{ "Tag": [], "Problem": "$ln(x^{2}+5x)=2ln(x+1)$\r\n\r\nsolve for x", "Solution_1": "hello, write the equation in the following form:\r\n$ln\\left(x^{2}+2x\\right)=ln\\left(x+1\\right)^{2}$. From this we get:\r\n$x^{2}+5x=x^{2}+2x+1$ with $x=\\frac{1}{3}$.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, write the equation in the following form:\n$ln\\left(x^{2}+2x\\right)=ln\\left(x+1\\right)^{2}$. From this we get:\n$x^{2}+5x=x^{2}+2x+1$ with $x=\\frac{1}{3}$.\nSonnhard.[/quote]\r\n\r\nhow did you cancel out the logs? what law did you use?\r\n\r\nthanks for the help", "Solution_3": "hello, i did use that $exp\\left(ln(x)\\right)=x$ for $x>0$.\r\nSonnhard.", "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, i did use that $exp\\left(ln(x)\\right)=x$ for $x>0$.\nSonnhard.[/quote]\r\n\r\nDO you mean?\r\n\r\n$e^\\left(ln(x)\\right)$=$x$", "Solution_5": "hello, yes. Sonnhard.", "Solution_6": "[hide=\" Step by step\"]\nFirst $2\\ln(x+1)$ can be written $\\ln(x+1)^{2}$ by the properties of logs. Then $e^{\\ln(x^{2}+5x)}= e^{\\ln(x+1)^{2}}$.\nThen we have \n$x^{2}+5x = (x+1)^{2}$\n$\\implies x^{2}+5x = x^{2}+2x+1$\n$\\implies 3x = 1 \\implies x = \\frac{1}{3}$[/hide]", "Solution_7": "[quote=\"bos1234\"]$ln(x^{2}+5x)=2ln(x+1)$\n\nsolve for x[/quote]\r\n\r\n[hide=\"Full explanation\"]Recall that $\\ln(a)+\\ln(b) = \\ln(a\\cdot b)$\n[hide=\"If you want, here is a proof\"]Let $x = \\ln(a)$, $y = \\ln(b)$, and $z = \\ln(ab)$. We want to show that $x+y = z$.\n\nFrom the definition of logs, $e^{x}= a$, $e^{y}= b$, $e^{z}= a b$. Now it is relatively obvious that $e^{x}\\cdot e^{y}= e^{z}$. From a property of exponentials, $e^{x+y}= e^{z}$, so $x+y = z$, as desired.[/hide]\n\nLet $a=b$ in that. Then:\n\n$\\ln(a)+\\ln(a) = \\ln(a^{2}$\n$2\\ln(a) = \\ln(a^{2})$\n\nReplace $a$ with $x+1$:\n$2\\ln(x+1) = \\ln((x+1)^{2}) = \\ln(x^{2}+2x+1)$\n\nNow, we may substitute this equation into the given:\n$\\ln(x^{2}+5 x) = \\ln(x^{2}+2x+1)$\n\nThus, $x^{2}+5x = x^{2}+2x+1$. So: $3x = 1 \\Rightarrow \\boxed{x = \\frac{1}{3}}$.[/hide]", "Solution_8": "Then $e^{\\ln(x^{2}+5x)}= e^{\\ln(x+1)^{2}}$\r\n\r\nI know that the law exists, but how can you make the logs indices? What permits us to do that?" } { "Tag": [ "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "Let E be ther error when the Taylor Polynomial T(x) = x - [(x^3)/(3!)] is used to approximate f(x) = sinx at x = 5. Which of the following is true?\r\n\r\n|E| < 0.0001 ******statement 1\r\n0.0001 < |E| < 0.0003 *****statement 2\r\n0.0003 < |E| < 0.005 *******statement 3\r\n0.005 < |E| < 0.007 *******statement 4\r\n0.07 < |E| ********statement 5", "Solution_1": "Shouldn't you be asking your calculator this question?\r\n\r\nAnyway, the polynomial evaluates to -95/6, while sin is always between -1 and 1, so it's pretty clear that the error is enormous.", "Solution_2": "which would be satisfied by statement 5. Thank you.", "Solution_3": "Remark: There's of course no need for calculators:). Just look at the remainder (in Cauchy's or Lagrange's form) and estimate it." } { "Tag": [ "puzzles" ], "Problem": "What are the next five terms in the sequence:\r\n\r\nO, T, T, F, F, S, ...\r\n\r\npretty easy", "Solution_1": "[hide=\"answer\"]one, two, three, four...so \"S\" for seven[/hide]", "Solution_2": "Try this one.\r\n\r\nFill in the missing term:\r\n\r\n...G, M, k, h, D, 1, d, c, m, _, n...\r\n\r\n[hide=\"Hint\"]\nIf you didn't know, I could give you 52 tries and you probably couldn't guess it.\n[/hide]", "Solution_3": "umm...[hide]Does it have to do with the metric system? \nbecause it looks like the first letter of each measurement but I am not sure what comes between a millimeter and a nanometer.[/hide]", "Solution_4": "[hide=\"answer\"]$\\mu$\nThe letters are the prefixes for the metric system (giga, mega, kilo, hecto, deca, no prefix, deci, centi, milli, micro, nano)\n[/hide]", "Solution_5": "Grrrrrrrr...\r\n\r\nVery good. :thumbup:", "Solution_6": "[quote=\"math92\"]What are the next five terms in the sequence:\n\nO, T, T, F, F, S, ...\n\npretty easy[/quote]\r\n[hide]\nOne, Two, Three, Four\n\nThe next letter would be S in correspondence to 7.[/hide]" } { "Tag": [ "calculus", "derivative", "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Show that the successive derivatives of an analytic function at a point can never satisfy $ |f^{(n)}(z)| > n!n^n$.\r\nFormulate a sharper theorem of same kind.", "Solution_1": "Let $ z_0$ be the point. By Taylor's Theorem, $ f$ can be represented as\r\n\r\n$ f(z) \\equal{} \\sum_{j \\equal{} 0}^{\\infty} \\frac {f^{n}(z_0)}{n!}(z \\minus{} z_0)^n$\r\n\r\nin a small neighborhood of $ z_0$. However, $ \\left|\\frac {f^{n}(z_0)}{n!}\\right| > n^n$, so by the Cauchy\u2013Hadamard theorem, the radius of convergence is $ 0$, a contradiction." } { "Tag": [], "Problem": "Solve the equation $ |y\\minus{}6| \\plus{} 2y \\equal{} 9$.", "Solution_1": "If $ (y\\minus{}6) \\leq 0$, $ y \\leq 6$ and the initial equation is $ 6 \\minus{} y \\plus{} 2y \\equal{} 9$ or $ y \\equal{} 3$.\r\nIf $ 0 \\leq (y\\minus{}6)$, $ 6 \\leq y$ and the initial equation is $ y \\minus{} 6 \\plus{} 2y \\equal{} 9$ or $ y \\equal{} 5$, which isn't agreed upon the initial condition.\r\nTherefore, the answer is $ y\\equal{}3$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: R\\minus{}>R$ such that $ f(x)f(x\\plus{}y)\\plus{}2f(x\\plus{}2y)\\plus{}f(2x\\plus{}y)f(y)\\equal{}x^{4}\\plus{}y^{4}\\plus{}x^{2}\\plus{}y^{2}$.", "Solution_1": "There is no such function.\r\n\r\nAssume, for a contradiction, that $ f$ does satisfy the functional equation.\r\nThen, for $ x\\equal{}y\\equal{}0$, it leads to $ f(0)\\equal{}0$ or $ f(0)\\equal{}\\minus{}1$.\r\n\r\nFor $ x\\equal{}a,y\\equal{}0$, it gives $ [f(a)]^2\\plus{}2f(a)\\plus{}f(2a)f(0)\\equal{}a^4\\plus{}a^2$. (1)\r\nFor $ x\\equal{}0,y\\equal{}a$, it gives $ [f(a)]^2\\plus{}2f(2a)\\plus{}f(a)f(0)\\equal{}a^4\\plus{}a^2$.\r\nThus, for all real $ a$, we must have $ 2f(a)\\plus{}f(2a)f(0)\\equal{}2f(2a)\\plus{}f(a)f(0)$.\r\nSince $ f(0) \\neq 2$, it follows that $ f(a)\\equal{}f(2a)$ for all $ a$. (2)\r\n\r\nBack to (1), we deduce that $ [f(a)]^2\\plus{}[2\\plus{}f(0)]f(a)\\equal{}a^4\\plus{}a^2$.\r\n\r\nUsing (2), it follows that for each integer $ n$, we have\r\n$ [f(a)]^2\\plus{}[2\\plus{}f(0)]f(a) \\equal{} [f(2^na)]^2\\plus{}[2\\plus{}f(0)]f(2^na)\\equal{}(2^na)^4\\plus{}(2^na)^2$.\r\nFor $ a\\equal{}1$, the LHS is constant and the RHS is not. A contradiction.\r\n\r\nPierre." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": ":( :( :(", "Solution_1": "the orthocenter H is the symmetric of O wrt N. The circle $ \\Gamma$ with center O and radius OA is the circumcircle. $ P: AH \\cap \\Gamma$. As well-know the side BC is the perpendicolar axes of HP that meet $ \\Gamma$ on B and C.", "Solution_2": ":clap: :clap:" } { "Tag": [ "geometry", "AMC", "USA(J)MO", "USAMO", "AIME", "area of a triangle", "Heron\\u0027s formula", "\\/closed" ], "Problem": "Hi, I took Honors/Pre-AP Geometry several years ago and performed reasonably well. However, I could use a refresher. I also want to study geometry much more in depth than I did in high school. Would you recommend the \"Introduction to Geometry\" textbook by AoPS? I'm not particularly prepping for any competitions; it's mostly for enrichment. Thanks in advance.", "Solution_1": "For someone who completed Honors/Pre-AP Geometry, Introduction to Geometry may be too easy for you. Although you are using it for enrichment, it would be no good to waste 50 dollars on something you know everything of already, though, granted, you may not know everything completely and may pick up on one or two things the whole book although that would still be a waste of money. \r\n\r\nMy recommendation would be that you incorporate other subjects into your study of mathematics in addition to geometry. Think of bodybuilding. If you train nothing but your arms, you would look awkward, not get any awards, and overall be phsyically limited to your weakest link. The same principle applies to math. For you, perhaps an appropriate level book would be ACoPS (the Art and Craft of Problem Solving).", "Solution_2": "[quote=\"chopinftw\"]Hi, I took Honors/Pre-AP Geometry several years ago and performed reasonably well. However, I could use a refresher. I also want to study geometry much more in depth than I did in high school. Would you recommend the \"Introduction to Geometry\" textbook by AoPS? I'm not particularly prepping for any competitions; it's mostly for enrichment. Thanks in advance.[/quote]\r\n\r\nI think Introduction to Geometry might still have some concepts an ordinary school course couldn't give you, so it might be worth to take a look at it if you've money to spare. However, it might be better to look at \"Geometry Revisited\" (there are a bunch of free copies floating around the Internet), because that still has a problem solving aspect like Intro to Geo, but covers advanced topics that would probably be better suited for you.", "Solution_3": "I am kind of in the same place as chpinftw, but I have also completed Intro to Geo. I am also taking precalculus (through AoPS) right now and my goal is to make USAMO this year. What are some recommendations? I was thinking of these three options. I would also be doing AIME and maybe a few Olympiad problems before the AIME for all 3 of these options\r\n\r\n[u]Option A[/u]\r\nFinish AoPS volume 1 and do Volume 2\r\n\r\n[u]Option B[/u]\r\n Do ACoPS and maybe parts of Geometry Revisited\r\n\r\n[u]Option C[/u]\r\n Geometry Revisited and practice AIME's for the rest of the time\r\n\r\nIf you have any other suggestions, feel free to give me your input. Thank You.", "Solution_4": "Um... I don't think you should do Vol.2 until you do the subjects in detail... Vol. 2 is more contest prep oriented", "Solution_5": "[quote=\"AIME15USAMO\"]I am kind of in the same place as chpinftw, but I have also completed Intro to Geo. I am also taking precalculus (through AoPS) right now and my goal is to make USAMO this year. What are some recommendations? I was thinking of these three options. I would also be doing AIME and maybe a few Olympiad problems before the AIME for all 3 of these options\n\n[u]Option A[/u]\nFinish AoPS volume 1 and do Volume 2\n\n[u]Option B[/u]\n Do ACoPS and maybe parts of Geometry Revisited\n\n[u]Option C[/u]\n Geometry Revisited and practice AIME's for the rest of the time\n\nIf you have any other suggestions, feel free to give me your input. Thank You.[/quote]He doesn't want to prepare for any contests. He just wants to learn geometry in general.\r\n\r\nAs for a good way to learn, I hear this book: http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20 \r\nis good.\r\n\r\nAlso, if you ever wanna try a different type of problem, go here:\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/Category:Intermediate_Geometry_Problems", "Solution_6": "[quote]He doesn't want to prepare for any contests. He just wants to learn geometry in general. [/quote]\r\n\r\nIf by he, you mean me, then I am preparing for a contest, like I mentioned in my post. Also, I have finished the Intro Series (except for Intro Geo, which I'm almost done with), and I am doing precalc and Interm Alg right now.", "Solution_7": "Ihatepie meant chopinftw by he", "Solution_8": "[quote=\"Ihatepie\"]As for a good way to learn, I hear this book: http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20 \nis good.[/quote]\r\nI have just gotten that book (Challenging Problems in Geometry). Basically, it is just a problem book (as suggested by its name). But I think you actually have to know some of concepts there in order to solve those problems. The book has you do the problems, some of which proves a theorem. For instance, in one of the problems, it asks you to prove Heron's Formula, etc. And there are other problems that deal with Heron's Formula. It is a great problem book though! (The book also has solutions, so you can always learn the theorems that way too.)", "Solution_9": "I wonder what Mr. Rusczyk thinks. :maybe:", "Solution_10": "[quote=\"chopinftw\"]I wonder what Mr. Rusczyk thinks. :maybe:[/quote]\r\n\r\nSorry I didn't answer sooner; I was on vacation. I suspect the book would mostly be review, but you can take a look at the diagnostic tests here:\r\n\r\nhttp://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204\r\n\r\nIf you can't do most of the problems on the Post-test, then the book would likely be helpful." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all $ x\\in Q$ satisfy\r\n$ \\{x^{2}\\}\\plus{}\\{x\\}\\equal{}1$", "Solution_1": "[hide=\"Does the usual trick work?\"]$ x\\equal{} z\\plus{}\\frac{a}{b}$ with $ \\{x\\}\\equal{}\\frac{a}{b}$ and $ z\\in\\mathbb Z$ then $ x^{2}\\equal{}z^{2}\\plus{}z q\\plus{}q^{2}$ \n$ \\{\\frac{z a b\\plus{}a^{2}}{b^{2}}\\}\\equal{}\\frac{b\\minus{}a}{b}$[hide][/hide][/hide]", "Solution_2": "Er...\r\n[hide]\n$ x \\equal{}\\frac{p}{q}$, where $ (p, q) \\equal{} 1$\n\n$ x^{2}\\plus{}x\\in\\mathbb{Z}$\n$ \\frac{p(p\\plus{}q)}{q^{2}}$\nClearly this is irreducible. Hence $ q \\equal{} 1$. And that yields $ \\{x^{2}\\}\\plus{}\\{x\\}\\equal{}0$, so there are no solutions.[/hide]\r\n\r\nSomething wrong here?", "Solution_3": "No it's right there are no solutions.", "Solution_4": "[quote=\"K81o7\"]\n$ \\{x^{2}\\}\\plus{}\\{x\\}\\equal{} 0$, so there are no solutions.\n[/quote]\r\n\r\n\r\nTry $ x\\equal{}\\frac{1}{2}$ so $ \\{x^{2}\\}\\plus{}\\{x\\}\\equal{}\\frac{3}{4}$ :(", "Solution_5": "[quote=\"quangpbc\"][quote=\"K81o7\"]\n$ \\{x^{2}\\}\\plus{}\\{x\\}\\equal{} 0$, so there are no solutions.\n[/quote]\n\n\nTry $ x \\equal{}\\frac{1}{2}$ so $ \\{x^{2}\\}\\plus{}\\{x\\}\\equal{}\\frac{3}{4}$ :([/quote]\r\n\r\n\r\nWhat do you mean? :huh: Isn't what you're saying only confirming Krishanu's result?" } { "Tag": [], "Problem": "Hi, i am having alot of trouble on these problems, could someone please please help me, ill be grateful!\r\n\r\n[b]1.[/b]\r\n\r\nIn triangle [i]ABC[/i], [i]D[/i] and [i]F[/i] are points on [i]BC[/i], and [i]E[/i] is a point on [i]AB[/i] such that [i]AD[/i] is parallel to [i]EF[/i], $\\angle CEF \\cong \\angle ACB$, [i]AD[/i]=$15$ m, [i]EF[/i] = $8$ m and [i]BF[/i] = $8$ m. [b]Find the length of BC.[/b]\r\n\r\n[b]2.[/b]\r\n\r\nThe vertices [i]A[/i] and [i]C[/i] of a triangle [i]ABC[/i] lie on the circumference of a circle. The circle cuts the sides [i]AB[/i] and [i]CB[/i] at points [i]X[/i] and [i]Y[/i] respectively. If [i]AC = XC[/i] and $\\angle YXC = 40^{\\circ}$, [b]Find the size of $\\angle ABC$[/b]\r\n\r\n\r\n\r\n[b]3. [/b]\r\n\r\nIn triangle [i]ABC[/i], $\\angle ABC = 112^{\\circ}$ and [i]AB = BC[/i]. [i]D[/i] is a point on the side [i]AB [/i]and [i]E[/i] is a point on the side [i]AC[/i] such that [i]AE = ED = DB[/i]. [b]Find the size of $\\angle BCD$[/b]\r\n\r\nI look forward to the dreaded solutions! Thanks alot :) and good luck with your maths :D \r\n\r\nkeshin\r\n\r\n[i][Edit: Re - texed to make clearer.][/i]", "Solution_1": "number one looks like... triangle ECF is isoceles. if EF is 8, then FC is 8 too.\r\nand... FC and BF are colinear, so...\r\nAt least, thats how it appears from how i drew it", "Solution_2": "1.\r\n$\\triangle ADB \\sim \\triangle EFB$\r\n$\\frac{15}{DB} = \\frac{8}{8}$\r\n$DB = 15$, $DF = 7$\r\n$\\angle CDA = \\angle CFE$ (by congruent corresponding angles)\r\nand the given angle proves by AA that $\\triangle CDF \\sim \\triangle EFC$\r\nusing corresponding sides, $\\frac{8}{CD} = \\frac{7+CD}{15}$\r\n$(CD)^{2} + 7(CD) - 8*15 = 0$\r\n$(CD - 8)(CD + 15) = 0$\r\n$CD = 8$\r\n$8+7+8 = \\boxed{23}$", "Solution_3": "I think #2 is 40, can anyone verify?", "Solution_4": "2.\r\nsince an inscribed angle is half the arc it intercepts, \r\n$arc YC = 2*40 = 80$\r\nsince AC = XC, \\angle CAX = \\angle CXA\r\nsince those angles are congruent, the arcs they intercept are congruent, so $arc AC = arc XYC$\r\nby the secant-secant angle formula, \r\n$\\angle ABC = \\frac{arc AC - arc XY}{2} = \\frac{arc AC - (arc AC - 80)}{2} = \\boxed{40}$", "Solution_5": "Hm for #3 woudln't AE be longer than AB?\r\n\r\nOr are the points allowed to be on extensions?\r\n\r\nEdit: whoops misread, thought E was on BC\r\n\r\nIt should be 17, here's my solution:\r\n\r\nAED is similar to ABC, so its angles are 34-34-112 also.\r\n\r\nSince the external angle of BDE is 34, and BCE is also 34, quadrilateral DECB is concylic. Since ADE is 34, angle EDB is 146, and since EDB is isosceles, DEB is 17. Since DECB is concylic, BCD=DEB=17.", "Solution_6": "3. \r\nsince $\\triangle ABC$ is isosoles, $\\angle BAC$ = $\\angle BCA = 34$\r\nsince $\\triangle AED$ is isosoles, $\\angle DAE = \\angle ADE = 34$, then $\\angle DEA = 112$\r\nsince all of the angles are congruent, $\\triangle ABC \\sim \\triangle AED$\r\nby proportional sides, $\\frac{AE}{AD} = \\frac{AB}{AC}$ since $AB = BC$, and $AE = BD$\r\n\r\n$\\frac{BD}{AD} = \\frac{BC}{AC}$\r\n\r\n$\\frac{BD}{BC} = \\frac{AD}{AC}$\r\nthat shows the angle biscector theorem, so $\\angle BCD = \\angle ACD$\r\nso $\\angle BCD = \\frac{34}{2} = \\boxed{17}$" } { "Tag": [ "probability", "geometry", "3D geometry", "tetrahedron", "algebra", "function", "domain" ], "Problem": "Here are some moderate to hard ciphering questions \r\nComments are appreciated \r\n\r\n1. The expression (1+i)^k is a real number when k is a multiple of what number? (i is defined as the square root of \u20131) \r\n\r\n2. x, y, and z are distinct real numbers that sum to 0. What is the maximum possible value of \r\n(xy + xz + yz)/(x^.5 + y^.5 + z^.5) \r\n\r\n3. You are given a 4ft board. You go to the shop and receive 2 boards, each of which has a maximum possible length of 6ft. What is the probability that you can construct a triangle from the three boards? \r\n\r\n4. xyz=1 , x+y+z=0 , and xy+yz+xz=0. Exactly one of x, y, and z is a real number. Find this value. \r\n\r\n5. Two men are betting money on a game of tic-tac-toe. A man pays 3. On average, how much can the player expect to win or lose in dollars? \r\n\r\n6. Find sqrt(3 + sqrt(3 + sqrt(3 + \u2026.))). \r\n\r\n7. In triangle ABC, D lies on side BC and AB=6, BD=3, DC=7 and AC=5. The length of the segment AD can be written in the form sqrt(a/b) where a/b is a fraction in lowest terms. What is a+b? \r\n\r\n8. The sum of the coefficients of the expansion (a+b+c+d)^9 can be written in the form 4^k where k is an integer. Find k. \r\n\r\n9. Find the ordered 5-tuple (a,b,c,d,e) that satisfies the following system of equations: \r\n4a + b + c + d + e = 8 \r\na + 4b + c + d + e = 14 \r\na + b + 4c + d + e = 24 \r\na + b + c + 4d + e = 34 \r\na + b + c + d + 4e = 40 \r\n\r\n10. Find the volume of a regular tetrahedron with side length 1 \r\n\r\n11. The rational numbers a,b,c, and d have the property that (ax+b)/(cx+d)=1 has no solution in x. What is the value of (a^2)/(a^2 +c^2)? \r\n\r\n12. The number X20191817\u2026321 is divisible by 11 if and only if X equals what positive integer? (note that X is a digit of the number; this is not multiplication) \r\n\r\n13. x/(y-6)=y/(z-8)=z/(x-10)=3. What is the value of x+y+z? \r\n\r\n14. Find the sum \r\n\r\n(sqrt(1+1) \u2013 sqrt(1))/(sqrt(1^2 +1) + (sqrt(2+1) \u2013 sqrt(2))/(sqrt(2^2 +2) + \r\n(sqrt(3+1) \u2013 sqrt(3))/(sqrt(3^2 +3) + \u2026 + (sqrt(15+1) \u2013 sqrt(15))/(sqrt(15^2 +15) \r\n\r\n15. Find the largest n such that (2004!)! is divisible by ((n!)!)! \r\n\r\n16. Compute \r\n\r\n(1*1! + 2*2! +3*3! + \u2026 +7*7! +1)/(6!) \r\n\r\n17. Simplify \r\n\r\n1(nC1) + 2(nC2) + 3(nC3) + \u2026 + n(nCn) \r\n\r\n18. A teacher wants to divide un-graded papers into a collection of several stacks, each consisting of the same number of papers. He knows that there is somewhere between 1000 and 2000 papers. He tries dividing the stack into groups of 2, 3, 4, 5, 6,7, and 8, but repeatedly gets 1 paper left over. How many papers are there? \r\n\r\n19. Find the sum of the series \r\n\r\n1+ 2(1/5) + 3(1/5)^2 + 4(1/5)^3 + \u2026 \r\n\r\n20. What is the smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 2^2^2 and 2^3^3.(note: a^b^c=(a^b)^c ) \r\n\r\n21. What is the remainder when 2^100 is divide by 15? \r\n\r\n22. Find one value of a for which the triangle with vertices (20,30), (30,60), and (a,50) has area 20. \r\n\r\n23. In triangle ADC, angle bisector DB is drawn. If AB=3, AD=6, and CD=8, find DB. \r\n\r\n24. If f(x/(x-1)) for all x not equal to 0 or 1, and 05 and x-f(f(x+3)) for x :le: 5. \r\nfind f(1)\r\n\r\n7. tennyson is building a circular corral for his pet dinosaur. to form the corral he pounds 20 posts equally spaced around the circumference. each post contributes 6 inches to the circumference and the fencing between adjacent posts contributes 8 ft to the circumference. to the nearest whol number, what is the number of square feet in the area of the corral? express your answer to the nearest integer.\r\n\r\n8. find the value of m, express as an improper fraction, so that this relation is not a function. {(-2m+1, -4), (-6m+8, 0)}\r\n\r\n9. a + 1/(b+1/(c+(1/d))) = 181/42. find a+b+c+d.\r\n\r\n10. tom took five math tests and got an integer score on each of them. he never scored higher than 90 and his lowest score was on the fourth test. if his avg score, rounded to the nearest integer was 82, what is the lowest possible score he could have earned on the fourth test?\r\n\r\n11. a science test has 10 questions worth 5 points each, 7 questions worth 6 points each, and 4 questions worth 2 points each. none of these questions will be given partial credit. how many scores between 0 and 100 are impossible to score?\r\n\r\n12. a blank videotape has space for exactly two hours in SP mode, four hours in LP mode, and six hours in EP mode. vino has recorded three one hour episodes of party of five on the same tape: the first one in SP the second in LP and the third in EP. what is the number of min in SP mode that remain on the tape?", "Solution_1": "1. [hide]The number that he sells (15+x)(400-16x)=-16x^2+160x+6000=-16(x^2-10x+25)+6400=-16(x-5)^2+6400. So it is maximum when x=5, or when he sells his peaches for 20 cents each.[/hide]\n\n\n\n2. [hide]Since A and R are symmetric around the point P, that means that P is the midpoint of points A and R. So (9+a)/2=5 and (14+b)/2=3. So a=1, b=-8 and the point is (1, -8)[/hide]\n\n\n\nBTW check #3 there may be something wrong with it", "Solution_2": "4. [hide]The lowest point on the graph is 3. When you reflect it across the line y=-2, the highest point on the graph would be -7.[/hide]\n\n\n\nCould you rephrase #5??", "Solution_3": "#2 (9,14) is right 4 and up 11 of (5,3), so the new point is the opposite: it is [b](1,-8)[/b]\r\n\r\n#3 Old vertex: (1,-1); New vertex: (-1,-1). So the graph moves[b] 2 units left[/b]\r\n\r\n#4 Assuming you mean \"greatest value of y\": \r\n\r\nThis will occur at the vertex of the new graph. Translating (0,3) across the line, we get (0,-7), so the answer is [b]-7[/b]\r\n\r\n#5 From a Venn diagram, we see that the value of the sums of the non-Z zones is [b]37%[/b].\r\n\r\n#6 If you play with the function enough, you see that it [b]never resolves itself[/b]. Is that a good answer?", "Solution_4": "#10 Let us say that the 4 other tests were all 90's. Then, we have:\r\n\r\nx+360/5=82\r\n\r\nx+360=410\r\nx=[b]50[/b]", "Solution_5": "7. [hide]There are twenty posts and twenty spaces between the posts which makes for 20*(8.5)=170 feet in the circumference. So there are 85/pi feet in the radius, which makes for 7225/(pi^2) or about 732 square feet.[/hide]\n\n\n\n8. [hide]Since it can't be a function, -2m+1=-6m+8 (in a function, a value of x cannot have two different values for y) That gies 7/4=1 3/4[/hide]\n\n\n\n9. [hide]The largest integer smaller than 181/42 is 4, so a=4. Then 1/(b+1/(c+(1/d)))=13/42, and b+1/(c+(1/d))=42/13. The greatest integer smaller than that is 3, so b=3. Then 1/(c+(1/d))=3/13 and c+(1/d)=13/3. So c=4 and d=3. a+b+c+d=4+3+4+3=14[/hide]", "Solution_6": "could someone explain how to draw the graphs for 3 and 4? and how to do 6 and 11? thanks!\r\nheres two more: \r\n\r\n1. What is the probability that a randomly chosen factor of 15^90 is a multiple of 15^65? Express your answer as a common fraction.\r\n\r\n2. If a motorist increases her speed by 8 mph she will cover a distance of 120 miles in 30 fewer min. What is her original rate of speed in mph?", "Solution_7": "[quote=\"topper\"]#6 If you play with the function enough, you see that it [b]never resolves itself[/b]. Is that a good answer?[/quote]\r\n\r\nI'm pretty sure it does resolve itself.. how about you post some working and we can see what happens..", "Solution_8": "I'll get to it next week, the musical I'm in is this week. Thanks.", "Solution_9": "whats #11??", "Solution_10": "[hide=11]Because everything is literally just a reflection of another number, the only impossible numbers are 1, 3, 97, and 99. Use symmetry.[/hide]", "Solution_11": "#12. I believe the answer is 10.", "Solution_12": "10 is wrong\n\n[hide]its 48[/hide]\n\n[hide]Average is 81.5[/hide]" } { "Tag": [ "logarithms" ], "Problem": "I've done question 1, trying to do question 2 at the moment, but I don't think there is enough information.\r\n\r\nIt says:\r\n\r\n\"In a certain figure, triangle ABC is inscribed in a circle O with AB = 4, BC = 8 and AC = 9. If segment BY bisects AC, find DC.\"\r\n\r\nThanks,\r\n\r\n~~Simba\r\n\r\nEdit: Also question 6 looks to be a bit dodgy.\r\n\r\n\"In the diagram below, points B and C are centers of two tangent circles, points E and F are points of tangency, BF + 2, and BD = 10. Lines BD and BF intersect at A. Find AE.\"", "Solution_1": "[quote=\"Simba\"]I've done question 1, trying to do question 2 at the moment, but I don't think there is enough information.\n\nIt says:\n\n\"In a certain figure, triangle ABC is inscribed in a circle O with AB = 4, BC = 8 and AC = 9. If segment BY bisects AC, find DC.\"\n\nThanks,\n\n~~Simba\n\nEdit: Also question 6 looks to be a bit dodgy.\n\n\"In the diagram below, points B and C are centers of two tangent circles, points E and F are points of tangency, BF + 2, and BD = 10. Lines BD and BF intersect at A. Find AE.\"[/quote]I just went through both of the problems and they both have enough info. I'll edit problem 2 to make it clearer though, without changing the4 problem. :) \r\n\r\nThanks for pointing that out.", "Solution_2": "What about question 7?\r\n\r\nDoes the last part mean log to base 16 of (a + b), or log to base 16 of a, then plus b? Is the 'b' including in the logarithm?\r\n\r\nThanks,\r\n\r\n~~Simba\r\n\r\nReply: It's log 16 of (a+b). I'll fix that one too. :roll: :)", "Solution_3": "[quote=\"shobber\"]I think that in problem 6, $BF=2$.[/quote]\r\n\r\nOk, thanks :) ...\r\n\r\n(Problem Fixed. :roll: )", "Solution_4": "for question 3 is simplest form mean factored, expanded or something else", "Solution_5": "For simplest form, I just want it to have the least number of terms. :)", "Solution_6": "For question 9, are you referring to 'a' and 'b' as the roots of the equation?\r\n\r\nFor instance, if the roots were x = 7 and x = 9, you would want 2(7 + 9) = 2 * 16 = 32?\r\n\r\nThanks,\r\nSimba\r\n\r\nReply: A and B are the two unknown coefficients of the problem. You need to find those given the conditions. :)", "Solution_7": "[quote=\"Simba\"]Reply: A and B are the two unknown coefficients of the problem. You need to find those given the conditions. :)[/quote]\r\n\r\nSo in the question, 'A' and 'B' are the same as 'a' and 'b'?\r\n\r\nThanks,\r\n\r\n~~Simba", "Solution_8": "Revive" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "The circle $ C_1$ meets a circle $ C_2$ at $ A$ and $ B$ .The tangent of $ C_2$ at $ A$ meets $ C_1$ at $ C$ and the tangent of $ C_1$ at $ A$ meets $ C_2$ at $ D$. Let $ (d)$ be a line pass through $ A$ and $ C$ the circum-circle of $ \\Delta ACD$. $ (d)$ meets $ C_1,C_2$ and $ C$ at $ M,N$ and $ P$ respectively. Prove that $ AM\\equal{}NP$", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=311308[/url]\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=313905[/url]\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=246834[/url]\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=187699[/url]" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "let $n = 4m^2 + 3$, where m is an integer not divisible by 3. show that there exists a prime p | n with p = 7 mod 12", "Solution_1": "all primes dividing it are bigger than three\r\n\r\nusing quadratic reciprocity you can prove that $4 m^2+3\\equiv 0\\bmod{p}$ implies $p\\equiv 1\\bmod{3}$\r\n\r\nnow as $4 m^2+3\\equiv 3 \\bmod{4}$ there has to be a prime p that is $p\\equiv 3\\bmod{4}$\r\n\r\nthis prime will then satisfy $p\\equiv 7\\bmod{12}$" } { "Tag": [ "induction", "inequalities proposed", "inequalities" ], "Problem": "Let $x_1, x_2,...,x_n$ be the positive real numbers and $x_i=|=x_j$.Prove that:\r\n $x_1^3+x_2^3+...+x_n^3\\geq(x_1+x_2+...+x_n)^2$", "Solution_1": "the degree is different!", "Solution_2": "I think the inequlity is right.\r\nand induction works", "Solution_3": "He has forgotten to tell that $x_1$, $x_2$, ..., $x_n$ should be [b]integers[/b]. The problem was solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=54151 and probably elsewhere, too (I remember seeing it a number of times).\r\n\r\n darij" } { "Tag": [ "function", "integration", "algorithm", "real analysis", "real analysis unsolved" ], "Problem": "Let $C$ be a real number. Let $f_{1},f_{2},f_{3}$ be continuous functions defined on $[0,1]$ with $\\int_{0}^{1}f_{i}= C$ for each $i$. Prove that there exist mutually disjoint subsets $P_{1},P_{2},P_{3}$ of $[0,1]$ such that $P_{1}\\cup P_{2}\\cup P_{3}= [0,1]$ and $\\int_{P_{i}}f_{i}\\ge C/3$ for each $i$.", "Solution_1": "[b][i]Very Informal Idea.[/i][/b] We use a \"greedy algorithm\": assign each point $x$ to the set $P_{i}$ for which $f_{i}(x)$ is maximal, e.g. if $f(x_{1}) \\geq f(x_{2})$ and $f(x_{1}) \\geq f(x_{3})$ we add $x$ to $P_{1}$. Once one of the sets $P_{i}$ satisfies the desired condition:\r\n\\[\\int_{P_{i}}f_{i}(x) dx = C/3,\\]\r\nwe stop adding new points to it.\r\n\r\n[b][i]Formal proof.[/i][/b] Define sets $A_{1}$, $A_{2}$ and $A_{3}$ as follows\r\n\\[A_{1}= \\{x: f_{1}(x) \\geq f_{2}(x) \\text{ and }f_{1}(x) \\geq f_{3}(x)\\}\\]\r\n\\[A_{2}= \\{x: f_{2}(x) > f_{1}(x) \\text{ and }f_{2}(x) \\geq f_{3}(x)\\}\\]\r\n\\[A_{3}= \\{x: f_{3}(x) > f_{1}(x) \\text{ and }f_{3}(x) > f_{2}(x)\\}.\\]\r\nAssume without loss of generality that \r\n\\[\\int_{A_{1}}f_{1}(x) dx\\geq \\int_{A_{2}}f_{2}(x) dx \\geq \\int_{A_{3}}f_{3}(x) dx.\\]\r\nSince\r\n\\[\\int_{A_{1}}f_{1}(x) dx+\\int_{A_{2}}f_{2}(x) dx+\\int_{A_{3}}f_{3}(x) dx = \\\\ \\int_{0}^{1}\\max(f_{1}(x), f_{2}(x), f_{3}(x)) dx \\geq C,\\]\r\nwe have \r\n\\[I_{1}\\equiv \\int_{A_{1}}f_{1}(x) dx \\geq C/3.\\]\r\nNote that since the continuous function\r\n\\[t\\mapsto \\int_{A_{1}\\cap [0, t]}f_{1}(x) dx\\]\r\nvaries from $0$ to $I_{1}$, there exists $t_{0}$ s.t.\r\n\\[I_{1}= \\int_{P_{1}}f_{1}(x) dx = C/3\\]\r\nwhere $P_{1}= A_{1}\\cap [0, t_{0}]$. Now let \r\n\\[B_{2}= \\{x\\notin P_{1}: f_{2}(x) \\geq f_{3}(x) \\}\\]\r\n\\[B_{3}= \\{x\\notin P_{1}: f_{3}(x) > f_{2}(x)\\}.\\]\r\nAssume that \r\n\\[\\int_{B_{2}}f_{2}(x) dx \\geq \\int_{B_{3}}f_{3}(x) dx.\\]\r\nThen \r\n\\[\\int_{B_{2}}f_{2}(x) dx+\\int_{B_{3}}f_{3}(x) dx = \\int_{[0,1]\\setminus P_{1}}\\max(f_{2}(x), f_{3}(x))\\leq \\\\ \\int_{0}^{1}\\max(f_{1}(x), f_{2}(x), f_{3}(x)) dx-\\int_{P_{1}}f_{1}(x) dx \\geq 2/3\\, C.\\]\r\nTherefore,\r\n\\[\\int_{B_{2}}f_{2}(x)dx \\geq C/3.\\]\r\nLet $P_{2}$ be a subset of $B_{2}$ s.t.\r\n\\[\\int_{P_{2}}f_{2}(x) dx= C/3.\\]\r\nLet $P_{3}= [0,1]\\setminus P_{1}\\setminus P_{2}$. We finally have\r\n\\[\\int_{P_{3}}f_{3}(x)dx = \\int_{0}^{1}f_{3}(x) dx-\\int_{P_{1}\\cup P_{2}}f_{3}(x) dx \\geq \\\\ \\int_{0}^{1}f_{3}(x) dx-\\int_{P_{1}}f_{1}(x) dx-\\int_{P_{2}}f_{2}(x) dx \\geq C/3\\]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "I can't seem to figure out how to solve this one:\r\n\r\nIf $abc=1$, then prove\r\n\r\n$\\frac{1}{a+ab}+\\frac{1}{b+bc}+\\frac{1}{c+ca} \\ge \\frac{3}{2}$.\r\n\r\nHints or solutions are welcome!", "Solution_1": "By http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 , we have\r\n\r\n$ \\left( 1+abc\\right) \\left(\\frac{1}{a\\left( 1+b\\right) }+\\frac{1}{b\\left( 1+c\\right) }+\\frac{1}{c\\left( 1+a\\right) }\\right) \\geq 3$.\r\n\r\nBut since abc = 1, we have 1 + abc = 2, and thus\r\n\r\n$ 2 \\left(\\frac{1}{a\\left( 1+b\\right) }+\\frac{1}{b\\left( 1+c\\right) }+\\frac{1}{c\\left( 1+a\\right) }\\right) \\geq 3$,\r\n\r\nso that\r\n\r\n$ \\frac{1}{a\\left( 1+b\\right) }+\\frac{1}{b\\left( 1+c\\right) }+\\frac{1}{c\\left( 1+a\\right) }\\geq\\frac32$.\r\n\r\nThis is, of course, exactly your inequality.\r\n\r\n Darij", "Solution_2": "Thanks! That is indeed a very nice inequality you proved in the other thread.", "Solution_3": "See\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=inequalities&t=19469\r\n\r\nalso", "Solution_4": "Can someone please give a link, where Po-Shen Loh's Notes on Inequalities could be found. Thank you very much.", "Solution_5": "Alternatively, take $a = \\frac{x}{y}$, $b = \\frac{y}{z}$, and $c = \\frac{z}{x}$. We get\r\n\r\n$\\sum_{cyc} \\frac{1}{a+ab} = \\sum_{cyc} \\frac{1}{\\frac{x}{y} + \\frac{x}{z}} = \\sum_{cyc} \\frac{yz}{xy + xz}$\r\n\r\nBut adding three to both sides, we find the equivalent\r\n\r\n$\\left(xy+yz+zx\\right)\\left(\\frac{1}{xy+yz} + \\frac{1}{yz+zx} + \\frac{1}{zx+xy}\\right) \\ge \\frac{9}{2}$\r\n\r\nwhich follows from Cauchy-Schwartz or AM-HM.", "Solution_6": "[quote=\"Xixas\"]Can someone please give a link, where Po-Shen Loh's Notes on Inequalities could be found. Thank you very much.[/quote]\r\n\r\n[url=http://www.cco.caltech.edu/~po/docs/math/2-inequalities-solns.pdf]http://www.cco.caltech.edu/~po/docs/math/2-inequalities-solns.pdf[/url] -- it's mentioned in the Inequalities Theorems and Formulas forum." } { "Tag": [ "AMC", "AIME" ], "Problem": "Can you solve in constants k,j,n,m?\r\n\r\n$(x-\\frac{j^{2}-k^{2}}{4x})^{2}+(y-\\frac{n^{2}-k^{2}}{4y})^{2}= k^{2}\\\\ \\\\ (x+\\frac{j^{2}-k^{2}}{4x})^{2}+(y-\\frac{n^{2}-k^{2}}{4y})^{2}= j^{2}\\\\ \\\\ (x-\\frac{j^{2}-k^{2}}{4x})^{2}+(y+\\frac{n^{2}-k^{2}}{4y})^{2}= n^{2}\\\\ \\\\ (x+\\frac{j^{2}-k^{2}}{4x})^{2}+(y+\\frac{n^{2}-k^{2}}{4y})^{2}= m^{2}$", "Solution_1": "What is the exact problem here? To find the relation between $j,k,m,n$? Or to find $x,y$ in terms of $j,k,m,n$?", "Solution_2": "Express $x$ and $y$ in terms of the other four. That [i]is[/i] messy. :ninja:", "Solution_3": "That particular feat can't be accomplished, as there's actually only one independent equation.\r\n\r\nOn the other hand, the relation between $j,k,m,n$ can be found easily: $k^{2}+m^{2}=j^{2}+n^{2}$", "Solution_4": "Use the fact that the form of all of the equations is a circle.\r\nOr...\r\nIf you expand for example, the first term in the first equation, you get:\r\n\r\nx^2 - (j^2 - k^2)/8 - (j^2 - k^2)/16x^2\r\n\r\nThen, do this to all of them and put all of the variable free terms, such as (j^2 - k^2)/8, on one side and it is pretty striaghtforward to solve through subbing in, adding equations, and seperating variables and constant.\r\n\r\n[quote=\"Farenhajt\"]That particular feat can't be accomplished, as there's actually only one independent equation.\n\nOn the other hand, the relation between $j,k,m,n$ can be found easily: $k^{2}+m^{2}=j^{2}+n^{2}$[/quote]\r\n\r\nAnd you can use what Farenhajt said as one of the \"subs\".", "Solution_5": "i believe this is similar to AIME II 2006 #15, try and find a geometric interpretation", "Solution_6": "[quote=\"Altheman\"]i believe this is similar to AIME II 2006 #15, try and find a geometric interpretation[/quote]\r\n\r\nThanks for repeating:\r\n\"Use the fact that the form of all of the equations is a circle. \r\nOr... \" -neelnal\r\n\r\n :roll: :wink:", "Solution_7": "Umm, actually I don't think they are equations of circles. Look at them more carefully; there are $x$ and $y$ terms in the denominators.", "Solution_8": "After expanding the squares, we get\r\n\r\n\\begin{eqnarray*}x^{2}-\\frac{j^{2}-k^{2}}{2}+\\left(\\frac{j^{2}-k^{2}}{4x}\\right)^{2}+y^{2}-\\frac{n^{2}-k^{2}}{2}+\\left(\\frac{n^{2}-k^{2}}{4y}\\right)^{2}&=& k^{2}\\\\ x^{2}+\\frac{j^{2}-k^{2}}{2}+\\left(\\frac{j^{2}-k^{2}}{4x}\\right)^{2}+y^{2}-\\frac{n^{2}-k^{2}}{2}+\\left(\\frac{n^{2}-k^{2}}{4y}\\right)^{2}&=& j^{2}\\\\ x^{2}-\\frac{j^{2}-k^{2}}{2}+\\left(\\frac{j^{2}-k^{2}}{4x}\\right)^{2}+y^{2}+\\frac{n^{2}-k^{2}}{2}+\\left(\\frac{n^{2}-k^{2}}{4y}\\right)^{2}&=& n^{2}\\\\ x^{2}+\\frac{j^{2}-k^{2}}{2}+\\left(\\frac{j^{2}-k^{2}}{4x}\\right)^{2}+y^{2}+\\frac{n^{2}-k^{2}}{2}+\\left(\\frac{n^{2}-k^{2}}{4y}\\right)^{2}&=& m^{2}\\end{eqnarray*}\r\n\r\nSubtracting the second, the third and the fourth equation from the first we get\r\n\r\n\\begin{eqnarray*}x^{2}-\\frac{j^{2}-k^{2}}{2}+\\left(\\frac{j^{2}-k^{2}}{4x}\\right)^{2}+y^{2}-\\frac{n^{2}-k^{2}}{2}+\\left(\\frac{n^{2}-k^{2}}{4y}\\right)^{2}&=& k^{2}\\\\-(j^{2}-k^{2}) &=& k^{2}-j^{2}\\\\-(n^{2}-k^{2}) &=& k^{2}-n^{2}\\\\-(j^{2}-k^{2})-(n^{2}-k^{2}) &=& k^{2}-m^{2}\\end{eqnarray*}\r\n\r\nTherefore, the second and the third equations are redundant, since they reduce to $0=0$, and the fourth equation can be valid if and only if $-j^{2}+k^{2}-n^{2}+k^{2}=k^{2}-m^{2}\\iff k^{2}+m^{2}=j^{2}+n^{2}$. In that case, the fourth equation also becomes redundant, and the system is reduced to the first equation:\r\n\r\n$x^{2}+\\left(\\frac{j^{2}-k^{2}}{4x}\\right)^{2}+y^{2}+\\left(\\frac{n^{2}-k^{2}}{4y}\\right)^{2}= \\frac{n^{2}+j^{2}}{2}$\r\n\r\nwhich obviously has infinitely many solutions." } { "Tag": [ "blogs", "Mafia", "puzzles" ], "Problem": "(The ghost of) r15s11z55y89w21 presents |-|!dden wol2cl (if you can't figure this out, maybe you should consider something else...no offense intended). I give a picture, you figure out which word (or sometimes number) is hidden in the picture! Sometimes the word would be something you wouldn't find in a dictionary, like \"macromedia\". [u]All words used are at least four letters long, and all numbers are at least three-digit numbers. [/u]PM me (me=Hopkinsmathclub) your answer, [color=red][size=134][u][b]do not post your answers or anything that might give a hint towards the answer on the thread[/b][/u][/size][/color]! Those who answer correctly receive 5 points. If there are more incorrect submissions than correct submissions, the first two to submit incorrectly will receive 1 point each, and if there are no correct submissions, the first to submit will receive 3 points. If your submission contains inappropriate language, you will receive -15 points. None of the answers are inappropriate words. 420, 666, and 1488 will not be considered inappropriate numbers. On submission PMs, title your submission [color=green]\"Hidden #[insert number]: [insert word you're submitting]\"[/color] for example [color=blue]\"Hidden #11: bourgeoisie\"[/color]. [i]Please[/i] do not share answers with others. [i]Please[/i] don't sabotage this game by deleting PMs before r15s11z55y89w21 reads them if you know the password to this account; I am, after all, working in your favor because I'm working towards Hopkinsmathclub-being-able-to-have-a-blog-ness. [i]Please[/i] play this game fairly. All right, enough begging for playing it fair, enjoy the game!\r\n\r\n[size=200]Number One[/size]\r\n\r\nDifficulty: Moderately Easy\r\n\r\nFind the hidden word.", "Solution_1": "[u]Results[/u]\r\n\r\nSubmissions: 4\r\nCorrect Answers: 4 ([color=green][b][i]100.00%[/i][/b][/color])\r\nScore changes: [b]Dojo [color=green]+5[/color], dragon96 [color=green]+5[/color], jjfun1 [color=green]+5[/color], bbgun34 [color=green]+5[/color][/b]\r\n[i]Extra point for jjfun1 for noticing an error![/i]\r\nCurrent Top Scores: \r\njjfun1 6\r\nDojo 5\r\ndragon96 5\r\nbbgun34 5\r\n\r\nAnswer: [b][color=brown]Hidden[/color][/b]\r\n\r\n[size=200]Number Two[/size]\r\n\r\nDifficulty: Moderate", "Solution_2": "[u]Results[/u]\r\n\r\nSubmissions: 2\r\nCorrect Answers: 0 ([color=red][b][i]0.00%[/i][/b][/color])\r\nScore changes: [b]rd5493 [color=green]+3[/color], dragon96 [color=green]+1[/color][/b]\r\nCurrent Top Scores: \r\ndragon96 6\r\njjfun1 6\r\nDojo 5\r\nbbgun34 5\r\nrd5493 3\r\n\r\nAnswer: [b][color=blue]Green[/color][/b]\r\n\r\n[size=200]Number Three[/size]\r\n\r\nDifficulty: Moderate", "Solution_3": "[u]Results[/u]\r\n\r\nSubmissions: 7 (2: Careful, 3: Cape, 1: Chapel, 1: Carpool)\r\nCorrect Answers: 2 ([color=orange][b][i]28.57%[/i][/b][/color])\r\nScore changes: [b]Yoshi [color=green]+5[/color], bbgun34 [color=green]+5[/color], dragon96 [color=orange]+2[/color], vahalla [color=orange]+2[/color], Dojo [color=orange]+2[/color], PowerOfPi [color=orange]+2[/color], westiepaw [color=orange]+2[/color][/b]\r\nCurrent Top Scores: \r\nbbgun34 10\r\ndragon96 8\r\nDojo 7\r\njjfun1 6\r\nYoshi 5\r\nrd5493 3\r\nvahalla 2\r\nPowerOfPi 2\r\nwestiepaw 2\r\n\r\nAnswer: [b][color=orange]Careful[/color][/b]\r\n\r\n[size=200]Number Four[/size]\r\n\r\nDifficulty: Easy", "Solution_4": "[u]Results[/u]\r\n\r\nSubmissions: 5 (3: Easy, 1: Ask, 1: Fog)\r\nCorrect Answers: 3 ([color=indigo][b][i]60.00%[/i][/b][/color])\r\nScore changes: [b]PowerOfPi [color=green]+5[/color], vahalla [color=green]+5[/color], dragon96 [color=green]+5[/color][/b]\r\nNotes: I think I'll start taking off points for not following the submission format. \r\nCurrent Top Scores: \r\ndragon96 13\r\nbbgun34 10\r\nDojo 7\r\nvahalla 7\r\nPowerOfPi 7\r\njjfun1 6\r\nYoshi 5\r\nrd5493 3\r\nwestiepaw 2\r\n\r\nAnswer: [b][color=olive]Easy[/color][/b]\r\n\r\n[size=200]Number Five[/size]\r\n\r\nDifficulty: Moderately Hard", "Solution_5": "[size=200]Number Six[/size]\r\n\r\nJust giving it ahead of time. Number five submissions are still accepted. \r\n\r\nDifficulty: Moderate", "Solution_6": "[u]Results[/u]\r\n\r\nSubmissions: 10 (10: Truth)\r\nCorrect Answers: 10 ([color=green][b][i]100.00%[/i][/b][/color])\r\nScore changes: [b]james4l [color=green]+5[/color], vahalla [color=green]+5[/color], PowerOfPi [color=green]+5[/color], dragon96 [color=green]+5[/color], Dojo [color=green]+5[/color], math_explorer [color=green]+5[/color], bluecarneal [color=green]+5[/color], Yoshi [color=green]+5[/color], meewhee009 [color=green]+4[/color], bbgun34 [color=green]+4[/color][/b]\r\nNotes: As I have warned, I took off a point for not following the submission format. From now on, I'll take off two points. \r\nCurrent Scores: \r\ndragon96 18\r\nbbgun34 14\r\nDojo 12\r\nvahalla 12\r\nYoshi 10\r\nPowerOfPi 7\r\njjfun1 6\r\njames4l 5\r\nmath_explorer 5\r\nbluecarneal 5\r\nmeewhee009 4\r\nrd5493 3\r\nwestiepaw 2\r\n\r\nAnswer: [b][color=Blue]Truth[/color][/b]\r\n\r\n[color=red][u][b]Number Six is still open. [/b][/u][/color]", "Solution_7": "Sorry, but how is the ansewr truth? Doesn't make much sense.", "Solution_8": "Lower-right hand corner :wink:", "Solution_9": "[quote=\"andersonw\"]Lower-right hand corner :wink:[/quote]\r\n\r\nA green square?", "Solution_10": "You're looking at the wrong puzzle. Puzzle 5 has the answer \"truth\".", "Solution_11": "[quote=\"AIME15\"]You're looking at the wrong puzzle. Puzzle 5 has the answer \"truth\".[/quote]\r\n\r\nOh, I see. Thanks!", "Solution_12": "[size=200]Number Seven[/size]\r\n\r\nDifficulty: Hard\r\n\r\nNumber Six Results coming soon.", "Solution_13": "[u]Results[/u]\r\n\r\nSubmissions: 16 (16: Sight)\r\nCorrect Answers: 16 ([color=green][b][i]100.00%[/i][/b][/color])\r\nScore changes: [b]BOGTRO, vahalla, Yoshi, PowerOfPi, jjfun1, james4l, meewhee009, cf249, dragon96, zapi2007, math_explorer, westiepaw, bluecarneal, gaussintraining [color=green]ALL +5[/color], bbgun34 [color=green]+3[/color], rd5493 [color=green]+3[/color][/b]\r\nNotes: Please follow the format! You only get 60% of the score if you don't. \r\nCurrent Scores: \r\ndragon96 23\r\nbbgun34 17\r\nvahalla 17\r\nPowerOfPi 17\r\nYoshi 15\r\nDojo 12\r\njjfun1 11\r\njames4l 10\r\nmath_explorer 10\r\nbluecarneal 10\r\nmeewhee009 9\r\nrd5493 7\r\nwestiepaw 7\r\nBOGTRO 5\r\ncf249 5\r\nzapi2007 5\r\ngaussintraining 5\r\n\r\nAnswer: [b][color=Violet]Sight[/color][/b]\r\n\r\n[color=red][u][b]Number Seven is still open. [/b][/u][/color]", "Solution_14": "[hide=\"Results for Puzzle 4\"]\n[quote=\"Hopkinsmathclub\"][u]Results[/u]\n\nSubmissions: 5 (3: Easy, 1: Ask, 1: Fog)\nCorrect Answers: 3 ([color=indigo][b][i]60.00%[/i][/b][/color])\nScore changes: [b]PowerOfPi [color=green]+5[/color], vahalla [color=green]+5[/color], dragon96 [color=green]+5[/color][/b]\nNotes: I think I'll start taking off points for not following the submission format. \nCurrent Top Scores: \ndragon96 13\nbbgun34 10\nDojo 7\nvahalla 7\n[b]PowerOfPi 7[/b]\njjfun1 6\nYoshi 5\nrd5493 3\nwestiepaw 2[/quote][/hide]\n[hide=\"Results for Puzzle 5\"]\n[quote=\"Hopkinsmathclub\"][u]Results[/u]\n\nSubmissions: 10 (10: Truth)\nCorrect Answers: 10 ([color=green][b][i]100.00%[/i][/b][/color])\nScore changes: [b]james4l [color=green]+5[/color], vahalla [color=green]+5[/color], PowerOfPi [color=green]+5[/color], dragon96 [color=green]+5[/color], Dojo [color=green]+5[/color], math_explorer [color=green]+5[/color], bluecarneal [color=green]+5[/color], Yoshi [color=green]+5[/color], meewhee009 [color=green]+4[/color], bbgun34 [color=green]+4[/color][/b]\nNotes: As I have warned, I took off a point for not following the submission format. From now on, I'll take off two points. \nCurrent Scores: \ndragon96 18\nbbgun34 14\nDojo 12\nvahalla 12\nYoshi 10\n[b]PowerOfPi 7[/b]\njjfun1 6\njames4l 5\nmath_explorer 5\nbluecarneal 5\nmeewhee009 4\nrd5493 3\nwestiepaw 2[/quote][/hide]\r\nYou forgot to add 5 to my score (see my bolded score).", "Solution_15": "[quote=\"Hopkinsmathclub\"]Correct Answers: 9 ([color=green][b][i]100.00%[/i][/b][/color])[/quote]\r\nYou forgot to change that.", "Solution_16": "Oh, whoops. LOL. \r\n\r\n[size=200]Number Sixteen[/size]\r\n\r\nMade by [b][color=green]fortenforge[/color][/b]\r\n\r\nDifficulty: Moderate\r\n\r\n[color=red][u][b]Number Fifteen still open. [/b][/u][/color]", "Solution_17": "[u]Results[/u]\r\n\r\nSubmissions: 6 (3: Exit, 1: Edit, 1: Fist, 1: Fact)\r\nCorrect Answers: 3 ([color=orange][b][i]50.00%[/i][/b][/color])\r\nScore changes: [b]fortenforge [color=green]+5[/color], math_explorer [color=green]+5[/color], PowerOfPi [color=green]+5[/color], zapi2007 [color=green]+3[/color][/b]\r\nCurrent scoreboard:\r\n[b]james4l 45[/b]\r\n[b]vahalla 42[/b]\r\n[i]math_explorer 39[/i]\r\n[i]dragon96 37[/i]\r\n[i]PowerOfPi 37[/i]\r\n[i]zapi2007 28[/i]\r\nBOGTRO 22\r\nbbgun34 17\r\nDojo 15\r\nYoshi 15\r\nwestiepaw 12\r\njjfun1 11\r\nfortenforge 10\r\ncf249 10\r\nbluecarneal 10\r\nmeewhee009 9\r\nrd5493 7\r\ngaussintraining 5\r\n\r\nAnswer: [b][color=blue]Exit[/color][/b]\r\n\r\n[color=red][u][b]Number Sixteen is still open. [/b][/u][/color]", "Solution_18": "[size=200]Number Seventeen[/size]\r\n\r\nDifficulty: Moderately Easy\r\n\r\n[u][b][color=red]Number Sixteen is still open. [/color][/b][/u]", "Solution_19": "[u]Results[/u]\r\n\r\nSubmissions: 3 (2: Symbol, 1: Uyombs) <--Lol?\r\nCorrect Answers: 2 ([color=indigo][b][i]66.67%[/i][/b][/color])\r\nScore changes: [b]james4l [color=green]+5[/color], math_explorer [color=green]+5[/color], bluecarneal [color=green]+1[/color][/b]\r\nCurrent scoreboard:\r\n[b]james4l 50[/b]\r\n[b]math_explorer 44[/b]\r\n[b]vahalla 42[/b]\r\n[i]dragon96 37[/i]\r\n[i]PowerOfPi 37[/i]\r\n[i]zapi2007 28[/i]\r\nBOGTRO 22\r\nbbgun34 17\r\nDojo 15\r\nYoshi 15\r\nfortenforge 14\r\nwestiepaw 12\r\nbluecarneal 11\r\njjfun1 11\r\ncf249 10\r\nmeewhee009 9\r\nrd5493 7\r\ngaussintraining 5\r\n\r\nAnswer: [b][color=black]Symbol[/color][/b]\r\n\r\n[color=red][u][b]Number Seventeen is still open. [/b][/u][/color]", "Solution_20": "[size=200]Number Eighteen[/size]\r\n\r\nDifficulty: Hard\r\n\r\nThe answer is a number. Bonus point if you can point out the significance of the number.", "Solution_21": "[u]Results[/u]\r\n\r\nSubmissions: 2 (1: Oversize, 1: Pursue) <--Lol?\r\nCorrect Answers: 1 ([color=orange][b][i]50.00%[/i][/b][/color])\r\nScore changes: [b]math_explorer [color=green]+5[/color], james4l [color=green]+1[/color][/b]\r\nCurrent scoreboard:\r\n[b]james4l 51[/b]\r\n[b]math_explorer 49[/b]\r\n[i]vahalla 42[/i]\r\n[i]dragon96 37[/i]\r\n[i]PowerOfPi 37[/i]\r\nzapi2007 28\r\nBOGTRO 22\r\nbbgun34 17\r\nDojo 15\r\nYoshi 15\r\nfortenforge 14\r\nwestiepaw 12\r\nbluecarneal 11\r\njjfun1 11\r\ncf249 10\r\nmeewhee009 9\r\nrd5493 7\r\ngaussintraining 5\r\n\r\nAnswer: [b][color=red]Oversize[/color][/b]\r\n\r\n[color=red][u][b]Number Eighteen is still open. [/b][/u][/color][/quote]", "Solution_22": "[u]Results[/u]\r\n\r\nSubmissions: 4 (3: 5040, 1: 504) (I just realized I still forgot to delete the <--Lol label from ages ago)\r\nCorrect Answers: 3 ([color=green][b][i]75.00%[/i][/b][/color])\r\nScore changes: [b]PowerOfPi [color=green]+5,+1[/color], westiepaw [color=green]+5[/color], math_explorer [color=green]+5,+1[/color], james4l [color=green]+3[/color][/b]\r\nCurrent scoreboard:\r\n[color=orange][b]math_explorer 55[/b]\n[b]james4l 54[/b][/color]\r\n[i]PowerOfPi 43[/i]\r\n[i]vahalla 42[/i]\r\ndragon96 37\r\nzapi2007 28\r\nBOGTRO 22\r\nwestiepaw 17\r\nbbgun34 17\r\nDojo 15\r\nYoshi 15\r\nfortenforge 14\r\nbluecarneal 11\r\njjfun1 11\r\ncf249 10\r\nmeewhee009 9\r\nrd5493 7\r\ngaussintraining 5\r\n\r\nAnswer: [b]5040[/b]\r\n\r\nSignificance of the number: 7!, ALSO ACCEPTABLE: according to Plato, \"the ideal population of a city,\" due to easy divisibility when dividing into groups", "Solution_23": "[size=200]Number Nineteen[/size]\r\n\r\nDifficulty: Moderately Easy\r\n\r\nThe answer is a number. Bonus point if you can point out the significance of the number. Some of you will immediately recognize this number, while others might have a really tough time. It's one of those numbers that are life-crucial in some occupations and totally unimportant in others.", "Solution_24": "[u]Results[/u]\r\n\r\nSubmissions: 6 (6: 69314718)\r\nCorrect Answers: 6 ([color=green][b][i]100.00%[/i][/b][/color])\r\n\r\nCurrent scoreboard:\r\n[color=orange][b]math_explorer 60[/b]\n[b]james4l 59[/b][/color]\r\nPowerOfPi 48\r\ndragon96 42\r\nvahalla 42\r\nzapi2007 33\r\nBOGTRO 22\r\nwestiepaw 17\r\nbbgun34 17\r\ncf249 15\r\nDojo 15\r\nYoshi 15\r\nfortenforge 14\r\nbluecarneal 11\r\njjfun1 11\r\nmeewhee009 9\r\nrd5493 7\r\ngaussintraining 5\r\n\r\nAnswer: [b]69314718[/b]\r\n\r\nSignificance of the number: The first few decimal places in the decimal representation of ln 2 and ln (1/2). This is used in banking and accounting as a dividend for an expression to calculate doubling time. (Credit only given for both parts.)\r\n\r\n[size=200]Number Twenty[/size]\r\n\r\nNow for something fun to test your guessing skills! Match the person with their body text message for their submission to Number Six! \r\n\r\n[color=white]Hint: Twenty is an even number. [/color]\r\n\r\nBOGTRO\r\nvahalla\r\nYoshi\r\nPowerOfPi\r\njjfun1\r\njames4l\r\nmeewhee009\r\ncf249\r\ndragon96\r\nzapi2007\r\nmath_explorer\r\nwestiepaw\r\nbluecarneal\r\ngaussintraining\r\nbbgun34\r\nrd5493\r\n[hide=\" \"]-2[/hide]\r\nSight\r\nSight\r\nHidden #6: sight\r\nloldarn\r\nHidden #6: Sight\r\nHidden #6: sight\r\nHidden #6: Sight\r\n*insert filler text here*\r\nNot sure if you received it on FTW.\r\nhey.\r\nsight\r\nNo way that was moderate. I got it in 10 seconds.\r\nHello. Am I supposed to put anything of value in the body of this message?\r\n :lol: \r\nHidden #6: SIGHT\r\nUnless, of course, I missed a huge very important part of the picture.", "Solution_25": "You should make people PM PhireKalk6781 instead since I accidentally found out who you are and each time I log in to HMC, I see this PM pop up thing.\r\n\r\nEDIT: Also, what do we put as our title?", "Solution_26": "[21:27] PhireKaLk6781: Just put Hidden #20", "Solution_27": "What's going on?", "Solution_28": "[u]Results[/u]\r\n\r\nSubmissions: 2 (james4l [color=green][b]+2[/b][/color], dragon96 [color=green][b]+4[/b][/color])\r\n\r\nFinal scoreboard:\r\n[color=blue]\n[b]james4l 61[/b] :winner_first:\n[b]math_explorer 60[/b] :winner_second:\n[b]PowerOfPi 48[/b] :winner_third:[/color]\r\n[i]dragon96 46\nvahalla 42[/i]\r\nzapi2007 33\r\nBOGTRO 22\r\nwestiepaw 17\r\nbbgun34 17\r\ncf249 15\r\nDojo 15\r\nYoshi 15\r\nfortenforge 14\r\nbluecarneal 11\r\njjfun1 11\r\nmeewhee009 9\r\nrd5493 7\r\ngaussintraining 5\r\n\r\n[u]Answers[/u]\r\nBOGTRO is Hidden #6: SIGHT\r\nvahalla is Unless, of course, I missed....of the picture. \r\nFor everyone else, just move two up on the list. This solution could have been derived by quoting. \r\n\r\nFuture hidden puzzles, if any, will be posted on PhireKaLk6781's blog.", "Solution_29": "Very nice waste of your ti... er, I mean, very amusing thing to do when bored." } { "Tag": [ "ratio", "geometry" ], "Problem": "1. How many postive integers less than 36 are equal to 4 times an odd integer?\r\n\r\n2. A long distance phone call costs 1 dollar and 80 cents for the first 3 minutes, and 40 cents for each additional minute. If and $x$ minute phone call costs 4 dollars and 20 cents , then $x$=?\r\n\r\n3.After $1/8$ of a ribbon is thrown away, the rest is cut into 2 pieces that are of the ratio 4:5. If 9 inches were thrown away, then what is the length of the shorter piece?\r\n\r\n4. What's the least number of squares of side length 2 inches required to cover, without overlapping, a larger square of a side length of 8 inches?\r\n\r\n5.In the games that the Bengals played the Lions, the Bengals won $2/3$ of their games,and lost $3/4$ of the other games. If the teams tied in 2 games, how many games did the Bengals PLAY the Lions?", "Solution_1": "[quote=\"nonie\"]1. How many postitve integers less than 36 are equal to 4 times an odd integer?\n\n2. A long distance phone call costs 1.80 for the first 3 minutes, and 40 cents for each additional minute. If an $x$ minute phone call costs 4.20, then $x$=?\n\n3.After $1/8$ of a ribbon is thrown away, the rest is cut into 2 pieces that are of the ratio 4:5. If 9 inches were thrown away, then what is the length of the shorter piece?\n\n4. What's the least number of squares of side length 2 inches required to cover, without overlapping, a larger square of a side length of 8 inches?\n\n5.In the games that the Bengals played the Lions, the Bengals won $2/3$ of their games,and lost $3/$ of the othergames. If the teams tied in 2 games, how many games did the Bengals tie the Lions?[/quote]\r\n\r\n[hide=\"number 1\"]1, 3, 5, and seven, each multiplied by four give us 4, 12, 20, and 28. Four integers. [/hide]\n\n[hide=\"number 2\"]After three minutes we have 4.20 - 1.80 0r 2.40. That gives us 2.40/.40 = 6 minutes left. 3+6=9=x. [/hide]\n\n[hide=\"number 3\"]The original ribbon must've been 72 inches. There was 63 inches left after 1/8 was taken away. The smaller piece was 4/9 or this, or 28 inches. [/hide]\n\n[hide=\"number 4\"]the area of the smaller square is four sq.in. The area of the big square is 64 sq.in. 64/4=16. [/hide]\r\n\r\nNumber 5: The bengals lost 3/what of the other games?", "Solution_2": "Those are all right, and as for number 5 sorry it was 3/4 i edited a few minutes ago.", "Solution_3": "[hide=\"number 5\"]2 games - it's right in the problem :rotfl: I think the question is wrong... [/hide]", "Solution_4": "[quote=\"236factorial\"][hide=\"number 5\"]2 games - it's right in the problem :rotfl: I think the question is wrong... [/hide][/quote]\r\n\r\nOMG i was copying the question from my notebook and my wrinting was really messy so i must have not realized what i was writing... It is supposed to be how many games did they PLAY.. So i guess I'll change that now.", "Solution_5": "[hide=\"new answer\"]They tied in 1/12 of the games. So they played 24 games. [/hide]", "Solution_6": "Yeah that's right.", "Solution_7": "[quote=\"nonie\"]1. How many postive integers less than 36 are equal to 4 times an odd integer?\n\n2. A long distance phone call costs 1 dollar and 80 cents for the first 3 minutes, and 40 cents for each additional minute. If and $x$ minute phone call costs 4 dollars and 20 cents , then $x$=?\n\n3.After $1/8$ of a ribbon is thrown away, the rest is cut into 2 pieces that are of the ratio 4:5. If 9 inches were thrown away, then what is the length of the shorter piece?\n\n4. What's the least number of squares of side length 2 inches required to cover, without overlapping, a larger square of a side length of 8 inches?\n\n5.In the games that the Bengals played the Lions, the Bengals won $2/3$ of their games,and lost $3/4$ of the other games. If the teams tied in 2 games, how many games did the Bengals PLAY the Lions?[/quote]\r\n[hide]\n1)1, 3, 5, 7 are odd integers that can be multiplied by 4 but are stilll less than 36.\n1x4=4, 3x4=12, 5x4=20, 7x4=28\nThe numbers are 4, 12, 20, 28.\n\n2)(4.20-1.80)/.40=each minute over 3 minutes.\n=2.4/.40=6 minutes over 3, (3+6) so the call was 9 minutes long\n\n3)if 9=1/8 of the total, the total was 72 inches. After being cut, the remainder was 72-9=63 inches. 4/9 of 63=28 inches\n\n4)the small square has area 4 sq in. The large square is 64 sq in. 64/4=16. It requires 16 small squares.\n\n5)2/3x + 3/12 x +2 = x\n11/12x+2=x\n1/12x=2\nx=24[/hide]", "Solution_8": "[hide]\n[quote]\n3)if 9=1/8 of the total, the total was 72 inches. After being cut, the remainder was 72-9=63 inches. 4/9 of 63=36 inches\n[/quote]\n\nb4k4ka,\nYour answersfor 1,2,4,+5 are correct\nbut for number 3 you made a simple computation error 4/9ths of 63 is 28 not 36[/hide]", "Solution_9": "[quote=\"nonie\"][hide]\n[quote]\n3)if 9=1/8 of the total, the total was 72 inches. After being cut, the remainder was 72-9=63 inches. 4/9 of 63=36 inches\n[/quote]\n\nb4k4ka,\nYour answersfor 1,2,4,+5 are correct\nbut for number 3 you made a simple computation error 4/9ths of 63 is 28 not 36[/hide][/quote]\r\noh oops, sorry i'll change that right away" } { "Tag": [ "Putnam", "algebra", "polynomial", "logarithms", "quadratics", "college contests" ], "Problem": "A [i]repunit[/i] is a positive integer whose digits in base $ 10$ are all ones. Find all polynomials $ f$ with real coefficients such that if $ n$ is a repunit, then so is $ f(n).$", "Solution_1": "There are a two-dimensional family of polynomials with this property. For nonnegative integers $ m$ and integers $ r\\ge 1\\minus{}m$, the polynomial $ f_{m,r}(x)\\equal{}\\frac{10^r(9x\\plus{}1)^m\\minus{}1}{9}$ takes the $ k$-digit repunit $ \\frac{10^k\\minus{}1}{9}$ to the $ (mk\\plus{}r)$-digit repunit $ \\frac{10^{mk\\plus{}r}\\minus{}1}{9}$. As it happens, these polynomials have integer coefficients when $ r$ is nonnegative.\r\n\r\nVerifying that these polynomials work is simple arithmetic. We must also show that these are the only such polynomials.\r\nFirst, note that all constant polynomials $ f(x)\\equal{}\\frac{10^r\\minus{}1}{9}$ with $ r\\ge 1$ work. We assume that $ f$ has degree at least $ 1$ in the following.\r\nFor a general polynomial $ f(x)\\equal{}a_mx^m\\plus{}a_{m\\minus{}1}x^{m\\minus{}1}\\plus{}\\cdots$, we have $ \\log(f(x))\\equal{}m\\log x\\plus{}\\log a_m\\plus{}o(1)$ as $ m\\to\\infty$. Applying $ f$ to the $ k$-digit repunit $ x_k\\equal{}\\frac{10^k\\minus{}1}{9}$, $ \\log(f(x_k))\\equal{}m(\\log(10^k)\\minus{}\\log(9))\\plus{}\\log(a_m)\\plus{}o(1)$\r\n$ \\equal{}mk\\minus{}m\\log(9)\\plus{}\\log(a_m)\\plus{}o(1)\\equal{}mk\\plus{}\\log(9^{\\minus{}m}a_m)\\plus{}o(1)$ as $ k\\to\\infty$ (We specialize to the base 10 logarithm here for convenience).\r\n\r\nWe want this to be the logarithm of some $ g(k)$-digit repunit, which is $ g(k)\\minus{}\\log(9)\\plus{}o(1)$. For large $ k$, we therefore have $ mk\\plus{}\\log(9^{\\minus{}m}a_m)\\minus{}g(k)\\plus{}\\log(9)\\plus{}o(1)\\equal{}mk\\plus{}\\log(9^{1\\minus{}m}a_m)\\minus{}g(k)\\plus{}o(1)\\equal{}0$. Since the possible values of $ mk\\plus{}\\log(9^{1\\minus{}m}a_m)\\minus{}g(k)$ differ by integers, they must eventually be equal to zero, and thus $ g_k\\equal{}mk\\plus{}r$ for $ r\\equal{}\\log(9^{1\\minus{}m}a_m)$. Two polynomials which are equal for infinitely many values (large enough repunits) are equal, and $ f(x)\\equal{}\\frac{10^r(9x\\plus{}1)^m\\minus{}1}{9}$ for some positive $ m$ and some $ r$. For the values to actually be repunits all the way down to $ f(1)$, we must have $ mk\\plus{}r\\ge 1$ for all $ k\\ge 1$, and thus $ r\\ge 1\\minus{}m$.", "Solution_2": "Hm...this problem was given at MOP this year...what was it doing on the Putnam? Am I right in thinking that the Putnam occurs after MOP? (i.e. it occurred sometime in the Fall of 2007, as opposed to the Spring of 2007?)", "Solution_3": "The Putnam happens at the beginning of December. I doubt the question-writers for the Putnam know anything about the questions used at MOP.", "Solution_4": "Oops, looks like I was mistaken.\r\n\r\nThe MOP problem was as follows:\r\n\r\nFind all quadratic polynomials with integer coefficients that transform repunits into repunits.\r\n\r\nMy apologies.", "Solution_5": "[hide=Solution]. Let $f$ be such a polynomial and observe that from the hypothesis it follows that there exists a sequence $\\left(a_{n}\\right)_{n \\geq 1}$ of positive integers such that $f\\left(\\frac{10^{n}-1}{9}\\right)=\\frac{10^{a n}-1}{9} .$ But this sequence $\\left(a_{n}\\right)_{n \\geq 1}$ cannot be really arbitrary:\nactually we can find precious information from an asymptotic study. Indeed, suppose that $\\operatorname{deg}(f)=d \\geq 1$. Then there exists a nonzero number $A$ such that $f(x) \\approx A x^{d}$ for large values of $x .$ Therefore $f\\left(\\frac{10^{n}-1}{9}\\right) \\approx \\frac{A}{9^{d}} \\cdot 10^{n d} .$ Thus $10^{a_{n}} \\approx \\frac{A}{9^{d-1}} \\cdot 10^{n d} .$ This shows that the sequence $\\left(a_{n}-n d\\right)_{n \\geq 1}$ converges to a limit $l$ such that $A=9^{d-1} \\cdot 10^{l}$. Because this sequence consists of integers, itbecomes eventually equal to the constant sequence $l$. Thus from a certain point we have $f\\left(\\frac{10^{n}-1}{9}\\right)=\\frac{10^{n d+l}-1}{9}$. If $x_{n}=\\frac{10^{n}-1}{9},$ we deduce that the equation $f(x)=\\frac{(9 x+1)^{d} \\cdot 10^{6}-1}{9}$ has infinitely many solutions, so $f(X)=\\frac{(9 X+1)^{d} \\cdot 10^{t}-1}{9}$\n\n\n", "Solution_6": "[quote=duongbgbg32][hide=Solution]. Let $f$ be such a polynomial and observe that from the hypothesis it follows that there exists a sequence $\\left(a_{n}\\right)_{n \\geq 1}$ of positive integers such that $f\\left(\\frac{10^{n}-1}{9}\\right)=\\frac{10^{a n}-1}{9} .$ But this sequence $\\left(a_{n}\\right)_{n \\geq 1}$ cannot be really arbitrary:\nactually we can find precious information from an asymptotic study. Indeed, suppose that $\\operatorname{deg}(f)=d \\geq 1$. Then there exists a nonzero number $A$ such that $f(x) \\approx A x^{d}$ for large values of $x .$ Therefore $f\\left(\\frac{10^{n}-1}{9}\\right) \\approx \\frac{A}{9^{d}} \\cdot 10^{n d} .$ Thus $10^{a_{n}} \\approx \\frac{A}{9^{d-1}} \\cdot 10^{n d} .$ This shows that the sequence $\\left(a_{n}-n d\\right)_{n \\geq 1}$ converges to a limit $l$ such that $A=9^{d-1} \\cdot 10^{l}$. Because this sequence consists of integers, itbecomes eventually equal to the constant sequence $l$. Thus from a certain point we have $f\\left(\\frac{10^{n}-1}{9}\\right)=\\frac{10^{n d+l}-1}{9}$. If $x_{n}=\\frac{10^{n}-1}{9},$ we deduce that the equation $f(x)=\\frac{(9 x+1)^{d} \\cdot 10^{6}-1}{9}$ has infinitely many solutions, so $f(X)=\\frac{(9 X+1)^{d} \\cdot 10^{t}-1}{9}$[/quote]bruh why 13 year bump\n\n", "Solution_7": "It's encouraged to post complete solutions to past problems. There is nothing wrong with bumping this thread. " } { "Tag": [ "probability", "LaTeX" ], "Problem": "Suppose that you are playing blackjack against a dealer. In a freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack? (blackjack means one of the card is an ace, and the other is either ten, jack, queen, or king)", "Solution_1": "[hide]The chances that one of you will get it is 1/13*4/13*2=8/169, so the probability that you won't get it is 161/169, so that squared is [b]25921/28561[/b][/hide]", "Solution_2": "but after I get 2 cards, the dealer will only have 50 cards from the deck to choose from?", "Solution_3": "Oh yeah.\r\n\r\nHint\r\n[hide]Use choose numbers[/hide]", "Solution_4": "but when calculating the probability of the 2nd person getting a black jack somewhat depends on what 2 cards the frist person got. Like the first person can get an Ace and something else and not form a black jack, or he can get two Aces, or he can get two Tens,jacks,queens,kings etc....I am confused", "Solution_5": "I am not sure if this is right:\r\n[hide] There are 52 choose 2 ways to randomly pick 2 cards out of 52 (I will just refer to a choose b as aCb, sorry I don't know how to Latex \"choose\"). Now, there are 4 aces in a deck and 16 other royals (including 10). Thus, there are 64 ways for the dealer to give the first person a blackjack. Therefore, the chance is 64/52C2 that the first person will recieve a blackjack. Assume that the first person was not given a blackjack. This chance is 1-64/52C2.\nSince there are 50 cards left and you want to give the second person 2 of them, there are 50C2 ways to do this. Now, there are still 4 aces and 16 other royals (including 10), thus there are 64 ways to give the second person a blackjack. Thus, the chance that the second person recieves a blackjack is 64/50C2, so the chance that he won't is 1-64/50C2. Therefore, the chance that none of them will recieve a blackjack is (1-64/50C2)(1-64/52C2). [/hide]", "Solution_6": "Another hint\r\n\r\n[hide]Find the probability that the [b]four[/b] cards dealt are the \"right \" cards[/hide]", "Solution_7": "by \"right \" cards I suppose you mean all four cards are not either ace, ten,jack,queen,king. But there is always a possibility that you get an Ace and a two, or a ten and a ten?", "Solution_8": "Call the 10 Jack, Queen King face cards. You need 2 face cards and 2 aces. Find the probability of that happening", "Solution_9": "ok, that takes care of probability of bothing getting a black jack..but the question is neither of them getting a black jack", "Solution_10": "subtract each of the individual probabilities from 1, and then multiply them together!", "Solution_11": "what if only one of them gets a black jack?" } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove or disprove that : If $ A$ are matrices with rational numbers, and $ A^5\\equal{}I$, and $ 1$ is not eigenvalue of $ A$, then the order of $ A$ must be divisible by $ 4$. Thanks in advance.", "Solution_1": "The characteristic polynomial $ p_A(x)$ has rational coefficients. Since we have ruled out $ 1$ as an eigenvalue, that means that all roots of $ p_A$ must also be roots of $ x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1.$ But that polynomial is irreducible over the rationals. Hence, we must have that $ p_A(x) \\equal{} (x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1)^k$ for some $ k\\in\\mathbb{N},$ from which we see that $ A$ is a $ 4k\\times 4k$ matrix.\r\n\r\nSecond question: does any such $ A$ actually exist? Can we find a $ 4\\times 4$ rational matrix with $ x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1$ as its characteristic polynomial?", "Solution_2": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28984]here[/url] for an almost identical problem. The solution there also works here.", "Solution_3": "Ah, yes. Should have thought of that.\r\n\r\n$ A\\equal{}\\begin{bmatrix}0&1&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\minus{}1&\\minus{}1&\\minus{}1&\\minus{}1\\end{bmatrix}$ works as an example.", "Solution_4": "$ A$ is the (transpose of the) [url=http://en.wikipedia.org/wiki/Companion_matrix]companion matrix[/url] of $ x^4 \\plus{} x^3 \\plus{} x^2 \\plus{} x \\plus{} 1 \\equal{} 0$. (A companion matrix is its own rational canonical form.)" } { "Tag": [ "calculus", "integration", "function", "derivative", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [0,\\infty]\\rightarrow \\mathbb{R}$ a derivable function, convex, with $ f(0)\\equal{}0$. \r\nProve that $ \\int_{0}^{x}f(t)dt \\leq \\frac{x^2}{2}f'(x)$, $ \\forall x \\in [0, \\infty)$.", "Solution_1": "I think it will be sufficient to show $ \\int\\limits_0^x {f(t)dt} \\le \\int\\limits_0^x {tf'(t)dt}$, which is a consequence of the hypotheses. It's not that difficult; the main thing being tested is whether you know about convex functions.", "Solution_2": "[quote=\"\u00a7outh\u00a7tar\"] $ \\int\\limits_0^x {f(t)dt} \\le \\int\\limits_0^x {tf'(t)dt}$[/quote]\r\n\r\nOk. I proved that, but it isn't very clear for me why that is a consequence of the hypotheses :P something to enlight me would be great. Thanks :)", "Solution_3": "Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.", "Solution_4": "[quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\r\n\r\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac{x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt\\minus{}\\int_{0}^{x}f(t)dt \\geq 0$. (1)\r\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\r\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P", "Solution_5": "[quote=\"Aalexandru\"][quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\n\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac {x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt \\minus{} \\int_{0}^{x}f(t)dt \\geq 0$. (1)\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]\r\n\r\nTo prove that $ \\int\\limits_0^x {tf'(t)dt} \\le \\frac {x^2}{2}f'(x) \\forall x \\in [0, \\infty)$.\r\n\r\nLet $ f(x)$denote $ rhs \\minus{} lhs$. then \r\n$ f'(x) \\equal{} \\frac {x^2}{2}f''(x) \\plus{} f'(x)x \\minus{} xf'(x) \\ge 0 \\forall x \\in [0, \\infty)$ since $ f$ is convex.\r\nthus $ f(x) \\ge f(0) \\equal{} 0$.", "Solution_6": "[quote=\"Aalexandru\"][quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\n\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac {x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt \\minus{} \\int_{0}^{x}f(t)dt \\geq 0$. (1)\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]\r\nThere is a proof already but I wanna say your argument here is incorrect", "Solution_7": "@bogdanno: Yes, you are right :oops: \r\n@ith_power: Your solution is quite simple and natural, but you haven't used the fact that $ f(0)\\equal{}0$. The $ f$ you are refering to in your proof is not the same with the $ f$ from the hypotheses. Is there a problem or that condition is not necessary?", "Solution_8": "[quote=\"ith_power\"][quote=\"Aalexandru\"][quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\n\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac {x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt \\minus{} \\int_{0}^{x}f(t)dt \\geq 0$. (1)\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]\n\nTo prove that $ \\int\\limits_0^x {tf'(t)dt} \\le \\frac {x^2}{2}f'(x) \\forall x \\in [0, \\infty)$.\n\nLet $ f(x)$denote $ rhs \\minus{} lhs$. then \n$ f'(x) \\equal{} \\frac {x^2}{2}f''(x) \\plus{} f'(x)x \\minus{} xf'(x) \\ge 0 \\forall x \\in [0, \\infty)$ since $ f$ is convex.\nthus $ f(x) \\ge f(0) \\equal{} 0$.[/quote]\r\n\r\nRemind me again why f is twice differentiable :?:", "Solution_9": "[quote=\"\u00a7outh\u00a7tar\"][/quote][quote=\"ith_power\"][quote=\"Aalexandru\"][quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\n\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac {x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt - \\int_{0}^{x}f(t)dt \\geq 0$. (1)\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]\n\nTo prove that $ \\int\\limits_0^x {tf'(t)dt} \\le \\frac {x^2}{2}f'(x) \\forall x \\in [0, \\infty)$.\n\nLet $ f(x)$denote $ rhs - lhs$. then \n$ f'(x) = \\frac {x^2}{2}f''(x) + f'(x)x - xf'(x) \\ge 0 \\forall x \\in [0, \\infty)$ since $ f$ is convex.\nthus $ f(x) \\ge f(0) = 0$.[/quote][quote=\"\u00a7outh\u00a7tar\"]\n\nRemind me again why f is twice differentiable :?:[/quote]\r\n\r\nI sensed that something was wrong with ith_power's proof :) \r\nOk. Other ideas?:P ( I've noticed that the function $ f(x)=x$ holds the equality. Does that help in some way?)", "Solution_10": "I showed you what the first step is. Next you note that the derivative is nondecreasing and use corresponding properties of the integral to obtain your upper bound.", "Solution_11": "[quote=\"ith_power\"][quote=\"Aalexandru\"][quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\n\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac {x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt \\minus{} \\int_{0}^{x}f(t)dt \\geq 0$. (1)\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]\n\nTo prove that $ \\int\\limits_0^x {tf'(t)dt} \\le \\frac {x^2}{2}f'(x) \\forall x \\in [0, \\infty)$.\n\n[b]Let $ f(x)$denote $ rhs \\minus{} lhs$. then [/b]\n$ f'(x) \\equal{} \\frac {x^2}{2}f''(x) \\plus{} f'(x)x \\minus{} xf'(x) \\ge 0 \\forall x \\in [0, \\infty)$ since $ f$ is convex.\nthus $ f(x) \\ge f(0) \\equal{} 0$.[/quote]\r\n\r\nwhy can [b]you let f=lhs-rhs[/b] ?", "Solution_12": "[quote=\"1234567a\"][/quote][quote=\"ith_power\"][quote=\"Aalexandru\"][quote=\"BlueVelvet\"]Well, how did you prove it? You must have used some information about $ f$, and therefore it is a consequence of the hypothesis.[/quote]\n\nLike this: $ \\int_{0}^{x}f(t)dt \\leq \\frac {x^2}{2}f'(x) \\Leftrightarrow \\int_{0}^{x}tf'(x)dt - \\int_{0}^{x}f(t)dt \\geq 0$. (1)\nBut $ f'$ is increasing, due to the fact that $ f$ is convex, so we have $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$.\nAnd, from (1), we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$. Ok, I used the hypothesis, but what do I do from here?:P[/quote]\n\nTo prove that $ \\int\\limits_0^x {tf'(t)dt} \\le \\frac {x^2}{2}f'(x) \\forall x \\in [0, \\infty)$.\n\n[b]Let $ f(x)$denote $ rhs - lhs$. then [/b]\n$ f'(x) = \\frac {x^2}{2}f''(x) + f'(x)x - xf'(x) \\ge 0 \\forall x \\in [0, \\infty)$ since $ f$ is convex.\nthus $ f(x) \\ge f(0) = 0$.[/quote][quote=\"1234567a\"]\n\nwhy can [b]you let f=lhs-rhs[/b] ?[/quote]\n\nI am very sorry , actually it was a typo. :oops: assume $ g(x)$ denote rhs-lhs.. then\n$ g'(x) = \\frac {x^2}{2}f''(x) + f'(x)x - xf'(x) \\ge 0 \\forall x \\in [0, \\infty)$ since $ f$ is convex.\nthus $ g(x)\\ge g(0) = 0$\n\n[quote=\"\u00a7outh\u00a7tar\"]\nRemind me again why f is twice differentiable :?:[/quote]\nsince $ f$ is convex, i used the fact that $ f''(x) \\ge 0$. :maybe: \n\n[quote=\"Aalexandru\"]\n@ith_power: Your solution is quite simple and natural, but you haven't used the fact that $ f(0) = 0$. The $ f$ you are refering to in your proof is not the same with the $ f$ from the hypotheses. Is there a problem or that condition is not necessary?[/quote]\r\nis it really necessary to have $ f(0)=0$? :maybe:", "Solution_13": "Question: is it true for a convex function $ f$ the inequality: $ f(t) \\leq t\\frac{f(x)}{x}$, $ \\forall t \\in [0,x]$ and $ x \\in [0, \\infty]$? ( I saw this result in another problem, but I don't know how to prove it).\r\nIf it's true, I think I have a solution for the problem, finally :)\r\nOk, so I want to prove that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$.\r\n$ f$ convex $ \\Rightarrow f(t) \\leq t\\frac{f(x)}{x} \\Rightarrow \\int_{0}^{x}f(t)dt \\leq \\frac{f(x)}{x}\\int_{0}^{x}tdt\\equal{}\\frac{xf(x)}{2}$ [b] (1) [/b] $ \\Rightarrow \\minus{}\\int_{0}^{x}f(t) \\geq \\minus{}\\frac{xf(x)}{2}$ \r\nOn the other hand $ \\int_{0}^{x}tf'(t)dt\\equal{}tf(t)|_{0}^{x} \\minus{}\\int_{0}^{x}f(t)dt \\geq xf(x)\\minus{}\\frac{xf(x)}{2}\\equal{}\\frac{xf(x)}{2}$ [b] (2) [/b]\r\nFrom (1) and (2) we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$\r\nBecause $ f$ is convex $ f'$ is non decreasing so $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$ and from here think it's obvious.", "Solution_14": "[quote]since $ f$ is convex, i used the fact that $ f''(x) \\ge 0$. :maybe: [/quote]\r\n\r\nhow do you know f'' exists?", "Solution_15": "@southstar [quote=\"Aalexandru\"]Let $ f: [0,\\infty]\\rightarrow \\mathbb{R}$ a [b]derivable[/b] function, convex .[/quote]", "Solution_16": "[quote=\"Aalexandru\"]@southstar [quote=\"Aalexandru\"]Let $ f: [0,\\infty]\\rightarrow \\mathbb{R}$ a [b]derivable[/b] function, convex .[/quote][/quote]\r\nYes but its not aparent that $ f$ is twice differentiable. :(", "Solution_17": "[quote=\"Aalexandru\"]Question: is it true for a convex function $ f$ the inequality: $ f(t) \\leq t\\frac {f(x)}{x}$, $ \\forall t \\in [0,x]$ and $ x \\in [0, \\infty]$? ( I saw this result in another problem, but I don't know how to prove it).\nIf it's true, I think I have a solution for the problem, finally :)\nOk, so I want to prove that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$.\n$ f$ convex $ \\Rightarrow f(t) \\leq t\\frac {f(x)}{x} \\Rightarrow \\int_{0}^{x}f(t)dt \\leq \\frac {f(x)}{x}\\int_{0}^{x}tdt \\equal{} \\frac {xf(x)}{2}$ [b] (1) [/b] $ \\Rightarrow \\minus{} \\int_{0}^{x}f(t) \\geq \\minus{} \\frac {xf(x)}{2}$ \nOn the other hand $ \\int_{0}^{x}tf'(t)dt \\equal{} tf(t)|_{0}^{x} \\minus{} \\int_{0}^{x}f(t)dt \\geq xf(x) \\minus{} \\frac {xf(x)}{2} \\equal{} \\frac {xf(x)}{2}$ [b] (2) [/b]\nFrom (1) and (2) we have that $ \\int_{0}^{x}tf'(t)dt \\geq \\int_{0}^{x}f(t)dt$\nBecause $ f$ is convex $ f'$ is non decreasing so $ \\int_{0}^{x}tf'(x)dt \\geq \\int_{0}^{x}tf'(t)dt$ and from here think it's obvious.[/quote]\r\n\r\nI ask again if this solution is correct ( my proof doesn't rely on $ f$ being twice differentiable) :)", "Solution_18": "Obviously, given a point $ x > 0$, there exists $ \\xi\\in[0,x]$ such that\r\n\\[ \\int_0^x tf'(t) dt \\equal{} f'(\\xi) \\int_0^x t dt \\equal{} \\frac {x^2}2 f'(\\xi).\r\n\\]\r\nSince $ f'$ is increasing, $ f'(\\xi) \\le f'(x)$, yielding\r\n\\[ \\int_0^x tf'(t) dt \\le \\frac {x^2}2 f'(x).\r\n\\]\r\nThat's all." } { "Tag": [ "LaTeX" ], "Problem": "Hi, I noticed some of these people's on the site are very good at making their own avatars. My avatar is also given as a present by one of users of this site and I wonder if there's any site that lets you to draw these comic figures.\r\n\r\nI'm asking this because I have a biology exam and in this project, I need to draw a comic strip on terms of DNA. But problem is, my drawing is -10^9999 worse than my $\\LaTeX$.\r\n\r\nThanks.", "Solution_1": "if you have this thing called paint, i think you can draw you own thing", "Solution_2": "Paint is kind of hard to use. My photoshop trial version expired, but at least some computers at our school have it.", "Solution_3": "You coud try [url=http://www.gimp.org/]The GIMP[/url] though it's kinda hard to use and takes a bit of practice (probably not so good if you need it for something immediate), but it seems incredibly powerful. It's almost like a GNU clone of Photoshop but a little less user friendly and more powerful.", "Solution_4": "This took about three seconds :P .", "Solution_5": "lol... :D", "Solution_6": "I draw my frames in MSPAINT and compile them with a horrible device :P . See my signature for shiny animations.", "Solution_7": "I didn't make this, but look! [img]http://img199.exs.cx/img199/2918/newbie0tr.gif[/img]" } { "Tag": [ "geometry" ], "Problem": "In the diagram, $ OA$ is a radius of the large circle and a diameter of the smaller circle. What is the number of square centimeters in the area inside the larger circle but outside the smaller circle? Express your answer in terms of $ \\pi$.\n[asy]pair O,A; A = (2,0); draw(O--A); draw(Circle(O,2)); draw(Circle((1,0),1));\nlabel(\"O\",O,W); label(\"A\",A,E); label(\"4 cm\",(1,0),N);[/asy]", "Solution_1": "The area of the larger circle is $ 16\\pi$, and the area of the smaller circle is $ 4\\pi$. Thus we have $ 16\\pi\\minus{}4\\pi\\equal{}\\boxed{12\\pi}$." } { "Tag": [ "analytic geometry", "vector" ], "Problem": "Prove that it will always take even number of moves for a knight to move back to it's original position).\r\n\r\nEDIT-oops! I meant to post this in the Intermediate Forum", "Solution_1": "Extension:\r\n\r\nProve that if lattice points on the coordinate plane are made into squares, and the knight starts at (0, 0), then the sum of the coordinates of a square and the number of moves it takes to get there will be of the same parity.", "Solution_2": "This is really easy, since each move changes the color of the square the knight is on, if the knight starts on square of color \"a\", next move will be to a square of color \"b\", and then back to \"a\", so the knight will be on a space of the same color only after an even number of moves. Therefore, it can only get back to the original square after an even number of moves.", "Solution_3": "probability1.01 wrote:Extension:\n\nProve that if lattice points on the coordinate plane are made into squares, and the knight starts at (0, 0), then the sum of the coordinates of a square and the number of moves it takes to get there will be of the same parity.\n\n\n\nHere's my solution:\n\n\n\n[hide]A knight's move can be represented as the vector (+-1, +-2) or the vector (+-2, +-1). Either way you're either adding +-1 or +-3 to the sum of the knight's coordinates each move. Thus, each move, the parity of the sum of the coordiantes changes. Both the number of moves and the sum of the coordinates start off at the same parity (even). Each move, the parity of the number of moves changes and so does the parity of the sum of the coordinates.[/hide]", "Solution_4": "Yasha's solution is correct (and surprisingly, the first one I thought of, despite having been a chess player for 4-5 years), and Demon's answer is the one that chess coaches would give (and it also proves the extension). Good job!", "Solution_5": "Interesting!\r\nHeres another one:\r\n\r\nHow many different ways are there to arrange all the chess pieces on the 8x8 board:\r\n\r\n1. Not following the rules (like a lone bishop, two kings adjacent, all pawns on a rank, etc.)\r\n\r\n2. (very hard) Yes following the rules.\r\n\r\n(I thought i had the first one already figured out, but i realized i was slightly off)", "Solution_6": "The first is [hide]just 64!/32! then divide by permutations of equivalent placements (e.g., bishop and bishop have two ways of being placed).[/hide]\n\n\n\nThe second is incredibly hard, even rather challenging to program for a computer to do.", "Solution_7": "The first one is very big: [hide]64!/32!/(8!*2!*2!*2)^2 (8 for the pawns, 2 for the each of the rooks, knights, and bishops, this is for each color).[/hide]\n\n\n\nThis is equal to [hide]46,347,266,955,878,096,411,920,459,823,285,670,400,000.[/hide]\n\n\n\nGo TI-89!\n\n\n\n\n\nI think you would have to calculate the entire game tree in order to calculate the second one. I don't think it's possible to be able to quickly recognize every impossible situation. Some impossible situations are obvious, like two white bishops both on white spaces, but there are probably others for which it is harder to tell.", "Solution_8": "[quote=\"Hexaditidom\"]Interesting!\nHeres another one:\n\nHow many different ways are there to arrange all the chess pieces on the 8x8 board:\n\n1. Not following the rules (like a lone bishop, two kings adjacent, all pawns on a rank, etc.)\n\n2. (very hard) Yes following the rules.\n\n(I thought i had the first one already figured out, but i realized i was slightly off)[/quote]\r\n\r\nIs it all the chess pieces for a single side (16 white pieces) or ALL the pieces, black and white (32 pieces)?", "Solution_9": "Whoops! I made a mistake in the previous post. I divided by the equivalent placements of only one color of pieces. I'll go edit it.", "Solution_10": "Yeah. You would have discount everything with the pawns in the way, like the b and d pawns must be moved before white queens bishop can move...etc, situations with stalemate or checkmate, situations where both kings are in check, situations with pawns on the back rows...there are just so many. It would be quite hard." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find last 2 digits of $1234^23$ i think its 04 but i'm not sure :blush: \r\nPS: the power is 23 ...", "Solution_1": ":?: :?: :?: :( \r\nJust tell me if it's correct or not... plz", "Solution_2": "Yes, it's correct.", "Solution_3": "10x" } { "Tag": [ "floor function", "induction", "More Sequences" ], "Problem": "The infinite sequence of 2's and 3's \\[\\begin{array}{l}2,3,3,2,3,3,3,2,3,3,3,2,3,3,2,3,3, \\\\ 3,2,3,3,3,2,3,3,3,2,3,3,2,3,3,3,2,\\cdots \\end{array}\\] has the property that, if one forms a second sequence that records the number of 3's between successive 2's, the result is identical to the given sequence. Show that there exists a real number $r$ such that, for any $n$, the $n$th term of the sequence is 2 if and only if $n = 1+\\lfloor rm \\rfloor$ for some nonnegative integer $m$.", "Solution_1": "We claim that $ m$ th $ 2$ appears as $ [(2 \\plus{} \\sqrt {3})(m \\minus{} 1)] \\plus{} 1$th term is this sequence\r\n\r\nProve by induction,the foundation is trivial,and for every positive integer $ \\le m$ is true.\r\n\r\nLet's take a look at thr case of $ m \\plus{} 1$\r\n\r\n$ \\exists$ $ r,r\\in N$ s.t. $ [(2 \\plus{} \\sqrt {3})(r \\minus{} 1)] \\plus{} 1\\le m < [(2 \\plus{} \\sqrt {3})r] \\plus{} 1$\r\n\r\nAfter some simple calculation.We obtain $ r \\equal{} [\\frac {m}{2 \\plus{} \\sqrt {3}}] \\plus{} 1$\r\n\r\nThis yields the number of $ 2$ $ 3$ in the first $ m$ terms are $ r$ and $ m \\minus{} r$ respectively\r\n\r\nSo the position of $ m \\plus{} 1$th $ 2$ is $ 2r \\plus{} 3(m \\minus{} r) \\plus{} m \\plus{} 1 \\equal{} 4m \\minus{} r \\plus{} 1 \\equal{} [(2 \\plus{} \\sqrt {3})m] \\plus{} 1$ which complete yhe \r\n\r\ninduction.\r\n\r\nQED", "Solution_2": "This is sequence [url=http://www.research.att.com/~njas/sequences/A007538]A007538[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url]." } { "Tag": [ "Vieta", "AMC", "USA(J)MO", "USAMO" ], "Problem": "There are no rational numbers so that:\r\n\r\n$x^{2005}-3x-1=0$", "Solution_1": "Here's a start maybe?:\r\n[hide]\nThrough Vieta's formulae:\n\n$(r_{1}+r_{2}+r_{3}+...+r_{2005}) = 1$\n\n$(r_{1}r_{2}r_{3}r_{4}...r_{2005}) = 1$\n\nTherefore\n$r_{1}+r_{2}+r_{3}+...+r_{2005}= r_{1}r_{2}r_{3}r_{4}...r_{2005}$\n\nand then somehow that fact should prove that none of $r_{i}$ are rational?\n[/hide]\r\n\r\nEDIT: The fact that I jumped to this conclusion before thinking of the rational root theorem.... oh man that is not good.", "Solution_2": "If there exists a rational root $x=\\frac{p}{q},\\ \\gcd (p,q)=1$, then $p|(-1)\\wedge q|1$. These conditions do not lead to roots. And the conclusion follows.", "Solution_3": "rational root theorem...", "Solution_4": "[quote=\"max_tm\"]Therefore\n$r_{1}+r_{2}+r_{3}+...+r_{2005}= r_{1}r_{2}r_{3}r_{4}...r_{2005}$\n\nand then somehow that fact should prove that none of $r_{i}$ are rational?[/quote]\r\n\r\nActually, a variant of that was a USAMO problem last year. This fact alone is insufficient; consider\r\n\r\n$r_{1}= r_{2}= ... = r_{2003}= 1, r_{2004}= a, r_{2005}= b$\r\n\r\nThen we merely require that $2003+a+b = ab \\implies (a-1)(b-1) = 2004$ which has infinite rational (not to mention several integer!) solutions, although none of them satisfy the original two conditions.", "Solution_5": "You are right.", "Solution_6": "[hide]From the Rational Roots Theorem, the only possible rational roots are $\\pm1$. Neither work, so there are no rational number solutions.[/hide]" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Help me to calculate the following integral: $ \\int_{\\minus{}\\frac \\pi 2 }^{\\frac \\pi 2}\\frac{\\sin{\\theta}\\sin{2 \\theta}}{\\sqrt{R^2\\plus{}|x|^2\\minus{}2R|x|\\cos{\\theta}}}d \\theta$", "Solution_1": "Maybe I'm thinking about this in the complete wrong way but this reminds me of Poisson's integral formula.", "Solution_2": "hello, here can you read somthing about the Poisson's integral formula:\r\nhttp://msvlab.hre.ntou.edu.tw/paper/tmes/ijmest2006.pdf\r\nSonnhard.", "Solution_3": "I've seen two derivations of it: one manipulating Cauchy's integral formula with complex conjugates and another one exploiting properties of Fourier series." } { "Tag": [], "Problem": "Let us call P(n) - the product of all digits of number n (in decimal notation).\r\nFor example, P(1243)=1*2*4*3=24; P(198501243)=0.\r\nLet us call n to be a good number, if (p(n)<>0) and (n mod P(n)=0).\r\nLet us call n to be a perfect number, if both n and n+1 are good numbers.\r\n\r\nYou are to write a program, which, given the number K(k<=100000), counts all such\r\nnumbers n that n is perfect and n contains exactly K digits in decimal notation. Time limit 0.2 sec", "Solution_1": "If K>1, then both P(n) and P(n+1) are multiples of (n with the last digit removed)..what does that tell you about all of ns digits except the last?\r\nAfter that, its just checking all possibilities for the last digit." } { "Tag": [ "\\/closed" ], "Problem": "whenever i use the forums now, all the text is magnified really big... how do i make it regular again\r\nand sometimes when i click on a link the text will be ez to read for a sec\r\n\r\nthx", "Solution_1": "what skin and what browser do you use?", "Solution_2": "erm.. internet explorer most of the time \r\nim not too sure i know what u mean by skin?", "Solution_3": "Click [url=http://www.mathlinks.ro/Forum/usercp.php]here[/url]. The first option (the dropdown menu) is the Board Layout option. You can choose between MathLinks and AoPS. A picture would really help at this point to understand what's going on ... :? (use the Prt Scr button on your keyboard to take a picture of your screen).", "Solution_4": "erm.. the print screen didnt work... \r\nmaybe its an ie problem not an aops problem", "Solution_5": "[quote=\"sheepwarrior\"]erm.. the print screen didnt work... \nmaybe its an ie problem not an aops problem[/quote]What skin do you use? In IE 6.0 go to View -> Text Size and try to find the appropriate measure ;)", "Solution_6": "oh thx!! it worked! but still whenever i click a link the text shrinks...\r\n\r\n\r\nedit: wait nvm \r\n\r\nthx again! :D\r\n\r\n2nd edit: wait.... it still shrinks when i click a link.. it goes back to normal once the page comes up but still...\r\nallnd when i click reveal hidden content all teh text spreads apart or overlaps ?/???" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Hi I'm having a hard time finding the solution to the following problem. In triangle ABC, point D is on AC such that AB=AD. If the measure of angle ABC - measure of angle ACB = 30, find the measure of CBD.\r\n(Note that the triangle is NOT drawn to scale.)\r\n\r\n[img]http://i13.photobucket.com/albums/a260/ContraForce/Triangle.jpg[/img]\r\n\r\n\r\nSince AB=AD then by Isosceles Triangle Theorem mADB = mABD.\r\nThen using Euclidean Parallel Postulate, I construct line EA such that EA is parallel to CB and EC is perpendicular to CB, and construct AG in the same fashion so that AG is perpendicular to CB. Then a quadralateral is created with sides CG, AG,EA,EC, and diagonal AC. Using the converse to the alternate interior angles theorem (CAIA) we see that mCEA = mACG, and since mECG = mGAE = 90 this implies that mACG=mACE by angle addition axiom. Then by AIA , EC is parallel to AG. WE can show trianlge EAC is congruent to triangle GCA using Angle Side Angle. At this point the work I'm doing doesn't seem to be leading me to a solution for the question. I've spent a lot of time on this problem and its driving me crazy so any help is greatly appreciated. Please note that I am restricted to certain Theorems, axioms, etc.. ( i.e. I cannot use things we haven't learned in class as a justification.)", "Solution_1": "I'm sorry, I meant to post this in the unsolved section, I do know that there is a solution I just can't seem to get to it. Can a mod please move this thread? Thanks", "Solution_2": "No extra lines, no nothing... \r\nPut $\\angle CBA=\\beta$ and $\\angle ACB=\\gamma$. $\\angle BDA=\\angle ABD=\\delta$, and suppose that position of points on line $AC$ is $A-D-C$, it's the same for other position. Then $\\delta = \\gamma+\\beta-\\delta$ (from triangle $CBD$), and $\\beta-\\gamma = 30^{o}$. From this $\\beta-\\delta =\\angle CBD =15^{o}$.\r\nBye", "Solution_3": "Thank you very much :)" } { "Tag": [], "Problem": "Find all natural numbers $x,y$ such that\r\n\\[6(x!+3)=y^2+5\\]", "Solution_1": "[hide] Rewrite the equation as 6x!+13=y^2. Starting with x=5, 6x!+3 ends in 3. However, the last digits of a square can only be 0, 1, 4, 5, 6, and 9. So, x must be 4 or less. We can simply check x=1, 2, 3, and 4, and we see that the solutions are (2, 5) and (3,7). [/hide]", "Solution_2": "[quote=\"s0mp\"][hide] Rewrite the equation as 6x!+13=y^2. Starting with x=5, 6x!+3 ends in 3. However, the last digits of a square can only be 0, 1, 4, 5, 6, and 9. So, x must be 4 or less. We can simply check x=1, 2, 3, and 4, and we see that the solutions are (2, 5) and (3,7). [/hide][/quote]\r\n\r\nCouple of typos, but I think that's it. $6x!+18$, not $6x!+13$ or $6x!+3$ and so it ends in $8$, which also works with your argument. Nice solution.", "Solution_3": "[quote=\"paladin8\"][quote=\"s0mp\"][hide] Rewrite the equation as 6x!+13=y^2. Starting with x=5, 6x!+3 ends in 3. However, the last digits of a square can only be 0, 1, 4, 5, 6, and 9. So, x must be 4 or less. We can simply check x=1, 2, 3, and 4, and we see that the solutions are (2, 5) and (3,7). [/hide][/quote]\n\nCouple of typos, but I think that's it. $6x!+18$, not $6x!+13$ or $6x!+3$ and so it ends in $8$, which also works with your argument. Nice solution.[/quote]\r\ns0mp is correct, paladin8 you got it wrong because $6(x!+3)=6x!+18=y^2+5$ which means that $6x!+13=y^2$. and not only that he is doing $\\mod 10$ so $6x!+13\\equiv 6x!+3$", "Solution_4": "[quote=\"Palytoxin\"][quote=\"paladin8\"][quote=\"s0mp\"][hide] Rewrite the equation as 6x!+13=y^2. Starting with x=5, 6x!+3 ends in 3. However, the last digits of a square can only be 0, 1, 4, 5, 6, and 9. So, x must be 4 or less. We can simply check x=1, 2, 3, and 4, and we see that the solutions are (2, 5) and (3,7). [/hide][/quote]\n\nCouple of typos, but I think that's it. $6x!+18$, not $6x!+13$ or $6x!+3$ and so it ends in $8$, which also works with your argument. Nice solution.[/quote]\ns0mp is correct, paladin8 you got it wrong because $6(x!+3)=6x!+18=y^2+5$ which means that $6x!+13=y^2$. and not only that he is doing $\\mod 10$ so $6x!+13\\equiv 6x!+3$[/quote]\r\n\r\nOh, I see. Yeah I kept the $5$ on the other side when using the $6x!+18$. I thought he had a typo because if you use $6x!+13 = y^2+5$ you have no solutions for $x \\ge 5$ either. My bad.", "Solution_5": "we can work (mod 8) , if x>3 we have y\u00b2=5(mod 8) witch is impossible ,so x is 0,1,2 or 3, we check that 2 and 3 works..." } { "Tag": [ "Columbia", "AMC 10", "AMC" ], "Problem": "Is there an improvement award for the AMCs ?", "Solution_1": "Yes, but the award goes to schools based on team score, not to individuals.\r\nRecall that the school team score is the sum of the top 3 scores from a school.\r\n\r\nFrom the award letter that goes to schools:\r\n\"The American Mathematics Competitions program of the Mathematical Association of America annually sponsors an pair of challenging mathematics contests at the high school level, the AMC 10 and the AMC 12. Your school and mathematics teachers have participated in these challenging contests for at least the last two years. As a result of your school's improvement in performance from the 2006 AMC contests to the 2007 AMC contests, your school is the State/Provincial Winner of the 2007 Pedagoguery Award. One award goes to to the school in each state or province with the best difference in team score on the AMC 10 and also to to the school in each state or province with the best difference in team score on the AMC 12 from 2006 to 2007. The actual award is a set of die-cast aluminum mathematical models of the five platonic solids and three additional truncated platonic solids. This award is provided courtesy of Pedagoguery Software Inc. of 4446 Lazelle Avenue, Terrace, British Columbia, Canada V8G 1R8. Through the generosity of Pedagoguery Software, Inc, the American Mathematics Competitions is making this award to stimulate continued year-to-year interest in the AMC contests and to reward teachers who coach their teams to greater levels of achievement on the contests and in mathematics generally. \"\r\n\r\nA brief description of the award is on page 8 of the AMC 10/AMC 12 Teachers' Manual in the section on awards.\r\n\r\nNo, we do not have an award for individual students.\r\n\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_2": "Which is a shame, because next year I have a shot at winning the Most Unimproved award with a score of -150." } { "Tag": [ "inequalities", "algebra", "IMO Shortlist" ], "Problem": "If $a,b,c$ are the sides of a triangle, prove that\n\\[\\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}+\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}}+\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\leq 3 \\]\n\n[i]Proposed by Hojoo Lee, Korea[/i]", "Solution_1": "[quote=\"mattilgale\"]If $ a$, $ b$, $ c$ are the sides of a triangle, prove that\n\\[ \\sum_{\\mbox{cyc}}\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\leq 3 \\]\n[/quote]\r\n$ \\sum_{\\mbox{cyc}}\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\leq 3\\Leftrightarrow\\sum_{cyc}\\left(1-\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{\\sqrt{ab}-\\sqrt{c(a+b-c)}}{\\sqrt a+\\sqrt b-\\sqrt c}\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{(a-b)(a-c)}{(\\sqrt{bc}+\\sqrt{a(b+c-a)})(\\sqrt b+\\sqrt c-\\sqrt a)}\\geq0.$\r\nLet's $ a\\geq b\\geq c.$ Then \r\n$ \\frac{1}{(\\sqrt{bc}+\\sqrt{a(b+c-a)})(\\sqrt b+\\sqrt c-\\sqrt a)}\\geq\\frac{1}{(\\sqrt{ac}+\\sqrt{b(a+c-b)})(\\sqrt a+\\sqrt c-\\sqrt b)}$\r\nsince, $ \\sqrt{ac}+\\sqrt{b(a+c-b)}\\geq\\sqrt{bc}+\\sqrt{a(b+c-a)}$ and $ \\sqrt a+\\sqrt c-\\sqrt b\\geq\\sqrt b+\\sqrt c-\\sqrt a.$\r\nThus, your inequality is proved. :)", "Solution_2": "My solution is the same with yours :wink:", "Solution_3": "I don't understand this one [quote]$\\sum_{cyc}\\left(1-\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\right)\\geq0 \\Leftrightarrow\\sum_{cyc}\\frac{\\sqrt{ab}-\\sqrt{c(a+b-c)}}{\\sqrt a+\\sqrt b-\\sqrt c}\\geq 0$[/quote]", "Solution_4": "[quote=\"zibi\"]I don't understand this one [quote]$\\sum_{cyc}\\left(1-\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\right)\\geq0 \\Leftrightarrow\\sum_{cyc}\\frac{\\sqrt{ab}-\\sqrt{c(a+b-c)}}{\\sqrt a+\\sqrt b-\\sqrt c}\\geq 0$[/quote][/quote]\r\nYou are right:\r\n$\\sum_{cyc}\\left(1-\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\right)\\geq0 \\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{\\sqrt{ab}-\\sqrt{c(a+b-c)}}{\\left(\\sqrt a+\\sqrt b-\\sqrt c\\right)\\left(\\sqrt a+\\sqrt b+\\sqrt c+\\sqrt{a+b-c}\\right)}\\geq 0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{(a-b)(a-c)}{(\\sqrt{bc}+\\sqrt{a(b+c-a)})(\\sqrt b+\\sqrt c-\\sqrt a)\\left(\\sqrt a+\\sqrt b+\\sqrt c+\\sqrt{b+c-a}\\right)}\\geq0.$\r\nBut $\\sqrt a+\\sqrt b+\\sqrt c+\\sqrt{a+c-b}\\geq\\sqrt a+\\sqrt b+\\sqrt c+\\sqrt{b+c-a}$ true too. Thank you. :)", "Solution_5": "What was the idea behind the first (mistaken) simplification?", "Solution_6": "[quote=\"mattilgale\"]If a, b, c are sides of a triangle prove that\n\\[\\sum_{\\mbox{cyc}}\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\leq 3 \\]\n[/quote]\r\n\r\n$\\sum \\frac{\\sqrt{-a+b+c}}{-\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}$\r\n$\\leq \\sqrt{3\\sum \\frac{-a+b+c}{(-\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^{2}}}$ by Cauchy's\r\n$=\\sqrt{3\\sum \\frac{2x^{2}+2xy+2xz-2yz}{4x^{2}}}$\r\n(where $x=\\frac{-\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{2}$, and so on)\r\n$=\\sqrt{3\\sum \\frac{x^{2}+xy+xz-yz}{2x^{2}}}$\r\n$=\\sqrt{3 \\frac{3x^{2}y^{2}z^{2}+\\sum_{sym}x^{3}y^{2}z-\\sum_{cyc}x^{3}y^{3}}{2x^{2}y^{2}z^{2}}}$\r\n$\\leq \\sqrt{3 \\frac{6x^{2}y^{2}z^{2}}{2x^{2}y^{2}z^{2}}}=3$ by Schur's for $xy,yz,zx$.", "Solution_7": "Let me post the solution I had for this problem (I was hoping that I would get to coordinate this problem if it were chosen in the competition last year :) )\r\n\r\nSince $ a,b,c$ are the sides of a triangle, it is clear that $ x=\\sqrt a, y=\\sqrt b, z= \\sqrt c$ are also the sides of a triangle (so the numbers $ -x+y+z$, $ x-y+z$ and $ x+y-z$ are positive). We first apply a Cauchy to get rid of the square roots in the numerators of the fractions \r\n\\[ \\left(\\sum \\frac{ \\sqrt{-x^{2}+y^{2}+z^{2}}}{-x+y+z}\\right)^{2}\\leq 3 \\sum \\frac{-x^{2}+y^{2}+z^{2}}{(-x+y+z)^{2}}.\\] Therefore what we want to prove is \r\n\\[ \\sum \\frac{-x^{2}+y^{2}+z^{2}}{(-x+y+z)^{2}}\\leq 3. \\]\r\nMoving all into one side (and distributing the 3 to each of the fractions equally) we get\r\n\\[ 0 \\leq \\sum \\frac{ x^{2}+yz-xy-xz }{ (-x+y+z)^{2}}= \\sum \\frac 1{(-x+y+z)^{2}}(x-y)(x-z) , \\]\r\nwhich of course is nothing else than Vornicu-Schur, because if $ x \\geq y\\geq z$ then $ \\frac 1{(-x+y+z)^{2}}\\geq \\frac 1{(x-y+z)^{2}}\\geq \\frac 1{(x+y-z)^{2}}$ and we are done.", "Solution_8": "Who is the proposer of that problem ? It's from Korea according to IMO compedium so Hojoo Lee ?", "Solution_9": "[quote=\"silouan\"]Who is the proposer of that problem ? It's from Korea according to IMO compedium so Hojoo Lee ?[/quote]\r\n\r\nAccording to http://ultrametric.googlepages.com/proposals.pdf\r\nthis problem indeed has been constructed by Hojoo Lee.", "Solution_10": "Is It True ? \r\n\r\n[tex]$ \\frac{\\sqrt[n]{a\\plus{}b\\minus{}c}}{\\sqrt[n]{a}\\plus{}\\sqrt[n]{b}\\minus{}\\sqrt[n]{c}}\\plus{} \\frac{\\sqrt[n]{b\\plus{}c\\minus{}a}}{\\sqrt[n]{b}\\plus{}\\sqrt[n]{c}\\minus{}\\sqrt[n]{a}}\\plus{} \\frac{\\sqrt[n]{c\\plus{}a\\minus{}b}}{\\sqrt[n]{c}\\plus{}\\sqrt[n]{a}\\minus{}\\sqrt[n]{b}}\\leq 3$ ? :D", "Solution_11": "Put $ a\\equal{}x\\plus{}y, b\\equal{}x\\plus{}z, c\\equal{}y\\plus{}z$. Then we should prove that $ \\sum\\frac{\\sqrt{2x}}{\\sqrt{x\\plus{}y}\\plus{}\\sqrt{x\\plus{}z}\\minus{}\\sqrt{y\\plus{}z}}\\le 3$.\r\n\r\nBy Cauchy, we get that\r\n\r\n$ LHS\\le \\frac{(\\sqrt[4]{2x}\\plus{}\\sqrt[4]{2y}\\plus{}\\sqrt[4]{2z})^2} {\\sqrt{x\\plus{}y}\\plus{}\\sqrt{x\\plus{}z}\\plus{}\\sqrt{y\\plus{}z}}\\le 3\\frac{\\sqrt{2x}\\plus{}\\sqrt{2y}\\plus{}\\sqrt{2z}} {\\sqrt{x\\plus{}y}\\plus{}\\sqrt{x\\plus{}z}\\plus{}\\sqrt{y\\plus{}z}}\\equal{} 3\\frac{2\\sqrt{x}\\plus{}2\\sqrt{y}\\plus{}2\\sqrt{z}} {\\sqrt{2x\\plus{}2y}\\plus{}\\sqrt{2x\\plus{}2z}\\plus{}\\sqrt{2y\\plus{}2z}}$.\r\n\r\n$ \\sqrt{x}\\plus{}\\sqrt{y}\\le \\sqrt{2x\\plus{}2y}$, because its squared is true: $ x\\plus{}y\\plus{}2\\sqrt{xy}\\le x\\plus{}y\\plus{}x\\plus{}y$ (we have positive numbers here), so numerator$ \\le$denominator. Thus\r\n\r\n$ 3\\frac{2\\sqrt{x}\\plus{}2\\sqrt{y}\\plus{}2\\sqrt{z}} {\\sqrt{2x\\plus{}2y}\\plus{}\\sqrt{2x\\plus{}2z}\\plus{}\\sqrt{2y\\plus{}2z}}\\le 3$.\r\n\r\nWe're done.", "Solution_12": "Uh.. sorry, my solution above is wrong..", "Solution_13": "we take $ a\\equal{}x^{2}$ $ b\\equal{}y^{2}$ $ c\\equal{}z^{2}$\r\nthen $ x,y,z$ are also side of triangle\r\nso there are exist uniqe triple such that $ x\\equal{}a'\\plus{}b' y\\equal{}b'\\plus{}c' z\\equal{}c'\\plus{}a'$\r\nso only condition for triples $ (a',b',c')$ should be $ a',b',c' \\in R^\\plus{}$ \r\nthe we rewrite our inequality\r\n$ \\sum\\frac{\\sqrt{4a'^2\\minus{}2(a'\\minus{}b')(a'\\minus{}c')}}{2a'}$\r\nbut we have $ 4a'^2\\minus{}2(a'\\minus{}b')(a'\\minus{}c')$ is negatif for example triple $ (1,10,2)$ \r\nwhere is my fault? can anyone tell me", "Solution_14": "I find my mistake.I use Ozandro's Lemma for this solution but it is not true", "Solution_15": "[quote=\"mestav\"]I find my mistake.I use Ozandro's Lemma for this solution but it is not true[/quote]\r\nwhat is Ozandro's Lemma?", "Solution_16": "other solution:\nWe have that \n\\[ \\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} =\\sqrt{1-2\\frac{(\\sqrt{c}-\\sqrt{a})(\\sqrt{c}-\\sqrt{b})}{(\\sqrt{a}+\\sqrt{b}-\\sqrt{c})^2}} \\le 1-\\frac{(\\sqrt{c}-\\sqrt{a})(\\sqrt{c}-\\sqrt{b})}{(\\sqrt{a}+\\sqrt{b}-\\sqrt{c})^2} .\\]\nSo we prove that if $x=\\sqrt{b}+\\sqrt{c}-\\sqrt{a}>0$, $y=\\sqrt{a}+\\sqrt{c}-\\sqrt{b}>0$, $z=\\sqrt{a}+\\sqrt{b}-\\sqrt{c}>0$ then \\[ \\sum x^{-2}(x-y)(x-z) \\ge 0. \\]\nIt is obviously true by Schur's inequality.", "Solution_17": "Hey mathuz Schur's Inequality tells $r>0$ but you have $r=-2$", "Solution_18": "[quote=\"TripteshBiswas\"]Hey mathuz Schur's Ineuality tells r>0 but you have r=-2[/quote]\nSee here the proof when $r$ is negative:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=415872\nthe last post", "Solution_19": "thanks mudok it was not known to me when r is negative", "Solution_20": "[b]Lemma[/b]:\n$\\sum_{cyc} \\frac {(x-z)(x-y)}{x^2}\\ge 0$. Since $2(x-y)(x-z)=(x-y)^2+(x-z)^2-(y-z)^2$ the inequality is reduced to the inequality $\\sum_{cyc} (x-y)^2(\\frac {1}{x^2}+\\frac {1}{y^2}-\\frac{1}{z^2})$ where we deonte $S_x=\\frac {1}{x^2}+\\frac {1}{y^2}-\\frac{1}{z^2}$ this is indeed true by the SOS method just assume WLOG $x\\ge y\\ge z$ and put $x-y=a,y-z=b$ the inequality becomes $a^2(S_z+S_y)+b^2(S_x+S_y)+2abS_y$ but we can easily see that $S_z+S_y,S_x+S_y,S_y$ are all greater than $0$ so our inequality is proven.\n[b]Back to the problem[/b]\nDenote by $S_a,S_b,S_c$ those expressions in the inequality we need to prove. Now notice that the inequality $\\sum_{cyc}(1-(S_a)^2)\\ge 0$ reduces to the lemma for numbers $\\sqrt{a}+\\sqrt{b}-\\sqrt{c},\\sqrt{a}-\\sqrt{b}+\\sqrt{c},-\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$.\nThis gives $\\sum_{cyc} (S_a)^2\\le 3$ and from here we easily get the desiret result", "Solution_21": "My solution is easy\n\nSince $a,b,c$ are the sides of a triangle, we can set\n$a=(q+r)^2$, $b=(p+r)^2$, $c=(p+q)^2$\n\nThus the inequality can be rewritten as \n$\\sum_{cyclic} \\frac{\\sqrt{2p^2+2pr+2pq-2qr}}{p} \\le 6$\n\nThus is it is sufficient to show that\n$\\sum_{cyclic} \\frac{2p^2+2pr +2pq -2qr}{p^2} \\le 12$\n$\\iff \\sum_{cyclic} \\frac{pr+pq-qr}{p^2} \\le 3$\n$\\iff \\sum_{cyclic} q^2r^2(pr+pq-qr) \\le 3p^2q^2r^2$\n\nThis is third degree Schur's inequality for $pq,qr,rp$\n\nHence proved", "Solution_22": "Let [b]a,b,c>0[/b],and [b]a+b+c=3[/b] prove that:\n${\\frac {a\\sqrt {\\sqrt {b}+\\sqrt {c}}}{\\sqrt {a+\\sqrt {bc}}}}+{\\frac {b\n\\sqrt {\\sqrt {c}+\\sqrt {a}}}{\\sqrt {b+\\sqrt {ac}}}}+{\\frac {c\\sqrt {\n\\sqrt {a}+\\sqrt {b}}}{\\sqrt {c+\\sqrt {ab}}}}\\leq 3.$\nhttp://www.artofproblemsolving.com/community/c6h1098581p4979124", "Solution_23": "[quote=mattilgale]If $a,b,c$ are the sides of a triangle, prove that\n\\[\\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}+\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}}+\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}}\\leq 3 \\][i]Proposed by Hojoo Lee, Korea[/i][/quote]\n[url=http://www.artofproblemsolving.com/community/c6h149013p842206]TST PERU - IMO 2007[/url]", "Solution_24": "my solution:\nconsider a function $ f(x)=\\sqrt{x} $ and three point A(a,$ \\sqrt{a} $),B(b,$ \\sqrt{b} $),C(c,$ \\sqrt{c} $)\nchoose D,E,F s.t. ABDC,ABCE,and AFBC are all parallelogram\nchoose $ G_{D} $,$ G_{E} $,and $ G_{F} $ on f(x) s.t. $ DG_{D} $,$ EG_{E} $,and $ FG_{F} $ are plumb lines\nnow,we only show that $ \\frac{y_{G_{D}}}{y_{D}}+\\frac{y_{G_{E}}}{y_{E}}\\leq 2 $\nthat is $ \\frac{DG_{D}}{y_{D}}\\geq \\frac{EG_{E}}{y_{E}} $\nit is obious because $ y_{D}\\geq y_{E} $\u4e14$ DG_{D}\\leq EG_{E} $\n", "Solution_25": "goooooooood!", "Solution_26": "Solution from [i]Twitch Solves ISL[/i]:\n\nTo deal with the long denominators, let $x = \\sqrt b + \\sqrt c - \\sqrt a$, etc.\\ for brevity. Note that $x > 0$, and similarly for the others. Then we have that \\[ 0 < b+c-a = \\left( \\frac{x+y}{2} \\right)^2 \t+ \\left( \\frac{x+z}{2} \\right)^2 - \\left( \\frac{y+z}{2} \\right)^2 \t= x^2+xy+xz-yz \\] Thus the problem asks us to prove that \\[ \\sum_{\\text{cyc}} \\frac{\\sqrt{x^2+xy+xz-yz}}{x} \\le 3\\sqrt 2 \\] for positive real numbers $x$, $y$, $z$ satisfying the extra condition that all radicals are positive.\nBy Cauchy-Schwarz though, we have that \\[ \t\\left( \\sum_{\\text{cyc}} \\frac{\\sqrt{x^2+xy+xz-yz}}{x} \\right)^2 \t\\le 3 \\sum_{\\text{cyc}} \\frac{x^2+xy+xz-yz}{x^2} \\] and expanding fully, we see it suffices to show \\[ \\sum_{\\text{sym}} xy^2z^3 \\le \t(xy)^3 + (yz)^3 + (zx)^3 + 3(xyz)^2 \\] which is Schur's inequality applied to $xy$, $yz$, $zx$.", "Solution_27": "[quote=v_Enhance]Solution from Twitch stream:\n\nThen we have that \\[ 0 < b+c-a = \\left( \\frac{x+y}{2} \\right)^2 \t+ \\left( \\frac{x+z}{2} \\right)^2 - \\left( \\frac{y+z}{2} \\right)^2 \t= x^2+xy+xz-yz \\] [/quote]\nSir, it would be $b+c-a=\\left(\\frac{x+y}{2}\\right)^2+\\left(\\frac{z+x}{2}\\right)^2-\\left(\\frac{y+z}{2}\\right)^2=\\frac{2x^2+2xy+2zx-2yz}{4}=\\frac{x^2+xy+xz-yz}{2}$", "Solution_28": "Solved with [b]eisirrational[/b], [b]goodbear[/b], [b]nukelauncher[/b], [b]Th3Numb3rThr33[/b].\n\nLet \\(a=(q+r)^2\\), \\(b=(r+p)^2\\), \\(c=(p+q)^2\\), so that the desired inequality is \\[\\sum_\\mathrm{cyc}\\frac{\\sqrt{p(p+q+r)-qr}}p\\le3\\sqrt2.\\]\n\nThen observe that \\begin{align*} \\left(\\sum_\\mathrm{cyc}\\frac{\\sqrt{p(p+q+r)-qr}}p\\right)^2 &\\le3\\sum_\\mathrm{cyc}\\frac{p(p+q+r)-qr}{p^2}\\\\ &=3\\left(3+\\sum_\\mathrm{sym}\\frac pq-\\sum_\\mathrm{cyc}\\frac{qr}{p^2}\\right), \\end{align*} so it suffices to show \\[3+\\sum_\\mathrm{cyc}\\frac{qr}{p^2}\\ge\\sum_\\mathrm{sym}\\frac pq.\\] This is just Schur's inequality on \\(p^{-2/3}q^{1/3}r^{1/3}\\), \\(p^{1/3}q^{-2/3}r^{1/3}\\), \\(p^{1/3}q^{1/3}r^{-2/3}\\): \\begin{align*} 3\\prod\\sqrt[3]{\\frac{qr}{p^2}}+\\sum_\\mathrm{cyc}\\frac{qr}{p^2} &\\ge\\sum_\\mathrm{cyc}\\sqrt[3]{\\frac{qr}{p^2}}\\sqrt[3]{\\frac{rp}{q^2}}\\left(\\sqrt[3]{\\frac{qr}{p^2}}+\\sqrt[3]{\\frac{rp}{q^2}}\\right)\\\\ &=\\sum_\\mathrm{cyc}\\sqrt[3]{\\frac{r^2}{pq}}\\left(\\frac{q\\sqrt[3]r+r\\sqrt[3]q}{\\sqrt[3]{p^2q^2}}\\right)\\\\ &=\\sum_\\mathrm{cyc}\\frac{qr+rp}{pq}=\\sum_\\mathrm{sym}\\frac pq. \\end{align*}", "Solution_29": "This inequality is just rationalization and schur. ", "Solution_30": "Let $a=(x+y)^2$, $b=(y+z)^2$, $c=(z+x)^2$, $a$, $b$, $c\\ge0$. It suffices to show that \n\\[S=\\sum_{\\text{cyc}}{\\frac{\\sqrt{x^2+xy+xz-yz}}{x} }\\le 3\\sqrt{2}\\]\nbut by Schur's, $\\sum_{\\text{cyc}}{(xy)^2(yz+zx-xy)}\\le 3x^2y^2z^2$ so \\[S^2 \\le (1+1+1)\\left(\\sum_{cyc}{\\frac{z^2}{z^2}+\\frac{yz+zx-xy}{z^2}}\\right)\\le (1+1+1)(3+3)=18\\] which gives the desired result.", "Solution_31": "Let $x = \\sqrt b + \\sqrt c - \\sqrt a$ and cyclic permutations. The inequality boils down to $$\\sum \\frac{\\sqrt{x^2+xy+xz-yz}}x \\leq 3\\sqrt 2.$$ But by Cauchy $$\\left(\\sum \\frac{\\sqrt{x^2+xy+xz-yz}}x\\right)^2 \\leq 3\\sum \\frac{x^2+xy+xz-yz}{x^2} \\leq 18.$$ The last inequality $$\\sum \\frac{xy+xz-yz}{x^2} \\leq 3$$ expands to Schur with $r=2$.", "Solution_32": "Substitute $x = \\sqrt{b} + \\sqrt{c} - \\sqrt{a}$ and so on. This then simplifies as \\[ \\frac13 \\sum_{\\text{cyc}} \\frac{\\sqrt{x^2 + xz + xy - yz}}{x} \\le \\sqrt{2} \\]\nBy Power-Mean and Schur on $r = -2$ it follows that \\[ \\frac13 \\sum_{\\text{cyc}} \\frac{\\sqrt{x^2 + xz + xy - yz}}{x} \\le \\left(\\frac13 \\sum_{\\text{cyc}} \\frac{x^2 + xz + xy - yz}{x^2}\\right)^{\\frac 12} \\le \\left(\\frac 13 (3 + 3) \\right)^{\\frac 12}. \\]", "Solution_33": "We can first make the substitution $x = \\sqrt b + \\sqrt c - \\sqrt a$, and cyclically for $y,z$, noting all of these variables are also positive. Then the left hand side can be expressed as\n\\[\\sum_{\\text{cyc}} \\sqrt{\\frac{x^2+xy+xz-yz}{2x^2}} \\leq \\sqrt{3 \\cdot \\sum_{\\text{cyc}} \\frac{x^2+xy+xz-yz}{2x^2}}\\]\n\nusing Cauchy. Hence it suffices to show\n\\[\\sum_{\\text{cyc}} \\frac{x^2+xy+xz-yz}{x^2} \\leq 6\\]\n\\[\\iff (xy)^3 + (yz)^3 + (zx)^3 + 3(xy)(yz)(zx) \\ge \\sum_{\\text{sym}} (xy)^2(yz),\\]\n\nwhich is indeed just Schur's for $t=1$ on $xy,yz,zx$. $\\blacksquare$", "Solution_34": "Make the substitution $x = \\sqrt{b} + \\sqrt{c} - \\sqrt{a}$, with $y$ and $z$ defined cyclically. Now note,\n\\begin{align*}\n\\sqrt{b+c-a} = \\sqrt{\\frac{x^2+xy+xz-yz}{2}}\n\\end{align*}\nThus we wish to prove,\n\\begin{align*}\n \\sum_{cyc} \\frac{\\sqrt{x^2+xy+xz-yz}}{x} \\leq 3\\sqrt{2}\\\\\n\\left( \\sum_{cyc} \\frac{\\sqrt{x^2+xy+xz-yz}}{x} \\right)^2 \\leq 18\n\\end{align*}\nHowever Cauchy gives,\n\\begin{align*}\n\\left( \\sum_{cyc} \\frac{\\sqrt{x^2+xy+xz-yz}}{x} \\right)^2 \\leq 3\\left(\\sum_{cyc} \\frac{x^2+xy+xz-yz}{x^2} \\right)\n\\end{align*}\nSubstituting this into our original expression we find,\n\\begin{align*}\n\\sum_{cyc} \\frac{x^2+xy+xz-yz}{x^2} &\\leq 6\\\\\n\\end{align*}\nwhich is just Schurs upon expansion. ", "Solution_35": "Let $x=\\sqrt{b}+\\sqrt{c}-\\sqrt{a}>0$ (positive by the triangle inequality), etc., so $y+z=2\\sqrt{a}$ and the LHS becomes\n$$\\sum_{\\mathrm{cyc}} \\frac{\\sqrt{(\\frac{x+y}{2})^2+(\\frac{x+z}{2})^2-(\\frac{y+z}{2})^2}}{x}=\\sum_{\\mathrm{cyc}} \\frac{\\sqrt{x^2+xy+xz-yz}}{x\\sqrt{2}}.$$\nIt thus suffices to prove that\n$$\\sum_{\\mathrm{cyc}} \\sqrt{1+\\frac{y}{x}+\\frac{z}{x}-\\frac{yz}{x^2}} \\leq 3\\sqrt{2}.$$\nBy Jensen's, it suffices to show that\n$$1+\\frac{1}{3}\\left(\\sum_{\\mathrm{cyc}} \\frac{y}{x}+\\frac{z}{x}-\\frac{yz}{x^2}\\right) \\leq 2 \\iff \\sum_{\\mathrm{cyc}} \\frac{y}{x}+\\frac{z}{x} \\leq 3+\\sum_{\\mathrm{cyc}} \\frac{yz}{x^2} \\iff 3x^2y^2z^2+3x^3y^3+3y^3z^3+3z^3x^3 \\geq x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3 \\iff [3,3,0]+[2,2,2] \\geq 2[3,2,1].$$\nThis is just first-degree Schur on $xy,yz,zx$, so we're done. $\\blacksquare$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Given $ a, b, c \\geq\\ 0.$ Prove that:\r\n$ (\\frac {a}{b \\plus{} c})^3 \\plus{} (\\frac {b}{c \\plus{} a})^3 \\plus{} (\\frac {c}{a \\plus{} b})^3 \\plus{} \\frac {abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {1}{2}. (\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca})^2$\r\nIt's stronger than a well-known:\r\n$ (\\frac {a}{b \\plus{} c})^3 \\plus{} (\\frac {b}{c \\plus{} a})^3 \\plus{} (\\frac {c}{a \\plus{} b})^3 \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$ \r\nPS: With this old ineq, we are easily to solve it by Am-Gm and... \r\nNotice that: $ abc(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca) \\geq\\ 0$\r\n :)", "Solution_1": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0.$ Prove that:\n$ (\\frac {a}{b \\plus{} c})^3 \\plus{} (\\frac {b}{c \\plus{} a})^3 \\plus{} (\\frac {c}{a \\plus{} b})^3 \\plus{} \\frac {abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {1}{2}. (\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca})^2$\nIt's stronger than a well-known:\n$ (\\frac {a}{b \\plus{} c})^3 \\plus{} (\\frac {b}{c \\plus{} a})^3 \\plus{} (\\frac {c}{a \\plus{} b})^3 \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$ \nPS: With this old ineq, we are easily to solve it by Am-Gm and... \nNotice that: $ abc(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca) \\geq\\ 0$\n :)[/quote]\r\nI haven't tried with AM-GM as you said but the inequality is actually weak and we can prove it using Cauchy Schwarz + Schur as follow:\r\nApplying Cauchy Schwarz Inequality, we have\r\n\\[ \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b} \\ge \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)}\\]\r\nIt follows that\r\n\\[ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca} \\le 2\\left( \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b} \\minus{}1\\right)\\]\r\nAnd we can deduce our inequality to\r\n\\[ \\frac{a^3}{(b\\plus{}c)^3}\\plus{}\\frac{b^3}{(c\\plus{}a)^3}\\plus{}\\frac{c^3}{(a\\plus{}b)^3} \\plus{}\\frac{abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\ge 2\\left( \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}\\minus{}1\\right)^2\\]\r\nSetting $ x\\equal{}\\frac{2a}{b\\plus{}c},y\\equal{}\\frac{2b}{c\\plus{}a},z\\equal{}\\frac{2c}{a\\plus{}b}$, then $ xy\\plus{}yz\\plus{}zx\\plus{}xyz\\equal{}4$ and our inequality becomes\r\n\\[ x^3\\plus{}y^3\\plus{}z^3\\plus{}xyz \\ge 4(x\\plus{}y\\plus{}z\\minus{}2)^2\\]\r\nNow, we denote $ p\\equal{}x\\plus{}y\\plus{}z,q\\equal{}xy\\plus{}yz\\plus{}Zx,r\\equal{}xyz,$ then $ q\\plus{}r\\equal{}4$ and our inequality is equivalent to\r\n\\[ p^3\\minus{}3pq\\plus{}4r \\ge 4(p\\minus{}2)^2\\]\r\n\\[ p^3 \\minus{}3p(4\\minus{}r)\\plus{}4r \\ge 4(p\\minus{}2)^2\\]\r\n\\[ (p\\minus{}4)(p^2\\plus{}4) \\plus{}(3p\\plus{}4)r \\ge 0\\]\r\nIf $ p \\ge 4$, it is trivial. If $ 4 \\ge p \\ge 3$, applying Schur's Inequality, we obtain $ r \\ge \\frac{p(4q\\minus{}p^2)}{9}$, hence\r\n\\[ 4\\equal{}q\\plus{}r \\ge q \\plus{}\\frac{p(4q\\minus{}p^2)}{9}\\]\r\nIt follows that\r\n\\[ q \\le \\frac{p^3\\plus{}36}{4p\\plus{}9}\\]\r\nand we obtain\r\n\\[ r\\equal{}4\\minus{}q \\ge 4\\minus{}\\frac{p^3\\plus{}36}{4p\\plus{}9}\\equal{}\\frac{p(16\\minus{}p^2)}{4p\\plus{}9}\\]\r\nWe have to prove\r\n\\[ (p\\minus{}4)(p^2\\plus{}4) \\plus{}(3p\\plus{}4)\\cdot \\frac{p(16\\minus{}p^2)}{4p\\plus{}9} \\ge 0\\]\r\n\\[ p(p\\plus{}4)(3p\\plus{}4)\\minus{}(4p\\plus{}9)(p^2\\plus{}4) \\ge 0\\]\r\n\\[ 7p^2\\minus{}p^3\\minus{}36 \\ge 0\\]\r\n\\[ (p\\minus{}3)(12\\plus{}4p\\minus{}p^2) \\ge 0\\]\r\nwhich is obviously true because $ 4 \\ge p \\ge 3$.\r\nThis completes our proof. Equality holds if and only if $ a\\equal{}b\\equal{}c$ or $ a\\equal{}b,c\\equal{}0$ and its cyclic permutations. :)", "Solution_2": "[quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0.$ Prove that:\n$ (\\frac {a}{b \\plus{} c})^3 \\plus{} (\\frac {b}{c \\plus{} a})^3 \\plus{} (\\frac {c}{a \\plus{} b})^3 \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$ \n :)[/quote]\r\nBut with this old ineq, your method can't solve it. I tried.\r\nI also see a long proof for it by SOS in Pham Van Thuan's book. But using Am-Gm, we have a very short solution :) \r\nAnd it's seems not weaker than mine :oops:", "Solution_3": "[quote=\"nguoivn\"][quote=\"nguoivn\"]Given $ a, b, c \\geq\\ 0.$ Prove that:\n$ (\\frac {a}{b \\plus{} c})^3 \\plus{} (\\frac {b}{c \\plus{} a})^3 \\plus{} (\\frac {c}{a \\plus{} b})^3 \\plus{} \\frac {5abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$ \n :)[/quote]\nBut with this old ineq, your method can't solve it. I tried.\nI also see a long proof for it by SOS in Pham Van Thuan's book. But using Am-Gm, we have a very short solution :) \nAnd it's seems not weaker than mine :oops:[/quote]\r\nI haven't read Thuan's book, so I dont know that solution. Anyways, the following inequality is stronger than yours ;)\r\n\\[ \\left( \\frac{a}{b\\plus{}c}\\right)^3 \\plus{}\\left( \\frac{b}{c\\plus{}a}\\right)^3 \\plus{}\\left( \\frac{c}{a\\plus{}b}\\right)^3 \\plus{}\\frac{9abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)} \\ge \\frac{a}{b\\plus{}c}\\plus{}\\frac{b}{c\\plus{}a}\\plus{}\\frac{c}{a\\plus{}b}.\\]\r\n;)", "Solution_4": "Yes, that's it :P \r\nYour proof is same to me :oops: \r\nI also used a nice lenma:\r\n$ \\sum\\ \\frac{a}{b\\plus{}c} \\geq\\ \\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca} \\plus{} \\frac{4abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}$\r\nAfter expand, it's become: $ abc(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca) \\geq\\ 0$ (as I said)\r\nAnd I think it's an useful lenma because notice that it's stronger than:\r\n$ \\sum\\ \\frac{a}{b\\plus{}c} \\geq\\ \\frac{(a\\plus{}b\\plus{}c)^2}{2(ab\\plus{}bc\\plus{}ca)}$", "Solution_5": "[quote=\"nguoivn\"]Yes, that's it :P \nYour proof is same to me :oops: \nI also used a nice lenma:\n$ \\sum\\ \\frac {a}{b \\plus{} c} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\plus{} \\frac {4abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}$\nAfter expand, it's become: $ abc(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca) \\geq\\ 0$ (as I said)\nAnd I think it's an useful lenma because notice that it's stronger than:\n$ \\sum\\ \\frac {a}{b \\plus{} c} \\geq\\ \\frac {(a \\plus{} b \\plus{} c)^2}{2(ab \\plus{} bc \\plus{} ca)}$[/quote]\r\nWe dont need to expand here, nguoi vn ;)\r\nWe rewrite it as follow:\r\n\\[ \\sum \\frac {a[a(b \\plus{} c) \\plus{} bc]}{b \\plus{} c} \\ge a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} \\frac {4abc(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}\r\n\\]\r\n\r\n\\[ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} abc\\sum \\frac {1}{b \\plus{} c} \\ge a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} \\frac {4abc(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}\r\n\\]\r\n\r\n\\[ \\sum \\frac {1}{b \\plus{} c} \\ge \\frac {4(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}\r\n\\]\r\nwhich is obviously true by Cauchy Schwarz because\r\n\\[ \\sum \\frac {1}{b \\plus{} c} \\ge \\frac {9}{2(a \\plus{} b \\plus{} c)}\r\n\\]\r\nand\r\n\\[ \\frac {4(ab \\plus{} bc \\plus{} ca)}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\le \\frac {9}{2(a \\plus{} b \\plus{} c)}\r\n\\]\r\n;) :)", "Solution_6": "I meant:\r\n$ \\sum\\ \\frac {a}{b \\plus{} c} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca} \\plus{} \\frac {4abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}$\r\n<=> $ \\frac {(a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2) \\minus{} abc}{(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)} \\geq\\ \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{ab \\plus{} bc \\plus{} ca}$ $ (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2)(ab \\plus{} bc \\plus{} ca) \\minus{} abc \\sum\\ ab \\geq\\ [(a \\plus{} b \\plus{} c)(ab \\plus{} bc \\plus{} ca) \\minus{} abc](a^2 \\plus{} b^2 \\plus{} c^2)$\r\n<=> $ abc(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ca) \\geq\\ 0$\r\nI think both of ways also short and nice :)" } { "Tag": [ "inequalities", "calculus", "calculus computations" ], "Problem": "[url=http://bildr.no/view/489513][img]http://bildr.no/thumb/489513.jpeg[/img][/url]\r\n[url=http://bildr.no/view/489514][img]http://bildr.no/thumb/489514.jpeg[/img][/url]\r\n[url=http://bildr.no/view/489526][img]http://bildr.no/thumb/489526.jpeg[/img][/url]\r\n4 pics are continual . because the forum only allowed 3 images. the last image is on the first reply post.\r\nI can not understand the lines denoted by red line. \r\n\r\nthanks", "Solution_1": "continued \r\n[url=http://bildr.no/view/489516][img]http://bildr.no/thumb/489516.jpeg[/img][/url]", "Solution_2": "continued \r\n[url=http://bildr.no/view/489516][img]http://bildr.no/thumb/489516.jpeg[/img][/url]\r\n\r\n\r\nI have understood it ." } { "Tag": [], "Problem": "What is the sum of the reciprocals of the first three positive even integers?", "Solution_1": "[hide]\nthe first three positive even integers are 2, 4, and 6.\nThe recipricols are:\n$\\frac{1}{2} + \\frac{1}{4} +\\frac{1}{6}=\\frac{14}{12} =\\frac{7}{6}=1 \\frac{1}{6}$[/hide]", "Solution_2": "it is not over one ;)", "Solution_3": "[hide]\nis it $\\frac{11}{12}$[/hide]", "Solution_4": "[quote=\"math92\"][hide]\nis it $\\frac{11}{12}$[/hide][/quote]\r\n\r\nThat's better :)" } { "Tag": [ "geometry", "function", "integration", "logarithms", "calculus", "real analysis", "real analysis unsolved" ], "Problem": "$y=f(x)$ is a differantiable function in $R$. $a0$. Arc of this function from $a$ to $b$ is equal to area between $x=a , x=b , y=0 , y=f(x)$ Find all $y=f(x)$.", "Solution_1": "[quote=\"mayhem\"]$y=f(x)$ is a differantiable function in $R$. $a0$. Arc of this function from $a$ to $b$ is equal to area between $x=a , x=b , y=0 , y=f(x)$ Find all $y=f(x)$.[/quote]\r\n$f: [a,b]\\to R^{2}$ ? \r\nIf I understand Everything right then ( if arc is $\\int_{a}^{b}\\sqrt{1+f'^{2}(x)}dx$) then \r\n$(b-a)f(x)=\\int_{a}^{b}\\sqrt{1+f'^{2}(x)}dx$ and therefore\r\n$f(x)=\\frac{\\int_{a}^{b}\\sqrt{1+f'^{2}(x)}dx}{(b-a)}$", "Solution_2": "A little correction to Extremal: \r\n$\\int_{a}^{b}f(x)dx=\\int_{a}^{b}\\sqrt{1+f'^{2}(x)}dx$\r\n\r\nAlso, in tis question asked for all functions $f(x)$.", "Solution_3": "Oh, Then I was wrong. \r\nI can represent some runctions for wich it will be true :\r\n\r\nIn first cours in my universuty we still not begin to study How solve differential equations therefore in my solution maybe will some simpl mistake but I'll try :oops: \r\nif $f(x)=\\sqrt{1+f'^{2}(x)}$ then our equation will be true\r\nlet's consider this case.\r\n\r\n$f^{2}(x)-1=f'^{2}(x)$\r\nnote that $f'(x)$ can be $>0$ and $<0$\r\ntherefore consider two case\r\n1. $f'>0$\r\n$\\sqrt{f^{2}-1}=f'$ $f' <>0$\r\n$1=\\frac{f'}{f^{2}-1}$\r\ntherefore\r\n$\\int=\\int\\frac{f'}{f^{2}-1}$\r\n$x+C=\\ln{(f+\\sqrt{f^{2}-1})}$\r\n$e^{x+C}=f+\\sqrt{f^{2}-1}$\r\ntherefore\r\n$f=\\frac{e^{2x+2C}+1}{e^{x+C}}$\r\nlet check our conditions ( we do it for find constant $c$)\r\n\r\n$(e^{x+c}+e^{-x-c})|_{a}^{b}=\\int_{a}^{b}\\sqrt{e^{2x+2c}+e^{-2x-2c}-1}dx$\r\nyou only need to find \r\n$\\int_{a}^{b}\\sqrt{e^{2x+2c}+e^{-2x-2c}-1}dx$\r\n=${\\int_{a}^{b}\\sqrt{2\\cosh{(2x+2c)-1}}dx=\\int_{2a-2c}^{2b-2c}\\sqrt{2\\cosh(y)-1}}dy$\r\nAnd (if I remember right) last integral was discused in this forum few times ago)", "Solution_4": "extremal, i think that there is a small error here :) ${\\frac{d\\log \\left(f(x)+\\sqrt{f(x)^{2}-1}\\right)}{dx}=\\frac{f'(x)}{\\sqrt{f(x)^{2}-1}}}$", "Solution_5": "[quote=\"yagaron\"]extremal, i think that there is a small error here :) ${\\frac{d\\log \\left(f(x)+\\sqrt{f(x)^{2}-1}\\right)}{dx}=\\frac{f'(x)}{\\sqrt{f(x)^{2}-1}}}$[/quote]\r\nyes really I rewrite very fast \r\nwhen I wrote \r\n$f'=\\sqrt{f^{2}-1}$\r\nthen we have\r\n$1=\\frac{f'}{\\sqrt{f^{2}-1}}$\r\nthanks for correction yagaron\r\n :wink:", "Solution_6": "no extremal. what I meant is that your solution is wrong.\r\n\r\nYou comit some mistakes. (actualy forgot to divide by 2 :oops: )\r\n\r\nand we can solve this ODE in a very easy way\r\n\r\n[quote]\n${\\frac{d\\cos (x)}{dx}=-\\sin (x)}$,${\\frac{d\\sin (x)}{dx}=\\cos (x)}$\n\n${\\left(\\cos^{2}(x)+\\sin^{2}(x)=1 \\right)\\Rightarrow \\left( \\cos^{2}(x)+\\left(\\frac{d\\cos (x)}{dx}\\right)^{2}=1\\right)}$\n\n${\\frac{d\\cosh (x)}{dx}=\\sinh (x)}$,${\\frac{d\\sinh (x)}{dx}=\\cosh (x)}$\n\n${\\left(\\cosh^{2}(x)-\\sinh^{2}(x)=1\\right)\\Rightarrow \\left(\\cosh^{2}(x)-\\left(\\frac{d\\cosh (x)}{dx}\\right)^{2}=1\\right)}$\n\n${\\left(\\sqrt{f'(x)^{2}+1}=f(x)\\right)\\Rightarrow \\left(f(x)^{2}-f'(x)^{2}=1\\right)}$\n\nSo, we can conclude ${f(x)=\\cosh \\left(x+c_{1}\\right)}$ is a specific result.\n\nwe just need to proof uniqueness with some theorems, but i dont want to make this post more than it is\n[/quote]" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Circles $C_{1},C_{2}$meet at points $A,B$,A line through A is parrallel to the line through the centres of $C_{1},C_{2}$ and meets$C_{1},C_{2}$ again $C and D$respectively.Circle $C_{3}$ having $CD$as its diameter meets $C_{1},and C_{2}$again at $P$ and $Q$ respectively. Prove that lines $CP,DQ,AB$ are concurent and $CQ,DP,AB$ are concurent.", "Solution_1": "What an easy problem. Proof:\r\n\r\nIt's simple to see that $C, O_{1}, B$ line on the same line and so do $D, O_{2}, B$.\r\n\r\nLet $I$ is the midpoint of the segment $CD$. We have: $IO_{2}$ is parallel to $CB$ [color=red](1)[/color]\r\n\r\nAs $Q$ lies on the circle whose diameter is $CD$, $QC$ is perpendicular to $QD$. On the other hand, $(I)$ and $(O_{2})$ intersect each other at $D$ and $Q$, thus $IO_{2}$ is perpendicular to $QD$. Therefore, $IO_{2}$ is parallel to $CQ$ [color=red](2)[/color]\r\n\r\nFrom (1) and (2) we have $B$ lies on $CQ$.\r\n\r\nSimilarly, $B$ lies on $DP$" } { "Tag": [ "geometry", "3D geometry", "sphere", "Asymptote" ], "Problem": "is there a way to extrude curved text (labelpath or texpath) ? \r\ni'd be grateful for any tips \r\nbest \r\nzb", "Solution_1": "[code]\nimport solids;\nviewportmargin=(10,10);\nsize(200);\nrevolution sphere=sphere(0.75);\nsurface s=surface(sphere);\nstring text=\"some text\";\nreal uoffset=-1.2;\nreal voffset=7.5;\nreal h=0.05;\ndraw(surface(xscale(0.4)*yscale(0.08)*text,s,uoffset,voffset,h),yellow);\ndraw(s.uequals(uoffset),red+dashed);\ndraw(s.vequals(voffset),blue+dashed);\ndraw(surface(s),palegray);\n\n[/code]", "Solution_2": "THANK YOU!\r\n\r\nzb" } { "Tag": [ "inequalities", "algebra", "polynomial", "algorithm", "inequalities unsolved" ], "Problem": "I composed\r\nIf a,b,c >0 then\r\n$a^{20}b^{4}+b^{20}c^{4}+c^{20}a^{4}+3a^{8}b^{8}c^{8}\\geq 2a^{2}b^{2}c^{2}(a^{10}b^{8}+b^{10}c^{8}+c^{10}a^{8})$\r\n\r\n[hide]I think it very good\n[hide] L\u00ea_V\u0103n_Ch\u00e1nh[/hide][/hide]", "Solution_1": "Is it not just Am-Gm: \\[a^{20}b^{4}+b^{20}c^{4}+c^{20}a^{4}+3a^{8}b^{8}c^{8}\\geq 2a^{2}b^{2}c^{2}(a^{10}b^{8}+b^{10}c^{8}+c^{10}a^{8}) =2\\sum_{cyc}a^{12}b^{10}c^{2}\\] but \\[a^{20}b^{4}+a^{8}b^{8}c^{8}\\geq 2a^{14}b^{6}c^{4}\\geq 2a^{12}b^{10}c^{2}\\] where the first is am-gm, the second is muirhead (am-gm in disguise)\r\n\r\nso the result follows by taking cyclic sum", "Solution_2": "[quote=\"me@home\"]Is it not just Am-Gm: \\[a^{20}b^{4}+b^{20}c^{4}+c^{20}a^{4}+3a^{8}b^{8}c^{8}\\geq 2a^{2}b^{2}c^{2}(a^{10}b^{8}+b^{10}c^{8}+c^{10}a^{8}) =2\\sum_{cyc}a^{12}b^{10}c^{2}\\] but \\[a^{20}b^{4}+a^{8}b^{8}c^{8}\\geq 2a^{14}b^{6}c^{4}\\geq 2a^{12}b^{10}c^{2}\\] where the first is am-gm, the second is muirhead (am-gm in disguise)\n\nso the result follows by taking cyclic sum[/quote]\r\nIt is easy!!!", "Solution_3": "Any type of problem in the form you posted can just as easily be solved by muirhead/am-gm, I am right?\r\nWhat other problems have you \"composed\"? edit: this is my post #1000!!!!", "Solution_4": "[quote=\"me@home\"]Any type of problem in the form you posted can just as easily be solved by muirhead/am-gm, I am right?\n[/quote]\r\nThat is not in fact true. There was a user here not long ago [b]fjwxcsl[/b] who posted some solutions to symmetric polynomial inequalities that looked like pasted computer output. This is Shengli Chen of the Chinese Academy of Science, and he is the coauthor of a paper about inequalities of symmetric polynomials of three variables. In the paper, they present an algorithm which will efficiently split a polynomial into Schur-like forms:\r\n$\\sum_{cyc}x^{k}(x-y)(x-z)$\r\n$\\sum_{cyc}(yz)^{k}(x-y)(x-z)$\r\n$\\sum_{cyc}x^{k}(y+z)(x-y)(x-z)$\r\n$\\sum_{cyc}(yz)^{k}(y+z)(x-y)(x-z)$\r\nAll of these are positive; therefore, such a decomposition will always enable you to check the validity of a polynomial inequality. (For details about that, see their paper.)\r\n\r\nHowever, I don't think it is true that any symmetric polynomial inequality can be decomposed in such a way that it is killed by AM-GM and Schur alone (certainly not AM-GM alone). See for example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107851]here[/url]." } { "Tag": [ "induction", "number theory", "relatively prime" ], "Problem": "Prove that there are no integers $m,n \\geq 1$ satisfying\r\n\r\n\r\n$\\sqrt{m+\\sqrt{m+\\sqrt{m+\\ldots+\\sqrt{m}}}}=n$\r\n\r\nwhere there are $2006$ square root signs.", "Solution_1": "[hide=\"Hint\"]\ngeneralize (2006 isn't special)\n[/hide]\n\n[hide=\"Solution\"]\nInstead of solving this problem for 2006, we will solve it for $k$ nested radicals where $k$ is a natural number greater than one. We will proceed by induction.\n\nThe base case is where $k=2$, or:\n\n$\\sqrt{m+\\sqrt{m}}=n$\n\nSquaring both sides and subtracting by $m$ yields:\n\n$\\sqrt{m}=n^{2}-m \\rightarrow m=(n^{2}-m)^{2}$\n\nTherefore, $m$ is a perfect square. Let $m=x^{2}$:\n\n$x=n^{2}-x^{2}\\rightarrow n=\\sqrt{x(x+1)}$\n\nSince $x$ and $x+1$ are relatively prime, $n$ is not an integer, which is a contradiction, which closes the base case.\n\nNow we will assume that the case with $k$ nested radicals does not yield an integer value for $n$ in order to prove the result for the case with $k+1$ nested radicals. After squaring both sides and subtracting $m$ from both sides in the equation with $k+1$ nested radicals, we find that:\n\n$\\sqrt{m+\\sqrt{m+\\sqrt{m+\\ldots+\\sqrt{m}}}}=n^{2}-m$\n\nwhere the nested radical appears $k$ times. From our inductive hypothesis, the left side is not an integer and the right side is by the given information (when $m$ is a natural number), so we have a contradiction and induction is complete.\n[/hide]", "Solution_2": "If you want to make your induction nice, do something like this:\r\n\r\nLet $f^{n+1}(x)=f(f^{n}(x))$ for $n\\in \\mathbb{N}^{+}$ and $f(x)=\\sqrt{m+x}$, $f^{1}(x)=\\sqrt{m}$." } { "Tag": [ "inequalities" ], "Problem": "$x,y,z>1$ and $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2$ prove:\r\n$\\sqrt{x+y+z}\\ge\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1}$", "Solution_1": "I could be wrong, but \\[\\sqrt{x-1}+\\sqrt{y-1}+\\sqrt{z-1}\\ \\geq\\ \\sqrt{x-1}-\\sqrt{y-1}-\\sqrt{z-1}.\\] Therefore, I think the inequality posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=533927#533927]here[/url] implies this one.", "Solution_2": "There's probably a typo." } { "Tag": [ "function" ], "Problem": "Decide whether or not there is a function $f$ defined for all positive integers and taking positive integer values such that \r\n\r\n$f(f(1)) = 5$\r\n\r\n$f(f(2)) = 6$\r\n\r\n$f(f(3)) = 4$\r\n\r\n$f(f(4)) = 3$ \r\n\r\n$f(f(n)) = n+2$ for $n\\geq5$", "Solution_1": "[hide]Let $y=f(1)$. Then $f(y)=5$. $y$ must be less than 5 or else $f(y)=y+2>y\\geq 5$. Also $y>1$ because $y=1\\Rightarrow f(y)=1$. \n\nThen $y$ is 2, 3, or 4. But in either case, $f(y)=5$ so $f(5)$ is 6, 4, or 3 depending on whether $y$ is 2, 3, or 4. Since this is false, no such function can exist.[/hide]", "Solution_2": "I got what I think is a solution, even though the argument's shitty.\r\n\r\nWe're looking at a function $f$ which applied twice to $x$ gives $x+2$ for some values of $x$. What's more is that $f$ is defined for positive integers and have an image of positive integers. So all that is is $f(x) = x+1$. But that doesn't work for $x<5$. So, no, there isn't such a function.\r\n\r\nOr is that too obvious to be true?\r\n\r\n(Are you just going to post them all up Sen, or actually do some yourself? :) )", "Solution_3": "how do we prove that $f(f(n)) = n + 2 \\implies f(n) = n+1$ is the only function for f" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "There is a fly at point X on a vertex of a rectangular prism with height 6, length 4, and width 3 units. The fly wants to fly to point Y which is on the vertex farthest from point X. How far must the fly fly to get to point Y?\r\n\r\n[hide] hint: use a 3-4-5 triange to find the distance from X to Y [/hide]", "Solution_1": "the title says \"3-d pythagorean theorem.\" do u mean for us to find the space diagonal, but the fly can't go like that. otherwise, you would find the shortest distance on the surface of the prism from vertex to the other vertex.", "Solution_2": "[quote=\"fireemblem13\"]the title says \"3-d pythagorean theorem.\" do u mean for us to find the space diagonal, but the fly can't go like that. otherwise, you would find the shortest distance on the surface of the prism from vertex to the other vertex.[/quote]\r\n\r\nWhy can't it go along the space diagonal? It can fly...\r\n[hide]$\\sqrt{6^2+4^2+3^2}=\\sqrt{61}$[/hide]", "Solution_3": "isn't it like a box? :huh:", "Solution_4": "Well I assumed it was empty..." } { "Tag": [ "geometry", "rectangle", "search", "factorial", "area of a triangle" ], "Problem": "1. In rectangle [tex]ABCD[/tex], [tex]AD=10[/tex], and [tex]AB=8[/tex]. Point [tex]K[/tex] is chosen in [tex]DC[/tex] such that, when triangle [tex]ADK[/tex] is reflected in [tex]AK[/tex], the image of [tex]D[/tex] is on [tex]BC[/tex]. The length of [tex]DK[/tex] is\r\n\r\na) [tex]5[/tex] b) [tex]6[/tex] c) [tex]2\\sqrt{5}[/tex] d)[tex]3\\sqrt{3}[/tex] e)[tex]4\\sqrt{2}[/tex]\r\n\r\n2. If [tex]P = 64 * 63 * 62 * ... * 3 * 2 * 1[/tex], then the largest positive integer [tex]n[/tex], such that [tex]P[/tex] is divisible by [tex]12^n[/tex] is\r\n\r\na) 5 b) 21 c) 28 d) 30 e) 32\r\n\r\n3. The number of integers between [tex]500[/tex] and [tex]600[/tex] whose digits have a sum of [tex]12[/tex] is\r\n\r\na) 4 b) 5 c) 6 d) 7 e) 8\r\n\r\n4. If the product of [tex](4^5)(5^{13})[/tex] is written as an integer, then the number of digits in the integer is\r\n\r\na) 12 b) 13 c) 18 d) 19 e) 24\r\n\r\n5. [tex]ABC[/tex] is a trangle with sides [tex]a,b,c[/tex]. If [tex]\\measuredangle C = 90^\\circ$[/tex], [tex]c= 4[/tex] and [tex]a+b=\\sqrt{18}[/tex], then the area of triangle [tex]ABC[/tex] is\r\n\r\na) [tex]\\frac{1}{2}[/tex] b) [tex]1[/tex] c) [tex]2[/tex] d) [tex]4[/tex] e) [tex]18[/tex]", "Solution_1": "For the degree sign it would probably be better to use \\circ so you would have 90^\\circ instead of 90^0.\r\n\r\n$90^\\circ$\r\n\r\ninstead of \r\n\r\n$90^0$\r\n\r\nThe latter might be confused to be equal to 1.", "Solution_2": "3.[hide]We know the first digit has to be 5, so the last two digits add up to 7. So the questions is basically asking, how many pairs of integers (a, b), add up to 7 where a and b are both less than or euqal to 7, and less than or equall to 0.\n\n\n\nSince 534 is different from 543, order does matter. Let's list the pairs, since there is just a small number.\n\n\n\n7, 0\n\n6, 1\n\n5, 2\n\n3, 4\n\n4, 3\n\n2, 5\n\n1, 6\n\n0, 7\n\n\n\nSo the answer is e, 8.\n\n\n\nIf I'm correct, you can also use permutations.[/hide]\n\n\n\n4.\n\n[hide]\n\nThe product can be simpliefied to \n\n always has digits, so has digits.\n\n\n\n = 125 which has 3 digits which is about \n\n which has 13 digits[/hide]\n\n\n\nAnd for 5, is c supossed to be the hypotenuse?", "Solution_3": "Joml, thank you for the tip.\r\n\r\nHspotter, yes, c is the hypotenuse", "Solution_4": "5. \n\n[hide]a 2 + b 2 can be written as (a+b) 2 - 2ab. Thus, 18 2 - 2ab = 4 2 by Pythagorean Theorem. So 324 - 2ab = 16, and ab = 154. Area of a triangle will be ab/2, so ab/2 = 77.[hide][/hide][/hide]", "Solution_5": "2. If P=64*63*62*...*3*2*1 , then the largest positive integer n, such that P is divisible by 12^n is ..?\n\n\n\n[hide]64! = k.12^n\n\n\n\nThere are 5 factors of 12 from 1 to 64, so I'd say 5.[/hide]", "Solution_6": "4. If the product of (4^5)(5^13) is written as an integer, then the number of digits in the integer is ..?\n\n\n\n[hide](4^5)(5^13) = (4^5)(5^5)(5^8) = (20^5)(5^8) = (10^5)(2^5)(5^8)\n\n\n\n10^5 contributes 5 digits to (2^5)(5^8)\n\n\n\n(2^5)(5^8) = (10^5)(5^3)\n\n\n\n10^5 contributes another 5 digits to 5^3\n\n\n\n5^3 = 5*5*5 = 125\n\n\n\nSo the integer is 125*10^10\n\nwhich has 13 digits.[/hide]", "Solution_7": "For number 5, I made a typing mistake. It was [tex]a+b=18[/tex] when it should have been [tex]a+b=\\sqrt{18}[/tex].\r\n\r\nThe mistake has been corrected.", "Solution_8": "kool_dudy wrote:5. is a trangle with sides . If , and [tex]a+b=\\sqrt{18}[/tex], then the area of triangle [tex]ABC[/tex] is\n\na) [tex]\\frac{1}{2}[/tex] b) [tex]1[/tex] c) [tex]2[/tex] d) [tex]4[/tex] e) [tex]18[/tex]\n\n\n\n[hide]We know that , and that and can be the height and width of the triangle respectively, since this is a right triangle. That means that is the area of the triangle.\n\n\n\nWe can also figure out that and that . We can both of these things together using an equation - \n\n\n\n\n\n\n\n\n\n\n\nSince is also the are of the triangle, the are of the triangle is a, or [/hide]", "Solution_9": "mikewd wrote:2. If P=64*63*62*...*3*2*1 , then the largest positive integer n, such that P is divisible by 12^n is ..?\n\n[hide]64! = k.12^n\n\nThere are 5 factors of 12 from 1 to 64, so I'd say 5.[/hide]\n\n\n\nHow can you tell if something is a factor of 12? Could there be any ways that you could produce more factors 12 than just the number 12,24,36...? Heres a slight hint: 4! has one factor of 12. Try to see if you can work with that.", "Solution_10": "hmm... 12 as a product of primes is 2*2*3. so maybe the way to find more is to search for how many 2*2*3 s there are there. Or actually, how many 2*2 s and how many 3 s, one will be the limiting factor?\r\n\r\nOk. how many times is something multiplied by 2 in 1*2*3*4*5*......*61*62*63*64 ?\r\n\r\nWell at least 32 times, as there are 32 even numbers. so 64! can be written:\r\n\r\n(1*3*5*7*...*61*63)(1*2*3*4*5*....*31*32)(2)\r\n\r\nTHen the middle bracket again can be half-factorised to get more even numbers out (and more 2 factors out):\r\n\r\n32! = (1*3*5*7*...*31)(1*2*3*4*...*16)(2)\r\n\r\nNow it's quite obvious that every time, the factorial we are half-dividing (dividing only the even multiples) gets smaller... first we took 32 2s out, then 16, then 8 then 4 then 2. then none left. add these to make 32+16+8+4+2 = 62 2s, or 31 (2^2)s.\r\n\r\n Now to find how many 3s, to see whether the (2^2)s or the 3s are the \"weakest link\"... :lol: \r\n\r\nok.. how many 3s in 64!... I guess the same method would work.\r\n\r\n1*2*3*4*5*...*61*62*63*64 = (1*2*4*5*7*...*62*62*64)(3*6*9*...*60*63) = (1*2*4*5*7*...*62*62*64)(1*2*3*4...*20*21)(3)\r\n\r\nSucking the 3s out of the next part that has 3s (ones that used to be 9s, 18s, which have... oh yeah of course, we are just taking all 3s out and ounting them once, taking 9s out (but only counting them once because we've already counted them once in the previous step)\r\n\r\nso skipping some of that, the highest multiple of 3 below 64 is 63, giving 21*3. 21 3s here. 7*3 is 21, so there are 7 more 3s here. the highest multiple of 3 below 7 is 6, and thats 3*2, so there are 2 more 3s here. \r\n\r\nadding them up gives us 21+7+2 = 30. so there are 30 threes in 64!\r\n\r\nok now to go back and edit some working on the 2s.\r\n\r\nAs there are 31 2^2s, and 30 3s, the abundance is 2^2s and the limiting factor is the 3s. so it is therefore possible to divide 64! by 12 30 times, or divide it by 12^30, and n is 30.\r\n\r\ncrikey... thats probably the hardest maths I've ever had to actually think about. thanks for the hint :)" } { "Tag": [ "calculus", "integration", "geometry", "perimeter", "inequalities" ], "Problem": "[b]How many distinct isosceles triangles having sides of integral length and perimeter 113 are possible?[/b]\r\n\r\nHow do u approach this type of problem. Do we use an equation? i got 2x+y=113. but where do i go with tht? Do we use the fact tht the side of a triangle has to be smaller than the sum of the other two sides?\r\n\r\nPLz show me a way where you do not plug numbers in if you can.", "Solution_1": "Yes.\r\n\r\n$ 2x\\plus{}y\\equal{}113$\r\n\r\nAlso, $ 2x>y$\r\n\r\nNow, assuming that $ 2x\\equal{}y$ will give us the highest $ x$ which does not satisfy the conditions.\r\n\r\nSo, $ 4x\\equal{}113$, $ x\\equal{}\\frac{113}{4}\\equal{}28.25$\r\nThe smallest value of $ x$ that will satisfy the equation is $ 29$.\r\n\r\nThe largest is $ 56$.\r\n\r\n$ 29\\rightarrow56\\equal{}28$ numbers", "Solution_2": "how do u know that 2x > y?", "Solution_3": "Triangle Inequality: a+b>c\r\n\r\nso in this case, x+x>y" } { "Tag": [ "LaTeX" ], "Problem": "salam man taze in forume irano peida kardam vali hala mibinam ziad fa'aliat nadare. man ye soale rahat amma ghashang daram:\r\n\r\n$ \\frac {1}{\\sqrt {a^2 \\plus{} 1}} \\plus{} \\frac {1}{\\sqrt {b^2 \\plus{} 1}} \\plus{} \\frac {1}{\\sqrt {c^2 \\plus{} 1}}\\geq\\frac {1}{\\sqrt {ab \\plus{} 1}} \\plus{} \\frac {1}{\\sqrt {ac \\plus{} 1}} \\plus{} \\frac {1}{\\sqrt {bc \\plus{} 1}}$\r\n$ a,b,c\\geq1$\r\n\r\n[color=red][Moderator edit : LaTeX formula corrected.][/color]", "Solution_1": "[quote=\"Kamyar1991\"]\n$ \\frac{1}{\\sqrt{a^2+1}}+\\frac{1}{\\sqrt{b^2+1}}+\\frac{1}{\\sqrt{c^2+1}}\\geq\\frac{1}{\\sqrt{ab+1}}+\\frac{1}{\\sqrt{ac+1}}+\\frac{1}{\\sqrt{bc+1}}$\n\n$ a,b,c\\geq 1$ \n[/quote]\r\nino ghablanam post kardi,age kasi khas hallesho bebine bere inja:\r\n\r\n[hide][url]http://www.mathlinks.ro/viewtopic.php?p=1102834#1102834[/url][/hide]\r\n\r\nvase $ \\text{\\LaTeX}$ ham chize khassi lazem nis download koni,faghat kafie do tarafe ebaratet alamate dolar bezario ye seri chizaye payaro balad bashi ke inja tozih dade shode:\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?p=1135325#1135325[/url]", "Solution_2": "hi man koli ruye in soal dobare fekr kardam va farz bayad tabdil be $ a,b,c>3$ beshe az hamatun ham mazerat mikham.!! :blush:" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Does there exist the greatest value of the expression $\\boxed{\\sqrt{x_1x_2}+\\sqrt{x_2x_3}+\\cdots+\\sqrt{x_{n-1}x_n}}$, where $x_i >0, \\sum_{i=1}^nx_i=1, i=\\overline{1,n}$.", "Solution_1": "I think mixing variables, first for any consecutive pairs of $x_2,...,x_{n-1}$ and finally for $x_1,x_n$ shows that $max=\\frac{n-1}n$ :)" } { "Tag": [ "linear algebra" ], "Problem": "Yeah!\r\n\r\nSo, um, I'm wondering what courses I should take (I'm in Pure Maths) and where I should live (which residence). Anyone who goes there have any comments?\r\n\r\nThanks in advance.\r\n\r\nAnd if you're also going to Waterloo please share your excitement!", "Solution_1": "I see you just joined the UW forum (err, sorry for blowing your cover :P). So basically, you don't choose that major till second year. First year you would take two maths a CS and two electives. That means MATH 145 and 147 first semester (145 is a course in \"classical algebra\" and is more an introduction to formal mathematics than anything, and 147 is the calculus/analysis course) and 146 and 148 second semester (146 is Linear Algebra, 148 is the second half of 147). Then your choice of CS and electives. Matter of fact, you should be getting your first year in math book in a couple days, should answer most questions. \r\n\r\n\r\n\r\nAnd yes, I'm soo excited for Waterloo in September!", "Solution_2": "me going to waterloo too!\r\n\r\nsoft eng !!!", "Solution_3": "[quote=\"max_tm\"]I see you just joined the UW forum (err, sorry for blowing your cover :P). So basically, you don't choose that major till second year. First year you would take two maths a CS and two electives. That means MATH 145 and 147 first semester (145 is a course in \"classical algebra\" and is more an introduction to formal mathematics than anything, and 147 is the calculus/analysis course) and 146 and 148 second semester (146 is Linear Algebra, 148 is the second half of 147). Then your choice of CS and electives. Matter of fact, you should be getting your first year in math book in a couple days, should answer most questions. \n\n\n\nAnd yes, I'm soo excited for Waterloo in September![/quote]\r\nUhhh I've been trying to join that forum for a while but I haven't been able to register. Maybe I should sort that out.\r\n\r\nI guess I will see you there!", "Solution_4": "Wait, so you're not the Steven on the forum who got a 94 on the Euclid?", "Solution_5": "ya i'm having trouble joining that forum too, it keeps saying i have an invalid UW ID..ya im going to UW Nanotech", "Solution_6": "[quote=\"max_tm\"]Wait, so you're not the Steven on the forum who got a 94 on the Euclid?[/quote]\r\nI am Steven and I got a 94 on the Euclid, but I did not make that post! Weird!", "Solution_7": "I remebered you got 93, steven............", "Solution_8": "[quote=\"Scarfy\"]I remebered you got 93, steven............[/quote]\r\nDude no, that's the UBC score, I got 94 after remarking at Waterloo.", "Solution_9": "lol...i dont even kno wut i got...\r\nsteven...how did u get a promise of scholarship from ubc if ur going to waterloo?...that was funny...", "Solution_10": "I'm going there in September too! Stream-4 (pure) math. See you all soon ... :lol: \r\n\r\nAnd if anyone has done math there already, how much homework is advanced math? What are the cheesiest electives? What is the more useful comp-sci course?", "Solution_11": "[quote=\"Ethanol\"]\nAnd if anyone has done math there already, how much homework is advanced math? [/quote]\r\n\r\ni heard advanced math was hard..idk if its completely true", "Solution_12": "I would say for a typical AoPSer, advanced math is exactly what they've wanted in a math course their whole lives but were not able to get in high school (obviously I've yet to do the course... I'm just basing this on what I know of the program and the fact that the reference textbook in MATH 147 is Spivak and the reference textbook in MATH 137 is Stewart... see the difference?) Most people don't take advanced math or find it difficult because they're accustomed to high school math and need that year to get used to a different style of instruction, different problem sets, etc. Not to mention that quite a few programs in the math degree aren't based around rigorous mathematics, so some people just aren't interested.\r\n\r\n\r\nAs for comp-sci courses... :P :\r\n\r\nTwo choices, Java-based and Scheme based:\r\n\r\nJava-based, three entry points:\r\nCS 125 - For the person who has never programmed, starts from scratch in Java.\r\nCS 133 - For people with some programming experience, not necessarily in an OO language.\r\nCS 134 - For people with 'extensive' programming experience in Java. This is actually the second semester course for 125 and 133 but can be taken in first semester.\r\n\r\nScheme-based, one entry point:\r\nCS 135 - This course differs from the others in that it is accessible to any level (from never-evers to experienced programmer) because the focus is a little shifted; the course is centered on learning good programming principles and design, outside the context of a specific programming language (it uses Scheme, which is a functional programming language based off of Lisp, but the course isn't about 'Scheme programming' per-se). This course is relatively new, but it's been really successful and, starting fall '08, they're replacing the java based courses with variants of this course. IMO, this is the one to be in.", "Solution_13": "Mhm, I'm going to Waterloo. Yay!", "Solution_14": "[quote=\"max_tm\"]I would say for a typical AoPSer, advanced math is exactly what they've wanted in a math course their whole lives but were not able to get in high school (obviously I've yet to do the course... I'm just basing this on what I know of the program and the fact that the reference textbook in MATH 147 is Spivak and the reference textbook in MATH 137 is Stewart... see the difference?) Most people don't take advanced math or find it difficult because they're accustomed to high school math and need that year to get used to a different style of instruction, different problem sets, etc. Not to mention that quite a few programs in the math degree aren't based around rigorous mathematics, so some people just aren't interested.\n\n\nAs for comp-sci courses... :P :\n\nTwo choices, Java-based and Scheme based:\n\nJava-based, three entry points:\nCS 125 - For the person who has never programmed, starts from scratch in Java.\nCS 133 - For people with some programming experience, not necessarily in an OO language.\nCS 134 - For people with 'extensive' programming experience in Java. This is actually the second semester course for 125 and 133 but can be taken in first semester.\n\nScheme-based, one entry point:\nCS 135 - This course differs from the others in that it is accessible to any level (from never-evers to experienced programmer) because the focus is a little shifted; the course is centered on learning good programming principles and design, outside the context of a specific programming language (it uses Scheme, which is a functional programming language based off of Lisp, but the course isn't about 'Scheme programming' per-se). This course is relatively new, but it's been really successful and, starting fall '08, they're replacing the java based courses with variants of this course. IMO, this is the one to be in.[/quote]\r\nThanks! As for residence, I'm hoping to get into a suite at MKV. They seem nice. What rez do you recommend?", "Solution_15": "everyone join UW network on facebook.", "Solution_16": "I've joined that Adavanced Math Waterloo Facebook group, btu never bothered looking into it. :P\r\nI guess I'm see many of you in Kitchener this coming September eh? cheers!\r\n\r\nPS And since many german-speaking people are still living over there, all you need to know is to say \"ich weiss nicht\" when they try to talk to ye. hahaha.", "Solution_17": "Yeah, I just finished the first year of advance math at UW. If you're the kind of person who does math for fun, and shows up on aops, you'll have a blast in the 140's at UW. You /will/ work hard, but you'll enjoy the work." } { "Tag": [ "algebra", "polynomial", "function", "algebra proposed" ], "Problem": "Let $P(x)=2x^{2}+6x^{3}+....+n(n-1)x^{n}$ where n is a natural number $n\\geq 2$ .Find $P(3)$ as a function of $n$ .", "Solution_1": "$P(x)=x^{2}(1+x+x^{2}+...+x^{n})''=x^{2}(\\frac{x^{n+1}-1}{x-1})''$." } { "Tag": [ "LaTeX" ], "Problem": "I'm writing up some lecture notes and want to refer back to a previously-defined equation, while still labelling it by its original equation number. To show you what i'm aiming at, the general scheme i've tried is:\r\n\r\n[in the preamble]\r\n \\newcounter{keepeqtn}\r\n\r\n[in the text body]\r\n \\begin{equation}\r\n a = b+c\r\n \\label{eq:one}\r\n \\end{equation}\r\n\r\n[yadah yadah yadah]\r\n\r\n \\setcounter{keepeqtn}{\\value{equation}}\r\n \\setcounter{equation}{\\value{\\ref{eq:one}}}\r\n \\begin{equation}\r\n a = b+c\r\n \\end{equation}\r\n \\setcounter{equation}{\\value{keepeqtn}}}\r\n\r\nbut this breaks at\r\n \\setcounter{equation}{\\value{\\ref{eq:one}}}\r\nwith a \"missing \\endcsname\" error\r\n\r\nAny ideas?\r\n[For info: i'm using \\documentclass{report}, and \"\\input\"ing (not \"\\include\"ing) separate \"chapters\". I'm interested in the general case where i can refer back to equations in previous \"chapters\" - although the scheme above breaks even for equations previously defined in the same \"chapter\"]", "Solution_1": "Does \\setcounter{equation}{\\ref{eq:one}} instead of \\setcounter{equation}{\\value{\\ref{eq:one}}} work?", "Solution_2": "Thanks for the suggestion, but no, it doesn't.\r\n\r\nAn alternative approach would just be something like\r\n a = b+c \\hfill (\\ref{eq: one})\r\nbut of course \\hfill doesn't work in math mode so the alignment comes out all wrong (and \\mbox-ing the \\hfill doesn't work)", "Solution_3": "I've just realised what you want to do, in which case, providing that you have loaded amsmath (which you are likely to have done) you can do it with \\tag and there's no need to use a counter. [code]\\usepackage{amsmath}\n...\n\\begin{equation} \na = b+c \n\\label{eq:one} \n\\end{equation} \n\n[yadah yadah yadah] \n\n\\begin{equation} \na = b+c \\tag{\\ref{eq:one}}\n\\end{equation} [/code]", "Solution_4": "Steve - right on the button, thanks very much!" } { "Tag": [ "inequalities", "geometry", "circumcircle", "geometry unsolved" ], "Problem": "$ AD$,$ BE$,$ CF$ are cevians in triangle $ ABC$(not necessarily concurrent), intersecting the circumcircle of triangle $ ABC$ in the points $ P$,$ Q$,$ R$. \r\nProve that $ \\frac{AD}{DP} \\plus{} \\frac{BE}{EQ} \\plus{} \\frac{CF}{FR} \\ge 9$", "Solution_1": "The bisectors $ AA_1, BB_1, CC_1$ ($ A_1,B_1,C_1$ lie on the sides) of the triangle $ ABC$ meets the circumcircle at again $ A_2,B_2,C_2$ respectively. \r\nDenote $ H$ be the foot of the $ A$-altitude. We have that $ d(P;BC)\\le d(A_2;BC)\\Rightarrow \\frac {AD}{DP} \\equal{} \\frac {AH}{d(P;BC)}\\ge \\frac {AH}{d(A_2;BC)} \\equal{} \\frac {AA_1}{A_1A_2}$\r\nwhere $ d(X,l)$ is the distance from $ X$ to the line $ l$.\r\nSimilarly, $ \\frac {BE}{EQ}\\ge \\frac {BB_1}{B_1B_2}$, $ \\frac {CF}{FR}\\ge \\frac {CC_1}{C_1C_2}$.\r\nThe inequality $ \\frac {AA_1}{A_1A_2} \\plus{} \\frac {BB_1}{B_1B_2} \\plus{} \\frac {CC_1}{C_1C_2} \\ge 9$ can be get easily :wink: .", "Solution_2": "[quote=\"keyree10\"]$ AD$,$ BE$,$ CF$ are cevians in triangle $ ABC$(not necessarily concurrent), intersecting the circumcircle of triangle $ ABC$ in the points $ P$,$ Q$,$ R$. \nProve that $ \\frac {AD}{DP} \\plus{} \\frac {BE}{EQ} \\plus{} \\frac {CF}{FR} \\ge 9$[/quote]\r\n\r\n\r\nSee also:\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=308544" } { "Tag": [ "inequalities", "geometry", "3D geometry", "tetrahedron" ], "Problem": "Derive a necessary and sufficient condition for six positive reals $ a, b, c, d, e, f$ to be the side lengths of a tetrahedron.", "Solution_1": "I think the short answer is that this is complicated. I haven't studied the [url=http://mathworld.wolfram.com/Cayley-MengerDeterminant.html]Cayley-Menger determinant[/url] in great detail, but it's possible that the determinant is positive if and only if there actually is a tetrahedron with given edge-lengths, so in principle you could just run through the 30 distinct permutations of the edges to see if any of the results are positive.\r\n\r\nOn [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83093]this thread[/url] I asked some questions about when we can construct more than one tetrahedron of given side-lengths:\r\n\r\n[quote=\"JBL\"]1. Find constraints on positive reals $ a, b$ such that:\n(a) There are three distinct tetrahedra with sides $ a, a, a, b, b, b$\n(b) There are only two such tetrahedra\n(c) There is only one such tetrahedron\n(d) There are no such tetrahedra\n\n2. What about if the tetrahedron is to have sides of length $ a, a, a, a, b, b$?[/quote]" } { "Tag": [ "calculus", "derivative", "inequalities", "inequalities solved" ], "Problem": "Show that for any $x \\in [0,1]$ we have $ 1+ \\frac{x}{2}- \\frac{x^2}{8} \\leq \\sqrt{1+x} \\leq 1+ \\frac{x}{2}$.\r\n\r\ncheers! :D :D :cool:", "Solution_1": "sqrt(1+x)\\leq 1+x/2 is easy to proof.. just square both sides..\r\ncan we use derivatives for the left inequality?", "Solution_2": "For 0 \\leq x , we have \r\n\r\n\\sqrt (1+x) \\leq 1+x/2 <===> 1+x \\leq 1 + x + (x/2) 2 <===> 0 \\leq x\r\n\r\n1+x/2-x2/8 \\leq |1+x/2-x2/8| \\leq \\sqrt (1+x) <===> 1+x + x2/4 - (x2/4)*(1+x/2) + x 4/64 \\leq 1+x <===> x2/16 - x/2 \\leq 0 <===> x \\leq 8\r\nAnd for 8 \\leq x we have 1+x/2-x2/8 < 0 < \\sqrt (1+x)", "Solution_3": "[quote=\"liyi\"]can we use derivatives for the left inequality?[/quote]\r\n\r\nwell.. using derivates is kinda easy.. lets stick please to elementary methods!\r\n\r\ncheers! thank you very much!", "Solution_4": "(1+x/2-x^2/8)^2 = 1+x-x^3/8+x^4/64\r\nwe need only to show that\r\n-x^3/8+x^4/64 \\leq 0\r\ni.e.,\r\nx^4\\leq 8x^3\r\nthis holds obviously for x\\in [0,1]." } { "Tag": [ "\\/closed" ], "Problem": "All the links in my post\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=198223\r\nhave been changed to some bullshiit by you righteous great Moderator. And I am not allow to edit it now. \r\n\r\nHave you no shame?\r\n\r\nIs this what you called freedom of speech?\r\n\r\nI must thank you for moving my first post to \r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=198129\r\n\r\nSeems like you can not edit the post there. :rotfl:", "Solution_1": "I don't think I ever said we are practicing freedom of speech here. This is mainly a [b]math and sciences[/b] forum. [b]It is NOT[/b] a forum where you complain about political situations. \r\n\r\nNext post of yours in this tone will make me ban you." } { "Tag": [ "AMC", "AIME", "MATHCOUNTS", "FTW", "USA(J)MO", "USAMO" ], "Problem": "How does the AMC/AIME committee choose names for characters in their problems? Just curious...", "Solution_1": "Lock this. No relevance at all.", "Solution_2": "Just because it's irrelevant doesn't mean it isn't fun. This might belong in G+FF, though.", "Solution_3": "[quote=\"gauss1181\"]Lock this. No relevance at all.[/quote]\r\nRelevant to what? This is a new topic. :) \r\nSure, a mod can move it wherever it needs to go.", "Solution_4": "Personally, I would hope that the characters are chosen so that they have unique first letters. Beyond that I don't really care.\r\n\r\nCan you just imagine a problem like this?\r\n\r\nBilly-Bob, Bimbo-bob and billy-bill each have some number of cows. Billy-bob and billy-bill have seven more than billy-bill and Bimbo-bob combined... (etc.)", "Solution_5": "THey used \"Mr.Dunbar\" like last year for a character name.\r\n\r\nMaybe they use the names of people that work on the AMC/AIME\r\n\r\nOr they pruposely make names that start with the same letter, so I have to denote them with different variables.", "Solution_6": "Look at problem 11 on the AMC 10B this year. Steve and LeRoy...Doesn't that sound like the names of two people involved with AMC? Hmmm...I wonder who they are...\r\n\r\nHELLO!!!!! Steve as in Mr. Dunbar and LeRoy as in the other person on the back of the AMC cover...(sry, whoever you are, I forgot your full name).", "Solution_7": "there was this city in MC called IMAFREEZIN", "Solution_8": "[quote=\"abacadaea\"]there was this city in MC called IMAFREEZIN[/quote]\r\n\r\nI remember seeing that.. and also i've seen\r\n\r\nAlice, Bob, and Casey are ....\r\n\r\nand a few more like that.", "Solution_9": "Al lot of the times on Canadian contests they'll do something like Andrew, Betty, Carlos :lol: (ABC get it... okay not funny I know). Maybe this is sometimes done on the Aimes Amcs and stuff.\r\nI remember one question where the names spelled out MATH", "Solution_10": "[quote=\"hadasah\"]Look at problem 11 on the AMC 10B this year. Steve and LeRoy...Doesn't that sound like the names of two people involved with AMC? Hmmm...I wonder who they are...\n\nHELLO!!!!! Steve as in Mr. Dunbar and LeRoy as in the other person on the back of the AMC cover...(sry, whoever you are, I forgot your full name).[/quote]\r\nHmm... [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1052015#1052015[/url]\r\nDo you mean AMC 10A?", "Solution_11": "they go to the big fat book of [b]Choosing Names for Contest Problems[/b].", "Solution_12": "I like problems where they are called Albert, Bert, Carlos, etc. or something of the sort. I distinctly remember, however, on the AMC this year a Brenda and Bobby problem.", "Solution_13": "As a totally unrelated comment, when I name people, I tend to use my own name, and the names of the people receiving the test. I always go to multicultural names, too. Like take the MATHCOUNTS test I'm writing. On one of the problems, my four people are Ahmed, Bhat, Carl, and Delfonso.\r\n\r\nAs an even more unrelated comment, on English vocab or grammar quizzes, when we have to make up sentences, I always name my people with stereotypical names, like Muhammad, Jean-Phillipe, Mr. Nguyen, Hsin-pu, Toyoko, Mthomba Mb'dbonouoe (MP 06 FTW), etc.\r\n\r\nI can't wait for the day where Zuming is part of an AMC problem. If you want to find a winning strategy for one of Zuming and someone else on the USAMO, assume Zuming is going to win. :D", "Solution_14": "Hmm, I don't usually think about these things.\r\n\r\n@Catalyst: Lol, whenever I make names for my own problems, the only ones I ever use are Joe, Bob, and Samantha. Yes, I need more creativity. :D", "Solution_15": "[quote=\"randomdragoon\"]Personally, I would hope that the characters are chosen so that they have unique first letters. Beyond that I don't really care.\n\nCan you just imagine a problem like this?\n\nBilly-Bob, Bimbo-bob and billy-bill each have some number of cows. Billy-bob and billy-bill have seven more than billy-bill and Bimbo-bob combined... (etc.)[/quote]\r\nermm.. Bimbo-bob? ...\r\n\r\nMy names = Person A, B, C, etc. Amazing, I know", "Solution_16": "The best names to use are Albert, Joe, Bob, Tyler, Arlene, and Vladmir.\r\n\r\nHowever, I don't remember any of the above names being used.", "Solution_17": "Sometimes is seems like they pick names that are from different languages and ethnicities just to seem like they're all multiculturalized and everything....", "Solution_18": "Lol, on [i]some[/i] Canadian Contests (the COMC), the names are always something like Alphonse and Beryl... Beryl usually wins... :D", "Solution_19": "[quote=\"CatalystOfNostalgia\"]I can't wait for the day where Zuming is part of an AMC problem. If you want to find a winning strategy for one of Zuming and someone else on the USAMO, assume Zuming is going to win. :D[/quote]\r\n\r\n*facepalms* I would laugh if I saw that.. but i don't think that Mr. Feng is going to allow that.. :)", "Solution_20": "[quote=\"CatalystOfNostalgia\"]As a totally unrelated comment, when I name people, I tend to use my own name, and the names of the people receiving the test. I always go to multicultural names, too. Like take the MATHCOUNTS test I'm writing. On one of the problems, my four people are Ahmed, Bhat, Carl, and Delfonso.\n[/quote]\r\n\r\nYou mean the one about the murder and you have to figure out who committed it?" } { "Tag": [], "Problem": "Consider the following heterogeneous equilibrium:\r\n\r\n$NH_{2}CO_{2}NH_{4}(s) \\rightleftharpoons 2NH_{3}(g)+CO_{2}(g)$.\r\n\r\nDiscuss the effect that the addition of carbon dioxide at constant temperature and pressure might have on this equilibrium.", "Solution_1": "Problem still not solved.", "Solution_2": "[quote=\"Carcul\"]Consider the following heterogeneous equilibrium:\n\n$ NH_{2}CO_{2}NH_{4}(s)\\rightleftharpoons 2NH_{3}(g)\\plus{}CO_{2}(g)$.\n\nDiscuss the effect that the addition of carbon dioxide at constant temperature and pressure might have on this equilibrium.[/quote]\r\n\r\nSince no one tried this problem, then I'll give it a try. One way to look at this problem is to look at the gas equation PV = nRT. Since R is the universal constant and P and T are unchanging, the increase in the number of moles of carbon dioxide will increase the volume. According to La Chatelier's principle, when there is a change in the concentration, pressure, volume, or total pressure, the equilibrium will shift in such a way as to counteract this change. Thus, to decrease the volume, the equilibrium will shift to the reactant side with the solid. The amount of $ NH_{2}CO_{2}NH_{4}$ will increase, the amount of NH3 will decrease and the amount of CO2 will decrease but will still be more than before the additional CO2 was added.", "Solution_3": "That answer is only partially correct. On other words, that will only happen if a certain condition is met: otherwise, that will not happen.", "Solution_4": "well thsi may come handy the concentration of solid is 1 or nearly always the same", "Solution_5": "No, it doesn't have nothing to do with that.", "Solution_6": "[quote=\"Carcul\"]That answer is only partially correct. On other words, that will only happen if a certain condition is met: otherwise, that will not happen.[/quote]\r\n\r\nDo you mean that under constant temperture and pressure, the Kc would have to be more than 1 for this to happen. And that the number of molecules in the gas state on the product side would have to be more than the number of molecules in the gas state on the reactant side so the volume would be able to decrease.", "Solution_7": "At constant temperature and pressure Kc is constant. What I mean is, if a certain condition (to be found) is met, then adding carbon dioxide will shift the equilibrium to the right (producing more carbon dioxide!). If that condition is not met, then the usual prediction (the reaction will shift to the left) will happen. Now, what is that condition?", "Solution_8": "Just wondering, if you add more $ CO_{2}$ and keep the pressure constant, obviously you would need to increase the volume of the overall system. What if you increase the volume in such a way as to decrease the overall concentration of $ CO_{2}$ (i.e. you increase the volume so much that $ [CO_{2}]$ actually becomes smaller than it was originally)." } { "Tag": [], "Problem": "At one hospital, there are 100 patients, all of whom have at least one of the following ailments: a cold, the flu, or an earache. 38 have a cold, 40 have the flu, and some number have earaches. If 17 have both colds and earaches, and 7 have all three, how many have earaches?", "Solution_1": "patients having earache=46.\r\nI used Venn diagram to solve this. To see the diagram open the image file attached", "Solution_2": "hmm, I agree with varun...... for now.", "Solution_3": "Good job, that's correct.", "Solution_4": "yup ..............sets and stuff , venn 's diagram's that is ....... same what i did.......\r\n\r\nisnt it that the venn's diagram's stuff that we do has been derived from algebra, itself? cos these sums you can do algebraically( possibly arithmetically too ) only venn's diagrams makes it a whole lot easier :P", "Solution_5": "[hide]i get 29. i don't know what i'm doing wrong. Can somebody post a solution?[/hide]", "Solution_6": "I think that you counted some of the overlaps of the venn-diagram twice." } { "Tag": [ "\\/closed" ], "Problem": "In the \"Solved Reports\" forum, the description says that you can't make new topics, but in the forum, there is a link that makes a new topic. Is this supposed to happen?", "Solution_1": "Click on it and see what happens. :wink: \r\n\r\n[quote]Only users granted special access can post topics in this forum.[/quote]" } { "Tag": [ "function", "inequalities proposed", "inequalities" ], "Problem": "Let a,b,c be positive number \r\nProve that\r\n$ \\sum_{cyc}\\frac{a^2}{\\sqrt{a^4\\plus{}24b^3c^3}} \\geq \\frac{3}{5}$", "Solution_1": "[quote=\"8826\"]Let a,b,c be positive number \nProve that\n$ \\sum_{cyc}\\frac {a^2}{\\sqrt {a^4 \\plus{} 24b^3c^3}} \\geq \\frac {3}{5}$[/quote]\r\nIt's wrong. Try $ a\\equal{}b\\equal{}c\\equal{}2.$ :wink:", "Solution_2": "[quote=\"8826\"]Let a,b,c be positive number \nProve that\n$ \\sum_{cyc}\\frac {a^2}{\\sqrt {a^4 \\plus{} 24b^3c^3}} \\geq \\frac {3}{5}$[/quote]\r\nI suppose you mean:\r\n$ \\sum_{cyc}\\frac {a^2}{\\sqrt {a^4 \\plus{} 24b^2c^2}} \\geq \\frac {3}{5}$, or equivalently:\r\n$ P \\equal{} \\sum_{cyc}\\frac {a}{\\sqrt {a^2 \\plus{} 24bc}} \\geq \\frac {3}{5}$.\r\nThen H\u00f6lder gives us:\r\n$ \\left ( \\sum_{cyc} a^3\\plus{}24abc \\right )P^2 \\ge \\left ( \\sum_{cyc} a \\right )^3$, so it suffices to prove\r\n\r\n$ 25 \\left ( \\sum_{cyc} a \\right )^3 \\ge 9\\left ( \\sum_{cyc} a^3\\plus{}24abc \\right )$, which is obvious upon expanding...", "Solution_3": "If it is as as Mathias said...\r\nAnother solution is(dedicated to quykhtn-qa1):\r\n$ \\sum\\frac {a^2}{\\sqrt {a^4 \\plus{} 24b^2c^2}} \\equal{} \\sum\\sqrt {\\frac {a^4}{a^4 \\plus{} 24b^2c^2}} \\equal{} \\sum\\frac {1}{\\sqrt {1 \\plus{} \\frac {24a^2b^2c^2}{a^6}}}$\r\nWe consider the function\r\n$ f(x) \\equal{} \\frac {1}{\\sqrt {1 \\plus{} \\frac {24a^2b^2c^2}{x^6}}}$\r\n$ f''(x)\\geq 0$\r\nSo $ f(a) \\plus{} f(b) \\plus{} f(c) \\geq 3f(\\frac {a \\plus{} b \\plus{} c}{3})$\r\nSo the $ LHS \\geq 3\\sum\\frac {1}{\\sqrt {1 \\plus{} \\frac {3^6*24a^2b^2c^2}{(a \\plus{} b \\plus{} c)^6}}}$\r\nSo we have to prove that:\r\n$ \\sum\\frac {1}{\\sqrt {1 \\plus{} \\frac {3^6*24a^2b^2c^2}{(a \\plus{} b \\plus{} c)^6}}}\\geq \\frac {1}{5}\\Leftrightarrow(a \\plus{} b \\plus{} c)^6 \\geq 3^6 a^2b^2c^2$ which is true :)" } { "Tag": [ "puzzles" ], "Problem": "1.Take away my first letter; take away my second letter; take away \r\nall my letters, and I would remain the same. What am I?\r\n\r\n2.The more you take the more you leave behind.\r\n\r\nWhat are they?\r\n\r\n3.Where is the only place that yesterday always follows today?\r\n\r\n4.A certain large animal lives happily and thrives here on Earth. \r\nOne day, every single one of these critters is wiped out by a \r\nmysterious disease which affects only this particular animal. \r\nThere are none left anywhere on earth -- they are all gone. Dead. \r\nAbout a year or so later, they begin to reappear on Earth again. \r\nHow can this be?\r\n\r\nThe animals were not created artifically by humans", "Solution_1": "1.\r\n[hide]postal carrier[/hide]", "Solution_2": "3\r\n[hide=\"3\"]a dictionary[/hide]", "Solution_3": "[hide=\"2\"]Footprints[/hide]", "Solution_4": "[hide] Hint on the fourth puzzle: Try to find out what kind of animal it is[/hide]", "Solution_5": "[hide=\"4\"]\nare they mules? (mix between horses and donkeys, they can't actually breed among each other.)\n[/hide]", "Solution_6": "[quote=\"Hamster1800\"][hide=\"4\"]\nare they mules? (mix between horses and donkeys, they can't actually breed among each other.)\n[/hide][/quote]\n[hide]Mules is the animal I was thinking of. Correct answer[/hide]", "Solution_7": "[hide=\"1.\"] mailman?[/hide]", "Solution_8": "D3m0n ShaD0w, \r\n\r\nyour answer to the first problem \r\n\r\nis equivalent to the one that I \r\n\r\nALREADY gave immediately following\r\n\r\nthe original post.", "Solution_9": "[quote=\"Hamster1800\"][hide=\"4\"]\nare they mules? (mix between horses and donkeys, they can't actually breed among each other.)\n[/hide][/quote]\r\n\r\nwell don't ligers have more issues? not only can they not interbreed, but they are larger and fit the description", "Solution_10": "[quote=\"box of rocks\"]D3m0n ShaD0w, \n\nyour answer to the first problem \n\nis equivalent to the one that I \n\nALREADY gave immediately following\n\nthe original post.[/quote]\r\n\r\noh sorry i didnt know i dont read hides unless i myself give a brainteaser" } { "Tag": [ "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Prove that in locally compact spaces, a subspace is of dimension 0 iff it is totally disconnected.\r\n\r\n1. We talk about second countable metric spaces only.\r\n2. dimension 0 means for any $p \\in X$, there exists an arbitrary small neighborhood which is open and close.\r\n\r\nIn general, $\\Rightarrow$ is true, but how to prove the reverse direction?", "Solution_1": "Take $X$ a totally disconnected compact space and $p\\in X$. We must find arbitrary small open-closed neighborhoods, so take an open neighborhood $U$ of $p$. For each $y\\in X-U$, there is an open-closed set $V_{y}$ such that $y\\in V_{y}\\subset X-\\{p\\}$. Cover $X-U$ by a finite number of $V_{y}$ (a closed set of a compact space is compact), the complement of the union of that finite number works as an open-closed neighborhood of $p$ inside $U$.\r\n\r\n :D", "Solution_2": "Please note that I am talking about LOCALLY compact spaces." } { "Tag": [ "puzzles" ], "Problem": "61, 52, 63, 94, 46, ?\r\n\r\n[hide=\"Hint\"]Quite familiar numbers.\nDont' need any calculation.[/hide]", "Solution_1": "[hide]LOL it's the two-digit squares backwards! The next one is 18[/hide]", "Solution_2": "Right :lol:" } { "Tag": [], "Problem": "I'm reading this book called Catch-22 by Joseph Heller now. It seems like a good book and I read about 60 pages but there is one question: What is the plot?? I mean, I read 60 pages so I'm hoping to see some actions but there isn't really anything going on. Maybe I'm not reading this well so perhaps someone who read this book can give me general outline of the story (don't tell how it will end or stuff like that). I know it's about war and something about U.S.O. (not sure what this is as well).\r\n\r\nI appreciate this greatly.", "Solution_1": "[url=http://www.sparknotes.com/lit/catch22/]Sparknotes[/url] is your friend. As far as I remember, the plot was pretty much random throughout the whole book... but pretty funny anyway :P.", "Solution_2": "I've been meaning to read this book...", "Solution_3": "I'm reading the book too (about page 50ish) and it hilarious. You must remember that it's not one of those traditionally good books... at first it got very bad reviews until people realized the hilarity and greatness of the book." } { "Tag": [ "modular arithmetic", "absolute value", "AMC" ], "Problem": "1. The digits of a positive integer [i]n[/i] are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 37? \r\n\r\nso it's like 1000x+100(x-1)+10(x-2)+(x-3)??? but how can I find out the remainder?\r\n\r\n2. Set A consists of [i]m [/i]consecutive integers whose sum is 2[i]m[/i], and set B consists of 2[i]m[/i] consecutive integers whose sum is [i]m[/i]. The absolute value of the difference between the greatest element of A and the greatest element of B is 99. Find m.\r\n\r\nCan any of you give me any clue, thanks.", "Solution_1": "The integers are $ 9876$, $ 8765$,...$ 4321$. The remainder when $ 4321$ is divided by $ 37$ is $ 29$. Each successively larger such $ n$ is $ 1111$ plus the previous $ n$. $ 1111 \\equiv 1 \\pmod{37}$, so the sum of the remainders is\r\n\\[ 29 \\plus{} 30 \\plus{} 31 \\plus{} 32 \\plus{} 33 \\plus{} 34 \\equiv \\minus{} 8 \\minus{} 7 \\minus{} 6 \\minus{} 5 \\minus{} 4 \\minus{} 3 \\equiv \\minus{} 33 \\equiv \\boxed{4} \\pmod{37}.\r\n\\]\r\n\r\nFor number 2, let the smallest element of $ A$ be $ x$ and the smallest element of $ B$ be $ y$. Where can you go from there?", "Solution_2": "We want sum of the remainders, not mod 37 of sum of remainders :)\r\n\r\nSo answer is 189." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a, b, c$ be non-negative real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove or disprove that\r\n\r\n$ \\sum\\frac{a^4(a\\plus{}1) \\plus{} b^5}{a^3 \\plus{} b^3}\\ge \\frac{5}{6}$ .\r\n\r\n_____________\r\n :?:", "Solution_1": "Yes it is true!\r\nFollows from\r\n1) $ \\sum_{cyc}\\frac{a^4}{a^3\\plus{}b^3}\\geq \\frac{a\\plus{}b\\plus{}c}{2}$\r\n2) $ \\sum_{cyc}\\frac{a^k\\plus{}b^k}{a\\plus{}b}\\geq (a^{k\\minus{}1}\\plus{}b^{k\\minus{}1}\\plus{}c^{k\\minus{}1})$ ($ k\\geq 1$)", "Solution_2": "[quote=\"Albanian Eagle\"]Yes it is true!\nFollows from\n1) $ \\sum_{cyc}\\frac {a^4}{a^3 \\plus{} b^3}\\geq \\frac {a \\plus{} b \\plus{} c}{2}$\n2) $ \\sum_{cyc}\\frac {a^k \\plus{} b^k}{a \\plus{} b}\\geq (a^{k \\minus{} 1} \\plus{} b^{k \\minus{} 1} \\plus{} c^{k \\minus{} 1})$ ($ k\\geq 1$)[/quote]\r\n\r\nused technique is very nice. Thanks \r\n\r\nAspect further resolutions...." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Hello!!\r\n\r\nLet $a_{1},a_{2},\\cdots,a_{n}$ be $n$ positive real numbers such that $a_{1}\\cdot a_{2}\\cdots\\cdot a_{n}=1$. Prove that : \\[a_{1}^{n}(1+a_{1})+a_{2}^{n}(1+a_{2})+\\cdots+a_{n}^{n}(1+a_{n})\\geq{{n}\\over{2^{n-1}}}(1+a_{1})\\cdots(1+a_{n}).\\]", "Solution_1": "known fact (*) $\\prod a_{i}=1, 0a(k+2) \r\nIf a(k+2)>a(k+3) then we change a(k+2) with a(k+3) and we get S'\r\n\r\nWe see that S\u2264S'. Indeed:\r\nLet {a(k);a(k+1);a(k+2)}={a>b>c}\r\n If a(k+3)>b then S= a-2c+a(k+3) =S' \r\n If b>a(k+3)>c then S= a-2c+a(k+3)a(k+3) then S=a-a(k+3)1, then $ 2^m\\minus{}1|\\frac{2^n\\minus{}1}{3}$. But $ 2^m\\minus{}1\\equal{}7\\mod 8$ and exist prime divisor $ q\\equal{}3\\mod 4$, suth that $ q|2^m\\minus{}1$.\r\nTherefore second condition equavalent to $ n\\equal{}2^k$, first $ k\\ge 1$.", "Solution_2": "Thanks for your nice solution." } { "Tag": [ "induction" ], "Problem": "$ a_{1} \\equal{} 1$, $ a_{k} \\equal{} a_{k\\minus{}1} \\plus{} 3$ for every natural number $ k>1$. Find $ a_{n}$ in terms of $ n$.", "Solution_1": "[hide]$ a_{n}\\equal{}3n\\minus{}2$[/hide]", "Solution_2": "Thanks dgreenb801.\r\n\r\n[hide=\" Proof by mathematical induction:\"]\n\nBasis step:\n\nFor $ n\\equal{}1$, $ a_{n} \\equal{} a_{1} \\equal{} 1 \\equal{} 3(1)\\minus{}2$\n\nInductive step:\n\nIf $ a_{k} \\equal{} 3k\\minus{}2$, then $ a_{k\\plus{}1} \\equal{} a_{k} \\plus{} 3 \\equal{} 3k\\minus{}2 \\plus{} 3 \\equal{} 3(k\\plus{}1) \\minus{} 2$\n\n[/hide]", "Solution_3": "$ a_n \\equal{} a_1 \\plus{} \\sum_{k\\equal{}2}^n (a_k \\minus{} a_{k\\minus{}1}) \\equal{} 1 \\plus{} 3(n\\minus{}1) \\equal{} \\boxed{3n \\minus{} 2}$.\r\nBy [url=http://mathworld.wolfram.com/TelescopingSum.html]Telescoping Sum[/url]", "Solution_4": "Thanks Kouichi." } { "Tag": [ "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "A graph has $n$ vertix.Prove that:\r\n$\\chi(G).\\chi(\\overline{G}) \\geq n$", "Solution_1": "this is also trivial.since an independent set in $G$ becomes a clique in $\\overline{G}$ it follows that $\\alpha(G)\\leq\\chi(\\overline{G})$. the inequality follows since $\\chi(G)\\geq n/\\alpha(G)$." } { "Tag": [ "LaTeX", "inequalities", "continued fraction" ], "Problem": "Hello, everyone; I'm sorry to disturb you again, but: I found this formula/series which converges on $ \\sqrt{n}$ for some non-negative integer \"n\". The first term of the series should always be 1, and the recursive formula of the series should be $ (x\\plus{}ny)/(x\\plus{}y)$ (Sorry I'm bad at LaTex), or $ 1\\plus{}y(n\\minus{}1)/(x\\plus{}y)$. The idea is that the series should slowly converge to $ \\sqrt{n}$. I tested this for a few examples (not many, just a few), and I was wondering whether someone here would be kind enough to either prove or disprove this (Probably this might be something simple, so I'm sorry if this is too easy for you)", "Solution_1": "What are $ x$ and $ y$?", "Solution_2": "Oh yeah, sorry :( . x/y should be the term that precedes the new term (I'm sorry if the language is confusing, as in the first term of the series for $ \\sqrt5$ is 1, so $ (1\\plus{}5(1))/(1\\plus{}1)$, or $ 6/2$ is the next term (well, it's faster if you don't simplify the fractions), and the term after that is $ (6\\plus{}5(2))/(6\\plus{}2)\\equal{}16/8$, etc.", "Solution_3": "So you've got a sequence $ a_k$ defined by $ a_1 \\equal{} 1, a_{k\\plus{}1} \\equal{} \\frac{a_k \\plus{} n}{a_k \\plus{} 1}$. The fixed points of the transformation $ q \\mapsto \\frac{q \\plus{} n}{q \\plus{} 1}$ are in fact $ \\pm \\sqrt{n}$, so if the sequence converges at all it must converge to one of these two numbers.\r\n\r\nThere are a couple of ways to do it from here.\r\n\r\n[hide=\"Method 1\"] $ a_{k\\plus{}1} \\equal{} 1 \\plus{} \\frac{n\\minus{}1}{1 \\plus{} a_k}$ gives a (non-simple) [url=http://en.wikipedia.org/wiki/Continued_fraction]continued fraction[/url], and there are some standard convergence theorems for these. [/hide]\n[hide=\"Method 2\"] Figure out what bound you need on $ a_k$ to apply the [url=http://en.wikipedia.org/wiki/Banach_fixed_point_theorem]contraction principle[/url] to $ f(x) \\equal{} \\frac{x \\plus{} n}{x \\plus{} 1}$. [/hide]\n[hide=\"Method 3\"] Explicitly show that the quantity $ \\left| \\sqrt{n} \\minus{} a_{k\\plus{}1} \\right|$ is decreasing to zero, for example by computing it as $ \\left| \\sqrt{n} \\minus{} \\frac{a_k \\plus{} n}{a_k \\plus{} 1} \\right|$. [/hide]", "Solution_4": "Thank you, though I can't really understand this because I'm not advanced (well probably, it is easy, but I have barely begun to even understand math yet). Can you please simplify this (is there any basic algebra-style method for this). Thanks for all your help.", "Solution_5": "Method 3 is probably the most elementary way to do it, but the argument doesn't seem to carry through unless you apply the recursion twice; this is because the sequence $ a_k$ approaches $ \\sqrt {n}$ alternately from the top and bottom, and the error in the approximation isn't monotonic.\r\n\r\n[hide=\"Details\"] I claim that the inequality\n\n$ \\left| \\sqrt {n} \\minus{} a_{k \\plus{} 2} \\right| < \\left| \\sqrt {n} \\minus{} a_k \\right| \\left( \\frac {n}{n \\plus{} 1} \\right)$\n\nholds. Since $ \\frac {n}{n \\plus{} 1} < 1$, this implies convergence as $ k \\to \\infty$.\n\nThe proof is \"basic algebra-style\": first write $ a_{k \\plus{} 2} \\equal{} \\frac {a_{k \\plus{} 1} \\plus{} n}{a_{k \\plus{} 1} \\plus{} 1} \\equal{} \\frac { (n \\plus{} 1) a_k \\plus{} 2n}{ 2 a_k \\plus{} (n \\plus{} 1) }$ by applying the recurrence twice, then compute\n\n$ \\left| \\sqrt {n} \\minus{} \\frac {(n \\plus{} 1) a_k \\plus{} 2n}{2a_k \\plus{} (n \\plus{} 1)} \\right| \\equal{} \\left| \\frac {(a_k \\minus{} \\sqrt {n})(\\sqrt {n} \\minus{} 1)^2}{2a_k \\plus{} (n \\plus{} 1)} \\right| \\equal{} \\left| \\sqrt {n} \\minus{} a_k \\right| \\frac { (\\sqrt {n} \\minus{} 1)^2}{2a_k \\plus{} (n \\plus{} 1)}$\n\nand then note that $ (\\sqrt {n} \\minus{} 1)^2 < n, 2a_k \\plus{} (n \\plus{} 1) > (n \\plus{} 1)$. The result follows. [/hide]\r\n\r\nI should also mention that it is actually possible to write down closed forms for sequences of the form $ s_{k \\plus{} 1} \\equal{} \\frac {as_k \\plus{} b}{cs_k \\plus{} d}$ using linear algebra. The nature of these closed forms provides a good justification for why these convergence results are true of fractional linear transformations.\r\n\r\nAlso, Method 2 probably doesn't work.", "Solution_6": "$ \\frac{q\\plus{}n}{1\\plus{}q} \\minus{} q \\equal{} \\frac{n\\minus{}q^2}{1\\plus{}q}$. So, for $ \\minus{}1 < q < \\sqrt{n}$, $ q$ increases through the transformation, and for $ \\sqrt{n} < q$, the transformation reduces $ q$.\r\n\r\n[u]Here's a question I haven't tried:[/u] For what $ m$ does $ a_{n\\plus{}1} \\mapsto \\frac{ma_n\\plus{}n}{a_n\\plus{}m}$ with initial value $ a_0$ converge to $ \\sqrt{n}$?", "Solution_7": "Oh thank you. Now I understand it.", "Solution_8": "[quote=\"TZF\"]for $ \\minus{} 1 < q < \\sqrt {n}$, $ q$ increases through the transformation, and for $ \\sqrt {n} < q$, the transformation reduces $ q$.[/quote]\nThis only proves boundedness, not convergence.\n\n[quote=\"TZF\"][u]Here's a question I haven't tried:[/u] For what $ m$ does $ a_{n \\plus{} 1} \\mapsto \\frac {ma_n \\plus{} n}{a_n \\plus{} m}$ with initial value $ a_0$ converge to $ \\sqrt {n}$?[/quote]\r\nAgain the fixed points are $ \\pm \\sqrt{n}$, so if the sequence converges at all it converges to one of the square roots. The eigenvalues of $ \\left[ \\begin{array}{cc} m & n \\\\ 1 & m \\end{array} \\right]$ are $ m \\pm \\sqrt{n}$, so convergence clearly can't occur for $ m \\equal{} 0$ but should occur for all real $ m > 0$. Things get iffy for negative $ m$, since the denominator can blow up." } { "Tag": [], "Problem": "a convex $ 2m$-gon has vertecies; $ A_1,A_2,A_3,...A_{2m}$. If a point, P, lies inside the $ 2m$-gon and does not lie on a diagonal, show that it lies inside an even number of triagnles, with vertecies at $ A_1,A_2...A_{2m}$.", "Solution_1": "Cute!\r\n\r\n[hide]Choose a point very close* to edge $ A_{2m}A_1$. Then this point is contained exactly in triangles $ A_{2m}A_1A_i$ for $ i \\in \\{2, 3, \\ldots, 2m \\minus{} 1\\}$, an even number. Now, it's clear that if we move a point around without crossing any diagonals, we don't change the number of triangles in which it lies. So we just need to show that if we move from one side of a diagonal (and very close to that diagonal) to the other side (and still very close to the diagonal), we don't change the parity. But it's easy to see that when a point is very close to diagonal $ A_iA_j$ and then crosses the diagonal, it stays in every triangle it was in except those of the form $ A_iA_jA_k$ and it stays out of every triangle it was out of except those of the form $ A_iA_jA_k$. But since there are a total of $ 2m \\minus{} 2$ triangles of this form, we're done.\n\n* \"Very close\" means \"close enough so there are no diagonals between the point and the line segment in question.\"\n\nThis method is essentially what Po-Shen Loh described as \"brutal force\" when he taught it at MOP, lo those many years ago.[/hide]" } { "Tag": [ "geometry", "MATHCOUNTS", "3D geometry", "number theory", "prime numbers" ], "Problem": "Hard puzzeling math problems that make you think out of the box.", "Solution_1": "Do you want us to post \"puzzle problems\"?", "Solution_2": "[quote=\"goldendomer\"]Do you want us to post \"puzzle problems\"?[/quote]\r\n\r\nYes, that is correct.", "Solution_3": "I thought we couldn't create marathons...", "Solution_4": "[quote=\"i_like_pie\"]I thought we couldn't create marathons...[/quote]\r\n\r\nI thought so too.", "Solution_5": "Marathon #2 has been goin on and i dont think a mod or administrator has said anything.", "Solution_6": "[quote=\"indianamath\"]Marathon #2 has been goin on and i dont think a mod or administrator has said anything.[/quote]\r\n\r\nAre you talking about Marathon 2 in the MATHCOUNTS forum? If you are, it's the only going in the MC forum. #1 was locked because it was way too big.", "Solution_7": "Here is your fist problem:\r\n\r\nThe is a planet named Zoobop.\r\nZoobog is a peculiar planet, in the sence that it has two \"hemispheres\" that are shaped like right-circular cones.\r\nThe two cones are congruent, and are joined at their circular bases. The height of the cones are 3000 miles each, and they have a base radius of 4000 miles.\r\nTwo cities are located on the equator of Zoobop, 180 degrees apart.\r\n\r\nWhat is the shortest distance between the two cities?", "Solution_8": "its 4000pi", "Solution_9": "[quote=\"hairymonkey2\"]its 4000pi[/quote]\r\n\r\nNope.\r\n\r\n[hide]Hint: unfold the cone[/hide]", "Solution_10": "the answer is : 0 pi haha\r\nsphericon", "Solution_11": "[quote=\"8parks11\"]the answer is : 0 pi haha\nsphericon[/quote]\r\n\r\nWrong. Hint: The answer does not include pi.", "Solution_12": "isnt it 0?\r\nthe pi was a joke", "Solution_13": "isnt it just\r\n[hide]8000[/hide]", "Solution_14": "[quote=\"8parks11\"]isnt it 0?\nthe pi was a joke[/quote]\r\n\r\nThe question is basically asking...", "Solution_15": "i understand haha\r\ni thought this was a brain teaser\r\nso i just said 0\r\n\r\nshortest would be to take the triangular arc?\r\nor the circumference? or the arc of the half of the circle?", "Solution_16": "where did you get the name \"Zoboop\" anyway :huh: ?", "Solution_17": "[quote=\"8parks11\"]i understand haha\ni thought this was a brain teaser\nso i just said 0\n\nshortest would be to take the triangular arc?\nor the circumference? or the arc of the half of the circle?[/quote]\r\n\r\nunfold ONE cone into a pie-like section with a piece missing. The draw a straight line between the two points on the section.\r\nThis is the shortest distance", "Solution_18": "What is this question asking?", "Solution_19": "Isn't it just 8000 miles?\r\nOr possibly 2000 pi?", "Solution_20": "[quote=\"hrzdlake\"]Isn't it just 8000 miles?\nOr possibly 2000 pi?[/quote]\r\nif they go through it is 8000", "Solution_21": "[quote=\"funcia\"][quote=\"hrzdlake\"]Isn't it just 8000 miles?\nOr possibly 2000 pi?[/quote]\nif they go through it is 8000[/quote]\r\nHe says that there is no pi in the answer...doesn't there have to be though?", "Solution_22": "[quote=\"snipefreak\"][quote=\"funcia\"][quote=\"hrzdlake\"]Isn't it just 8000 miles?\nOr possibly 2000 pi?[/quote]\nif they go through it is 8000[/quote]\nHe says that there is no pi in the answer...doesn't there have to be though?[/quote]\r\nnot if dig hole to china", "Solution_23": "[quote=\"hrzdlake\"]Isn't it just 8000 miles?\nOr possibly 2000 pi?[/quote]\n\n[quote=\"pn12345\"]The answer does not include pi.[/quote]", "Solution_24": "[quote=\"buzzer11\"][quote=\"hrzdlake\"]Isn't it just 8000 miles?\nOr possibly 2000 pi?[/quote]\n\n[quote=\"pn12345\"]The answer does not include pi.[/quote][/quote]\r\nspam?!\r\nno, that person didn't read question", "Solution_25": "[quote=\"pn12345\"][quote=\"8parks11\"]i understand haha\ni thought this was a brain teaser\nso i just said 0\n\nshortest would be to take the triangular arc?\nor the circumference? or the arc of the half of the circle?[/quote]\n\nunfold ONE cone into a pie-like section with a piece missing. The draw a straight line between the two points on the section.\nThis is the shortest distance[/quote]\r\nthe shortest distance between two points is a direct straight line. it does not have to be above the surface. it can be just a direct line. the question did not ask for shortest traveling distance, just shortest distance.", "Solution_26": "Question 1: is it 10000 miles...oops, I didnt see the posts about 8000 miles, and explanations about... would 10000 miles be the shortest traveling distance, though?\r\nQuestion 2:52?", "Solution_27": "[quote=\"pn12345\"]Here is your fist problem:\n\nThe is a planet named Zoobop.\nZoobog is a peculiar planet, in the sence that it has two \"hemispheres\" that are shaped like right-circular cones.\nThe two cones are congruent, and are joined at their circular bases. The height of the cones are 3000 miles each, and they have a base radius of 4000 miles.\nTwo cities are located on the equator of Zoobop, 180 degrees apart.\n\nWhat is the shortest distance between the two cities?[/quote]\r\nWouldn't you just double the height, giving you 8000 miles?", "Solution_28": "[quote=\"erdogankerem123\"]Question 1: is it 10000 miles...oops, I didnt see the posts about 8000 miles, and explanations about... would 10000 miles be the shortest traveling distance, though?\nQuestion 2:52?[/quote]\r\nFor question 2 I get 52 too.", "Solution_29": "[quote=\"pn12345\"]Here is your fist problem:\n\nThe is a planet named Zoobop.\nZoobog is a peculiar planet, in the sence that it has two \"hemispheres\" that are shaped like right-circular cones.\nThe two cones are congruent, and are joined at their circular bases. The height of the cones are 3000 miles each, and they have a base radius of 4000 miles.\nTwo cities are located on the equator of Zoobop, 180 degrees apart.\n\nWhat is the shortest distance between the two cities?[/quote]\r\n\r\nI think it is the square root of ((1-cos144)(2)(5000^2)) which is about 9510.56516. This is traveling in an arc across the lateral surface of the cone. If you unfold the cone to get a section of a circle, you can compare the radius or the circle 5000 (lateral hight of the cone) to the radius of the base of the cone (4000, given) to find that the section represents 4/5 of the circle, which is 288 degrees. one city is in the middle of the arc and one is on the end, so by dropping a line from the center of the circle to the second city and connecting the two cities with a strait line we form a triangle with two sides of 5000 and an included angle of 144 degrees. Using law of cos we find the final side.", "Solution_30": "Here's a new problem:\r\n3.The sequence 2, 3, 5, 6, 7, 10, \u2026 consists of all natural\r\nnumbers which are neither perfect squares nor perfect\r\ncubes. Find the 75th term of this sequence.", "Solution_31": "[hide]The 75th number is 75, but 1^2, 2^2, 2^3, 3^2, 4^2, 5^2, 3^2, 6^2, 7^2, and 8^2 (or 4^3) are counted. This is 10 squares or cubes, so 75 is actually the 65th term in the sequence. 80 is the 70th term, then we have: 82, 83, 84, 85, 86. 86 is the 75th term.[/hide]", "Solution_32": "[quote=\"funcia\"][quote=\"buzzer11\"][quote=\"hrzdlake\"]Isn't it just 8000 miles?\nOr possibly 2000 pi?[/quote]\n\n[quote=\"pn12345\"]The answer does not include pi.[/quote][/quote]\nspam?!\nno, that person didn't read question[/quote]\r\n\r\ni did NOT write that... :maybe:", "Solution_33": "[hide]86. Wasn't this on a previous MOEMS test?[/hide]", "Solution_34": "[hide=\"Simple.\"]86[/hide]\r\nThis is a long thread.", "Solution_35": "$a$ is a prime number. In the set ${2a,3a,5a}$, how many prime numbers are there?\r\nBTW, this thread isn't long if you compare it to the problem solving thread in the Mathcounts Forum (>90 pages)", "Solution_36": "[hide=\" :arrow: :roll: :wink: \"]That was weird. None, of course.[/hide]", "Solution_37": "[quote=\"vamathletes\"]$a$ is a prime number. In the set ${2a,3a,5a}$, how many prime numbers are there?\nBTW, this thread isn't long if you compare it to the problem solving thread in the Mathcounts Forum (>90 pages)[/quote]\r\n[hide]Obviously, each will be divisible by $a$, and the coefficient, so no prime numbers.[/hide]", "Solution_38": "[quote=\"xpmath\"][quote=\"vamathletes\"]$a$ is a prime number. In the set ${2a,3a,5a}$, how many prime numbers are there?\nBTW, this thread isn't long if you compare it to the problem solving thread in the Mathcounts Forum (>90 pages)[/quote]\n[hide]Obviously, each will be divisible by $a$, and the coefficient, so no prime numbers.[/hide][/quote]\n\nYou should hide things in smilies.\n[hide=\":roll:\"]It's cool.[/hide]\r\n\r\nNew problem?", "Solution_39": "what is the ones digit of 2007^2007?\r\nobviously, you dont have a calculator with a screen big enough to fit that number.", "Solution_40": "[quote=\"Nerd_of_the_Ages\"]what is the ones digit of 2007^2007?\nobviously, you dont have a calculator with a screen big enough to fit that number.[/quote]\r\n[hide]$2007^{2007}\\equiv2007^{2007\\mod4}\\equiv2007^{3}\\equiv7^{3}\\equiv\\boxed{3}\\mod{10}$[/hide]", "Solution_41": "Bobby runs on a track at a two feet per second. If an object at his starting place is falling from 2 miles up starting at .1 feet a second (abnormally slow, I know), how many quarters of a lap will Bobby have run before the object hits the ground, to the nearest lap.", "Solution_42": "Are we assuming each lap is one mile? If so\r\n[hide]Since the object falls .1 feet each second, and 1 mile is 5280 feet, it will take 10*5280*2= 105600 seconds. \nThen, since Bobby travels 2 feet per second, he travels 211200 feet, which is 40 miles. So if each lap is one mile, he needs to do 40 laps?[/hide]\r\n\r\nI'm really tired right now, so I probably messed up.", "Solution_43": "Are we assuming the object is falling at a constant rate, because in real life, there is acceleration, which changes almost the whole problem.", "Solution_44": "Well, I said [b]starting[/b] at that speed, implying acceleration." } { "Tag": [ "algebra", "binomial theorem" ], "Problem": "Using the Binomial Theorem Part 3\n\nWe use our understanding of the Binomial Theorem to quickly find specific terms of expansions of powers of binomials, without finding the full expansion!", "Solution_1": "Thank you for the non-existent video" } { "Tag": [ "quadratics", "calculus", "integration" ], "Problem": "On a harder problem, i have everything simplified down to a 2 by 2 quadratic, solving for x: Please help:\r\n\r\n$ \\minus{} ax^2 \\plus{} 2x \\plus{} a \\equal{} 1$\r\n$ a^2x^2 \\minus{} 2ax \\plus{} a^2 \\equal{} 0$", "Solution_1": "The \"obvious\" thing to do is solve the first equation for $ a$ and substitute it in to the second equation. There're one \"nice\" and three \"ugly\" solutions (the roots of a cubic with integral coefficients; one irrational and two complex with irrational real and imaginary parts).", "Solution_2": "er... now i need help solving a 6th power:\r\n\r\n$ a^6\\plus{}2a^4\\minus{}6a^3\\plus{}a^2\\minus{}3\\equal{}0$", "Solution_3": "I thought you were solving for $ x$... :maybe:\r\n[hide=\"Too lazy to show the numbers, sorry.\"]\nFactor the second equation to get $ a(ax^2\\plus{}a\\minus{}2)\\equal{}0$. Now we have $ a\\equal{}0$ or $ ax^2\\plus{}a\\minus{}2\\equal{}0$. The first case has $ x\\equal{} \\dfrac 12$. (Substitute $ a\\equal{}0$ in the first equation.) The other case, $ ax^2\\plus{}a\\minus{}2\\equal{}0$, when added with the first equation, yields $ \\minus{}(x\\plus{}\\dfrac 12)\\equal{}a$. Plugging that in the first equation eventually gives the cubic $ 2x^3 \\plus{} x^2 \\plus{}2x \\minus{}2\\equal{}0$, which has the \"ugly\" irrational and complex roots.\n[/hide]", "Solution_4": "[quote=\"unimpossible\"]On a harder problem, i have everything simplified down to a 2 by 2 quadratic, solving for x: Please help:\n\n$ \\minus{} ax^2 \\plus{} 2x \\plus{} a \\equal{} 1$\n$ a^2x^2 \\minus{} 2ax \\plus{} a^2 \\equal{} 0$[/quote]\r\nWait, I'm confused... \r\n[hide=\"Does this work?\"]\nWe multiply the first equation by $ a$ to get that $ \\minus{} a^2x^2 \\plus{} 2ax \\plus{} a^2 \\equal{} a$. Adding this to the second equation gives us that $ a^2 \\plus{} a^2 \\equal{} a\\implies 2a^2 \\equal{} a\\implies a \\equal{} \\frac {1}{2}, 0$. Then plug them in to get:\n\n[b]Case 1: $ a \\equal{} 0$[/b]\n$ 2x \\equal{} 1$ and $ 0 \\equal{} 0$, so one solution is $ x \\equal{} \\frac {1}{2}$\n\n[b]Case 2: $ a \\equal{} \\frac {1}{2}$[/b]\nThe first equation yields $ \\minus{} \\frac {x^2}{2} \\plus{} 2x \\minus{} \\frac {1}{2} \\equal{} 0\\implies x^2 \\minus{} 4x \\plus{} 1 \\equal{} 0\\implies x \\equal{} \\frac {4 \\pm \\sqrt {16 \\minus{} 4}}{2} \\equal{} 2\\pm \\sqrt {3}$. The second equation is now essentially the same thing as the first, so it will give the same solutions. \n\nSo, $ x \\equal{} \\boxed{\\frac {1}{2}, 2\\pm \\sqrt {3}}$\n[/hide]", "Solution_5": "The problem was originally written differently. Your solution seems correct for the new problem." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a,b,c are non-negative numbers such that $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$. Prove that\r\n$ \\sum_{sym}\\frac {1}{2\\minus{}3ab}\\le 3$", "Solution_1": "[quote=\"thegod277\"]Let a,b,c are non-negative numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$. Prove that\n$ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab}\\le 3$[/quote]\r\nShould it be $ \\sum_{cyc}\\frac {1}{2 \\minus{} 3ab}\\le 3$? $ a \\equal{} b \\equal{} c \\equal{} \\sqrt{\\frac{1}{3}}$ gives:\r\n$ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} 6 \\frac{1}{2\\minus{}1} \\equal{} 6 > 3$", "Solution_2": "No.$ \\sum_{sym}\\frac {1}{2\\minus{}3ab}\\equal{}\\frac {1}{2\\minus{}3ab}\\plus{}\\frac {1}{2\\minus{}3bc}\\plus{}\\frac {1}{2\\minus{}3ca}$. So, if $ a\\equal{}b\\equal{}c\\equal{}\\frac {1}{\\sqrt 3}$ we have $ \\sum_{sym}\\frac {1}{2\\minus{}3ab}\\equal{}3$", "Solution_3": "[quote=\"thegod277\"]No.$ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}$. So, if $ a \\equal{} b \\equal{} c \\equal{} \\frac {1}{\\sqrt 3}$ we have $ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} 3$[/quote]\r\nMaybe $ \\sum_{cyc}\\frac {1}{2 \\minus{} 3ab} \\equal{} \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}$ $ ?$ :wink:\r\nBy the way, $ \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}\\leq3$ is wrong. \r\nTry $ a\\equal{}\\frac{1}{\\sqrt{10}}$ and $ b\\equal{}c\\equal{}\\frac{3}{\\sqrt{10}}.$ :wink:", "Solution_4": "[quote=\"arqady\"][quote=\"thegod277\"]No.$ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}$. So, if $ a \\equal{} b \\equal{} c \\equal{} \\frac {1}{\\sqrt 3}$ we have $ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} 3$[/quote]\nMaybe $ \\sum_{cyc}\\frac {1}{2 \\minus{} 3ab} \\equal{} \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}$ $ ?$ :wink:\nBy the way, $ \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}\\leq3$ is wrong. \nTry $ a \\equal{} \\frac {1}{\\sqrt {10}}$ and $ b \\equal{} c \\equal{} \\frac {3}{\\sqrt {10}}.$ :wink:[/quote]\r\nArgady, I think when $ a \\equal{} \\frac {1}{\\sqrt {10}}$ and $ b \\equal{} c \\equal{} \\frac {3}{\\sqrt {10}}.$\r\nwe have $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1.9>1$??? :maybe: :maybe:", "Solution_5": "[quote=\"thegod277\"][quote=\"arqady\"][quote=\"thegod277\"]No.$ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}$. So, if $ a \\equal{} b \\equal{} c \\equal{} \\frac {1}{\\sqrt 3}$ we have $ \\sum_{sym}\\frac {1}{2 \\minus{} 3ab} \\equal{} 3$[/quote]\nMaybe $ \\sum_{cyc}\\frac {1}{2 \\minus{} 3ab} \\equal{} \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}$ $ ?$ :wink:\nBy the way, $ \\frac {1}{2 \\minus{} 3ab} \\plus{} \\frac {1}{2 \\minus{} 3bc} \\plus{} \\frac {1}{2 \\minus{} 3ca}\\leq3$ is wrong. \nTry $ a \\equal{} \\frac {1}{\\sqrt {10}}$ and $ b \\equal{} c \\equal{} \\frac {3}{\\sqrt {10}}.$ :wink:[/quote]\nArgady, I think when $ a \\equal{} \\frac {1}{\\sqrt {10}}$ and $ b \\equal{} c \\equal{} \\frac {3}{\\sqrt {10}}.$\nwe have $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1.9 > 1$??? :maybe: :maybe:[/quote]\r\nIt can be $ a \\equal{} \\frac {1}{\\sqrt {19}}$ and $ b \\equal{} c \\equal{} \\frac {3}{\\sqrt {19}}.$ :D", "Solution_6": "[quote=\"arqady\"][/quote][quote=\"thegod277\"][quote=\"arqady\"][quote=\"thegod277\"]No.$ \\sum_{sym}\\frac {1}{2 - 3ab} = \\frac {1}{2 - 3ab} + \\frac {1}{2 - 3bc} + \\frac {1}{2 - 3ca}$. So, if $ a = b = c = \\frac {1}{\\sqrt 3}$ we have $ \\sum_{sym}\\frac {1}{2 - 3ab} = 3$[/quote]\nMaybe $ \\sum_{cyc}\\frac {1}{2 - 3ab} = \\frac {1}{2 - 3ab} + \\frac {1}{2 - 3bc} + \\frac {1}{2 - 3ca}$ $ ?$ :wink:\nBy the way, $ \\frac {1}{2 - 3ab} + \\frac {1}{2 - 3bc} + \\frac {1}{2 - 3ca}\\leq3$ is wrong. \nTry $ a = \\frac {1}{\\sqrt {10}}$ and $ b = c = \\frac {3}{\\sqrt {10}}.$ :wink:[/quote]\nArgady, I think when $ a = \\frac {1}{\\sqrt {10}}$ and $ b = c = \\frac {3}{\\sqrt {10}}.$\nwe have $ a^2 + b^2 + c^2 = 1.9 > 1$??? :maybe: :maybe:[/quote][quote=\"arqady\"]\nIt can be $ a = \\frac {1}{\\sqrt {19}}$ and $ b = c = \\frac {3}{\\sqrt {19}}.$ :D[/quote]\r\noh, yes. I am wrong. Thank you argady, I will find a condition best for it.thank you. :) \r\nPS: A inequality $ \\frac {1}{1-ab}+\\frac {1}{1-bc}+\\frac {1}{1-ca}\\le \\frac 92$ is right. So what is genelization inequality? :huh:" } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra solved" ], "Problem": "Let A,B be two matrices in M_2(R). Prove that det(I_2+ (2AB+3BA)/5)=det(I_2+ (3AB+2BA)/5)", "Solution_1": "Let P(x) = det(I2+xAB+(1-x)BA). Since it's well-known that det(!2+AB)=det(I2+BA) then P(0)=P(1) (*). P has degree 2, so P(x)=ax^2+bx+1. From (*) we get a+b=0, so P(x)=ax^2-ax+1. It's easy to check now that P(x)=P(1-x) and you apply this for x=3/5.", "Solution_2": "where can i find some documentation about the polinomyal form of a determinant??" } { "Tag": [ "geometry", "calculus", "calculus computations" ], "Problem": "I'm having difficulty with these problems - please help!!!\r\n\r\n1.) y = (4x)^(1/2) from x = 0 to 8 about the x-axis\r\nso in terms of x this is:\r\nx = ((y-1)^(1/2))/2 from y = 1 to 257\r\nso in the formula this comes out to:\r\n2pi [int from 1 to 257] ((y-1)^(1/2)) / 2 * (1 + (1 /( 4(y - 1)^(1/2))) ^2) ^(1/2)\r\nwhich comes out to be:\r\n2pi* (1/192) (y-1)^(3/2) ((16y - 15) / (y - 1))^(3/2)\r\nwhich is about 142\r\nwhich is wrong! what have i done incorrectly?\r\n\r\nthe other is:\r\n2.) x = 2e^2y from y = 0 to y = 1 about y-axis\r\nin formula:\r\n2pi * [Int from 0-1] 2e^(2y) (1 + (4e^(2y))^2)^(1/2)\r\nwhich comes out to be:\r\n2pi (1/8) ln((1 + 16e^(4))^(1/2)) + 4e^(2) + (1/2)e^(2) (1 + 16e^(4))^(1/2)\r\nwhich is about 142\r\nwhich is wrong! what have i done incorrectly?", "Solution_1": "And would you know how to solve:\r\n\r\nFind the area of the surface obtained by rotating the curve:\r\n$ y \\equal{} 2e^{2x}$ from $ x \\equal{} 0$ to $ x \\equal{} 1$ about the $ x$ axis.", "Solution_2": "[quote=\"Beachic109\"]I'm having difficulty with these problems - please help!!!\n\n1.) y = (4x)^(1/2) from x = 0 to 8 about the x-axis\nso in terms of x this is:\nx = ((y-1)^(1/2))/2 from y = 1 to 257[/quote]\r\n\r\nThat doesn't hang together.\r\n\r\nIf $ y\\equal{}\\sqrt{4x}$ for $ 0\\le x\\le 8,$ then $ x\\equal{}\\frac{y^2}4$ for $ 0\\le y\\le 2\\sqrt{2}.$\r\n\r\nIf $ x\\equal{}\\frac{\\sqrt{y\\minus{}1}}2$ for $ 1\\le y\\le 257,$ then $ y\\equal{}1\\plus{}4x^2$ for $ 0\\le x\\le 8.$\r\n\r\nI can't even tell what problem you're working, and if I can't tell that, there's no way I can find what else might be wrong." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Prove that:\r\n[tex] \\sum{ \\sqrt{ \\frac{1}{(x+y)^2} + \\frac{1}{(y+z)^2} + \\frac{1}{(x+y)(y+z)}}} \\leq \\frac{1}{2} \\sum{ \\sqrt{ \\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{xy}}} [/tex]\r\n\r\nwith [tex]x,y,z>0[/tex]", "Solution_1": "Note that \\[{{\\sqrt{\\frac{1}{x^2}+\\frac{1}{y^2}+\\frac{1}{xy}}}+{\\sqrt{\\frac{1}{y^2}+\\frac{1}{z^2}+\\frac{1}{zy}}}\\ge{\\sqrt{(\\frac{1}{x}+\\frac{1}{y})^2+(\\frac{1}{y}+\\frac{1}{z})^2+(\\sqrt\\frac{1}{xy}+\\sqrt{\\frac{1}{yz}})^2}}\\ge4\\sqrt{\\frac{1}{(x+y)^2}+\\frac{1}{(y+z)^2}+\\frac{1}{(x+y)(y+z)}}}\\]Since ${(\\frac{1}{x}+\\frac{1}{y})^2\\ge\\frac{16}{(x+y)^2}}$ and ${{(\\sqrt\\frac{1}{xy}+\\sqrt{\\frac{1}{yz}})^2}}\\ge\\frac{16}{(x+y)(y+z)}$." } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra theorems" ], "Problem": "Please tell me what is erroneous about the following argument. \r\n\r\nSay we have some homomorphism phi from group X to group Y. For some x in X, we have the equality (phi(x))^n = phi(x^n). From this we see that if the order of x is n, then phi(x^n) = phi(1) = 1 = phi(x)^n, and hence the order of phi(x) is n. (Note: Showing that n is minimal is simple; assume there exists some m less than n such that phi(x)^m = 1. This implies phi(x^m) = 1, however this contradicts than n is the order of phi(x), thus by contradiction n in minimal such that phi(x)^n = 1.)\r\n\r\nHowever, this equality doesn't seem to make sense; for example if x comes from the kernel of phi, although the order of phi(x) is 1, that doesn't imply the order of x must be one. (A kernel can certainly contain more than just the identity.) So then what is wrong with the argument?", "Solution_1": "the argument uses its claim ...\r\n\r\nyou just can say, that the order of $ \\phi(x)$ divides the order of $ x$." } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "Let $(a_{n})$ be a decreasing positive sequence and the sequence $S_{n}=a_{1}+a_{2}+...+a_{n}$ is convergent. \r\nShow that $lim (na_{n})=0$", "Solution_1": "[hide=\"Hint\"]Consider $S_{2n}-S_{n}$ [/hide]", "Solution_2": "please write more details. I can't understand what you did", "Solution_3": "[quote=\"Ng\u00f4ng\"]please write more details. I can't understand what you did[/quote]\r\n\r\n\r\n$\\\\ S_{2n}-S_{n}= \\sum_{k=n}^{2n}a_{k}\\geq n a_{2n}>0$ since $a_{n}$ is decreasing.\r\n\r\nTherefore $\\lim_{n \\to \\infty }2n a_{2n}= 0$ from Squeeze theorem .\r\n\r\nBut since $a_{2n}\\geq a_{2n+1}$ we have \r\n\r\n$\\\\ \\frac{2n+1}{2n}2n a_{2n}\\geq (2n+1) a_{2n+1}\\Rightarrow \\lim_{n \\to \\infty }(2n+1) a_{2n+1}= 0$\r\n\r\nTherefore $\\lim_{n \\to \\infty }n a_{n}= 0$ .", "Solution_4": "I don't know Squeeze theorem .Please tell me more about it.", "Solution_5": "Suppose $(a_{n})$, $(b_{n})$, and $(c_{n})$ are three sequences of real numbers such that for all $i\\in \\mathbb{N}$ we have $a_{i}\\leq b_{i}\\leq c_{i}$. Suppose further that $\\exists x_{0}\\in\\mathbb{R}$ such that $\\lim_{n\\to \\infty}a_{n}=\\lim_{n\\to \\infty}c_{n}=x_{0}$, Then squeezing therem states that $\\lim_{n\\to \\infty}b_{n}=x_{0}$." } { "Tag": [ "email" ], "Problem": "Does anyone have the answers for the following:\r\nNational 1991-1995 sprint and target\r\n1996 target\r\n1998 sprint and target\r\n2003 sprint and target?\r\n\r\nThanks", "Solution_1": "[quote=\"anirudh\"]Does anyone have the answers for the following:\nNational 1991-1995 sprint and target\n1996 target\n1998 sprint and target\n2003 sprint and target?\n\nThanks[/quote]\r\nI have them somewhere. I need to dig them out.", "Solution_2": "Do you want us to pm, or should we post it like last time?\r\n\r\nIt seems like you want a lot of things.", "Solution_3": "It may be over the uploading quota, so just email them to me:\r\njdasarathy@yahoo.com", "Solution_4": "Well, I might as well just give the only thing I have.\r\n\r\nI will post it.\r\n\r\n\r\n[hide=\"Sprint Answers\"]1. $10$\n2. $\\$12$\n3. $\\frac{1}{4}$\n4. $252$\n5. $751$ toothpicks\n6. $17.5$\n7. $\\frac{47}{12}$\n8. $697$\n9. $\\frac{\\pi}{4}$\n10. Affirmed\n11. $25.5$ square units\n12. $6$ days\n13. $8192$\n14. $10$\n15. $144$\n16. $20$\n17. $38$ values\n18. $\\frac{1}{3}$\n19. $5$ arrangements\n20. $2$\n21. $\\frac{1}{4}$\n22. $24$\n23. $4.5$ miles per hour\n24. $\\frac{5}{6}$\n25. $4a+a\\sqrt{5}$ units\n26. $\\frac{18}{295}$\n27. $90$ (4's)\n28. $116+4\\sqrt{26}$\n29. $10$ regions\n30. $\\frac{1}{8}$[/hide]\n\n[hide=\"Target Answers\"]1. $6$ pairs\n2. $22.21$ inches\n3. $12$ squares\n4. $22.5$\n5. $4.14$ inches\n6. $24$\n7. $\\frac{83}{48}$\n8. $157.5$ degrees[/hide]\r\n\r\nThis is the only one I have", "Solution_5": "That's for year 2003 btw.", "Solution_6": "yes, that is for year 2003.\r\n\r\nThank You mad_skillz_aops." } { "Tag": [], "Problem": "Basicly, this is just tug of war, but now instead of moving left and right, you can move in 2 dimensions: left, right, up and down. Each move increments or deincrements the x or the y value in the last position by 1. Let's start:\r\n\r\n[b](0,0)[/b]", "Solution_1": "uhhh ok apparently ccy is too tired to do this right.\r\nso ignore the \"let's start\" part. we need teams anyways\r\nSo just sign up here I guess...", "Solution_2": "Not yet friends. Hold on a bit more! Tiem is not ripe. I'll decide the right moment for this. And I also have great plans ...\r\n\r\nKeep guessing!", "Solution_3": "wtf bubka\r\nI need more excuses so I can procrastinate on homework. Might as well start a 2d tow now", "Solution_4": "OMFG GUYS I BUMPED THE OLD ONE UP NO NEED TO HAVE ANOTHER THREAD", "Solution_5": "OMFG YOU BUMPED THAT ONE AFTER THIS GUY MADE THIS", "Solution_6": "OMFG SRRY BOUT TAT THE OLD ONE IS LOCKED NOW SO ALL IS GOOD\r\n\r\n*BLAH*", "Solution_7": "BLAH NOW WHAT", "Solution_8": "YES..........", "Solution_9": "[quote=\"mustafa\"]YES..........[/quote]\r\n\r\nUM IT WASN'T A YES OR NO QUESTION", "Solution_10": "YO GUYS WE NEED A MORE ORGANIZED START\r\n\r\nSIGNUPS OVER HERE:\r\n\r\n1)akalra01\r\n2)ccy\r\n3)junggi\r\n4)bubka", "Solution_11": "UM WHY ARE YOU THE 1) GUY TO SIGN UP", "Solution_12": ":roll: :roll:", "Solution_13": "[quote=\"Ignite168\"]:roll: :roll:[/quote]\r\nDOES THAT IMPLY THAT YOU SIGNED UP?", "Solution_14": "People you are doing nothing but turning this 2D tug of war into spam. Just be a bit more patient. I am making preparations.\r\n\r\nAnd someone lock this.", "Solution_15": "What preperations do you need to make a tug of war game? :huh:", "Solution_16": "LOTS\r\n[color=red]\nWATCH [/color]AND LEARN" } { "Tag": [ "trigonometry", "geometry proposed", "geometry" ], "Problem": "Prove that there exists $1977$ triangles $ABC$ satisfying the conditions:\r\n$\\frac{\\sin{A}+\\sin{B}+\\sin{C}}{\\cos{A}+\\cos{B}+\\cos{C}}=\\frac{12}{7}$ and $\\sin{A}\\sin{B}\\sin{C}=\\frac{12}{25}$.", "Solution_1": "We have $\\frac{4\\prod cos\\frac{A}{2}}{1+4\\prod sin\\frac{A}{2}}=\\frac{12}{7}$ and $8\\prod sin\\frac{A}{2}=\\frac{12}{25}$.Thus $\\prod cos\\frac{A}{2}=0,6$ and $\\prod \\sin\\frac{A}{2}=0,1$.\r\nNotice that $sin\\frac{C}{2}=cos\\frac{A}{2}cos\\frac{B}{2}-sin\\frac{A}{2}sin\\frac{B}{2}$ $\\Leftrightarrow$ $sin^2\\frac{C}{2}cos\\frac{C}{2}=0,6sin\\frac{C}{2}-0,1cos\\frac{C}{2}$.\r\nLet $cosC=t$,we have \r\n$(1-t^2)^2=0,6\\sqrt{1-t^2}-0,1t$\r\n$\\leftrightarrow$ $11t-10t^3=6\\sqrt{1-t^2}$\r\nSolved this equation,we have $cos\\frac{C}{2} \\in$ {$\\sqrt{\\frac{1}{2}},\\sqrt{\\frac{4}{5}},\\sqrt{\\frac{3}{10}}$}\r\ntherefore $sinC \\in$ {$1;0,8;0,6$} and from here the conclusion follows." } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Suppose that $ x\\equal{}cos\\angle{A}$, $ y\\equal{}cos\\angle{B}$, $ z\\equal{}cos\\angle{C}$, where $ \\angle{A},\\angle{B},\\angle{C}$ are the angles of an acute triangle $ ABC$. Prove that $ 2(xy\\plus{}yz\\plus{}zx)\\plus{}3\\geq{3(x\\plus{}y\\plus{}z)}$.", "Solution_1": "Sorry for my fault, the sign must be reversed. Hence, we need to prove $ 2(xy\\plus{}yz\\plus{}zx)\\plus{}3\\leq{3(x\\plus{}y\\plus{}z)}$." } { "Tag": [], "Problem": "Here's a pretty neat contest problem that i found the other day...\r\n\r\nIf $(x-a)(x-10)+3 = (x+b)(x+c)$ is true for all $x$ and if $a, b,$ and $c$ are integers, then the number of possible values for $a+b$ is....\r\n\r\nA) 0\r\nB) 2\r\nC) 3\r\nD) 4\r\nE) 6", "Solution_1": "[quote=\"revelation\"]Here's a pretty neat contest problem that i found the other day...\n\nIf $(x-a)(x-10)+3 = (x+b)(x+c)$ is true for all $x$ and if $a, b,$ and $c$ are integers, then the number of possible values for $a+b$ is....\n\nA) 0\nB) 2\nC) 3\nD) 4\nE) 6[/quote]\r\n\r\n[hide]Let $x = a$. That forces us to have $3 = (a+b)(a+c)$, which is the product of 2 integers. Since $3$ is prime, it only has two factors (which can be positive or negative). Therefore, the answer is D. [/hide]", "Solution_2": "And does it specify that $a+b$ must be positive? :wink:", "Solution_3": "just expand it, and equate coefficients....", "Solution_4": "[hide]Expanding and rearranging, we get two equations:\n$-10-a=b+c\\Longrightarrow a+b=-10-c$\n$10a+3=bc\\Longrightarrow 10a+ac+3=ac+bc\\Longrightarrow a(10+c)+3=(a+b)c$\n\nSubstituting in our value for $a+b$ and rearranging some more, we get:\n$(10+c)(a+c)=-3$. \nBecause both of these factors are integers, $(10+c)$ can only be $1,3,-1, or-3,$ and then $a+b$ is also $\\pm 1$ or $\\pm 3$ and the answer is D.[/hide]", "Solution_5": "[quote=\"white_horse_king88\"]And does it specify that $a+b$ must be positive? :wink:[/quote]\r\n\r\nWHY do i always miss small little details like this. :mad:" } { "Tag": [ "pigeonhole principle" ], "Problem": "If $ r$ is a real number, prove that there must be at least one term in the sequence $ r, 2r, \\cdots, 99r$ that is within $ \\frac {1}{100}$ an integer.", "Solution_1": "By \"within $ \\frac{1}{100}$, you include numbers which are exactly $ \\frac{1}{100}$ away from an integer right...\r\n\r\n[hide=\"hint\"]You know the difference between any two of them is also in the set. So assume otherwise: now construct $ 98$ pigeonholes from the interval between two integers not including the sections which correspond to the desired property.[/hide]" } { "Tag": [ "ratio" ], "Problem": "On side $ BC$ of $ \\triangle ABC$ is point $ D$, such that $ BD : DC \\equal{} 2: 1$. What is the ratio in which the median from point $ C$ to side $ AB$ divides the segment $ AD$?\r\n\r\nI'm also trying to figure it out without just looking at a solution, so a hint would be nice too. :)", "Solution_1": "First hint:\r\n\r\n[hide]Let the midpoint of AB be E. Draw the midsegment EF, where F is the midpoint of AC. Let the intersection of CE and AD be M, and let the intersection of EF and AD be N. See if you can find any interesting triangles? (and/or ratios involving the midsegment of the triangle?)\n[/hide]\n\nSecond hint:\n\n[hide]Indeed, triangles MNE and MDC are congruent. This means MN and MD are congruent. We can also note that EN is a midsegment of triangle ABD, so AN = ND.[/hide]\n\nWrapping it up:\n\n[hide]MN=MD\nMN+MD = ND\n2MD = ND\nMD = ND/2\n\nSo I'm not sure exactly what ratio you're looking for, but it would divide the median into parts with ratio 3:1[/hide]\r\n\r\nI think. ;)", "Solution_2": "I think for this problem it's easier to use mass points." } { "Tag": [ "calculus", "integration", "algebra", "function", "domain", "factorial", "algorithm" ], "Problem": "I saw posts about gaussian integer at\r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=179715[/url]\r\n\r\nbut I can't understand it\r\n\r\nSo I want to know about the number's definition & concept.\r\n\r\nAnd how to use it??", "Solution_1": "The set of Gaussian integers is $ \\mathbb{Z}[i]$. You can find lots of information about them on the internet, especially mathlinks. I suggest that you get a book on abstract algebra. :wink:", "Solution_2": "You may first want to look at my post http://www.mathlinks.ro/Forum/viewtopic.php?p=584714 for what rings are.\r\nThen see http://www.mathlinks.ro/Forum/viewtopic.php?p=935293 for some more terms.\r\n\r\nGaussian numbers $ \\mathbb Z[i] \\equal{} \\{ x \\plus{} iy \\in \\mathbb C | x,y \\in \\mathbb Z \\}$ are a special type of ring, namely they are:\r\na) commutative and unitary, thus $ ab \\equal{} ba$ and there is $ 1 \\in \\mathbb Z[i]$ such that $ 1 \\cdot a \\equal{} a$ for all $ a,b \\in \\mathbb Z[i]$.\r\nb) an integral domain: if $ ab \\equal{} 0$ then $ a \\equal{} 0$ or $ b \\equal{} 0$. Or the other way around: the product of nonzero numbers is a nonzero number.\r\nc) factorial: every such number can be written as a product of prime elements, and this is unique up to reordering and multiplication with units. See the second link.\r\n\r\nEven more:\r\nd) there is a norm: there is a mapping $ N: \\mathbb Z[i] \\to \\mathbb Z$ given by $ N(x \\plus{} iy) \\equal{} (x \\plus{} iy)\\overline{(x \\plus{} iy)} \\equal{} (x \\plus{} iy)(x \\minus{} iy) \\equal{} x^2 \\plus{} y^2$ for $ x \\plus{} iy \\in \\mathbb Z[i]$. It is multiplicative, thus $ N(a\\cdot b) \\equal{} N(a) \\cdot N(b)$.\r\ne) norm-eucidean: for $ a,b \\in \\mathbb Z[i]$, $ b \\neq 0$, we always find $ q,r \\in \\mathbb Z[i]$ such that $ a \\equal{} bq \\plus{} r$ and $ N(r) < N(b)$. This is what the euclidean algorithm is based upon.\r\n\r\nIt can be shown that a), b) and e) imply c).\r\nOne example of usage is the thread you linked at. Some more problems:\r\n- show that if $ x,y \\in \\mathbb Z$ are each sum of two perfect squares, then $ xy$ is a sum of two squares.\r\n- let $ p$ be prime; show that $ x^2 \\equiv \\minus{} 1 \\mod p$ has a solution iff $ p$ is a sum of two perfect squares. Show that this solution is unique (up to sign and permutation) and find all such $ p$.\r\n- find the number of ways to write some $ n$ as a sum of two squares.\r\n\r\n\r\nWell, and after concerning with Gaussian numbers, there are a lot of similar rings that deserve attention. Look for algebraic integers and number rings (number fields). I suggest some course on abstract algebra, then on algebraic number theory." } { "Tag": [ "symmetry", "geometry", "similar triangles", "geometry proposed" ], "Problem": "Let two circles (1),(2) intouches the circle (O). Two chords BC and DE of (O) are the external common tangents of (1),(2). A point A lies on the arc BC of (O), which does not contain D,E. AD,AE meets BC at M,N. Prove that if the A-median of triangle AMN touches (1),(2), then the second internal common tangent of (1),(2) passes through the midpoint of BC.", "Solution_1": "The first common internal tangent of the circles $ (1), (2)$ cuts $ BC \\equiv MN$ at $ P$ and $ (O)$ at $ A$ on the arc $ BC$ which does not contain $ D, E.$ The 2nd common internal tangent of $ (1), (2)$ cuts $ DE$ at $ F$ and the circle $ (O)$ at $ Z,$ again on the arc $ BC$ which does not contain $ D, E.$ By the parallel tangent theorem, $ AZ \\parallel BC \\equiv MN.$ The $ \\triangle AMN \\sim \\triangle ZED$ are oppositely similar, having equal angles: $ \\angle MAN \\equal{} \\angle DAE \\equal{} \\angle DZE$ and $ \\angle MNA \\equal{} \\angle MAZ \\equal{} \\angle EAZ \\equal{} \\angle EDZ.$ By symmetry, the angle $ \\angle APM \\equal{} \\angle APB$ formed by the 1st common pair of the common external / internal tangents $ BC, AP$ of $ (1), (2)$ is equal to the angle $ \\angle ZFE$ formed by the 2nd pair of their common external / internal tangents $ DE, ZF.$ Therefore $ P, F$ are corresponding points of the similar triangles $ \\triangle AMN \\sim \\triangle ZED.$ Consequently, $ P$ is the midpoint of $ MN$ $ \\Longleftrightarrow$ $ F$ is the midpoint of $ DE.$", "Solution_2": "[hide]With the above lemma, we are able to prove the problem at [url]http://www.mathlinks.ro/viewtopic.php?t=213165[/url][/hide]" } { "Tag": [], "Problem": "Yes, that old game. Here's the drill. Find what \"I\" in each riddle or passage refers to. All are written by me unless listed otherwise. \r\n\r\n[b][u]Submit your answer by private message[/u][/b]. On submission PMs, title (that means subject line) your submission [color=green]\"WAI #[insert number]: [insert answer you're submitting]\"[/color] for example [color=orange]\"WAI #22: Seventy-four\"[/color]. A scoreboard will be kept. [b][u]Do not, I repeat, do [i]not[/i], post your answers on the thread. [/u][/b]\r\n\r\nUse the most [u]specific[/u] answer you can think of that fits [u]all[/u] clues given. \r\n\r\nOkay, here's the first. \r\n\r\n[size=200]Puzzle One[/size]\r\n\r\nFor five points, find the person who would be the \"I\" below. \r\n\r\nA prodigy, a genius, a math wizard was me, \r\nWhen I traveled to Britain it was plain to see. \r\nI devised clever theorems when I was thirteen, \r\nAnd I cranked out numbers fast as a machine. \r\nI showed my peers math they never saw, \r\nAnd they all stared at me in awe, \r\nAlthough I must have went a little too askew, \r\nSince I died when I was thirty-two.", "Solution_1": "There were [b]six[/b] submissions for [b]Puzzle One[/b]. \r\n\r\n[u]Current Scoreboard[/u]\r\ntennis123 5\r\nmeewhee009 5\r\nmath_explorer 5\r\nwestiepaw 5\r\npacman2812 5\r\nk00lperson 5\r\n\r\n[size=200]Puzzle Two[/size]\r\n\r\nFor five points, find the geographical feature that would be the \"I\" below. \r\n\r\nOn Earth, I was a recent occurrence, \r\nI show early mankind's appearance, \r\nIn fossils inside, Where the desert can hide. \r\nAnd across my stretch over an entire continent, \r\nMaking a fault breaking ever so dissonant, \r\nAre all these lakes, \r\nSome are so great the seem fake, \r\nAnd one so dangerous it puts a million people at stake.", "Solution_2": "I forgot to post the answer to Puzzle One. :wallbash: anyways, it's Ramanujan. From now on, I will only accept submissions for the currently active puzzle. \r\nThe answer to Puzzle Two was the Great Rift Valley, specifically. \r\n\r\nThere were [b]three[/b] submissions for [b]Puzzle Two[/b]. \r\n\r\n[u]Current Scoreboard[/u]\r\nk00lperson 7\r\ntennis123 5\r\nmeewhee009 5\r\nmath_explorer 5\r\nwestiepaw 5\r\npacman2812 5\r\nPowerOfPi 5\r\nernie 2\r\nksun48 2\r\n\r\n[size=200]Puzzle Three[/size]\r\n\r\nFor five points, find the scientific phenomenon that would be the \"I\" below. \r\n\r\nI cause the west coasts to be warm, \r\nI cause the cyclones to be formed, \r\nI determine the winds that blow hilltops flat, \r\nI am part of what makes the earth fat, \r\nI impact climate (and maybe sensation), \r\nAnd I am the result of the Earth's rotation.", "Solution_3": "HURRAH! I'm winning..\r\ni probably am gonna start losing now.", "Solution_4": "can i still join?", "Solution_5": "Of course! You can.....\r\n\r\nLike just pm him saying the answers", "Solution_6": "There were [b]twelve[/b] submissions (including [b]two[/b] [color=green]correct[/color] ones! =P) for [b]Puzzle Three[/b]. The answer was the [b]Coriolis Effect[/b]. There was a diverse variety of submitted answers, as can can see from the list of the twelve submissions (in chronological order of submitted time) below. \r\n\r\nEl Nino, El Nino, Centrifugal Force, Heat, Equator, Coriolis Effect, Gravity, El Nino, Tide, Coriolis Effect, Heat, Ocean Currents\r\n\r\nCongratulations to [b]vahalla[/b] and [b]eqjj168[/b] who submitted correct answers! \r\n\r\n[u]Also, start following the format. I'll start to take off points for the wrong submission format. See the first post. [/u]\r\n\r\n[u]Current Scoreboard[/u]\r\nk00lperson 7\r\nmath_explorer 6\r\nvahalla 5\r\neqjj168 5\r\ntennis123 5\r\nmeewhee009 5\r\nwestiepaw 5\r\npacman2812 5\r\nPowerOfPi 5\r\nernie 3\r\nabcak 2\r\nksun48 2\r\nbluecarneal 1\r\ndragon96 1\r\n\r\nPuzzle Four will be up shortly.", "Solution_7": "darn! i knew it was going to be the coroilis effect!\r\n\r\nI learned that last year.", "Solution_8": "[size=200]Puzzle Four[/size]\r\n\r\nFor four points, find who or what the \"I\" in the following passage refers to. \r\n\r\nCompared to my siblings, I am so small; \r\nI can easily fit through the narrowest hall. \r\nAs to why I'm so small, well I don't know why, \r\nI was cloned to get why, and the still they all sigh, \r\nAs I still don't know, and why still don't so.", "Solution_9": "There were [b]four[/b] submissions for [b]Puzzle Four[/b]. Indeed, \"I\" is the letter \"I.\" \r\n\r\n[u]Three out of four people were still docked off for the wrong format, despite my warning. Again, put WAI #[number]: [submitted answer] in the subject line. I know I'm a little picky here, but chances are one of your future teachers will be highly anal about strict following of directions. That having been said, let's move on. [/u]\r\n\r\n[u]Current Scoreboard[/u]\r\nvahalla 8\r\nk00lperson 7\r\nmath_explorer 6\r\neqjj168 5\r\ntennis123 5\r\nmeewhee009 5\r\nwestiepaw 5\r\npacman2812 5\r\nPowerOfPi 5\r\naleph0 4\r\nernie 4\r\nksun48 3\r\nabcak 2\r\nbluecarneal 1\r\ndragon96 1\r\n\r\n[size=200]Puzzle Five[/size]\r\n\r\nFor five points, find the integer that fits the description of \"I\" below. The third line refers to the digits themselves, not the values of the digits. \r\n\r\nI'm not too large, I'm not too small, \r\nJust two digits, yup, that's all. \r\nThe ones is six more than the tens, \r\nAnd I am prime! Win, win win!", "Solution_10": "Gah. I put the format right, but I guessed a liger.", "Solution_11": "Contrariwise, ernie, the format was not right. :wink:", "Solution_12": "Oh right, I got lazy and put the WAI in the subject.", "Solution_13": "[quote=\"PhireKaLk6781\"]There were [b]twelve[/b] submissions for [b]Puzzle Five[/b], less than half of which had the right format. The answer is [b]seventeen[/b]. \r\n\r\n[u]Again, put WAI #[number]: [submitted answer] [b]in the subject line[/b] (without the brackets). It's for convenience reasons. [/u]\r\n\r\nYou are allowed to submit your own WAIs for use in this thread. In that case, you get the median score of the submissions. \r\n\r\n[u]Current Scoreboard[/u]\r\nvahalla 12\r\nmath_explorer 11\r\nwestiepaw 10\r\neqjj168 10\r\ntennis123 10\r\nk00lperson 7\r\nernie 7\r\nksun48 6\r\nbluecarneal 6\r\nmusicnmath 5\r\nmeewhee009 5\r\npacman2812 5\r\nPowerOfPi 5\r\nSabien5 5\r\naleph0 4\r\nandrewjjiang97 3\r\nmonkeygirl13 3\r\nabcak 2\r\ndragon96 1\r\n\r\n[size=200]Puzzle Six[/size]\r\n\r\nFor six points, find the musical form described as \"I\" below: \r\n\r\nI was popular ten scores ago, \r\nBut still a little today, \r\nMy form has many different sections, \r\nBetween repeated section A.", "Solution_14": "There were [b]six[/b] submissions for [b]Puzzle Five[/b]. The answer is [b]rondo[/b]. \r\n\r\n[u]Current Scoreboard[/u]\r\nvahalla 16\r\ntennis123 15\r\neqjj168 12\r\nmath_explorer 11\r\nwestiepaw 10\r\nernie 8\r\nk00lperson 7\r\nksun48 6\r\nbluecarneal 6\r\nzapi2006 5\r\nmusicnmath 5\r\nmeewhee009 5\r\npacman2812 5\r\nPowerOfPi 5\r\nSabien5 5\r\naleph0 4\r\nandrewjjiang97 3\r\nmonkeygirl13 3\r\nabcak 2\r\ndragon96 1\r\n\r\n[size=200]Puzzle Seven[/size]\r\n\r\nFor four points, find the contemporary meme described below: \r\n\r\nPronunciation goes out of hand, \r\nLike in a distant land. \r\nAnd yes it happens in funny verse, \r\nBecause it's in reverse. \r\nIt isn't annoying as a rickroll, \r\nBut funny as bear and bull, \r\nOverused? Maybe, but truth be said, \r\nGoogle has over 9000 yet.", "Solution_15": "There were [b]four[/b] submissions for [b]Puzzle Seven[/b]. The answer is [b]In Soviet Russia[/b]. \r\n\r\n[u]Current Scoreboard[/u]\r\ntennis123 16\r\nvahalla 16\r\nmath_explorer 12\r\neqjj168 12\r\nwestiepaw 10\r\nernie 8\r\nk00lperson 7\r\nksun48 6\r\nbluecarneal 6\r\nzapi2006 5\r\nmusicnmath 5\r\nmeewhee009 5\r\npacman2812 5\r\nPowerOfPi 5\r\nSabien5 5\r\naleph0 4\r\nandrewjjiang97 3\r\nmonkeygirl13 3\r\nabcak 2\r\ndragon96 1" } { "Tag": [ "integration", "calculus", "limit", "function", "complex analysis", "calculus computations" ], "Problem": "Evaluate\r\n\r\n$\\sum_{n=0}^{\\infty}\\left(\\frac{1}{5n+1}-\\frac{1}{5n+4}\\right)$", "Solution_1": "Well, I can express it as\r\n\r\n$\\int_{0}^{1}\\frac{1-x^{3}}{1-x^{5}}\\,dx= 1-\\int_{0}^{1}\\frac{x^{3}+x^{4}}{1+x+x^{2}+x^{3}+x^{4}}\\,dx.$\r\n\r\nWhether that does any good is a different question.\r\n\r\nActually, I think I'd be better off trying to find a Fourier series than a power series.", "Solution_2": "That integral can be done in elementary terms- it's just ugly.\r\n\r\nActually, I like the more symmetric form $\\lim_{N\\to\\infty}\\sum_{n=N}^{N+1}\\frac1{1+5n}$- it's a principal value.\r\n\r\nSee also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21316]here[/url].", "Solution_3": "[quote=\"Kent Merryfield\"]Well, I can express it as\n\n$\\int_{0}^{1}\\frac{1-x^{3}}{1-x^{5}}\\,dx= 1-\\int_{0}^{1}\\frac{x^{3}+x^{4}}{1+x+x^{2}+x^{3}+x^{4}}\\,dx.$\n\nWhether that does any good is a different question.\n\nActually, I think I'd be better off trying to find a Fourier series than a power series.[/quote]\r\nhow are u able to do that?", "Solution_4": "Let $f(x)=\\sum_{n=0}^{\\infty}\\left(\\frac{x^{5n+1}}{5n+1}-\\frac{x^{5n+4}}{5n+4}\\right).$\r\n\r\nDifferentiate to find $f'(x),$ then note that we can compute $f'(x)$ as a geometric series.", "Solution_5": "\\[\\sum_{n = 0}^\\infty{(\\frac{1}{{5n+1}}-\\frac{1}{{5n+4}})}= \\frac{\\pi }{5}\\cot \\frac{\\pi }{5}\\]", "Solution_6": "did you find that by just evaluating the integral?", "Solution_7": "An identity: $\\lim_{N\\to\\infty}\\sum_{n=-N}^{N}\\frac1{1+\\frac{n}{z}}=\\lim_{N\\to\\infty}\\sum_{n=-N}^{N}\\frac{z}{z+n}=\\pi z\\cot(\\pi z)$ whenever $z$ is not an integer. The proof uses complex analysis; if we divide by $z$, we get an obviously periodic function with poles of residue 1 at every integer. There are several more steps to pin it down exactly, but that is doable- see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41504]here[/url]." } { "Tag": [ "function", "inequalities", "inequalities proposed" ], "Problem": "$ a,b,c,d\\geq 0$ and at least two of them are small than 1.prove that \r\n\\[ (1\\minus{}a)(1\\minus{}b)(1\\minus{}c)(1\\minus{}d)\\plus{}a\\plus{}b\\plus{}c\\plus{}d\\geq 1\\]", "Solution_1": "Consider\r\n\\[ f(a) \\equal{} (1 \\minus{} a)(1 \\minus{} b)(1 \\minus{} c)(1 \\minus{} d) \\plus{} a \\plus{} b \\plus{} c \\plus{} d \\minus{} 1\\]\r\nas a linear function in which a is a variable lies in $ [0,1]$\r\n\r\nClearly, we can see that $ f(0)$ and $ f(1)$ are non-negative so we complete our proof here.", "Solution_2": "[quote=\"Tourish\"]$ a,b,c,d\\geq 0$ and at least two of them are small than 1.prove that\n\\[ (1 - a)(1 - b)(1 - c)(1 - d) + a + b + c + d\\geq 1\\]\n[/quote]\r\nIt's a very nice Inequality. \r\nHere my solution:\r\nExpanding the Inequality , we have:\r\n$ ab+bc+ca+bd+cd+ad+abcd\\geq 1 \\1$", "Solution_3": "[quote=\"Tourish\"]$ a,b,c,d\\geq 0$ and at least two of them are small than 1.prove that\n\\[ (1 \\minus{} a)(1 \\minus{} b)(1 \\minus{} c)(1 \\minus{} d) \\plus{} a \\plus{} b \\plus{} c \\plus{} d\\geq 1\\]\n[/quote]\r\nIt's a very nice Inequality. \r\n[b]Here my solution:[/b]\r\nWe may assume that $ a,b \\le 1$\r\nExpanding the Inequality , we have:\r\n$ ab \\plus{} bc \\plus{} ca \\plus{} ad \\plus{} bd \\plus{} cd \\plus{} abcd \\ge bc(a \\plus{} d) \\plus{} ad(b \\plus{} c)$\r\nor $ (a \\plus{} c)(b \\plus{} d) \\plus{} ac \\plus{} abcd \\plus{} bd \\ge ac(b \\plus{} d) \\plus{} bd(a \\plus{} c)$\r\n $ \\Leftrightarrow (b \\plus{} d)(a \\plus{} c \\minus{} ac) \\plus{} bd(ac \\minus{} a \\minus{} c) \\plus{} ac \\plus{} bd \\ge 0$\r\n $ \\begin{array}{l} \\Leftrightarrow (b \\plus{} d \\minus{} bd)(a \\plus{} c \\minus{} ac) \\plus{} ac \\plus{} bd \\ge 0 \\\\\r\n\\Leftrightarrow (b \\plus{} d(1 \\minus{} b))(a \\plus{} c(1 \\minus{} a)) \\plus{} ac \\plus{} bd \\ge 0 \\\\\r\n\\end{array}$\r\nWhich is true because $ a,b \\le 1$\r\nThe equality occurs when $ a \\equal{} b \\equal{} c \\equal{} d \\equal{} 0$" } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "reflection", "circumcircle", "geometry proposed" ], "Problem": "Prove that the product of a whole altitude $ AD$ of a triangle and the part of this altitude between its foot and the orthocentre $ HD$ is equal to the product of the two segments into which the altitude divides the side.", "Solution_1": "[color=darkblue]$ AD\\equal{}AC\\sin C\\equal{}2R\\sin B\\sin C$\n$ HD\\equal{}AD\\minus{}AH\\equal{}AD\\minus{}2R\\cos A\\equal{}2R(\\sin B\\sin C\\minus{}\\cos A)$\n\n$ BD\\equal{}AB \\cos B\\equal{}2R\\sin C \\cos B$\n$ CD\\equal{}AC \\cos C\\equal{}2R \\sin B \\cos C$\n\nSo, we need to prove\n\n$ \\sin B \\sin C(\\sin B \\sin C \\minus{}\\cos A)\\equal{}\\sin B \\sin C \\cos B \\cos C$\n\n$ \\sin B \\sin C\\minus{} \\cos B \\cos C\\equal{}\\cos A$\n\nWhich is clearly true, because\n\n$ \\cos A\\equal{} \\minus{}\\cos(B\\plus{}C)\\equal{}\\sin B \\sin C\\minus{} \\cos B \\cos C$.[/color]" } { "Tag": [ "function", "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "let $X$ be the compact topological space with some metric that induced this topology.If $f: X\\rightarrow X$ be isometry show that this isometry is surjective.\r\nfunction $f: X\\rightarrow X$ is isometry where $(X,d)$ is metric space if for every $x,y\\in X$ we have $d(x,y)=d(f(x),f(y))$.", "Solution_1": "$f(K)$ is a compact subset of $K$ since $f$ is continuous. $O=K-f(K)$ is an open set of $K$. Suppose there is some $x \\in O$. Then one can find some $r>0$ such that $K \\cap B(x,r)$ is included in $O$. Hence, for any $y \\in K$, $d(x,f(y)) > r$. But this is absurde: take for example the sequence $u_0=x$, $u_{n+1}=f(u_n)$. Then for any$ar$. Hence the sequence $u_n$ has no convergent subsequence: impossible since $K$ is compact." } { "Tag": [ "probability", "logarithms", "SFFT", "special factorizations" ], "Problem": "Two cards are drawn from a regular deck of 52 cards, without replacement.\r\nLet $x$ be the first number and let $y$ be the second number.\r\n($A=1$, $J=11$, $Q=12$ and $K=13$)\r\n\r\nWhat is the probability that $x$ and $y$ satisfy $\\log_3 (x+y) -\\log_3 x -\\log_3 y +1=0$?", "Solution_1": "[hide=\"hint\"]$\\log_3 (x+y) - \\log_3 (xy) =-1 \\Leftrightarrow \\log_3 \\frac{x+y}{xy} = -1 \\implies \\left( \\frac{xy}{x+y} \\right)=3$. Now use Simon's Favorite Factoring Trick to find possible $x$ and $y$.[/hide]", "Solution_2": "[hide]Which changes into $(x-3)(y-3)=9$\nSo $(x,y)=(4,12),(12,4)(6,6)$[/hide]", "Solution_3": "[hide=\"sweet trick\"]\n\n$\\log_3(x+y) - \\log_3 x - \\log_3 y + 1 = 0 = \\log_3 (\\frac{x+y}{xy}) + 1$\n\n$\\log_3(\\frac{x+y}{xy}) = -1$\n$\\frac{xy}{x+y} = 3$\n\n$xy - 3x - 3y + 9 = 9$\n$(x-3)(y-3) = 9$\n\n$9 = 3*3$ or $1*9$, excluding the negatives factor combinations as they will yield $0$ or a negative number for x or y.\n\nSo there are three xy ordered pairs: $(6, 6), (4, 12), (12, 4)$\n\n$\\frac{4*3}{52*51} + 2\\frac{4^2}{52*51} = \\boxed{\\frac{11}{663}}$[/hide]", "Solution_4": "That is some pretty sweet unparsable or dangerous code you got going all right!", "Solution_5": "Haha, yeah, its fixed now.", "Solution_6": "[quote=\"breez\"][hide=\"sweet trick\"]\n\n$\\log_3(x+y) -\\log_3 x -\\log_3 y + 1 =0 = \\log_3 (\\frac{x+y}{xy}) + 1$\n\n$\\log_3(\\frac{x+y}{xy}) = -1$\n$\\frac{xy}{x+y} = 3$\n\n$xy - 3x - 3y + 9 = 9$\n$(x-3)(y-3) = 9$\n\n$9 = 3*3$ or $1*9$, excluding the negatives factor combinations as they will yield $0$ or a negative number for x or y.\n\nSo there are three xy ordered pairs: $(6, 6), (4, 12), (12, 4)$\n\n$\\frac{4*3}{52*51} + 2\\frac{4^2}{52*51} = \\boxed{\\frac{11}{663}}$[/hide][/quote]\r\nNicely done! :)" } { "Tag": [ "modular arithmetic" ], "Problem": "What's the smallest positive value of $ 33^a \\minus{} 7^b$ where $ a$ and $ b$ are non-negative integers?", "Solution_1": "[hide=\"Solution?\"]Okay, so we have to solve $ 33^a \\minus{} 7^b \\equal{} n$ such that $ n$ is minimized. Re-write the expression as $ 33^a \\equal{} 7^b \\plus{} n$.\n\nNow obviously $ a\\ne0$ or $ n$ will be negative. For $ a\\ge 1$, $ 7^b \\plus{} n \\equiv 0 \\pmod{33}$. So, we have to find the smallest $ n$ such that $ 7^b \\plus{} n$ is a multiple of $ 33$. \n\nThe powers of $ 7 \\pmod {33}$ are $ 7, 16, 13, 25, 10, 4, 28, 31, 19, 1$. Choosing $ 7^b \\equiv 31\\pmod{33}$, the smallest $ n$ is $ 33 \\minus{} 31 \\equal{} \\boxed{2}$.[/hide]", "Solution_2": "You haven't shown that there actually exist $ a, b$ such that the minimum is achieved; \"power of $ 33$\" is a stronger condition than \"multiple of $ 33$.\"", "Solution_3": "[quote=\"RisingFlame\"][hide=\"Solution?\"]Okay, so we have to solve $ 33^a \\minus{} 7^b \\equal{} n$ such that $ n$ is minimized. Re-write the expression as $ 33^a \\equal{} 7^b \\plus{} n$.\n\nNow obviously $ a\\ne0$ or $ n$ will be negative. For $ a\\ge 1$, $ 7^b \\plus{} n \\equiv 0 \\pmod{33}$. So, we have to find the smallest $ n$ such that $ 7^b \\plus{} n$ is a multiple of $ 33$. \n\nThe powers of $ 7 \\pmod {33}$ are $ 7, 16, 13, 25, 10, 4, 28, 31, 19, 1$. Choosing $ 7^b \\equiv 31\\pmod{33}$, the smallest $ n$ is $ 33 \\minus{} 31 \\equal{} \\boxed{2}$.[/hide][/quote]\r\n\r\nCould you please let me know the related values of $ a$ and $ b$?" } { "Tag": [ "calculus", "integration", "analytic geometry", "calculus computations" ], "Problem": "Use a triple integral to define the volume under the surface z=8-2x\u00b2-2y\u00b2 and over the xy plane.\r\nI need to use this simple problem as a guide, to solve problems of this kind. Thanks", "Solution_1": "If you are familiar with cylindrical coordinates, that is a good way to go here. The curve along which $ z = 0$ is $ x^2 + y^2 = 4$, the circle centered at the origin with radius $ 2$. The surface itself in cylindrical coordinates is $ z = 8 - 2r^2$, and recall that the Jacobian is $ r$ (i.e. the 'volume element' is $ dV = rdzdrd\\theta$). Thus, our integral is\r\n\\[ \\int_0^{2\\pi}{\\int_0^2{\\int_0^{8 - 2r^2}{r dz dr d\\theta}}} = \\int_0^{2\\pi}{\\int_0^2{r(8 - 2r^2)drd\\theta}} = \\int_0^{2\\pi}{8d\\theta} = 16\\pi\r\n\\]\r\nAlternatively, if you didn't want to use cylindrical coordinates, you could set up the integral directly in rectangular coordinates. In this case it would be\r\n\\[{ \\int_{ - 2}^2{\\int_{ - \\sqrt {4 - x^2}}^{\\sqrt {4 - x^2}}{\\int_0^{8 - 2x^2 - 2y^2}{dzdydx}}}}\r\n\\]\r\nthe calculation would of course yield the same result $ 16\\pi$, but would be quite a bit more difficult to compute.", "Solution_2": "Thank you!" } { "Tag": [ "function" ], "Problem": "Here are some more (they're all kind of easy, but I liked them, that's why I'm posting them):\r\n\r\n1. Find f:N\\{0}->N s.t. f(1)=0 and f(2n+1)=f(2n)=2f(n)+1 for all n>=1.\r\n\r\n2. Find f:N\\{0}->N s.t. f(1)=0 and f(2n+1)=f(2n)-1=f(n) for all n>=1.\r\n\r\n3. Find f:N->N s.t. f(0)=0 and f(2n+1)=f(2n)+1=f(n)+1 for all n>=0.\r\n\r\n4. Find f:N\\{0}->N s.t. f(1)=0 and f(2n+1)=f(2n)-1=2f(n) for all n>=1.", "Solution_1": "These were all given in different contests! :D :D(I know all these,I have solved them some time ago :) ) \r\nthe 1st one was given in France and f(n)=2^[log_2 (n)] - 1\r\nwhere [x]=x-{x} \r\nthe 2nd one also in Estonia and f(n)=1+[log_2(x)]-n+[n/2]+[n/2^2]+... where x=[log_2(n)]\r\nthe 3rd one in Italy and f(n)=n-([n/2]+[n/2^2]+..)\r\nand the 4th one in SUA and f(n)=2^(x+1)-n-1 where x=[log_2(n)]\r\n\r\nSo I am going to post a solution only for the 2nd one as the others are done in the same way! :D :D\r\nthis is the solution given at the contest! :D \r\n\r\nFor a n natural lets put n=2[n/2]+r and r in {0,1}. so we consider the relations like so: f(n)=f([n/2])-r+1.\r\nlets consider now n=n0+n1*2+..nx*2^x with all n0,n1 and n_(x-1) in {0,1} and nx=1, we represented n in base 2 like so! :)\r\nwe now have f(n)=f(n0+n1*2+..nx*2^x)=..=1+x-(n0+n1+..+nx).(after some computation)\r\nwe know that nk=[n/2^k]-2[n/2^(k+1)] and x=[log_2 (n)] so we get the answer I gave!!\r\nthats all! :D :D", "Solution_2": "That's also what I did (mainly), but here are some ideas which might be useful for other similar ones :\r\n\r\n1. f(n)+1=the number of digits in the binary representation of n\r\n\r\n2. f(n)=the number of 0's in the binary representation of n\r\n\r\n3. f(n)=number of 1's in the binary representation of n\r\n\r\n4. I didn't see this one (I saw it in the 'Answers' section of the book) \r\nf(n) is obtained from n by interchanging 1's and 0's (this one's kind of smart, huh?; this might explain why I didn't see it... :D :D )" } { "Tag": [], "Problem": "Dear every body! I wan to download problems on this site, but I can't! I do not Why that! Could you show me, please!", "Solution_1": "do you mean AoPS?", "Solution_2": "Contest downloading is currently broken." } { "Tag": [ "blogs", "Olimpiada de matematicas" ], "Problem": "Me informaron que un integrante del equipo peruano (creo que CESAR CUENCA o Salas Hunman) fue el mejor oro de este a\u00f1o, otro oro (Anibal Veloso) se qued\u00f3 en el pa\u00eds sede, Chile; al igual que un bronce (Benjamin Baeza).\r\n\r\nEso es lo que pude extraer de una converzaci\u00f3n telef\u00f3nica, ma\u00f1ana cuando el ambiente est\u00e9 m\u00e1s calmado de seguro saldr\u00e1 la informaci\u00f3n oficial.\r\n\r\n\r\nSi pudieran aportar con mas informaci\u00f3n para el tema ser\u00eda vital.\r\n\r\nSaludos y felicidades a los medallistas.", "Solution_1": "Per\u00fa Campe\u00f3n!...\u00a1Per\u00fa Campe\u00f3n!...\r\nes el grito que repite la afici\u00f3n.\r\n\u00a1Per\u00fa Campe\u00f3n!...\u00a1Per\u00fa Campe\u00f3n!... :D \r\n\r\nPer\u00fa quedo 1ero con 3 Oros y 1 Plata.\r\nse que Per1 y Per2 tienen 2 oros... Salu2\r\n..Mas Info..me dijeron que Argentina tenia 3 platas y una bronce..Paraguay una bronce y un mension honrosa..de ahi no se mas..\r\nChaufa!\r\n\r\nPer\u00fa Campe\u00f3n!...\u00a1Per\u00fa Campe\u00f3n!...\r\nes el grito que repite la afici\u00f3n.\r\n\u00a1Per\u00fa Campe\u00f3n!...\u00a1Per\u00fa Campe\u00f3n!...\r\ndice en cada palpitar mi coraz\u00f3n. :D", "Solution_2": "Brasil 2 platas y 2 bronces.", "Solution_3": "[color=blue]\u00a1Maldita sea la prensa paraguaya, que siempre nos da la informaci\u00f3n con crasos errores!\n\nParaguay obtuvo una medalla de bronce, y tres menciones de honor.[/color]", "Solution_4": "Hola:\r\n\r\nCopio textualmente lo que escribio Jhon Cuya en la p\u00e1gina de http://selectivos-peru.blogspot.com/2008/05/cono-sur-2008.html:\r\n\r\n\"En general se repartieron 4 medallas de oro, 6 de plata, 7 de bronce y 4 menciones honrosas, donde la otra oro se la gan\u00f3 un participante chileno. Las cortes para las medallas fueron 43 puntos para la medalla de oro, 22 para la plata y 15 para la bronce.\"\r\n\r\nEntonces:\r\n\r\nOro (C\u00e9sar Cuenca, Fernando Manrique, Iv\u00e1n Mu\u00f1oz -Per\u00fa-, Anibal Veloso-Chile-)\r\nPlata(Eduardo Salas -Per\u00fa-, -Brasil-)\r\nBronce(Benjamin Baeza -Chile-, -Paraguay-, -Brasil- )\r\nMenci\u00f3n Honrosa (-Paraguay-)\r\n\r\nCalculo que para obtener el ranking por equipos suman el puntaje de los integrantes del mismo \u00bfverdad?\r\n\r\nSaludos \r\nOscar", "Solution_5": "[color=blue]Me llama la atenci\u00f3n que casi nunca se publican los datos de las olimpiadas del Cono Sur, as\u00ed como se hace con la Iberoamericana, o bien la Internacional.\n\u00bfPor qu\u00e9 sucede eso? No es dif\u00edcil, aunque sea en un blog, enviar los datos interesantes de cada olimpiada.[/color]", "Solution_6": "Hola:\r\n\r\nEl medallero en la pagina:\r\n\r\nhttp://www.olimpiadadematematica.cl/index.php?option=com_wrapper&Itemid=68\r\n\r\nSaludos\r\n\r\nOscar" } { "Tag": [ "MATHCOUNTS" ], "Problem": "I am new to mathcounts class. Where can I find today's mathcounts assignments and transcripts?", "Solution_1": "Weekly homework is found under the \"Online School\" section (and please ask related questions in the \"Classes Information\" forum.\r\n\r\nTranscripts can be found under \"My Classes\" and clicking on the MATHCOUNTS class." } { "Tag": [ "ratio", "geometry", "circumcircle", "rectangle", "geometry proposed" ], "Problem": "[color=darkred]Let $w$ be the circumcircle of the triangle $ABC$. For a point $M\\in (BC)$ denote : the ratio $\\frac{MB}{MC}=m$ ;\nthe second intersection $T$ of the line $AM$ with the circle $w$ ; the point $X\\in BC\\cap TT$. Prove that $\\boxed{\\frac{XB}{XC}=\\left(\\frac{b}{c}\\cdot m\\right)^{2}}$.[/color]", "Solution_1": "This is much easier than the previous problem.\r\n\r\nLet $BT=y$ and $CT=z$. \r\n\r\n$TBX\\sim\\triangle CTX\\implies\\frac{y}{z}=\\frac{BX}{TX}=\\frac{TX}{CX}$\r\n$\\frac{y^{2}}{z^{2}}=\\frac{BX}{TX}*\\frac{TX}{CX}=\\frac{BX}{CX}$ (1)\r\n\r\nSince $\\triangle CTM\\sim\\triangle ABM$, $\\frac{z}{CM}=\\frac{c}{AM}$. Likewise, $\\frac{y}{BM}=\\frac{b}{AM}$.\r\n\r\nHence: $\\frac{y}{z}=\\frac{b}{c}*\\frac{BM}{CM}=\\frac{bm}{c}$ (2)\r\n\r\nPlug (2) into (1) to get the desired result.\r\n\r\n--\r\n\r\nDid you have an interesting solution that uses more advanced tools? [like harmonic progressions, branchion's theorem, poles/polars, etc?] I am interested in applications of those theorems; I am making an effort to learn those geometry techniques this summer :) .", "Solution_2": "[quote=\"Virgil Nicula\"][color=darkred]Let $w$ be the circumcircle of the triangle $ABC$. For a point $M\\in (BC)$ denote : the ratio $\\frac{MB}{MC}=m$ ;\nthe second intersection $T$ of the line $AM$ with the circle $w$ ; the point $X\\in BC\\cap TT$. Prove that $\\boxed{\\frac{XB}{XC}=(\\frac{b}{c}\\cdot m)^{2}}$.[/color][/quote]\n[color=darkblue][b] O.K. Altheman ![/b] I\"ll present you two well-known simple lemmas.\n\n[b]Lemma I.[/b] Let $ABCD$ be convexe quadrilateral inscribed in the circle $w=C(O,R)$. Denote $I\\in AC\\cap BD$ and $\\{\\begin{array}{c}AB=a\\ ,\\ BC=b\\\\\\ CD=c\\ ,\\ DA=d\\\\\\ AC=e\\ ,\\ BD=f\\end{array}$. \nThen $\\frac{IA}{da}=\\frac{IB}{ab}=\\frac{IC}{bc}=\\frac{ID}{cd}=\\frac{e}{da+bc}=\\frac{f}{ab+cd}=\\sqrt{\\frac{-p_{w}(I)}{abcd}}$, where $p_{w}(I)\\equiv IO^{2}-R^{2}$ is the power of the point $I$ w.r.t. the circle $w$.\n\n[b]Lemma II.[/b] Let $w$ be the circumcircle of the triangle $ABC$. Denote $T\\in BC\\cap AA$. Then $\\frac{TB}{TC}=(\\frac{AB}{AC})^{2}$.\n\n[b]Proof of the proposed problem.[/b] Apply the lemma II to the triangle $BTC$ : $\\frac{XB}{XC}=(\\frac{TB}{TC})^{2}$.\nApply the lemma I to the quadrilateral $BACT$ : $\\frac{MB}{MC}=\\frac{BA\\cdot BT}{CA\\cdot CT}$ $\\implies$ $m=\\frac{c}{b}\\cdot\\frac{TB}{TC}$ $\\implies$ $\\frac{TB}{TC}=\\frac{b}{c}\\cdot m$ $\\implies$ $\\frac{XB}{XC}=(\\frac{b}{c}\\cdot m)^{2}$.\n\n[b][u]Here is an nice, difficult and interesting proposed problem.[/u][/b][/color]\n\n[quote][color=darkred]$O\\in AM$ and $\\{\\begin{array}{c}U\\in AB\\cap XO\\\\\\ V\\in AC\\cap XO\\end{array}$ $\\implies$ $OU=OV$. [/color] [/quote]", "Solution_3": "It suffices to prove that $AC//UT$ because if we say $UT$ intersects $C(O,R)$ again at $T'$, then $ACTT'$ is a rectangle, so $C$, $O$, $T'$ are collinear and we can apply simple triangle similarity.\r\n\r\nAdditional Problem: If we fix $A$, $C$, $O$, and we move $B$ around the circle, then the locus of $U$ is a line. [I suspect harmonic progressions can be used here.]\r\n\r\nI will further investigate tomorrow." } { "Tag": [ "AMC 10", "AMC 10 A" ], "Problem": "I was wondering if anyone was interested in trading math books? It could be really helpful. So let's talk about feasability, and books and competitions we are willing to trade.\r\n\r\nFeasability: I think we can trust most members, so the best thing to do would be just mail the books on a honor system. And we have a rule that we can't trade with anyone having less than 100 posts.\r\n\r\nBooks: Mathematics of Choice, 2005 AMC 10 A\r\n\r\nSorry, this is not a finished idea, but let's work on it.", "Solution_1": "Yep, trading's a great idea. I do that with some of my friends already, but it gets a little inconvenient when you have to mail them all over the place.", "Solution_2": "Anyone want Mathematical Mosaics?", "Solution_3": "I'd love to have it. :)", "Solution_4": "Cool! What do you have?", "Solution_5": "Mathematics of Choice.", "Solution_6": "Ooh, I want that! :) I'll PM you.", "Solution_7": "The honor system is good, but not unbreakable. Aops should not have any responsiblity for stolen books and the such.", "Solution_8": "[quote=\"white_horse_king88\"]The honor system is good, but not unbreakable. Aops should not have any responsiblity for stolen books and the such.[/quote]\r\nI also think the members should be responsible.", "Solution_9": "Make sure you trust or know the person that you'll be trading with to prevent something bad from happening in the first place. :D" } { "Tag": [ "geometry", "geometric transformation", "reflection", "trapezoid", "geometry proposed" ], "Problem": "Given $\\triangle ABC$ such that $\\measuredangle \\ CBA=90^\\circ$. Point $E$ lies from $\\overline{BC}$ such that $\\measuredangle \\ BAE=\\alpha$. Let $D$ and $F$ be the foot of the perpendicular from $B$ and $E$ to $\\overline{AC}$. If $\\measuredangle \\ ACB=2\\alpha$, then prove that $\\overline{BD}=\\frac{\\overline{AB}+\\overline{EF}}{2}$\r\n\r\n[hide=\"Read this before any reply\"]\nNot allowed use trig.[/hide]", "Solution_1": "Let $E' \\in BC$ extended be a reflection of E in AB and F' the foot of a normal from E' to AC. Then $BD = \\frac{EF+E'F'}{2}.$ AE'BF' is cyclic with right angles $\\angle ABE' = \\angle AF'E' = 90^\\circ.$ $\\angle AE'F' = 90^\\circ-[(90^\\circ-2 \\alpha)+\\alpha] = \\alpha = \\angle E'AB,$ so that BE' = AF', AE'BF' is a cyclic isosceles trapezoid with equal diagonals AB = E'F'.", "Solution_2": "Great!!\r\n\r\nIf someone want to give an alternative solution, since do it.", "Solution_3": "Draw the bisector of $\\angle{ACB}$ which intersects AB at G. Draw perpendicular to AC from G and the foot is B'. Extends B'G and intersect BC at A'. We constructed a right triangle $\\triangle{A'CB'}\\cong \\triangle{ACB}$\r\nIt's obvious now that $BD$ is the midline of the trapezoid$A'EFB'$,\r\n(${\\angle{AA'C}=\\angle{AEA'}=90^\\circ-\\alpha\\Rightarrow AA'=AE, BE=A'B=AB'}$)\r\n$BD=\\frac{EF+A'B'}{2}=\\frac{EF+AB}{2}$" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "let Pn(x)=n+(n+1)*x+(n+2)*x^2+...+2n*x^n and z1,z2,..,zn the n complex roots of Pn. Prove that there exist i<>j so that |zi-zj|<8/n\r\n\r\nhave fun ..", "Solution_1": "hint: the roots are between two circles ... :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "There are 32 spies working on the same project. Each spy has at least one piece of information that is not known to the others. At the end of the project they plan to phone each other to bring together all the information that has been collected. These calls are conference calls which include three spy. In each of the calls the three spies pass on to each other all the information that they have learnt up to that moment. At least, how many conference calls are required so that all the information is obtained by all the spies?", "Solution_1": "For the sake of completeness http://www.mathlinks.ro/Forum/viewtopic.php?t=31684" } { "Tag": [], "Problem": "1. integrate cos^8x with respect to x.\r\n\r\n2. solve for real x,sin3x+cos3x=2sinx * cosx\r\n\r\n3. prove, cotx*cot2x+cot2x*cot3x+2=cotx(cotx-cot3x)", "Solution_1": "Something smells like homework..." } { "Tag": [ "inequalities" ], "Problem": "\u0391\u03bd $a,b,c>0$ \u03bc\u03b5 $ab+bc+ca=1$ \u039d\u03b4\u03bf\r\n\r\n$\\sum\\frac{1-a^2}{1+a^2}\\leq \\frac{3}{2}$\r\n\r\n\r\nM\u03b9\u03b1 \u03b1\u03bb\u03b3\u03b5\u03b2\u03c1\u03b9\u03ba\u03ae \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03bb\u03c5\u03c3\u03b7!\r\n :)", "Solution_1": "... \u039a\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae:\r\n\r\n\u0391\u03bd\u03c4\u03b9\u03ba\u03b1\u03b8\u03b9\u03c3\u03c4\u03ce\u03bd\u03c4\u03b1\u03c2: $a = tan\\frac{A}{2}$, $b = tan\\frac{B}{2}$, $c = tan\\frac{C}{2}$, \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9:\r\n\r\n$tan\\frac{A}{2}tan\\frac{B}{2} + tan\\frac{B}{2}tan\\frac{C}{2} + tan\\frac{C}{2}tan\\frac{A}{2} = 1$ (1)\r\n\r\n\u03ba\u03b1\u03b9 \u03c9\u03c2 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc\u03bd, \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf ABC \u03bc\u03b5 \u03b3\u03c9\u03bd\u03af\u03b5\u03c2 $\\angle A$, $\\angle B$, $\\angle C$ \u03bf\u03b9 \u03bf\u03c0\u03bf\u03af\u03b5\u03c2 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03cd\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd (1), \u03ac\u03c1\u03b1 \u03b8\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 $A + B + C = \\pi$\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2, \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9:\r\n\r\n$\\frac{1-a^2}{1+a^2} = \\frac{1-tan^2\\frac{A}{2}}{1+tan^2\\frac{A}{2}} = \\cdot \\cdot \\cdot = cos^2\\frac{A}{2} - sin^2\\frac{A}{2} = cosA$\r\n\r\n\u0386\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9:\r\n\r\n$cosA + cosB + cosC \\leq \\frac{3}{2}$ \u03cc\u03c4\u03b1\u03bd $A + B + C = \\pi$\r\n\r\n\u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03c4\u03cc.", "Solution_2": "cmad, \u03ba\u03b1\u03bb\u03ce\u03c2 \u03ae\u03bb\u03b8\u03b5\u03c2!\r\n\r\n\u03a0\u03ce\u03c2 \u03b2\u03b3\u03ac\u03b6\u03b5\u03b9\u03c2 \u03c4\u03bf \u03c3\u03c5\u03bc\u03c0\u03ad\u03c1\u03b1\u03c3\u03bc\u03b1 \u03cc\u03c4\u03b9 $tan\\frac{A}{2}tan\\frac{B}{2} + tan\\frac{B}{2}tan\\frac{C}{2} + tan\\frac{C}{2}tan\\frac{A}{2} = 1$ ?\r\n\r\n\u0388\u03ba\u03b1\u03bd\u03b1 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03b1\u03c1\u03cc\u03bc\u03bf\u03b9\u03bf \u03bc\u03b5 \u03c3\u03c5\u03bd\u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c4\u03ad\u03bb\u03b7\u03be\u03b1 \u03c3\u03c4\u03bf $cos2A + cos2B + cos2C \\geq -\\frac{3}{2}$\r\n\r\n\r\n\u039b\u03bf\u03c5\u03ba\u03ac\u03c2", "Solution_3": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce\r\n\r\n\u0394\u03af\u03bd\u03b5\u03b9 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03cc\u03c4\u03b9 $ab + bc + ca = 1$, \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03b8\u03b9\u03c3\u03c4\u03ce, \u03ba\u03b1\u03b9 \u03c0\u03b1\u03af\u03c1\u03bd\u03c9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b9\u03c2 \u03b5\u03c6\u03b1\u03c0\u03c4\u03bf\u03bc\u03ad\u03bd\u03b5\u03c2.", "Solution_4": "\u03b1, \u03bf\u03ba! :)", "Solution_5": "\u039d\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5. \r\n\u0398\u03b5\u03c9\u03c1\u03ce \u03c4\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $f(x)=\\frac{1-x^2}{1+x^2}$ \u03bc\u03b5 \r\n\r\n\u03a3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03bf\u03af\u03bb\u03b7 \u03c3\u03c4\u03bf [0,1] \u03ac\u03c1\u03b1 \u03b1\u03c0\u03cc Jensen \r\n\r\n\u03c0\u03c1\u03ce\u03c4\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2 $\\leq 3\\frac{1-(\\frac{a+b+c}{3})^2}{1+(\\frac{a+b+c}{3})^2}$ \r\n\r\n\u0386\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af $3\\frac{1-(\\frac{a+b+c}{3})^2}{1+(\\frac{a+b+c}{3})^2}\\leq\\frac{3}{2}$ \r\n\r\n\u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 :lol:", "Solution_6": "[quote=\"silouan\"]\u039d\u03b1 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5. \n\u0398\u03b5\u03c9\u03c1\u03ce \u03c4\u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $f(x)=\\frac{1-x^2}{1+x^2}$ \u03bc\u03b5 $x\\in \\left[ 0,1\\right]$\n\n\u03a3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03bf\u03af\u03bb\u03b7 \u03c3\u03c4\u03bf [0,1] \u03ac\u03c1\u03b1 \u03b1\u03c0\u03cc Jensen \n\n\u03c0\u03c1\u03ce\u03c4\u03bf \u03bc\u03ad\u03bb\u03bf\u03c2 $\\leq 3\\frac{1-(\\frac{a+b+c}{3})^2}{1+(\\frac{a+b+c}{3})^2}$ \n\n\u0386\u03c1\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af $3\\frac{1-(\\frac{a+b+c}{3})^2}{1+(\\frac{a+b+c}{3})^2}\\leq\\frac{3}{2}$ \n\n\u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 :lol:[/quote]\r\n\r\n\r\ngiati $x\\in \\left[ 0,1\\right]$ ??? :?", "Solution_7": "\u03a0\u03b1\u03c1\u03b1\u03b8\u03ad\u03c4\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03b7 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7.\r\n\r\n\u0391\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03cc\u03c4\u03b9 \u03a3[1/(1+\u03b1^2)]<=9/4 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03a3[1/(\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1+\u03b1^2]<=9/4 \u03ae \u03a3{1/[(\u03b1+\u03b2)(\u03b1+\u03b3)]}<=9/4.\r\n\u039a\u03ac\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c1\u03b9\u03ba\u03bf\u03cd\u03c2 \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 \u03bc\u03b5 \u03b2\u03ac\u03c3\u03b7 \u03c4\u03b7\u03bd \u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1=1 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03c3\u03c4\u03b7\u03bd\r\n\u03b1+\u03b2+\u03b3>=9\u03b1\u03b2\u03b3 => 1/\u03b1\u03b2+1/\u03b2\u03b3+1/\u03b3\u03b1>=9 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd Andreescu :lol:\r\n\u03a3\u03c5\u03b3\u03b3\u03bd\u03ce\u03bc\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03c4\u03c1\u03cc\u03c0\u03bf \u03b3\u03c1\u03b1\u03c6\u03ae\u03c2, \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03b5\u03af\u03bc\u03b1\u03b9 \u03b5\u03be\u03bf\u03b9\u03ba\u03b5\u03b9\u03c9\u03bc\u03ad\u03bd\u03bf\u03c2 \u03bc\u03b5 \u03c4\u03bf Latex.", "Solution_8": "[quote=\"socrates\"]\n\ngiati $x\\in \\left[ 0,1\\right]$ ??? :?[/quote]\r\n\r\nAn x > 1 \u03b7 f<0 \u03ac\u03c1\u03b1 \u03c4\u03bf \u03b1\u03c1\u03b9\u03c3\u03c4\u03b5\u03c1\u03cc \u03bc\u03ad\u03bb\u03bf\u03c2 \u03b5\u03bb\u03bb\u03b1\u03c4\u03ce\u03bd\u03b5\u03c4\u03b1\u03b9 (\u03bd\u03bf\u03bc\u03af\u03b6\u03c9).", "Solution_9": "\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03c0\u03b1\u03c1\u03b1\u03b3\u03ce\u03b3\u03b7\u03c3\u03b5 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b4\u03b5\u03af\u03c2 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03bf [0, 1] \r\n :lol: \u03b1\u03bb\u03bb\u03ac \u03b1\u03bd \u03b4\u03b5\u03bd \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2 \u0384\u03ba\u03bf\u03af\u03c4\u03b1 \u03ba\u03ac\u03c4\u03b9\r\n\r\n $f(x) = \\frac{2}{1 +x^2} -1$ \u03c0\u03b1\u03c1\u03b1\u03b3\u03ce\u03b3\u03b7\u03c3\u03b5 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03b4\u03b5\u03af\u03c2 \u03b1\u03bb\u03bb\u03ac \u03b1\u03c6\u03bf\u03c5 \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b2\u03b3\u03ac\u03bb\u03b5\u03b9\u03c2 \u03ba\u03ac\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ba\u03b1\u03bc\u03c0\u03b7\u03bb\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03b7\u03c2 f \u03c4\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03c3\u03b7\u03bc\u03bf \u03c4\u03b7\u03c2 \u03bc\u03ae\u03c0\u03c9\u03c2 \u03b5\u03bd\u03bd\u03bf\u03b5\u03af\u03c2 f''(x) <0", "Solution_10": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03ce\u03c1\u03b1 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03bf\u03bb\u03bf\u03ba\u03bb\u03ae\u03c1\u03c9\u03c3\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7.\u03a0\u03c1\u03b9\u03bd \u03b5\u03af\u03c7\u03b1 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03b9\u03ba\u03ac \u03bb\u03b1\u03b8\u03b7 .\r\n\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b8\u03b5\u03c9\u03c1\u03ce \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $f(x)=\\frac{1-x^2}{1+x^2}$.\r\n\r\n\u0392\u03c1\u03af\u03c3\u03ba\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $f'(x)=\\frac{-4x}{(1+x^2)^2}<0$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 \u03c7 \u03b8\u03b5\u03c4\u03b9\u03ba\u03cc \u03ac\u03c1\u03b1 f \u03b3\u03bd\u03b7\u03c3\u03af\u03c9\u03c2 \r\n\r\n\u03c6\u03b8\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2.\r\n\r\n\u0395\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 $f''(x)=\\frac{4(3x^2-1)}{(1+x^2)^3}$ .H f'' \u03b1\u03bb\u03bb\u03ac\u03b6\u03b5\u03b9 \u03c0\u03c1\u03cc\u03c3\u03b7\u03bc\u03bf \u03c3\u03c4\u03bf \r\n\r\n$\\frac{\\sqrt 3}{3}$ \u03ac\u03c1\u03b1 f'' \u03ba\u03bf\u03af\u03bb\u03b7 \u03c3\u03c4\u03bf $\\left( 0, \\frac{\\sqrt 3}{3} \\right)$ .\u0386\u03c1\u03b1 \u03b1\u03c0\u03cc \r\n\r\nJensen \u03cc\u03c0\u03c9\u03c2 \u03ad\u03b4\u03b5\u03b9\u03be\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 A \u03bc\u03ad\u03bb\u03bf\u03c2 $\\leq\\frac{3}{2}$\r\n\r\n\u038c\u03bc\u03c9\u03c2 \u03b1\u03c6\u03bf\u03cd f \u03b3\u03bd\u03b7\u03c3\u03af\u03c9\u03c2 \u03c6\u03b8\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 \u03bf\u03b9 \u03c4\u03b9\u03bc\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7\u03c2 \u03b3\u03b9\u03b1 $x\\in \\left( 0, \\frac{\\sqrt 3}{3} \\right)$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5\u03b3\u03b1\u03bb\u03cd\u03c4\u03b5\u03c1\u03b5\u03c2 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ad\u03c2 \u03cc\u03c4\u03b1\u03bd $x\\in\\left( \\frac{\\sqrt 3}{3}, +\\infty\\right)$ \u03ac\u03c1\u03b1 \u03c4\u03b5\u03bb\u03b5\u03b9\u03ce\u03c3\u03b1\u03bc\u03b5.\r\n\u03a0\u03ce\u03c2 \u03c3\u03b1\u03c2 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b5 ?", "Solution_11": "\u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03b5 \u03bd\u03bf\u03bc\u03b9\u03b6\u03c9 \u03b5\u03af\u03c3\u03b1\u03b9 \u03bf\u03ba \u03bc\u03b5 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7!!\r\n :)", "Solution_12": "\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u0398\u03b1\u03bd\u03ac\u03c3\u03b7 \u03b8\u03b1 \u03c4\u03b1 \u03c0\u03bf\u03cd\u03bc\u03b5" } { "Tag": [ "calculus" ], "Problem": "Guessing Question\r\n--------------------\r\n\r\nOut Of 150 Students there is 6 teachers 4 of the teachers are failing students but in each class there are 5 honor students but 2 honor students are terrible in a subject. What is The Subject That Most Students are Failing On?", "Solution_1": "The problem is unclear. Please clarify.", "Solution_2": "Failing students means giving at least 1 student a failing grade (usually F, possibly D). It says terrible at a subject, not failing. Terrible can be A- for some people", "Solution_3": "Sorry if i was unclear but i meant that the honor students are only a honor student in a few subjects. This school wasnt exactly a great school so the students basically didnt care about their education. Failing Grades Meant: F's. \r\nThe failing possible subjects could be: Calculous, Biology, Language, and Music.\r\nThese are 10th graders." } { "Tag": [ "geometry", "geometric transformation", "number theory", "prime numbers" ], "Problem": "Well, some days ago a friend told me the so-called \"Lord Smurf's Theorem\"... I stared at it... Then realised..... :laugh: \r\n\r\nHere it is:\r\n\r\n\"If $ p$ and $ p^2\\plus{}2$ are prime numbers, then also $ p^3\\plus{}2$ is a prime.\"\r\n\r\n\r\nTry to prove it... (I'm sure you'll get an unexpected surprise... :yup: )", "Solution_1": "Not sure how to prove it, but I gotta say I love the name.", "Solution_2": "[quote=\"chetjan\"]I gotta say I love the name.[/quote]\nYeah, that's brilliant! :yup:\n\n[quote=\"chetjan\"]Not sure how to prove it[/quote]\r\nWell, here is just a little hint:\r\n[hide=\"hint\"]\nTry it for small values of $ n$ (let's say... up to 5 for example)\nThen... tell me what you think...\n[/hide]", "Solution_3": "I like it. How'd he get the name?\r\n\r\n[hide=\"proof\"]Take the 2nd prime mod 3. If $ p\\equiv\\pm1\\mod3$, then $ p^2\\plus{}2\\equiv 3\\mod3$ and so since its a prime it must be 3. But then $ p\\equal{}1$ which isn't a prime. Thus $ p\\equiv0\\mod3$ and so $ p\\equal{}3$. $ p^2\\plus{}2\\equal{}11$ is a prime, so $ p^3\\plus{}2\\equal{}29$ must be a prime too, and it is![/hide]", "Solution_4": "Well... the name is a translation from Italian to English, so I hope I chose it properly :) \r\n\r\nLord Smurf is the leader of the Smurfs, and can be easily distinguished from the other Smurfs by his red clothes and his bushy white mustache & beard. Everyone in the Smurf Village turns to Papa Smurf when things go awry. Papa Smurf is very skillful in making magical spells and potions. (thanks Wikipedia!)\r\n\r\n[img]http://www.cartonionline.com/gif/CARTOON/puffi/immagini/Puffi_02.jpg[/img]\r\n\r\n\r\n\r\nIn Italy he's very funny!" } { "Tag": [ "trigonometry" ], "Problem": "The angle of elevation of the tree from one position on a flat path from the tree is 40 degrees and from a second position 40 feet along this same path is 30 degrees. How high is the tree?", "Solution_1": "$ \\tan 40\\equal{}\\frac{h}{x}$\r\n\r\n$ \\tan 30\\equal{}\\frac{h}{x\\plus{}40}$\r\n\r\nCan you go from there?", "Solution_2": "Do I solve for x first or for h first?\r\n\r\ntan40 = h/x \r\n\r\nDoes x = tan40(h)\r\n\r\nDo I then plug the above value for x into\r\n\r\ntan30 = h/(x + 40)?", "Solution_3": "You could solve for either first, but it's obviously simpler if you just solve for h straight away. \r\n\r\nFrom equation 1:\r\n\r\n$ \\tan40^{\\circ}\\equal{}\\frac{h}{x}$\r\n\r\n$ \\Rightarrow x \\equal{}\\frac{h}{\\tan40^{\\circ}}$\r\n\r\nSubstituting in 2:\r\n\r\n$ \\tan30^{\\circ}\\equal{}\\frac{h}{\\frac{h}{\\tan40^{\\circ}}\\plus{}40}$\r\n\r\n$ \\frac{1}{\\sqrt{3}}\\equal{}\\frac{h}{\\left(\\frac{h\\plus{}40\\tan40^{\\circ}}{\\tan40^{\\circ}}\\right)}$\r\n\r\n$ \\frac{1}{\\sqrt{3}}\\equal{}\\frac{h\\tan40^{\\circ}}{h\\plus{}40\\tan40^{\\circ}}$\r\n\r\n$ h\\plus{}40\\tan40^{\\circ}\\equal{}\\sqrt{3}h\\tan40^{\\circ}$\r\n\r\n$ h\\minus{}\\sqrt{3}h\\tan40^{\\circ}\\equal{}\\minus{}40\\tan40^{\\circ}$\r\n\r\n$ h(1\\minus{}\\sqrt{3}\\tan40^{\\circ}) \\equal{}\\minus{}40\\tan40^{\\circ}$\r\n\r\n$ h \\equal{}\\frac{\\minus{}40\\tan40^{\\circ}}{1\\minus{}\\sqrt{3}\\tan40^{\\circ}}$\r\n\r\n$ h\\approx74$ feet" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "find all natural numbers such that; 0.5(a+b)(b+c)(c+a) +(a+b+c)^3=1-abc.", "Solution_1": "Do not name topics like this!\r\nThat name is useless, meaningless, senseless, etcless and helps noone recognicing the problem.\r\nPlease change the title.\r\n(And I think it was already posted)", "Solution_2": "$(a,b,c)\\in N^{3}$ ; $\\frac{1}{2}(a+b)(b+c)(c+a)+(a+b+c)^{3}=1-abc$\r\nbut $1-abc\\le 0$ and $\\frac{1}{2}(a+b)(b+c)(c+a)+(a+b+c)^{3}\\geq 0$\r\n$S=\\{nothing\\}$ :!: :!:", "Solution_3": "oh i dont know that 1-abc<0 . the numbers lie in Z :ninja:", "Solution_4": "[quote=\"fermat3\"] natural numbers [/quote]\r\nit's $N$\r\nZ: integer numbre", "Solution_5": "if$a,b,c \\in Z$ then the condition is equivlent to $(2a+b+c)(2b+c+a)(2c+a+b)=1$\r\nthe rest is trivial." } { "Tag": [], "Problem": "This problem was from Russia 2002 MO.\r\n\"Show that there are integers a,b such that n< a^2 +b^2 < n+3*(n)^(1/4) .\r\nFor every n >10000, n is integer\"", "Solution_1": "Possible outline of proof (offhand):\n\n\n\n[hide]\n\n A number k can be written in the form a^2 + b^2 iff it's squarefree part has no factors congruent to 3 (mod 4). \n\n\n\n There are at least 29 numbers in the range, for any n satisfying the conditions of the problem.\n\n\n\n[/hide]", "Solution_2": "Well, I had thought of that before but i didn't manage to get something that works.\r\nI have found a solution by representig n = k^2 +h, with 0<=h<=2k.\r\nThen i have put a=k and b=x+1, where x^2<=h<(x+1)^2 this works for all x>5 and x<3.\r\nIf x=3 or 4 you easily play with." } { "Tag": [ "geometry", "3D geometry", "algorithm", "Old" ], "Problem": "What are you guy's rubiks time. My record is 4:43", "Solution_1": "there's already lots of rubiks cube threads like [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40921[/url]", "Solution_2": "Those are old ones. i won't revive them", "Solution_3": "Then I will do that :P \r\nThere is no reason not to revive it...", "Solution_4": "cool 4 mintues... what method?", "Solution_5": "[hide]Rubik's cube is wicked hard, I am thinking of just putting all the stickers together again and starting over. I am completely stumped by it.[/hide]", "Solution_6": "[quote=\"SoxFan34\"][hide]Rubik's cube is wicked hard, I am thinking of just putting all the stickers together again and starting over. I am completely stumped by it.[/hide][/quote]\r\n\r\nturn the top side 45 degrees and pop the middle edge out.\r\nDont take off the stickers lol.", "Solution_7": "Well, my record is about 2 minutes (2:01 to be exact) using a tiny little key chain cube (the only one I have :P ).\r\nBut I've only learned how to solve the cube a little more than a week ago. Is that good?", "Solution_8": "The only cube that I have is the little one that I got at Mathcounts. I can do one face of it, but I have never actually been taught how to do it. I'm trying to teach myself......slowly..... :rotfl:", "Solution_9": "Teaching yourself doesn't work well at all.\r\n\r\nI taught myself and consistantly solved it in around 5 or 6 minutes, and after reading a guide I could solve it in roughly a minute 15 seconds.\r\n\r\nI only solved Rubik's cubes during this summer and last summer out of boredom, so I don't know if I can solve them anymore.", "Solution_10": "I can do the top face in like 1:30, but I don't feel like memorizing the algorithms for doing the rest.", "Solution_11": "[quote=\"Treething\"]Teaching yourself doesn't work well at all.\n[/quote]\r\n\r\nIt may not work as well, but it would be a lot more fun to teach myself than just memorize a bunch of algorithms. I just want to get a whole cube in under 10 minutes by myself, and then I will look at some algorithms. It just seems like cheating to memorize algorithms. If you had a math problem, would you figure it out yourself, or look at an answers booklet and let someone else figure it out for you? \r\n\r\nThe point of this forum, The [b]Art[/b] of Problem Solving is to do math creatively, not to just memorize a bunch of formulas, which is what most people seem to do to get good at rubiks cube. Just because it isn't math doesn't mean that your standards should suddenly change.", "Solution_12": "[quote=\"nebula42\"]Well, my record is about 2 minutes (2:01 to be exact) using a tiny little key chain cube (the only one I have :P ).\nBut I've only learned how to solve the cube a little more than a week ago. Is that good?[/quote]\r\n\r\nNeal, I know exactly where you got that cube. MC Nationals 2006! :D :D :D I have one too, except it keeps coming apart when I try to turn it.", "Solution_13": "I actually got my own method down to abuout 1:30 average with a crappy cube. Then I got a better cube, sprayed it, and modified my method dramatically (closer to Petrus), reducing that average to under 20 seconds over time. So teaching yourself can be good.", "Solution_14": "Moved from RT.", "Solution_15": "My record is 36.23 seconds.......u guys should learn f2l. The world record is 4.90 by Lucas Etter.", "Solution_16": "[quote=aleeshen]My record is 36.23 seconds.......u guys should learn f2l. The world record is 4.90 by Lucas Etter.[/quote]\n\nLol first post again in 10 years :rotfl: ", "Solution_17": "My record is 15 seconds but I average about 25 seconds. I really want to get sub-20!!! I also use CFOP", "Solution_18": "4:00 rec!!!!!", "Solution_19": "$1:23$\n :first: :winner_first: " } { "Tag": [], "Problem": "Which of the following methods of proving a geometric figure a locus is not correct?\r\n\r\n$ \\textbf{(A)}\\ \\text{Every point of the locus satisfies the conditions and every point not on the locus does not satisfy the conditions.}$\r\n$ \\textbf{(B)}\\ \\text{Every point not satisfying the conditions is not on the locus and every point on the locus does satisfy the conditions.}$\r\n$ \\textbf{(C)}\\ \\text{Every point satisfying the conditions is on the locus and every point on the locus satisfies the conditions.}$ \r\n$ \\textbf{(D)}\\ \\text{Every point not on the locus does not satisfy the conditions and every point not satisfying} \\\\\r\n\\text{the conditions is not on the locus.}$\r\n$ \\textbf{(E)}\\ \\text{Every point satisfying the conditions is on the locus and every point not satisfying the conditions is not on the locus.}$", "Solution_1": "[hide=\"Click for solution\"]\nTo prove that a geometric figure is a locus you must include all proper points and exclude all improper point. $ \\boxed{\\textbf{(B)}}$ does not say that every point satisfying the conditions lies on the locus.\n[/hide]", "Solution_2": "Note that choice B doesn't guarantee that there isn't any other point not on the locus but satisfies the conditions." } { "Tag": [ "modular arithmetic", "ratio", "number theory unsolved", "number theory" ], "Problem": "n is odd, composite and not a power of an integer. How to show that at least half the integer pairs x,y in [0,n) with x^2 = y^2 (mod n) necessarily have gcd(x-y,n) as a proper factor of n?", "Solution_1": "That's not quite true. \r\nFor $ n \\equal{} 15$, there are 45 pairs $ (x,y)$ in $ [0,14]$ with $ x^2\\equiv y^2\\pmod{15}$ but only 22 of them give $ \\gcd(x \\minus{} y,15)$ as a proper factor of $ n$.\r\n\r\nHowever, if we restrict our attention to $ x,y$ co-prime to $ n$, then the number of pairs delivering a proper factor of $ n$ is indeed at least a half of the total amount. To be precise, there are $ \\varphi(n)\\cdot 2^k$ such pairs overall (where $ k$ is the number of distinct prime factors of $ n$) while only $ 2\\varphi(n)$ of them give a non-proper factor. So, the success ratio is \r\n\\[ \\frac{\\varphi(n)\\cdot (2^k\\minus{}2)}{\\varphi(n)\\cdot 2^k} \\equal{} 1 \\minus{} \\frac{1}{2^{k\\minus{}1}} \\geq \\frac{1}{2}.\\]" } { "Tag": [ "real analysis", "calculus", "integration", "calculus computations" ], "Problem": "Show that if a real series $ \\sum_{i,j\\equal{}1}^\\infty a_{ij}$ converges absolutely, then \r\n\\[ \\sum_{i,j\\equal{}1}^\\infty a_{ij}\\equal{}\\sum_{i\\equal{}1}^\\infty\\sum_{j\\equal{}1}^\\infty a_{ij}\\equal{}\\sum_{j\\equal{}1}^\\infty\\sum_{i\\equal{}1}^\\infty a_{ij}\\]", "Solution_1": "this is related to fubini's theorem for sums which states, \r\nTheorem: If $ \\sum_{i,j \\in \\mathbb{N}} |f_{ij}| < \\infty$ then\r\n\\[ \\sum_{i,j \\in \\mathbb{N}} f_{ij} \\equal{} \\sum_{i\\equal{}1}^{\\infty} \\sum_{j\\equal{}1}^{\\infty} f_{ij} \\equal{} \\sum_{j\\equal{}1}^{\\infty} \\sum_{i\\equal{}1}^{\\infty} f_{ij}\\]\r\nWhich is Fubini's original theorem for lebesgue integrals except you use the counting measure $ \\mathbb{N}$ so I assume the proof of Fubini's theorem for lebesgue integrals will be similar to this proof. Here it is [url]http://planetmath.org/?op=getobj&from=objects&id=7189[/url]", "Solution_2": "Yes, this is one form of Fubini's theorem. Counting measure is simpler than general product measures, so the proof doesn't have too many complications.\r\n\r\nSee also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=174318]here[/url]." } { "Tag": [ "number theory", "prime numbers", "number theory unsolved" ], "Problem": "a,b and c are natural number N*\r\n\r\nprove that a=b=c if we have a^2/a+b, b^2/b+c, and c^2/a+c prime numbers\r\n :oops: :P", "Solution_1": "Let $x = a+b, y =b+c, z = c+a$. Then the quantities\r\n\r\n$\\frac{(x-y+z)^{2}}{4x}, \\frac{(x+y-z)^{2}}{4y}, \\frac{(-x+y+z)^{2}}{4z}$\r\n\r\nAre prime numbers, say, $p, q, r$. Then\r\n\r\n$(x-y+z)^{2}= 4xp$\r\n$(x+y-z)^{2}= 4yq$\r\n$(-x+y+z)^{2}= 4zr$\r\n\r\nAfter dividing out by the irrelevant factor of $4$, we clearly have that $p, q, r$ must divide a square, hence they must divide it twice. It follows that $p | x, q | y, r | z$. \r\n\r\nSo let $x = p \\alpha, y = q \\beta, z = r \\gamma$. Then\r\n\r\n$\\left( \\frac{p \\alpha-q \\beta+r \\gamma}{2p}\\right)^{2}= \\alpha$\r\n$\\left( \\frac{p \\alpha+q \\beta-r \\gamma}{2q}\\right)^{2}= \\beta$\r\n$\\left( \\frac{-p \\alpha+q \\beta+r \\gamma}{2r}\\right)^{2}= \\gamma$\r\n\r\nFor some integers $\\alpha, \\beta, \\gamma$. \r\n\r\n[b]To be continued...[/b]", "Solution_2": "assume the contrary, WLOG 0 p|c (1)\r\nc^2 /p =2c-k\r\nc/p = 2 \u2013 k/c\r\nfrom (1) k/c is a positive integer then k=0" } { "Tag": [ "trigonometry", "AMC", "AIME" ], "Problem": "This problem was on 4everwise's mock AIME 2 from 2006-2007:\r\n\r\nIf $ \\displaystyle \\tan 15^\\circ \\tan 25^\\circ \\tan 35^\\circ \\equal{}\\tan \\theta$ and $ \\displaystyle 0^\\circ \\le \\theta \\le 180^\\circ$, find $ \\displaystyle \\theta$.", "Solution_1": "You should be able to use the identity $ \\tan{a}\\tan{b}\\tan{c}\\equal{}\\frac{\\sin{2a}\\sin{2b}\\sin{2c}}{(\\cos{2a}\\plus{}1)(\\cos{2b}\\plus{}1)(\\cos{2c}\\plus{}1)}$\r\nand find that $ \\theta\\equal{}5^\\circ$", "Solution_2": "I can see why that identity is true, but why would somebody know that off the top of his head?", "Solution_3": "Stank, I can't really see how that identity is going to solve the problem, but we can use $ \\tan(60\\minus{}x)\\tan x\\tan(60\\plus{}x)\\equal{}\\tan3x$ with $ x\\equal{}25$", "Solution_4": "[quote=\"mathwizarddude\"]Stank, I can't really see how that identity is going to solve the problem, but we can use $ \\tan(60 \\minus{} x)\\tan x\\tan(60 \\plus{} x) \\equal{} \\tan3x$ with $ x \\equal{} 25$[/quote]\r\n\r\nHow did you think to use that identity? And did you just derive that identity while thinking about the problem? Is it a common identity?" } { "Tag": [ "trigonometry", "inequalities", "ARML", "geometry", "angle bisector", "congruent triangles", "Pythagorean Theorem" ], "Problem": "Here's what I did. Please correct me if I'm wrong. \r\n\r\n[hide=\"1\"]First off, we wish to minimize the digits. Clearly, the smallest partition of 2009 with digits from 0-9 is one 2 and 223 9's, yielding a total of 224 digits. \n\nIf the leading digit is 1, the sum of the other 223 digits must be 2008, which is impossible since $ 9 \\cdot 223 < 2008$. If the leading digit is 2, then we get $ 2 \\underbrace{99 \\ldots 9}_{223}$ satisfies all the conditions of the problems, so that is our answer.[/hide]\n\n[hide=\"2\"]Suppose $ n^2 \\plus{} 2009n \\equal{} m^2$. Then $ (2n \\plus{} 2009)^2 \\minus{} 2009^2 \\equal{} (2m)^2$, so $ (2n \\plus{} 2009 \\minus{} 2m)(2n \\plus{} 2009 \\plus{} 2m) \\equal{} 2009^2$. Letting $ u \\equal{} 2n \\plus{} 2009$, and $ v \\equal{} 2m$, we wish to find the maximal solution of \n$ u$ in $ (u \\minus{} v)(u \\plus{} v) \\equal{} 7^4 41^2$, where $ u$ is odd and $ v$ is a multiple of 4. If we let $ u \\plus{} v \\equal{} 41^2 7^3$ and $ u \\minus{} v \\equal{} 7$, we see that that $ u \\equal{} 288295$ and $ v \\equal{} 288288$ are odd and a multiple of 4, respectively, so we've found a solution. Clearly, since $ 41^2 7^3$ is the greatest factor of $ 2009^2$, this is the solution that maximizes $ u \\equal{} 288295$, so we get that $ n \\equal{} 143143$ is the greatest possible solution. [/hide]\n\n[hide=\"3\"]We're going to use a heuristic argument here. Suppose $ e_1, \\ldots, e_k$ are the powers of the prime factors. Then $ \\prod (e_i \\plus{} 1) \\equal{} 2009$. We guess that the $ e_i$'s that will minimize the product are $ e_1 \\equal{} 40$, $ e_2 \\equal{} 6$, and $ e_3 \\equal{} 6$. Clearly, the product is here minimized when it's $ 2^{40} 3^6 5^6$, so I suppose that's our answer. \n\nThere's probably a way to rigoroize (or correct =p) what I'm doing above. Any feedback would be appreciated. [/hide]\n\n[hide=\"4\"]Note that $ n \\equal{} 1! 2! \\ldots 120! \\equal{} 1^{120} 2^{119} \\ldots 120^1$. Note that if we divide out $ 2 \\cdot 4 \\cdot 8 \\cdot \\ldots \\cdot 120 \\equal{} 2^{60} 60!$, we'll end up with a perfect square (as all the odd exponents would become even.) However, if we divide out 60!, we'll get a perfect square divided by $ 2^{60}$, which itself is a perfect square. [/hide]\n\n[hide=\"5\"]Clearing denominators, we get that $ mn \\equal{} 2009m \\plus{} 2009n$, so $ (m \\minus{} 2009)(n \\minus{} 2009) \\equal{} 2009^2$. Note that if $ m, n < 2009$, since $ m$ and $ n$ are positive, $ |m \\minus{} 2009| < 2009$ and $ |n \\minus{} 2009| < 2009$, so their product is less than $ 2009^2$. Therefore, we only need consider the case in which $ m \\minus{} 2009$ and $ n \\minus{} 2009$ are both positive. There is a correspondence between these solutions and the number of positive solutions to $ xy \\equal{} 2009^2$, which is simply $ \\boxed{15}$. [/hide]\n\n[hide=\"6\"]A fun little problem. It's trivial if you apply law of cosines on triangles ABP, ACP, and ABC. There's another way to do it, though. \n\nConstruct a point P' such that BP'C is equilateral. Note that since $ \\angle BAC \\plus{} \\angle BPC \\equal{} 180^{\\circ}$, $ ABP'C$ is cyclic, so $ 60^{\\circ} \\equal{} m \\angle BAP' \\equal{} \\angle CAP'$, so $ P'$ lies on the angle bisector of $ \\angle CAB$. By Ptolemy's theorem, $ AP' \\equal{} AB \\plus{} AC$. Since the point $ X$ lying on the angle bisector of $ \\angle CAB$ such that $ AX \\equal{} AB \\plus{} AC$ is unique, we get that $ X \\equal{} P' \\equal{} P$, so triangle BPC is equilateral. \n\nThere was another way using a line extension and congruent triangles, nothing else, but I forgot it...\n[/hide]\n\n[hide=\"7\"]This is $ 4 \\times \\frac {10^{2n} \\minus{} 1}{9} \\minus{} 8 \\times \\frac {10^{n} \\minus{} 1}{9} \\equal{} \\frac {4 \\times 10^{2n} \\minus{} 8 \\times 10^n \\plus{} 4}{9} \\equal{} \\left ( \\frac {2 \\cdot 10^n \\minus{} 2}{3} \\right )^2$. [/hide]\n\n[hide=\"8\"]Haven't worked on this long enough. Muirhead fails me. :/ Overall, inequalities are probably my worst subject, and Muirhead can't help me here, so I'm stuck. [/hide]\n\n[hide=\"9\"]I haven't got the full solution, but here's my progress. For any $ x^2 a$, with $ x > 0$, if we can write $ a$ as a sum of 6 distinct squares $ s_1, \\ldots, s_6$, we can also write $ x^2 a$ as a sum of 6 distinct squares $ xs_1, \\ldots, x s_6$. \n\nI let $ x \\equal{} 2009^{2008}$, so we need to express $ 2009^2$ as a sum of 6 distinct squares. \n\nI haven't worked on this enough to get the full solution. I suppose we could reduce the problem to writing $ 1681 \\equal{} 41^2$ as a sum of 6 distinct squares and hope for the best. Or we could just write a program and bash it out, but oh well...\n[/hide]\n\n[hide=\"10\"]I actually made up a problem using the exact same concept, and was hoping to propose it to a random competition when I get old enough, but my hopes were crushed when I saw this problem. -_- On the plus side, it made this problem trivial for me. \n\nDraw BD and AC. We know $ m \\angle ABC \\equal{} m \\angle ACD \\equal{} 90^{\\circ}$. Therefore, by the Pythagorean theorem, $ BD \\equal{} \\sqrt {x^2 \\minus{} a^2}$ and $ AC \\equal{} \\sqrt {x^2 \\minus{} c^2}$. By Ptolemy's theorem, $ (bx \\plus{} ac)^2 \\equal{} (x^2 \\minus{} a^2)(x^2 \\minus{} c^2)$. Therefore, $ b^2x^2 \\plus{} 2abcx \\plus{} a^2c^2 \\equal{} x^4 \\minus{} a^2 x^2 \\minus{} c^2 x^2 \\plus{} a^2 c^2$. Rearranging, $ x^4 \\minus{} (a^2 \\plus{} b^2 \\plus{} c^2)x^2 \\minus{} 2abcx \\equal{} 0$, so $ x^3 \\minus{} (a^2 \\plus{} b^2 \\plus{} c^2)x \\minus{} 2abc \\equal{} 0$, as desired. [/hide]", "Solution_1": "Hmm, I didn't apply, but here's my solution to 9:\r\n\r\n[hide]Motivation: Writing something as 6 distinct squares can be done if the number is the product of a sum of 3 distinct squares and a sum of 2 distinct squares.\n\n2009^2010 = $ 7^{4020}$ multiplied by $ 41^{2010}.$ Now, $ 7^{4020} \\equal{} (7^{2009})^2 * 49 \\equal{} (7^{2009})^2*(9 \\plus{} 4 \\plus{} 36).$ We have succeeded in representing $ 7^{4020}$ as a sum of 3 squares.\n\nNow, we represent 41^2010 as a sum of 2 squares. 41 = 5^2+4^2, and (a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2, so we have 41^2 = (5^2-4^2)^2 +(2*4*5)^2 = 9^2+40^2. 41^2010 = (41^1004)^2 * (9^4+40^2), and we have succeeded in representing 41^2010 as the sum of 2 squares and we are done.[/hide]\n\nOutline for the inequality:\n\n[hide]Multiply numerator and denominator of a/a+2b by a, Cauchy-Schwarz Lemma, instant win.[/hide]", "Solution_2": "[quote]Cauchy-Schwarz Lemma[/quote]\nCould you remind me what that is again? \n\n[quote]Writing something as 6 distinct squares can be done if the number is the product of a sum of 3 distinct squares and a sum of 2 distinct squares. [/quote]\r\n$ (1^2 \\plus{} 2^2 \\plus{} 3^2)(4^2 \\plus{} 12^2) \\equal{} 4^2 \\plus{} 8^2 \\plus{} \\boxed{12^2 \\plus{} 12^2} \\plus{} 24^2 \\plus{} 36$ I'm aware that this isn't the only decomposition, but it's the one that's most consistent with what you said. Could you explain your reasoning a bit more? (I think your solution works anyway, though. )", "Solution_3": "[quote=\"Zhero\"][quote]Cauchy-Schwarz Lemma[/quote]\nCould you remind me what that is again? \n\n[quote]Writing something as 6 distinct squares can be done if the number is the product of a sum of 3 distinct squares and a sum of 2 distinct squares. [/quote]\n$ (1^2 \\plus{} 2^2 \\plus{} 3^2)(4^2 \\plus{} 12^2) \\equal{} 4^2 \\plus{} 8^2 \\plus{} \\boxed{12^2 \\plus{} 12^2} \\plus{} 24^2 \\plus{} 36$ I'm aware that this isn't the only decomposition, but it's the one that's most consistent with what you said. Could you explain your reasoning a bit more? (I think your solution works anyway, though. )[/quote]\r\n\r\nYeah that statement is false :oops: \r\n\r\nThe proof still gives the number as a sum of 6 squares, you just have to check that they are distinct. (And they indeed are)\r\n\r\nCauchy-Schwarz Lemma is a1^2/a2 + b1^2/b2 + c1^2/c2 >= (a1+b1+c1)^2/a2+b2+c2. (This is the 3 variable version.)", "Solution_4": "For #2, I got something different\r\n[hide]\nSetting $ n^2 \\plus{} 2009n \\equal{} (n \\plus{} 1004)^2$ gives $ n \\equal{} 1004^2$ as an answer.\nIf n is bigger, then $ (n \\plus{} 1004)^2 < n^2 \\plus{} 2009n < (n \\plus{} 1005)^2$\n[/hide]\n\nOn #4, it really went to 120? I thought it was 121! when I printed it out for some reason (at ARML, a friend and I managed to prove that it was impossible with 121). I need to check the test I printed out again.\n\nFor #8:\n[hide]\n$ (\\frac {a}{a \\plus{} 2b} \\plus{} \\frac {b}{b \\plus{} 2c} \\plus{} \\frac {c}{c \\plus{} 2a})(a(a \\plus{} 2b) \\plus{} b(b \\plus{} 2c) \\plus{} c(c \\plus{} 2a))\\ge (a \\plus{} b \\plus{} c)^2$ by Cauchy.\nIt turns out that $ (a \\plus{} b \\plus{} c)^2 \\equal{} a(a \\plus{} 2b) \\plus{} b(b \\plus{} 2c) \\plus{} c(c \\plus{} 2a)$, so we can divide that out and win.[/hide]\n\nFor #9:\n[hide]\nMy friend and I think we figured this out during ARML (after the deadline, of course). Basically, you can keep decomposing $ (7^{2010}*41^{1005})^2$ into 9-40-41 pythagorean triples a bunch of times. I think there might be something wrong, though, because that means that you can split $ 2009^{2010}$ into a sum of at least 1005 distinct squares.[/hide]\r\n\r\n\r\nEdit: I just checked the test I had printed out, and it does in fact go to 121!. So, they might have changed the pdf document after I had already saved and printed it.", "Solution_5": "For #2, you're probably right then (I was searching for a solution like that, and I was like meh whatever bashy time), and I thought up my solution was I was typing it up, so yeah..." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Find the greatest positive prime factor of 2^20 - 1.", "Solution_1": "[hide=\"Hint\"]$ 2^{20}\\minus{}1\\equal{}2^{20}\\minus{}1^{20}$.[/hide]", "Solution_2": "A repost of problem?\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=314061", "Solution_3": "[hide=\"Click here to reveal hidden content\"]\n\n\n2^20-1=(2^10-1)(2^10+1). 2^10=1024, so there is 1023 and 1025. 1023=3*11*31, and 1025=5^2*41. Thus the largest prime factor is [b]41[/b].[/hide]", "Solution_4": "${ 2^{20}-1=2^{20}-1^{20}=(2^{10}-1)(2^{10}+1}$\r\n$ (1023)(1025)=(11*3*33)(41*5*5)$ 41 prime factor", "Solution_5": "[hide][quote=\"sri340\"]Find the greatest positive prime factor of 2^20 - 1.[/quote]\n$ 2^{20} \\minus{} 1 \\equal{} 2^{20} \\minus{} 1^{20}.$\n\n$ (2^{10} \\plus{} 1^{10})(2^{10} \\minus{} 1^{10})$\n\n$ (2^{10} \\plus{} 1^{10})(2^5 \\plus{} 1^5)(2^5 \\minus{} 1^5)$\n\nThe prime factorization of $ 2^{10} \\plus{} 1$ is $ 5^2\\cdot41$.\n\nThe prime factorization of $ 33$ is $ 11\\cdot3$.\n\nThe prime factorization of $ 31$ is simply $ 31$ because 31 is prime. \n\nTherefore, the largest positive prime factor of $ 2^{20} \\minus{} 1$ is $ \\fbox{41}$.[/hide]" } { "Tag": [ "logarithms" ], "Problem": "Suppose that: $a^{\\log_{b}c}+b^{\\log_{c}a}=m$. Find the value of $c^{\\log_{b}a}+a^{\\log_{c}b}$", "Solution_1": "[hide]\r\n$a^{\\log_{b}c}+b^{\\log_{c}a}= a^\\frac{ \\log_{a}c}{\\log_{a}b}+b^\\frac{\\log_{b}a}{\\log_{b}c}= c^{\\frac{1}{\\log_{a}b}}+a^{\\frac{1}{\\log_{b}c}}= c^{\\log_{b}a}+a^{\\log_{c}b}=\\boxed{m}$" } { "Tag": [ "geometry", "similar triangles", "geometry unsolved" ], "Problem": "A circle through vertices $ A,B$ of $ \\Delta ABC$ meets the sides $ AC,BC$ at $ D,E$ respectively. The lines $ AB$ and $ DE$ meet at $ F$ and the lines $ BD$ and $ CF$ meet at $ M$. Show that $ M$ is the midpoint of $ CF$ iff $ MB.MD\\equal{}{MC}^2$", "Solution_1": "MB*MD = MC^2 is equivlanet to angle FCA = angle CBM (similar triangles)\r\n\r\nthe idea revovling around this one is:\r\n \r\n1. extend BM to BN such that MN = MB\r\n2. NCDF is cyclic by an angle chase", "Solution_2": "[quote=\"hell_ever\"]A circle through vertices $ A,B$ of $ \\Delta ABC$ meets the sides $ AC,BC$ at $ D,E$ respectively. The lines $ AB$ and $ DE$ meet at $ F$ and the lines $ BD$ and $ CF$ meet at $ M$. Show that $ M$ is the midpoint of $ CF$ iff $ MB.MD \\equal{} {MC}^2$[/quote]\r\nLook here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=336205#336205" } { "Tag": [ "inequalities" ], "Problem": "APMO 2004 Problem 1: Find all non-empty finite sets $ S$ of positive integers such that for all $ i,j\\in S$, we have $ \\dfrac{i\\plus{}j}{(i,j)}\\in S$.\r\n\r\nThere's a link to a solution in the thread found in the contest's section. But what if we allow for infinite sets as well?", "Solution_1": "[hide=\"what happens\"]\nThen the problem is actually pretty easy. If we have any odd number $ N$, we can generate all odd numbers greater than $ N$ by getting $ f(N,2) \\rightarrow N \\plus{} 2$. So any odd number generates an infinitude of odd numbers.\n\nBut then $ f(N,N \\plus{} 2k) \\rightarrow 2N \\plus{} 2k$. So we can generate all even numbers that are greater than or equal to $ 2N \\plus{} 2$.\n\nFor any prime $ p$, there must be a multiple of $ p$ not divisible by 2 in the interval $ [N,N \\plus{} 2p)$ and in the interval $ [N \\plus{} 2p,N \\plus{} 4p)$. WLOG let $ N \\plus{} k, k < 2p$ be divisible by $ p$ and not by 2. Then we can generate $ f(N \\plus{} k,N \\plus{} k \\plus{} 2p) \\rightarrow \\frac {2N \\plus{} 2k \\plus{} 2p}{p} \\equal{} \\frac {2N \\plus{} 2k}{p} \\plus{} 2 \\le \\frac {2N}{p} \\plus{} 6$. So if we choose $ p > 2N$, we must generate some number $ k \\le 7$. But we can't generate 2, since that only comes when $ i \\equal{} j$. \n\nIf we generate 3, then $ f(3,2) \\rightarrow 5, 7, 9, \\cdots$. But $ f(3,5) \\rightarrow 8$, and $ f(5,19) \\rightarrow 24$. Obviously $ f(8,24) \\rightarrow 4$. Similarly $ f(8,40) \\rightarrow 6$. So we generate $ \\mathbb{Z} \\plus{} ^\\backslash\\{1\\}$.\n\nIf we generate 4, then we have $ f(2,4) \\rightarrow 3$ and we proceed as above.\n\nIf we generate 5, then we have $ f(5,7) \\rightarrow 12$ and $ f(5,19)\\rightarrow 24$. But $ f(12,24) \\rightarrow 3$ and we proceed as above.\n\nIf we generate 6, we have $ f(2,6) \\rightarrow 4$ and we proceed as above.\n\nIf we generate 7, then we have $ f(7,9) \\rightarrow 16$ and $ f(7,35) \\rightarrow 6$ and we proceed as above.\n\nSo, if we have any odd number, we can generate the set $ S \\equal{} \\mathbb{Z}^ \\plus{} \\backslash\\{1\\}$. But the set $ S \\cup \\{1\\}$ also has the given property. Hence these are the two possibilities.\n\nEdit: Fixed LaTeX. Edit 2: Fixed more LaTeX. Edit 3: Fixed even more LaTeX. Edit 4: Fixed major mistake and hid response. Edit 5: Tightened inequality to equality.\n\n[/hide]", "Solution_2": "The case that $ S$ contains an odd number is a little easier than that.\r\n[hide=\"Discussion\"] I am assuming that it is possible for $ i$ to equal $ j$ in which case $ 2 \\in S$ whenever $ S$ is non-empty. If in addition $ a \\in S$ and $ a$ is odd, then all odd integers greater than $ a$ are in $ S$. In particular, $ (2k \\minus{} 1)a \\in S$ for all positive integers $ k$, so $ \\frac {a \\plus{} (2k \\minus{} 1)a}{a} \\equal{} 2k \\in S$ for all positive integers $ k$. And then $ \\frac {2 \\plus{} 2k}{2} \\equal{} k \\plus{} 1 \\in S$ for all positive integers $ k$. So we have $ S \\equal{} \\mathbb{N} \\minus{} \\{1\\}$ (if $ a > 1$) or $ S \\equal{} \\mathbb{N}$ (if $ a \\equal{} 1$). [/hide]\nEdit: [hide=\"A full solution\"] Aside from $ 2$, let $ n$ be the smallest integer with the property that some element of $ S$ is divisible by $ 2^n$ but not by $ 2^{n\\plus{}1}$. Let the smallest such integer be $ 2^n a \\neq 2$, where $ a$ is odd. We have handled the case $ n \\equal{} 0$. \n\n[b]Case:[/b] $ n \\equal{} 1$. Then $ \\frac {2 \\plus{} 2a}{2} \\equal{} a \\plus{} 1 \\in S$, which is even but smaller than $ 2a$ unless $ a \\equal{} 1$; contradiction.\n\n[b]Case:[/b] $ n > 1$. Then $ \\frac {2 \\plus{} 2^n a}{2} \\equal{} 1 \\plus{} 2^{n \\minus{} 1} a \\in S$, which is odd; contradiction. [/hide]" } { "Tag": [ "floor function", "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "fin all prime P such that$P/(1^{p-1}+2^{p-1}.....2004^{p-1})$", "Solution_1": "$k^{P-1}\\equiv 1 (mod P)$ if $(k,P) = 1$ and $k^{P-1}\\equiv 0 (mod P)$ if $P|k$. \r\nLet $n(P,2004)$ is the number of multiple of $P$ less or equal to $2004$. Then, $n(P,2004) = \\lfloor \\frac{2004}{P}\\rfloor$.\r\nWe have $\\sum_{i=1}^{2004}i^{P-1}\\equiv 2004-n(P,2004) (mod P)$, so it's sufficient to find $P$ prime, such that $\\lfloor \\frac{2004}{P}\\rfloor \\equiv 2004 (mod P)$.\r\n\r\nI don't know other way to find all $P$ beside trying all prime less than $2004$ one by one. Can someone help ? :maybe:", "Solution_2": "This problem is from Bulgarian Spring Math Competition 2004 for 9-10 grade. After you get $2004\\equiv\\left[\\frac{2004}{p}\\right]\\pmod{p}$ write $2004=pq+r,$ $r\\leq p-1.$ If $q>p-1$ then $p\\leq 43$ and after check ypu get $p=17$ as a solution. If $q 0$\r\nDone, the equality not holds." } { "Tag": [ "puzzles" ], "Problem": "Bob is babysitting 10 children. She has a jar with only 10 crackers in it. All the children want to have a cracker, but they also want Bob to leave one cracker in the jar. How is Bob able to give each child a cracker and still leave one in the jar?", "Solution_1": "Two options that I just made up\r\n\r\n[hide]Bob puts one of the kids in the jar![/hide]\n\n[hide]Bob has more crackers outside of the jar[/hide]", "Solution_2": "number 1, lol :lol: wrong though\r\n\r\nnumber 2, wrong", "Solution_3": "more guesses :D \r\n\r\n[hide]Bob breaks the crackers into pieces\n[/hide]\n[hide]SIAMESE TWINS!!!!!![/hide]", "Solution_4": "[hide]maybe bob leaves the cracker in the jar and gives both to one of the kids.[/hide]", "Solution_5": "Bob is one of the 10 children and is babysitting herself/himself(?) along with the 9 others, so the cracker left in the jar will be left for a child, which she/he would be", "Solution_6": "bob is a girl woohooo!!!!\r\nand bob has a car and a credit card therefore she is able to go but more crackers/or/ there is a hobo on the street who sells bob poisonous crackers...", "Solution_7": "mustafa, wrong again :lol: \r\n\r\nand bocks of rocks\r\n\r\nperfectnumber, I think ur right rephrase the sentence plz", "Solution_8": "[hide]bob gives 9 of the crackers to the first 9 kids, and leaves the other one in the jar. then she gives the jar with the cracker in it to the last kid.\n[/hide]", "Solution_9": "wow smart thinking :blush:", "Solution_10": "[quote=\"perfectnumber628\"][hide]bob gives 9 of the crackers to the first 9 kids, and leaves the other one in the jar. then she gives the jar with the cracker in it to the last kid.\n[/hide][/quote]\r\n\r\nyeah", "Solution_11": "perfectnumber628 is citing too specific of a case. In the instructions there is nothing required about the order of the \r\nchildren receiving the crackers.\r\n\r\nsee: \r\n[hide]In general, Bob could have removed nine crackers out of the jar. Then Bob could have picked one of the ten children at random to receive the jar with the one cracker inside. After that the remaining nine crackers could be passed out to the other nine children. Or by removing crackers, and holding on to the jar at a certain time, then the jar can be passed out at any\ntime and given to a random child (and the crackers without the jar fall will in place).[/hide]", "Solution_12": "[quote=\"box of rocks\"]perfectnumber628 is citing too specific of a case. In the instructions there is nothing required about the order of the \nchildren receiving the crackers.\n\nsee: \n[hide]In general, Bob could have removed nine crackers out of the jar. Then Bob could have picked one of the ten children at random to receive the jar with the one cracker inside. After that the remaining nine crackers could be passed out to the other nine children. Or by removing crackers, and holding on to the jar at a certain time, then the jar can be passed out at any\ntime and given to a random child (and the crackers without the jar fall will in place).[/hide][/quote]\r\ndude, stop being a gayass. really, stop.\r\nno one gives a bs about all these technical bs that you talk about.(and it's REALLY annoying too)\r\nAnyone who had a brain will know what perfect was talking about, and what else do you need to get out of the puzzle?", "Solution_13": "junggi,\r\n\r\nlearn how to argue:\r\n\r\n1) Typing \"gayass\" on here? -- Don't namecall on these boards.\r\n Also, I don't care to see your immaturity displayed.\r\n\r\n2) Puzzles are ALL about technicality, so DON'T EVEN\r\n try to bring that up. Don't read them if YOU don't want\r\n to see different leagues of people who can answer.\r\n\r\n3) \"Annoying\" is in the eye of the beholder, and you know what?\r\n You're changing the subject by bringing it up. If you can't\r\n DEAL with fine tunings of correctness in puzzles, math, etc.,\r\n then you're not a serious-minded puzzle solver or mathemat-\r\n ician.\r\n\r\n4) And no, perfectnumber628 was NOT to be given the benefit \r\n of the doubt. He knew PART of the picture. Just like a hasty\r\n person like you jumps to conclusions and tries to speak for\r\n everyone else. Well, you're not speaking for everyone else.\r\n Just stay out of a thread because you're non-specific, and\r\n you don't speak for another user.", "Solution_14": "[quote=\"box of rocks\"]junggi,\n\nlearn how to argue:\n\n1) Typing \"gayass\" on here? -- Don't namecall on these boards.\n Also, I don't care to see your immaturity displayed.\n\n2) Puzzles are ALL about technicality, so DON'T EVEN\n try to bring that up. Don't read them if YOU don't want\n to see different leagues of people who can answer.\n\n3) \"Annoying\" is in the eye of the beholder, and you know what?\n You're changing the subject by bringing it up. If you can't\n DEAL with fine tunings of correctness in puzzles, math, etc.,\n then you're not a serious-minded puzzle solver or mathemat-\n ician.\n\n4) And no, perfectnumber628 was NOT to be given the benefit \n of the doubt. He knew PART of the picture. Just like a hasty\n person like you jumps to conclusions and tries to speak for\n everyone else. Well, you're not speaking for everyone else.\n Just stay out of a thread because you're non-specific, and\n you don't speak for another user.[/quote]\r\n1) See some of Ignite's post. You'd realize that \"gayass\" is nothing.\r\n3) I'd say most people here aren't \"serious\" puzzle solvers, whatever you mean by that. At least, they're not \"serious\" enough to bring up all those technicalities. And where does math come in here? Did I ever say that you don't need rigor in math problems? Actually, you don't need complete rigor in math either, if what you're doing is completely informal, like answering puzzles here, or posting solutions to some problems in the middle school or high school forums, etc. And I don't even see what you're trying to argue here.\r\n4) Yep, you have brain damage.\r\n5) Do you have a second account that you used to spam my inbox with? :maybe:", "Solution_15": "[quote=\"box of rocks\"]perfectnumber628 is citing too specific of a case. In the instructions there is nothing required about the order of the \nchildren receiving the crackers.\n\nsee: \n[hide]In general, Bob could have removed nine crackers out of the jar. Then Bob could have picked one of the ten children at random to receive the jar with the one cracker inside. After that the remaining nine crackers could be passed out to the other nine children. Or by removing crackers, and holding on to the jar at a certain time, then the jar can be passed out at any\ntime and given to a random child (and the crackers without the jar fall will in place).[/hide][/quote]In my opinion, giving only one case is enough. We just have to find a way to solve the problem not all.", "Solution_16": "Junggi, \"Other people act like real jerks on the forum\" is not a good explanation for why [i]you[/i] should act like a real jerk on the forum. Don't flame, don't be offensive, and keep personal messages to PMs.", "Solution_17": "[quote=\"JBL\"]Junggi, \"Other people act like real jerks on the forum\" is not a good explanation for why [i]you[/i] should act like a real jerk on the forum. Don't flame, don't be offensive, and keep personal messages to PMs.[/quote]\r\nok sorry, I'll try to stop.(but it might be hard when he sends me some...um....very um strange pms)", "Solution_18": "So finally the answer is.... ????", "Solution_19": "[quote=\"Padmavathi\"]So finally the answer is.... ????[/quote]\r\nperfect's answer", "Solution_20": "[quote=\"junggi\"][quote=\"JBL\"]Junggi, \"Other people act like real jerks on the forum\" is not a good explanation for why [i]you[/i] should act like a real jerk on the forum. Don't flame, don't be offensive, and keep personal messages to PMs.[/quote]\nok sorry, I'll try to stop.(but it might be hard when he sends me some...um....very um strange pms)[/quote]\r\n\r\n\r\ni don't even want to know....\r\n\r\nand yes over the other boards I have seen you, junggi, flame other people for ridiculous and pointless things.", "Solution_21": "[quote=\"7h3.D3m0n.117\"][quote=\"junggi\"][quote=\"JBL\"]Junggi, \"Other people act like real jerks on the forum\" is not a good explanation for why [i]you[/i] should act like a real jerk on the forum. Don't flame, don't be offensive, and keep personal messages to PMs.[/quote]\nok sorry, I'll try to stop.(but it might be hard when he sends me some...um....very um strange pms)[/quote]\n\n\ni don't even want to know....\n\nand yes over the other boards I have seen you, junggi, flame other people for ridiculous and pointless things.[/quote]\r\nlike what?(something outside of GFF and something other than anirudh kind of thing flaming please)\r\nyou spamming?\r\nI'd say that was definitely not pointless.\r\n\r\np.s. I find your post very ironic" } { "Tag": [ "email" ], "Problem": "Does anyone know where the registration website is and when the festival will take place this year?", "Solution_1": "You didn't get the email? I'll just post it here:\r\nRegistration for email list: https://secure.msri.org/forms/jrmf/jrmfregistration \r\nRegistration for this year's event: https://secure.msri.org/forms/jrmf3/stanford" } { "Tag": [ "trigonometry", "geometry", "trig identities", "Law of Sines" ], "Problem": "Prove: $ \\frac{\\sin (A\\minus{}B)}{\\sin (A\\plus{}B)}\\equal{}\\frac{a^2\\minus{}b^2}{c^2}$\r\n\r\nFor any $ \\triangle ABC$", "Solution_1": "$ \\frac{a^2\\minus{}b^2}{c^2}$=$ \\frac{4R^2(sin^2A\\minus{}sin^2B)}{4R^2sin^2C}$ \r\n \r\n = $ \\frac{sin(A\\plus{}B)sin(A\\minus{}B)}{sin^2(180\\minus{}(A\\plus{}B)}$ \r\n \r\n =$ \\frac{sin(A\\minus{}B)}{sin(A\\plus{}B)}$", "Solution_2": "I'm just starting to learn trig concepts so I will just brush up on picture74's method with a few comments\r\n\r\n$ \\frac {a^{2} \\minus{} b^{2}}{c^{2}} \\equal{} \\frac {4R^{2}(\\sin^{2} A \\minus{} \\sin^{2} B)}{4R^{2}{\\sin^{2}} C}$\r\n\\* This is a consequence of the Law of sines: notice how $ \\frac {c}{\\sin c} \\equal{} 2r$ *\\\r\n\r\n$ \\frac {\\sin(A \\plus{} B)\\sin(A \\minus{} B)}{\\sin^{2}(180 \\minus{} (A \\plus{} B)}$ \\* This is just standard manipulation combined with a use of the $ \\angle A \\plus{} \\angle B \\plus{} \\angle C \\equal{} 180^{\\circ}$ in any triangle. *\\\r\n\r\n$ \\frac {\\sin(A \\minus{} B)}{\\sin(A \\plus{} B)}$ *\\ More simplifying and also utilizing that $ \\sin{\\theta} \\equal{} \\sin{180^{\\circ} \\minus{} \\theta}$, this can be derived quite easily from sketching a unit circle.", "Solution_3": "[hide=\"Different Way...\"]Look at the LHS only:\n\n$ \\equal{}\\frac{\\sin A \\cos B \\minus{} \\sin B \\cos A}{\\sin C}$\n\n$ \\equal{}\\frac{\\sin A \\cdot \\frac{a^2\\plus{}c^2\\minus{}b^2}{2ac} \\minus{} \\sin B \\cdot \\frac{b^2\\plus{}c^2\\minus{}a^2}{2bc}}{\\sin C}$\n\n$ \\equal{}\\frac{\\frac{a^2 \\sin A \\plus{} c^2 \\sin A \\minus{} b^2 \\sin A}{2ac} \\minus{} \\frac{b^2 \\sin B \\plus{} c^2 \\sin B \\minus{} a^2 \\sin B}{2bc}}{\\sin C}$\n\n$ \\equal{}\\frac{\\frac{(a^2 b \\sin A \\plus{} b c^2 \\sin A \\minus{} b^3 \\sin A) \\minus{} (a b^2 \\sin B \\plus{} a c^2 \\sin B \\minus{} a^2 \\sin B)}{2abc}}{\\sin C}$\n\n$ \\equal{}\\frac{\\frac{b \\sin A (a^2 \\plus{} c^2 \\minus{} b^2) \\minus{} a \\sin B (b^2 \\plus{} c^2 \\minus{} a^2)}{2abc}}{\\sin C}$\n\n$ \\equal{}\\frac{\\frac{b \\sin A(a^2 \\plus{} c^2 \\minus{} b^2 \\minus{} b^2 \\minus{} c^2 \\plus{} a^2)}{2abc}}{\\sin C}$\n\n$ \\equal{}\\frac{b \\sin A (2a^2 \\minus{} 2b^2)}{2abc \\sin C}$\n\n$ \\equal{}\\frac{2 \\sin A (a^2\\minus{}b^2)}{2 c^2 \\sin A}$\n\n$ \\equal{}\\frac{a^2\\minus{}b^2}{c^2}$\n\n\n\n[/hide]" } { "Tag": [ "conics", "ellipse", "geometry", "geometric transformation", "reflection", "ratio", "circumcircle" ], "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14549[/img]\r\n\r\n :(", "Solution_1": "The ellipse center is the triangle centrid $ G$ and it is tangent to $ BC, CA, AB$ at their midpoints $ A', B', C'.$ Reflections $ A'', B'', C''$ of $ A', B', C'$ in $ G$ give 3 more points.", "Solution_2": "and to show that G is the centre of the ellipse we consider a parallel projection making the triangle equilateral ( observe that parallel projetions coserve ratio of areas and collinear ratios) :D :lol:", "Solution_3": "I think the answer of yetty correspond to the previous post from jrrcb; if the ellipse is circumscribed, the solution is the ellipse which pass by the vertices A, B, C and which has tangents in these points parallel to the opposed sides, if I am not wrong.", "Solution_4": "Thanks for pointing it out, I flipped the replies. Answer to this problem is\r\n\r\nYou already have 3 ellipse points $ A, B, C.$ The ellipse center is the triangle centroid $ G$ and and reflections $ A', B', C'$ of $ A, B, C$ in $ G$ give 3 more points.", "Solution_5": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14565[/img]\r\n\r\n :thumbup:", "Solution_6": "How do we know that the ellipse with minimal area through the vertices of a fixed equilateral is a circle?", "Solution_7": "Let $ \\mathcal E$ be the smallest area ellipse circumcribed around an equilateral $ \\triangle ABC.$ The ratio $ \\frac{S(\\triangle ABC)}{S(\\mathcal E)}$ is largest possible. Assume $ \\mathcal E$ is different from its circumcircle $ \\mathcal O,$ i.e., not a circle. By a parallel projection $ \\mathbf P$, preserving area ratios, project $ \\mathcal E$ into a circle $ \\mathcal E'$ with radius $ R'.$ The equilateral $ \\triangle ABC$ inscribed in the ellipse $ \\mathcal E$ goes into a $ \\triangle A'B'C'$ inscribed in the circle $ \\mathcal E'.$ Its area is largest possible for triangles inscribed in $ \\mathcal E',$ $ \\frac{S'(\\triangle A'B'C')}{S(\\mathcal E')} \\equal{} \\frac{a'b'c'}{4\\pi R'^3} \\equal{} \\frac{2}{\\pi} \\sin A' \\sin B' \\sin C'$ is maximum $ \\Longrightarrow$ $ \\angle A' \\equal{} \\angle B' \\equal{} \\angle C'$ (for example, with Lagrange multipliers), the $ \\triangle A'B'C'$ is also equilateral $ \\Longrightarrow$ $ \\mathbf P$ is a translation or identity transformation $ \\Longrightarrow$ $ \\mathcal E$ is also a circle, contradiction.", "Solution_8": "Similarly, ellipse with maximum area incribed in an equilateral triangle is a circle. Combined with parallel projection, equivalent statement is that the triangle of minimum area circumscribed about a given circle is equilateral.\r\n\r\nLet $ (I)$ be a fixed circle with radius $ r$ and $ A$ arbitrary fixed point outside of it. Denote the segment length $ k = AI.$ Let $ b, c$ be tangents of $ (I)$ from $ A$ and $ a$ a tangent of $ (I)$ intresecting the tangent $ b, c$ at $ C, B,$ such that $ (I)$ is incircle of the resulting $ \\triangle ABC.$ Let $ X, Y, Z$ be its tangency points with $ BC, CA, AB.$ Since the inradius $ r$ is fixed, area $ S$ of the $ \\triangle ABC$ is proportional to its semiperimeter $ s,$ $ S = rs.$ The A-excirdcle $ (I_a),$ centered on the fixed angle bisector ray $ AI,$ is tangent to $ CA, AB$ at $ Y_a, Z_a$ and $ s = AY_a = AZ_a.$ Obviously, $ s$ is minimum, when $ (I_a)$ is smallest possible, tangent to $ (I)$ at $ X \\in BC$ and the $ \\triangle ABC$ is isosceles with $ AB = AC.$ The area $ S = rs$ is then also minimum. Thus we have to find the position of $ A$ corresponding to the minimum area only for isosceles triangles. In this case,\r\n\r\n$ S = AX \\cdot BX = AX \\cdot IZ \\cdot \\frac {AX}{AZ} = \\frac {r(k + r)^2}{\\sqrt {k^2 - r^2}}$\r\n\r\n$ \\frac {\\text dS}{\\text dk} = \\frac {r(k + r)}{(\\sqrt {k^2 - r^2})^3}\\ (k^2 - rk - 2r^2) = 0\\ \\Longrightarrow\\ k = -r$ or $ k = \\frac {r \\pm \\sqrt {r^2 + 8r^2}}{2} = - r$ or $ 2r.$\r\n\r\nOnly the last positive root is acceptable and isosceles triangle with $ AI = k = 2r$ means an equilateral triangle. We can easily verify that $ \\frac {\\text d^2S}{\\text dk^2}|_{k = 2r} = \\frac{3}{\\sqrt {3}} > 0,$ $ S$ has a minimum for $ k = 2r.$", "Solution_9": "[quote=\"yetti\"]Let $ \\mathcal E$ be the smallest area ellipse circumcribed around an equilateral $ \\triangle ABC.$ The ratio $ \\frac {S(\\triangle ABC)}{S(\\mathcal E)}$ is largest possible. Assume $ \\mathcal E$ is different from its circumcircle $ \\mathcal O,$ i.e., not a circle. By a parallel projection $ \\mathbf P$, preserving area ratios, project $ \\mathcal E$ into a circle $ \\mathcal E'$ with radius $ R'.$ The equilateral $ \\triangle ABC$ inscribed in the ellipse $ \\mathcal E$ goes into a $ \\triangle A'B'C'$ inscribed in the circle $ \\mathcal E'.$ Its area is largest possible for triangles inscribed in $ \\mathcal E',$ $ \\frac {S'(\\triangle A'B'C')}{S(\\mathcal E')} \\equal{} \\frac {a'b'c'}{4\\pi R'^3} \\equal{} \\frac {2}{\\pi} \\sin A' \\sin B' \\sin C'$ is maximum $ \\Longrightarrow$ $ \\angle A' \\equal{} \\angle B' \\equal{} \\angle C'$ (for example, with Lagrange multipliers), the $ \\triangle A'B'C'$ is also equilateral $ \\Longrightarrow$ $ \\mathbf P$ is a translation or identity transformation $ \\Longrightarrow$ $ \\mathcal E$ is also a circle, contradiction.[/quote]\r\n\r\nIt is an interesting proof but I don't really understand: $ \\frac {S(\\triangle ABC)}{S(\\mathcal E)}$ as large as possible (given that ABC is fixed) implies that $ \\frac {S'(\\triangle A'B'C')}{S(\\mathcal E')}$ is as large as possible given that $ \\mathcal{E}$ is fixed.\r\n\r\nI know that those ratios are equal, but I don't see how the as large as possible given constraint A implies as large as possible given constraint B\r\n\r\nIt makes sense intuitively, and I think it is the right method of proof, I'm just having trouble justifying to myself.", "Solution_10": "My last post in this thread is completely incorrect. I have to show that the largest equilateral triangle inscribed in a given ellipse is symmetrical with respect to the minor axis and I cannot.", "Solution_11": "[quote=\"yetti\"]My last post in this thread is completely incorrect. I have to show that the largest equilateral triangle inscribed in a given ellipse is symmetrical with respect to the minor axis and I cannot.[/quote]\r\nwell actually you can just see my previous post", "Solution_12": "[quote=\"jrrbc\"][img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14549[/img]\n\n :([/quote]\r\n\r\nI think that if AB = BC = CA\r\nthen it is the circle which A,B,C are on it.\r\nAffine transformation from K, let a triangle ABC become a triangular triangle\r\n\r\ncircle becomes ellispe\r\n\r\nMy English words are bad.I used Google translation.", "Solution_13": "[quote=\"gauravpatil\"][quote=\"yetti\"]My last post in this thread is completely incorrect. I have to show that the largest equilateral triangle inscribed in a given ellipse is symmetrical with respect to the minor axis and I cannot.[/quote]\nwell actually you can just see my previous post[/quote]\r\nYour previous post is completely insufficient, you made the same error I did.\r\n______________________________\r\n\r\nLemma: Let $ AB$ be an arbitrary chord of an ellipse $ \\mathcal K.$ A parallel projection $ \\mathbf P$ exists, which takes the ellipse into a circle and which keeps the points $ A, B$ in place.\r\n\r\nLet $ M$ be the midpoint the chord $ AB,$ let $ O$ be the ellipse center and let $ UV \\parallel AB$ be an ellipse diameter through its center $ O.$ Let $ OM$ cut the ellipse at $ X, Y.$ Then $ UV, XY$ is a pair of the ellipse conjugate diameters. When the ellipse is projected to a circle by a parallel projection, a pair of conjugate diameters goes to a pair of perpendicular diameters of the circle. Let $ n$ be the perpendicular to $ UV$ at $ O.$ A circle with center $ O$ and radius $ OU \\equal{} OV$ cuts this perpendicular at $ X', Y'.$ Let $ \\mathbf Q,$ be a parallel projection, which takes the line $ UV$ into itself and $ XY$ into $ X'Y'.$ This parallel projection takes the ellipse $ \\mathcal K$ into a circle $ \\mathcal K'$ with center $ O$ and radius $ OU \\equal{} OV \\equal{} OX' \\equal{} OY'.$ The ellipse chord $ AB$ goes to a circle chord $ A'B'.$ Since $ AB \\parallel UV,$ it follows that $ A'B' \\equal{} AB.$ Let $ \\mathbf T$ be a translation, which takes $ A'B'$ into $ AB$ and the circle $ \\mathcal K'$ into a circle $ \\mathcal K''.$ Translation is a parallel projection and composition of 2 parallel projections is again a parallel projection. Therefore $ \\mathbf P \\equal{} \\mathbf T \\circ \\mathbf Q$ is the desired parallel projection.\r\n\r\nLet $ \\mathcal K$ be the minimum area ellipse circumscribed around an equilateral $ \\triangle ABC.$ Assume that it is not identical with its circuircle and WLOG, assume that the triangle side $ AB$ is not parallel to either the ellipse major or minor axis (even though some other side might or might not be). Let $ \\mathbf P$ be the parallel projection, which takes the ellipse $ \\mathcal K$ into a circle $ \\mathcal K',$ while keeping the line $ AB$ in place. As the only pair of the ellipse conjugate diameters perpendicular to each other are the ellipse main axes, it follows that the projection $ C'$ of $ C$ is not on the perpendicular bisector of the side $ AB$ and $ AC' \\neq BC'.$ Let the perpendicular bisector of the side $ AB$ cut the projected circumcircle $ \\mathcal K'$ of the $ \\triangle ABC'$ at a point $ D$ on the same side of $ AB$ as $ C',$ so that $ AD \\equal{} BD.$ For areas of the $ \\triangle ABC', \\triangle ABD,$ we then have $ S(\\triangle ABC') < S(\\triangle ABD).$ Let $ \\mathbf S$ be a parallel projection, which keeps the line $ AB$ in place and which takes $ D$ back to $ C.$ In this, the common circumcircle $ \\mathcal K'$ of the $ \\triangle ABC', \\triangle ABD$ goes into an ellipse $ \\mathcal K''$ with one of the main axis parallel to $ AB.$ Since all area ratios are invariant in parallel projections,\r\n\r\n$ \\frac {S(\\mathcal K)}{S(\\triangle ABC)} \\equal{} \\frac {S(\\mathcal K')}{S(\\triangle ABC')} > \\frac {S(\\mathcal K')}{S(\\triangle ABD)} \\equal{} \\frac {S(\\mathcal K'')}{S(\\triangle ABC)}$\r\n\r\nand $ S(\\mathcal K) > S(\\mathcal K'').$ Thus we have a smaller area ellipse $ \\mathcal K''$ circumscribed around the equilateral $ \\triangle ABC$ with one main axis parallel to $ AB.$ As this can be done for any side of the equilateral $ \\triangle ABC$ not parallel to either main axis of the circumscribed ellipse, it follows that for the circumscribed ellipse with minimum area, all 3 sides $ AB, BC, CA$ have to be parallel to either main axis, which is possible only when the circumscribed ellipse with minimum area is a circle.", "Solution_14": "could you tell me the mistake Please :oops: :)", "Solution_15": "Proposition P (easy to prove): The maximum area triangle inscribed in a given circle is equilateral.\r\nConclusion Q: The minimum area ellipse circummscribed about a given equilateral triangle is a circle.\r\n\r\nEven though the conclusion Q is intuitively true, it does not follow from the proposition P, P => Q is incorrect. This is the error we both made. All that follows from P is that various (non-equilateral) triangles with maximum possible area, larger than the equilateral triangle area, can be inscribed in its circumellipse, not identical with its circumcircle. I appreciate [b]Altheman[/b] raising the question, questioning the incorrect proof, and the way he did it. :omighty:\r\n\r\nNote that statements about the minimum area of Steiner circumellipse and the maximum area of Steiner inellipse are followed by references at Mathworld. On the other hand, statements that under an affine transformation, the Steiner circumellipse and inellipse can be transformed into the circumcircle and incircle of an equilateral triangle are not followed by any references:\r\n[url]http://mathworld.wolfram.com/SteinerCircumellipse.html[/url]\r\n[url]http://mathworld.wolfram.com/SteinerInellipse.htm[/url]", "Solution_16": "although not directly but P does imply Q\r\nthis is because a parallel projection preserves ratios of areas\r\nso suppose for instance Q is not true then we use a parallel projection to make it a circle\r\nnow as this ratio of area could be made smaller (triangle is no longer equilateral)\r\nso our assertion was wrong\r\nQED\r\ni did not get the problem as yet\r\n\r\n( eDITED)", "Solution_17": "Supposing that Q is not true means that the minimum area ellipse around an equilateral triangle is not a circle. Assuming this, you are projecting an ellipse to another ellipse and the equilateral triangle to a general one - getting nowhere. If you project the minimum area ellipse around the equilateral triangle to a circle, you get a general triangle inscribed in a circle. You can find bigger ones inscribed in this circle, equilateral. This only tells you that the minimum area ellipse around the equilateral triangle has various larger general triangles inscribed in it. This does not contradict ~Q.", "Solution_18": "I HAVE EDITED MY MESSAGE \r\nI TAKE THE ELLIPSE TO A CIRCLE", "Solution_19": "I considered this possibility in my reply. Q is not true, the minimum area circumellipse of the equilateral triangle is not circle, until you get a contradiction. The fact that you can inscribe a multitude of larger area triangles in this ellipse does not contradict the asumption that this is the minimum area circumellipse of this particular (equilateral) triangle.", "Solution_20": "so my question is is my answer correct?", "Solution_21": "No, your answer is not correct. The question is not to find a larger area triangle in the minimum area circumellipse ellipse of the equilateral triangle (or demonstrate that there is such a triangle). The question is to find a smaller area circumellipse of the equilateral triangle than the supposed minimum area circumellipse (or demonstrate that there is such a circumellipse). A positive answer to the first question does not imply a positive answer to the second, different question. If you still think that it does, break your proof in small logical steps.", "Solution_22": "but both the problems are essentially the same after some knowledge in parllel projection\r\nor am i wrong" } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Find: $ \\int_{0}^{ln10}\\frac {{e^{x}\\sqrt {e^{x}\\minus{}1}}}{e^{x} \\plus{} 8}dx$", "Solution_1": "It looks like you have a typo- did you intend $ e^x\\minus{}1$ or $ e^{x\\minus{}1}$ inside that square root?", "Solution_2": "[quote=\"jmerry\"]It looks like you have a typo- did you intend $ e^x \\minus{} 1$ or $ e^{x \\minus{} 1}$ inside that square root?[/quote]\r\nI intended $ e^x \\minus{} 1$, corrected. :blush:", "Solution_3": "[quote=\"enb081\"]Find: $ \\int_{0}^{ln10}\\frac {{e^{x}\\sqrt {e^{x} \\minus{} 1}}}{e^{x} \\plus{} 8}dx$[/quote]\r\n\r\nHi :) !!\r\n\r\nwa put us $ \\sqrt {e^x \\minus{} 1} \\equal{} t$ than the integral become's:\r\n\r\n$ \\int_0^3 \\frac {2t^2}{t^2 \\plus{} 9}dt \\equal{} 6 \\minus{} 6 \\int_0^3 \\frac {\\frac {1}{3}}{\\left( \\frac {t}{3} \\right)^2 \\plus{} 1}dt$\r\n\r\nthan $ \\int_0^{ln(10)} \\frac {e^x \\sqrt {e^x \\minus{} 1}}{e^x \\plus{} 8}dx \\equal{} 6 \\minus{} \\frac {3 \\pi}{2}$\r\n\r\nand thinks" } { "Tag": [ "ratio", "geometry", "trapezoid", "similar triangles" ], "Problem": "Triangle $ABC$ in the figure has area $10$. Points $D$, $E$, and $F$, all distinct from $A$, $B$, and $C$, are on sides $AB$, $BC$, and $CA$ respectively, and $AD=2$, $DB=3$. If triangle $ABE$ and quadrilateral $DBEF$ have equal areas, then that area is?", "Solution_1": "[hide]\nWe know that triangle ABE and quadrilateral DBEF have equal areas. Let G be the point of intersection between EA and FD. Then we know that the area of triangle FEG is equal to that of triangle DAG. By this, we know that triangle FEG is congruent to triangle DAG. So DA = FE = 2.\n\nConsider the altitude from C drawn to AB. Because AB = 5 and the area of the triangle is 10, this altitude is 4.\n\nNow, consider the altitude from E to AB. Since FE = 2/5(AB), this altitude is 3/5 (1 - 2/5) of the altitude from C down to AB, or 3/5(4) = 12/5.\n\nNow we can find the area of triangle ABE (or quadrilateral DBEF):\n\nArea = 1/2 * AB * (altitude from E to AB)\nArea = 1/2(5)(12/5) = 6\n[/hide]", "Solution_2": "krsattack said:\r\n[quote][hide]We know that triangle ABE and quadrilateral DBEF have equal areas. Let G be the point of intersection between EA and FD. Then we know that the area of triangle FEG is equal to that of triangle DAG. By this, we know that triangle FEG is congruent to triangle DAG. So DA = FE = 2.\n\nConsider the altitude from C drawn to AB. Because AB = 5 and the area of the triangle is 10, this altitude is 4.\n\nNow, consider the altitude from E to AB. Since FE = 2/5(AB), this altitude is 3/5 (1 - 2/5) of the altitude from C down to AB, or 3/5(4) = 12/5.\n\nNow we can find the area of triangle ABE (or quadrilateral DBEF):\n\nArea = 1/2 * AB * (altitude from E to AB)\nArea = 1/2(5)(12/5) = 6 [/hide][/quote]\r\n\r\nkrsattack, just because triangles ADG and FEG have the same area, doesn't mean that they are congruent. It was not stated that FE was parallel to AD.", "Solution_3": "OK, here's my answer with proof.\r\n[hide=\"ANSWER!!!\"]\n [b]It is stated that quadrilateral FEBD and triangle AEB have the same area.[/b] If the intersection point of segments FD and AE is called G, then [b]triangles FEG and ADG have equal area[/b] (because FEBD and AEB have DBEG in common.)\n[i]Note: just because FEG and ADG have the same area doesn't mean that they are congruent! The only way that those two triangles are congruent is if FE and AD are parallel, and that is not stated in the problem.[/i]\n Now, you can say that [b]triangles FAE and FAD have equal area[/b] (because all you're doing is adding triangle FAG to FEG and ADG.)\n Since FAE and FAD have the same area [i]and[/i] the same base (FA), then [b]they must have the same height![/b] This shows that [b]if you draw ED, then it is parallel to FA![/b]\n Finally, something parallel! That is what we needed. Now, all we have to do is similar triangles! [b]Triangle DEB is similar to triangle ACB[/b] by angle-angle-angle similarity (both triangles have angle CBA in common, and the other two pairs of angles are congruent because DE is parallel to CA.) Since the area of triangle ACB is 10 and its base is 5, the height must be 4. Since the base of DEB is 3, then you can do a proportion to find the height of DEB. [b]This yields 12/5 as the height of triangle DEB.[/b]\n Now, we need to find the area of either AEB or FEBD. Since it is very difficult to find FEBD (because FE is not parallel to AB), then we must find the area of AEB. Because we found the height of DEB, we can use that as the height of AEB because they share that same height. (1/2)bh=(1/2)(5)(12/5)=[b]6[/b]\n[/hide]\r\nIf FEG was really congruent to DAG, then FE would be parallel to AD. Then that would mean that FEBD is a trapezoid, but then if you find the area of that, then it would be different than AEB. This shows that FE isn't parallel to AD.", "Solution_4": "I had originally read the question wrong. But I still think I made a [b]correct[/b] assumption (with, I'll admit, insufficient thought)...\r\n\r\nI might be wrong, but I believe that it turns out that EF is parallel to AD. In fact, I found the area of the trapezoid and here's what I got:\r\n\r\n[hide]The Area of a trapezoid is:\n$A = \\frac{1}{2}h(b_{1}+b_{2})$\nThus, since $b_{1}= 3$ and $b_{2}= 2$, and as I had found before, the height is $\\frac{12}{5}$, then\n$A = \\frac{1}{2}(\\frac{12}{5})(3+2) = 6$\nBut this is also the area of ABE!!!\n[/hide]\r\n\r\nAnyway...\r\nI am having trouble showing whether or not FE [b]MUST[/b] be parallel to AD, and perhaps I am totally wrong, but if there are any ideas as to proving this, I'd like to know.[/b]", "Solution_5": "[quote=\"krsattack\"]I had originally read the question wrong. But I still think I made a [b]correct[/b] assumption (with, I'll admit, insufficient thought)...\n\nI might be wrong, but I believe that it turns out that EF is parallel to AD. In fact, I found the area of the trapezoid and here's what I got:\n\n[hide]The Area of a trapezoid is:\n$A = \\frac{1}{2}h(b_{1}+b_{2})$\nThus, since $b_{1}= 3$ and $b_{2}= 2$, and as I had found before, the height is $\\frac{12}{5}$, then\n$A = \\frac{1}{2}(\\frac{12}{5})(3+2) = 6$\nBut this is also the area of ABE!!!\n[/hide]\n\nAnyway...\nI am having trouble showing whether or not FE [b]MUST[/b] be parallel to AD, and perhaps I am totally wrong, but if there are any ideas as to proving this, I'd like to know.[/b][/quote]\r\nWhoops, you're right. They are parallel! Sorry about the trapezoid thing. Yes, they are parallel, simply because the trapezoid works. Just understand that you got lucky on this problem. :wink:" } { "Tag": [ "trigonometry" ], "Problem": "More of a physics problem, but it's from a maths textbook.\r\n\r\nI cant seem to prove it and am banging my head against the wall.\r\n\r\n\r\n\r\nSnell's law states that light travels through a homogeneous medium in a straight line at a constant velocity dependent upon the medium. Let the velocity of light in air be v1 and the velocity of light in water be v2. Show that light will travel from a point in air to a point in water in the shortest possible time if sin(a1) / v1 = sin(a2) / v2 , where the light ray makes angles of a1 and a2 in air and water respectively with the normal to the surface.\r\n\r\n\r\nThanks in advanced", "Solution_1": "From Snell's law, $n_{1}\\sin a_{1}=n_{2}\\sin a_{2}$, where $n_{a}$ is the index of refraction of the medium, and for $n_{a}\\equiv \\frac{c}{v_{a}}\\text{, }\\frac{c\\sin a_{1}}{v_{1}}=\\frac{c\\sin a_{2}}{v_{2}}\\text{, }$\r\n$\\frac{\\sin a_{1}}{v_{1}}=\\frac{\\sin a_{2}}{v_{2}}$ :)", "Solution_2": "Read this proof. I tried to make it as simple as possible, but it does involve some differential calculus. Try to work through it step by step. This is a derivation similiar to Wolfram's Mathworld Snell's Derivation.", "Solution_3": "How can I put a PDF directly in my post? So you guys don't have to go through external sites (security)...", "Solution_4": "Attachment? :maybe:\r\nDiscover where to click the \"Add an Attachment\" in the reply window :P", "Solution_5": "How so? How do I attach something?\r\nSorry to sound like a forum noob :P", "Solution_6": "Read the proof and tell me what you think.", "Solution_7": "Thanks a million :) \r\n\r\nFantastic proof, couldn't have asked for more.\r\n\r\nCheers.", "Solution_8": "Hehe, it's funny. Your welcome. I just had to do the same project for physics." } { "Tag": [ "inequalities", "combinatorics proposed", "combinatorics" ], "Problem": "Let $A_1,A_2,\\ldots , A_n$ $(n\\geq 3)$ be finite sets of positive integers. Prove, that\r\n\\[ \\displaystyle \\frac{1}{n} \\left( \\sum_{i=1}^n |A_i|\\right) + \\frac{1}{{{n}\\choose{3}}}\\sum_{1\\leq i < j < k \\leq n} |A_i \\cap A_j \\cap A_k| \\geq \\frac{2}{{{n}\\choose{2}}}\\sum_{1\\leq i < j \\leq n}|A_i \\cap A_j| \\]\r\nholds, where $|E|$ is the cardinality of the set $E$", "Solution_1": "let there be an element $a$ that's in exactly $k$ sets. this element will add to the first sum $k$, to the second sum ${k}\\choose{3}$ and to the last sum ${k}\\choose{2}$. if we prove te inequality by doing it one element at a time, then at the end it'll be true (since with start with empty set, which gives $0 \\ge 0$). so we want to prove that:\r\n\r\n$\\frac{k}{n} + \\frac{{{k}\\choose{3}}}{{{n}\\choose{3}}} \\ge \\frac{2{{k}\\choose{2}}}{{{n}\\choose{2}}}$\r\n\r\n$\\frac{k}{n} + \\frac{k(k-1)(k-2)}{n(n-1)(n-2)} \\ge \\frac{2k(k-1)}{n(n-1)}$\r\n\r\n$1 + \\frac{(k-1)(k-2)}{(n-1)(n-2)} \\ge \\frac{2(k-1)}{n-1}$\r\n\r\n$(n-1)(n-2) + (k-1)(k-2) \\ge 2(k-1)(n-2)$\r\n\r\nlet $n = 2 + s$ with $s \\ge 1$ and $k = 1 + t$ with $t \\ge 0$\r\n\r\n$(s+1)s + t(t-1) \\ge 2ts$\r\n\r\n$(t-s)^2 \\ge t-s$\r\nwhich is true, since $t-s$ is an integer", "Solution_2": "Thats exactly the same solution I found ;)\r\nI just proved the last step directly without the substitution\r\n\r\nYimin" } { "Tag": [ "induction", "probability", "IMO 2009", "IMO", "algebra", "IMO Shortlist", "combinatorics" ], "Problem": "Let $ a_1, a_2, \\ldots , a_n$ be distinct positive integers and let $ M$ be a set of $ n \\minus{} 1$ positive integers not containing $ s \\equal{} a_1 \\plus{} a_2 \\plus{} \\ldots \\plus{} a_n.$ A grasshopper is to jump along the real axis, starting at the point $ 0$ and making $ n$ jumps to the right with lengths $ a_1, a_2, \\ldots , a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $ M.$\r\n\r\n[i]Proposed by Dmitry Khramtsov, Russia[/i]", "Solution_1": "Did they really consider this hard enough for a problem 6??", "Solution_2": "Well, as far as I know, many strong contestants did not solve it (maybe no italian, british, romanian, french...one german...) thus I guess the answer may be affirmative. :wink: \r\n\r\nPierre.", "Solution_3": "lol... looks like #3 and #6 are of similar flavor this year...\r\n\r\nFor this problem, can't you just\r\n\r\n[hide]induct? Let $ a_n$ be largest, and consider the largest element $ m$ of $ M$. In the case that $ s \\minus{} a_n$ is in $ M$ and $ m > s \\minus{} a_n$, note that for some $ 1 \\le i \\le n\\minus{}1$, we have that $ s \\minus{} a_i$ and $ s \\minus{} a_i \\minus{} a_n$ are both not in $ M$. Thus, apply the inductive hypothesis to the $ a_i$ excluding $ a_i$ and $ a_n$, and we are able to dodge the elements of $ M$, since at most $ n\\minus{}3$ elements of $ M$ are strictly between $ 0$ and $ s \\minus{} a_i \\minus{} a_n$.\n\nOtherwise, apply the inductive hypothesis to $ a_1$ through $ a_{n\\minus{}1}$ on the set $ M \\minus{} \\{ m \\}$, so that we make $ n\\minus{}1$ hops. If $ m$ is not reached after any of these hops, we can simply tag on a hop of length $ a_n$ at the end of the sequence. If $ m$ is reached after the $ k$th hop, then we revise the $ k$th hop to be of length $ a_n$ and move what was originally the $ k$th hop to the end.\n\nSeems a bit simple (although not totally trivial), so I might have missed something...[/hide]", "Solution_4": "I'm not sure if I understand this part of your argument:\r\n[quote=\"probability1.01\"]\n\n[hide]Otherwise, apply the inductive hypothesis to $ a_1$ through $ a_{n \\minus{} 1}$ on the set $ M \\minus{} \\{ m \\}$, so that we make $ n \\minus{} 1$ hops. If $ m$ is not reached after any of these hops, we can simply tag on a hop of length $ a_n$ at the end of the sequence. If $ m$ is reached after the $ k$th hop, then we revise the $ k$th hop to be of length $ a_n$ and move what was originally the $ k$th hop to the end.[/hide][/quote]\r\n\r\nOkay, so I have a sequence that doesn't reach $ M \\minus{} \\{m\\}$. But when I change one of the middle hops to be of length $ a_n$, why should I expect the sequence to keep that property? \r\n\r\n(EDIT: Ah, I missed that $ m$ was largest, as kdano pointed out.)", "Solution_5": "[quote=\"probability1.01\"]lol... looks like #3 and #6 are of similar flavor this year...\n\nFor this problem, can't you just\n\n[hide]induct? Let $ a_n$ be largest, and consider the largest element $ m$ of $ M$. In the case that $ s \\minus{} a_n$ is in $ M$ and $ m > s \\minus{} a_n$, note that for some $ 1 \\le i \\le n \\minus{} 1$, we have that $ s \\minus{} a_i$ and $ s \\minus{} a_i \\minus{} a_n$ are both not in $ M$. Thus, apply the inductive hypothesis to the $ a_i$ excluding $ a_i$ and $ a_n$, and we are able to dodge the elements of $ M$, since at most $ n \\minus{} 3$ elements of $ M$ are strictly between $ 0$ and $ s \\minus{} a_i \\minus{} a_n$.\n\nOtherwise, apply the inductive hypothesis to $ a_1$ through $ a_{n \\minus{} 1}$ on the set $ M \\minus{} \\{ m \\}$, so that we make $ n \\minus{} 1$ hops. If $ m$ is not reached after any of these hops, we can simply tag on a hop of length $ a_n$ at the end of the sequence. If $ m$ is reached after the $ k$th hop, then we revise the $ k$th hop to be of length $ a_n$ and move what was originally the $ k$th hop to the end.\n\nSeems a bit simple (although not totally trivial), so I might have missed something...[/hide][/quote]\r\n\r\nI was just about to type my solution that's basically the same, and i couldn't find a bug in it either.", "Solution_6": "[quote=\"probability1.01\"]For this problem, can't you just\n\n[hide]induct? Let $ a_n$ be largest, and consider the largest element $ m$ of $ M$. In the case that $ s \\minus{} a_n$ is in $ M$ and $ m > s \\minus{} a_n$, note that for some $ 1 \\le i \\le n \\minus{} 1$, we have that $ s \\minus{} a_i$ and $ s \\minus{} a_i \\minus{} a_n$ are both not in $ M$. Thus, apply the inductive hypothesis to the $ a_i$ excluding $ a_i$ and $ a_n$, and we are able to dodge the elements of $ M$, since at most $ n \\minus{} 3$ elements of $ M$ are strictly between $ 0$ and $ s \\minus{} a_i \\minus{} a_n$.\n\nOtherwise, apply the inductive hypothesis to $ a_1$ through $ a_{n \\minus{} 1}$ on the set $ M \\minus{} \\{ m \\}$, so that we make $ n \\minus{} 1$ hops. If $ m$ is not reached after any of these hops, we can simply tag on a hop of length $ a_n$ at the end of the sequence. If $ m$ is reached after the $ k$th hop, then we revise the $ k$th hop to be of length $ a_n$ and move what was originally the $ k$th hop to the end.\n\nSeems a bit simple (although not totally trivial), so I might have missed something...[/hide][/quote]\n\nIt seems like you can, if you missed something, so did I when reading your solution. Actually, I was trying along a similar line as you, but got stuck\n[hide]\nin the second case because I \"forgot\" to consider $ a_n$ as the largest, which makes the induction step work nicely in case you bump into $ m$ along the way... :blush: I found the first case relatively simple though...\n[/hide]\r\n\r\nI would like to know whether the shortlist contained any problems harder than #3 and #6, I guess I will have to wait for one year... In last year's shortlist, after looking through it, I consider problems A6 and C6 harder than the actual problems #3 and #6 from last year. I also think that A6 was quite original for a functional equation, and C6 a very beautiful problem, probably the most beautiful in the entire shortlist (working on it, haven't still got much). Anyway, I was surprised after seeing them in the shortlist that neither got selected for the contest... :?:", "Solution_7": "[quote=\"kevinatcausa\"]I'm not sure if I understand this part of your argument:\n\nOkay, so I have a sequence that doesn't reach $ M \\minus{} \\{m\\}$. But when I change one of the middle hops to be of length $ a_n$, why should I expect the sequence to keep that property?[/quote]\r\n\r\nNo, you change the second last hop to $ a_n$, thus increasing it (not touching $ m$ but a greater number). The hops until that don't change, and since $ m$ is the greatest of $ M$, it doesn't touch any number in $ M$.", "Solution_8": "[quote=\"kdano\"]\nI was just about to type my solution that's basically the same, and i couldn't find a bug in it either.[/quote]\r\n\r\nthat is really weird, I've done essentially the same stuff, but can not find any mistake in it either...", "Solution_9": "[quote=\"probability1.01\"]lol... looks like #3 and #6 are of similar flavor this year...\n\nFor this problem, can't you just\n\n[hide]induct? Let $ a_n$ be largest, and consider the largest element $ m$ of $ M$. In the case that $ s \\minus{} a_n$ is in $ M$ and $ m > s \\minus{} a_n$, note that for some $ 1 \\le i \\le n \\minus{} 1$, we have that $ s \\minus{} a_i$ and $ s \\minus{} a_i \\minus{} a_n$ are both not in $ M$. Thus, apply the inductive hypothesis to the $ a_i$ excluding $ a_i$ and $ a_n$, and we are able to dodge the elements of $ M$, since at most $ n \\minus{} 3$ elements of $ M$ are strictly between $ 0$ and $ s \\minus{} a_i \\minus{} a_n$.\n\nOtherwise, apply the inductive hypothesis to $ a_1$ through $ a_{n \\minus{} 1}$ on the set $ M \\minus{} \\{ m \\}$, so that we make $ n \\minus{} 1$ hops. If $ m$ is not reached after any of these hops, we can simply tag on a hop of length $ a_n$ at the end of the sequence. If $ m$ is reached after the $ k$th hop, then we revise the $ k$th hop to be of length $ a_n$ and move what was originally the $ k$th hop to the end.\n\nSeems a bit simple (although not totally trivial), so I might have missed something...[/hide][/quote]\r\n\r\nI also tried induction,but I havent finished it yet.", "Solution_10": "[hide=\"Kind of easy when saw what probability has done\"]\n\nInduction on n. Let it be true for k, now we prove it's true for k+1. We add $ m_k$, such that it's the largest of all m-s (if it isn't, then take all without greatest and add greatest to them in inductive step, as we assumed for all k number it's true). If it hits any point on which grasshoper lands, then just change that last jump before hitting $ m_k$ to some bigger jump... That's the whole problem... \n\nComments: I wouldn't probably do it without seeing the idea the probability did, but with it it's almost trivial... \n[/hide]", "Solution_11": "[quote=\"pbornsztein\"][quote=\"huang\"]Did they really consider this hard enough for a problem 6??[/quote]\n\nAs the coordinations take place, it turns to be the most difficult problem of the IMO for the last 50 years in terms of full solutions (maybe only 2, surely less than 5), and in terms of nonzero scores (probably less than 10 contestants will receive a nonzero score).\nSo, please stop with all these complaining about the level of difficulties of the IMO problems (each problem and each year), it becomes ridiculous.\n\nPierre.[/quote]\r\n\r\nSir, who were the two solved this problem? Could you please also check on the sketch posted by [b]probability 1.01[/b].", "Solution_12": "[quote=\"pbornsztein\"][quote=\"huang\"]Did they really consider this hard enough for a problem 6??[/quote]\n\nAs the coordinations take place, it turns to be the most difficult problem of the IMO for the last 50 years in terms of full solutions (maybe only 2, surely less than 5), and in terms of nonzero scores (probably less than 10 contestants will receive a nonzero score).\nSo, please stop with all these complaining about the level of difficulties of the IMO problems (each problem and each year), it becomes ridiculous.\n\nPierre.[/quote]\r\n\r\nWell the idea of induction with carefully chosen numbers was really hard, but I wanted to pointed out that the problem was not hard in technique of solving the problem.", "Solution_13": "That means absolutely nothing. 2007/3 certainly didn't require anything fancy.", "Solution_14": "[quote=\"Bugi\"]\nWell the idea of induction with carefully chosen numbers was really hard, but I wanted to pointed out that the problem was not hard in technique of solving the problem.[/quote]\r\n\r\nI might be overtrained in combinatorics, but I disagree with you. Induction was my second idea for this problem (after some Menger-related stuff, of which i had no knowledge last year, when i was in highschool). Of course one has to work with it to induct appropriately, but one gets promising results while trying to do so.\r\nI always wondered how hard people find combinatorics problems.. like i found 2008/5 to be the easiest of all 2008's questions, and what i did for 2007/3 seemed pretty straightforward to me (so straightforward, that i didn't notice that the only cases left don't exist :P).. :huh:", "Solution_15": "[REDACTED] Needs some checking I am trying to improve my older solutions, so wil update soon.", "Solution_16": "WLOG let $a_1 < a_2 < a_3\\ldots m_{n-1}$ and have it avoid all points in $M$. \n\n[color=#3D85C6][b]Case 2: [/b][/color]$m_{n-1} > s-a_n$.\n\nIf there exists a positive integer $i$ such that $s-a_i \\not\\in M$ and $s-a_i < m_{n-1}$, then we let $k$ be the index such that $m_k < s-a_i \\le m_{k+1}$. We can just apply the inductive hypothesis to $A \\setminus \\{a_i\\}$ and $M \\setminus m_k$, and then let the grasshopper jump $a_i$ to reach $s$ where it never touches a point in $M$. Therefore, assume otherwise. \n\nWe know there must exist an $i$ such that $s-a_i \\not\\in M$, and let this be maximal. We must have $N = \\{s-a_{i+1}, s-a_{i+2}, \\ldots s-a_n\\} \\subseteq M$, where $\\#(N) \\ge 2$. Therefore, because there must exist some $s-a_n-a_j \\not\\in M$ for $1 \\le j \\le i$ by the Pigeonhole Principle, we can use our inductive hypothesis on $A \\setminus \\{a_j, a_n\\}$ and $M \\setminus N$ to get the grasshopper to a point $p$ where $p < s-a_n$ and $p+a_n>s-a_{i+1}$, which we then can make it jump $a_n$ and then $a_j$ to reach $s$ without touching a point in $M$. \n\nTherefore, the inductive step is proven and we are done. $\\blacksquare$\n\n[color=#f00][b]Remarks: [/b][/color]I agree with people above that this problem has an easy solution (as I can do it lol), yet is one of the hardest IMO problems to date. Right after reading I instantly thought about induction and because I have unlimited time, I just tried it until I figured out that it worked. However this is a luxury that test-takers do not have and also with it being an \"intimidating IMO #6\" is probably the reason of its incredibly low solve rate. ", "Solution_17": "I am confused here, does all elements of M are less than S?", "Solution_18": "Bump, someone please help", "Solution_19": "[quote=abvk1718]I am confused here, does all elements of M are less than S?[/quote]\n\nYou can assume this if it helps your proof because clearly if an element is greater than $s$ then the grasshopper won't be able to jump on it. ", "Solution_20": "[quote=jeteagle][quote=abvk1718]I am confused here, does all elements of M are less than S?[/quote]\n\nYou can assume this if it helps your proof because clearly if an element is greater than $s$ then the grasshopper won't be able to jump on it.[/quote]\n\nGot it :coolspeak: ", "Solution_21": "We use the method of mathematical induction. Let our jump sizes be $a_1, a_2, \\dots a_n$ and the forbidden points be $p_1, p_2, \\dots, p_{n - 1}$. Assume $a_1 < a_2 < \\dots a_n$ and $p_1 < p_2 \\dots < p_n$. There are several cases:\n\n[b]Case 1: [/b] $a_1 + a_2 + \\dots + a_{n - 1} < p_{n - 1}.$\nIn this case, we can find some permutation of $a_{1}, a_2, \\dots a_{n -1}$ which avoids $p_1, p_2, \\dots p_{n - 2}$. It would also by default avoid $p_{n - 1}$, by our case 1 assumption. Then, we can jump $a_{n}$ to finish.\n\n[b]Case 2: [/b] $a_1 + a_2 + \\dots + a_{n - 1} \\ge p_{n - 1}.$\nIn this case, let's also use some permutation of $a_1, a_2, \\dots a_{n - 1}$ that avoids $p_1, p_2, \\dots p_{n - 2}$. If that ordering also avoids $p_{n -1 }$, we're done, so let's assume that it hits $p_{n - 1}$ at some point. At the point where it hits $p_{n - 1}$, swap $a_i$ with $a_n$. Then, the path will be valid.", "Solution_22": "Can someone please tell the motivation behind [hide=the soltion][img]https://i.imgur.com/KPw8mBn.png[/img]\n\n[img]https://i.imgur.com/rmiJ92a.png[/img]\n\n[img]https://i.imgur.com/3jwiQcm.png[/img][/hide] in the [url=https://www.imo-official.org/problems/IMO2009SL.pdf]official booklet[/url].\n\nPersonally, I was also looking at least $k$ for which $|M \\cap (0,T_k]| \\ge k$ (as that's sort of a generalization to the case when $M$ contains an element $\\le a_1$, which is easy to resolve by induction). \nBut I was unable to finish from there. ", "Solution_23": "We induct on $n$, with the base case $n=1$ obvious. Set $a_1 < a_2 < \\cdots < a_n$. For the inductive step, we split into four cases:\n\n[b]Case 1.[/b] $s-a_n \\in M$ is the maximal element of $M$. In this case, by the inductive hypothesis, there exists a sequence of jumps with lengths $a_1, a_2, \\cdots, a_{n-1}$ that can avoid the first $n-2$ elements of $M$. Let $a_k$ be the last jump in this sequence. Then, instead of jumping $a_k$ and then $a_n$, we can jump $a_n$ and then $a_k$, which avoids the last mine, so we are done.\n\n[b]Case 2.[/b] $s-a_n \\in M$, and there are elements of $M$ greater than $s-a_n$. First, notice that of the $n-1$ values $s-a_k$, $1 \\leq k \\leq n-1$, one of them must not be in $M$ because $M$ has at most $n-2$ elements in this range. Now, suppose there are $k$ elements of $M$ after $s-a_n$. Then, there are at least $n-k-1$ values $i$ such that $s-a_{n-1-i}$ is not in $M$. Because there are at most $n-k-2$ elements in $M$ before $s-a_n$, there must exist an index $i$ such that $s-a_{n-1-i} - a_n$ is not in $M$. Now, there are at most $n-3$ elements of $M$ left and we have $n-2$ jumps remaining, so it is possible to avoid all these points via a sequence of jumps.\n\n[b]Case 3.[/b] $s-a_n \\not \\in M$, and there are elements of $M$ greater than $s-a_n$. This case is obvious; simply make a mine-avoiding sequence of jumps to $s-a_n$, then jump $a_n$.\n\n[b]Case 4.[/b] $s-a_n \\not \\in M$, and all elements of $M$ are strictly less than $s-a_n$. Let $m$ be the maximal element of $M$. There exists a sequence of jumps in $\\{a_1, a_2, \\cdots, a_{n-1}\\}$ that avoids all elements of $M$ except for $m$. If it avoids $m$ as well, it lands on $s-a_n$, and we are done. If it lands on $m$, then on the jump right before it reaches $m$, change the grasshopper's jump length to $a_n$, which guarantees that it lands on a safe square as $m$ is maximal. Next, make the remaining jumps to reach $s$, as needed.\n\nThus, the induction is complete, and we are done.", "Solution_24": "Define $m=\\max M,$ $A=\\left \\{ a_1,a_2, \\dots ,a_n\\right \\}$ and WLOG $a_n=\\max a_i$. \nWe induct on $n;$ base case $n\\in \\left \\{ 1,2\\right \\}$ is trivial. For $n>2$ consider two cases.\n\n[b]Case 1.[/b] $s-a_n\\geq m.$\nBy inductive hypothesis there exist a sequence of jumps by lengths $A\\backslash a_n$ which avoid points from $M\\backslash m$ - add a jump by $a_n$ at the end. If the jump by $a_i$ leads to point $m,$ we swaps order of jumps by $a_i,a_n$ - clearly this works. \n\n[b]Case 2.[/b] $s-a_np_{n-2},...,p_1$ implies that jumping $a_k$ jumps past the points; in particular, we can move $a_n$ instead of $a_k$ at that move, since it will jump a further distance past all points.", "Solution_26": "solved with hints\n\nInduct on $n$. This is clearly true for $n=1,2$. \n\nDefine a critical point as a point that is the sum of $n-1$ out of the $a_i$. Define a trap as an element of $M$.\\\\\n\nConsider the smallest critical point, $c$.\\\\\n\nIf there is no trap on $c$ and no trap after $c$, then there are $n$ traps between 0 and $c$. Ignore the last trap before $c$ for now, by induction we can then find a path to $c$ since there are only $n-2$ traps ignoring that one and then we could finish and jump from $c$ to $s$. The only issue is that we might hit the last trap. However, if we do, make the largest possible move in place of the move that stepped on the last trap, which would guarantee skipping it as the largest move is not part of the road to $c$ as $c$ does not contain that term.\\\\\n\nIf there is no trap on $c$ and trap after $c$, we can directly induct to find a path to $c$, as there are at most $n-2$ traps in that area, so we would be done.\\\\\n\nIf there is a trap on $c$ but it is the furthest one along and there are no traps after $c$, take a path that would get you to $c$ except replace the last move with the largest possible move and then finish with whatever move is left.\\\\\n\nFinally, if there is a trap on $c$ and also a trap past $c$, consider the points that is one \"jump\" away from $c$, and the corresponding point that is that same jump away from $s$. There are $n-1$ such pairs, but with only $n-2$ traps other than the one at $c$, there must be a pair of these points that are both safe. Since there are at most $n-3$ traps before $c$, we can reach the point that is one jump away from $c$. Then, the last two jumps are used to get the corresponding point by making the largest jump and then finally jumping to the end.", "Solution_27": "Let's prove this statement by induction on $n$, the number of jumps the grasshopper makes.\n\nBase case: When $n = 2$, there is only one jump with length $a_1$ or $a_2$. Since $M$ contains $n - 1 = 1$ integer and cannot contain $s = a_1 + a_2$, the grasshopper can simply make the jump to the right with the remaining length and avoid landing on the point in $M$.\n\nInductive step: Assume the statement holds for $n = k$, where $k$ is some positive integer. We'll prove it for $n = k + 1$.\n\nLet the sequence of positive integers be $a_1, a_2, \\ldots, a_{k+1}$, and let $M$ be a set of $k$ positive integers not containing $s = a_1 + a_2 + \\ldots + a_{k+1}$. Consider $M'$, which is obtained by adding $s$ to each element of $M$, i.e., $M' = {m + s \\mid m \\in M}$.\n\nBy the induction hypothesis, there exists an order in which the grasshopper can jump $k$ times among $a_1, a_2, \\ldots, a_k$ without landing on any point in $M$. Let $p$ be the position after these $k$ jumps. Now, if $p + a_{k+1} \\notin M'$, the grasshopper can make the $k+1$-th jump to the right with length $a_{k+1}$ and avoid landing on any point in $M$ since $p + a_{k+1}$ cannot be in $M'$.\n\nIf $p + a_{k+1} \\in M'$, then $p + a_{k+1} = m + s$ for some $m \\in M$. But $p + a_{k+1} - s = p - (a_1 + a_2 + \\ldots + a_k) = p - s$, which means $p - s = m \\in M$. However, $p - s$ is the position after $k$ jumps, which, by the induction hypothesis, does not land on any point in $M$. This contradicts $p - s = m \\in M$. Therefore, $p + a_{k+1} \\notin M'$, and the grasshopper can make the $(k+1)$-th jump without landing on any point in $M$.\n\nBy induction, the statement holds for all positive integers $n$.", "Solution_28": "Proceed by induction on $n$. Let $a_1s-a_n$ and $s-a_n\\not\\in M$. Then, by the induction hypothesis, there exists a path from $0$ to $s-a_n$ that avoids all integers in $M$ that uses jumps $a_1$, $a_2$, $\\dots$, $a_{n-1}$. Then, once at $s-a_n$, the grasshopper can jump $a_n$ and win. If $\\max{M}\\le s-a_n$ then consider the path that the grasshopper would take if we removed $\\max{M}$ from $M$. If it doesn't hit $\\max{M}$, we're all good. If it does, then replace that move with a larger one. We can always do this since $a_n$ is available. The remaining case is $\\max{M}>s-a_n$ and $s-a_n\\in M$. Let $S'=\\{\\{s-a_n-a_i, s-a_i\\}\\mid 1\\le i\\le n-1\\}$. Note that $|S'|=n-1$ and $|M\\setminus \\{s-a_n\\}|=n-2$, there exists an element $k\\in S'$ such that $|M\\cap k|=0$. Note that $|M\\cap \\{1,2,\\dots,s-a_n-a_i-1\\}|\\le n-3$, so by the induction hypothesis, the grasshopper can jump to $s-a_n-a_i$ and then to $s-a_i$ and then to $s$.", "Solution_29": "Groupsolved with [b]cursed_tangent1434[/b].\nWe will use induction, and WLOG $a_1 < a_2 < a_3 < \\ldots < a_n$, and let the elements of set $M$ be $m_1 \\leq m_2 \\leq m_3 \\leq \\ldots \\leq m_{n-1}$.\n$\\newline$\nOur base case of $n = 1$ is clearly true, and so is our $n = 2$ case.\n$\\newline$\n\nNow assume that $n - 1$ works.\nWe then have to prove that for any choice of $m_{n-1}$ and choice of $a_n$ then it is possible to order the jumps in a way so that we land on $a_1 + a_2 + \\ldots + a_n = s$ without touching an element of $M$.\n$\\newline$\n\nCase $1$: $s - a_n < m_{n-1}$\nFrom here, we can simply add $a_n$ to $s - a_n$ to reach $s$.\nNote that it is possible to reach $s - a_n = a_1 + a_2 + \\ldots + a_{n-1}$ since our $n - 1$ case is true. \n$\\newline$\n\nCase $2$: $s - a_n \\geq m_{n-1}$\nIf $m_{n-1}$ does not coincide with the points the grasshopper jumps on, then we're done by simply adding $a_n$ to $s - a_n$.\nIf otherwise, then swap the distance that the grasshopper jumped to land on $m_{n-1}$ with $a_n$. Since $m_{n-1}$ is the maximum value in $M$ and that $a_n \\geq m_{n-1}$, this avoids all other elements of $M$, and leads the grasshopper to $s$.\nSo, our induction is done. $\\blacksquare$\n" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Show that if x1,x2,x3,x4>0 then:\r\n\r\nx1/(x2+x3)+x2/(x3+x4)+x3/(x4+x1)+x4/(x1+x2)>=2.", "Solution_1": "WLOG x1+x2+x3+x4=1\r\nf(x)=1/x convex we apply Jensen and we have that:\r\n\r\nx1/(x2+x3)+x2/(x3+x4)+x3/(x4+x1)+x4/(x1+x2)>=1/(x1x2+2x1x3+x1x4+x2x3+2x2x4+x3x4).\r\n\r\nwe have that:\r\n\r\n2(x1x2+2x1x3+x1x4+x2x3+2x2x4+x3x4)<=(x1+x2+x3+x4)^2=1 which is equivalent to\r\n(x1-x3)^2+(x2-x4)^2>=0\r\n\r\nThis is the like the ineq from Rusia 1980 (posted by Manlio)\r\n\r\ncheers! :D :D", "Solution_2": "Or we could amplify the i'h fraction with x_i for i from 1 to 4 and then use the result \r\n\\sum a_i2/x_i >= (\\sum a_i)2/ (\\sum x_i) for positive a_i and x_i, i from 1 to n. We continue as Lagrangia did." } { "Tag": [ "calculus", "integration", "trigonometry", "conics", "ellipse", "vector", "calculus computations" ], "Problem": "Calculate\r\n\\[\\int_\\gamma e^{x}(\\sin{(xy)}+y\\cos{(xy)})dx+xe^{x}(\\cos{(xy)})dy \\]\r\n\r\nwhere $\\gamma$ is the ellipse $4x^{2}+y^{2}=9$ oriented anticlockwise.", "Solution_1": "use Mr. Green! :D \r\n\r\n$\\int_{\\gamma}\\;f\\;dx\\;+\\;g\\;dy\\;=\\bigcirc \\hspace{-0.5cm}\\int \\hspace{-0.25cm}\\int_{R}\\; ( \\;-\\frac{\\partial f}{\\partial y}\\;+\\;\\frac{\\partial g}{\\partial x}\\; ) \\;d\\mathbb{A}$\r\n\r\n$\\;=\\;-\\;e^{x}\\; ( \\;x\\;\\cos{(xy)}\\;+\\;\\cos{(xy)}\\;-\\; ( \\;xy\\; ) \\sin{(xy)}\\; )\\\\ \\\\ \\hspace* \\;\\;\\;\\;\\;\\;\\; \\qquad-\\; ( \\;xy\\; ) \\;e^{x}\\;\\sin{(xy)}\\;+\\;x\\;e^{x}\\;\\cos{(xy)}\\;+\\;e^{x}\\;\\cos{(xy)}$\r\n\r\n$=\\;\\boxed{0}\\;$\r\n\r\n(circle or ellipse won't matter)", "Solution_2": "this vector is Path independence.\r\n\r\nSince the start point and the end point is equal, the integral is zero.\r\n\r\nsee ya http://en.wikipedia.org/wiki/Line_integral", "Solution_3": "[quote=\"misan\"]use Mr. Green! :D \n\n$\\int_{\\gamma}\\;f\\;dx\\;+\\;g\\;dy\\;=\\bigcirc \\hspace{-0.5cm}\\int \\hspace{-0.25cm}\\int_{R}\\; ( \\;-\\frac{\\partial f}{\\partial y}\\;+\\;\\frac{\\partial g}{\\partial x}\\; ) \\;d\\mathbb{A}$\n\n$\\;=\\;-\\;e^{x}\\; ( \\;x\\;\\cos{(xy)}\\;+\\;\\cos{(xy)}\\;-\\; ( \\;xy\\; ) \\sin{(xy)}\\; )\\\\ \\\\ \\hspace* \\;\\;\\;\\;\\;\\;\\; \\qquad-\\; ( \\;xy\\; ) \\;e^{x}\\;\\sin{(xy)}\\;+\\;x\\;e^{x}\\;\\cos{(xy)}\\;+\\;e^{x}\\;\\cos{(xy)}$\n\n$=\\;\\boxed{0}\\;$\n\n(circle or ellipse won't matter)[/quote]\r\nThanks :lol: \r\nI made that calculation too but initially got an $x$ too much but now i got it.", "Solution_4": "Err... I think is much more easy to see that this vector field is path independent\r\n\r\n${\\frac{\\partial \\left(e^{x}\\sin (x y)\\right)}{\\partial x}=e^{x}(y \\cos (x y)+\\sin (x y))}$\r\n${\\frac{\\partial \\left(e^{x}\\sin (x y)\\right)}{\\partial y}=e^{x}x \\cos (x y)}$", "Solution_5": "Yes it is also easy to find a potential for the field as you have shown above.\r\nThanks :lol:" } { "Tag": [], "Problem": "Put the best website you have ever been to and tell us why it is good. (try not to post the same ones again)\r\nI will make a list of them on this post.\r\n[url=http://www.miniclip.com]www.miniclip.com[/url]\r\n[url=http://www.gamesofgondor.com]www.gamesofgondor.com[/url]\r\n[url=http://www.onemorelevel.com]www.onemorelevel.com[/url]\r\n[url=http://www.artofproblemsolving.com]www.artofproblemsolving.com[/url]\r\n[url=http://mathworld.wolfram.com]mathworld.wolfram.com[/url]\r\n[url=http://www.braingle.com/index.php?ref=GarnetLover]www.braingle.com/index.php?ref=GarnetLover[/url]\\\r\n[url=http://www.albinoblacksheep.com/]www.albinoblacksheep.com/[/url]\r\n[url=http://www.gnu.org/]www.gnu.org/[/url]", "Solution_1": "Since I get the first post, I will say:\r\n\r\n[url=http://artofproblemsolving.com]www.artofproblemsolving.com!!!!!!![/url]\r\n\r\nThis is seriously the best site I've ever visited. I spend about 5x more time on this site than all the other put together. (Including classes) :)", "Solution_2": "I totally agree with 4everwise!!! This site's awesome!\r\n\r\nAlso, [u]http://mathworld.wolfram.com[/u] is a complete reference that any problem solver could want or need", "Solution_3": "Yeah, you guys should check out AoPS. It's seriously my favorite site on the whole internet. I'm not kidding! Check it out! http://www.artofproblemsolving.com :roll:", "Solution_4": "besides AoPS, http://www.braingle.com is awesome\r\n\r\nif u want to visit, plz click here: http://www.braingle.com/index.php?ref=GarnetLover thanx", "Solution_5": "My site! Well, actually its pretty terrible (haven't worked on it in about a year).. but i'm working on something new, so when i'm done with this new site ill put a post up.\r\n\r\nYeah, http://www.artofproblemsolving.com is pretty awesome. But I guess that makes me unoriginal.\r\n\r\nSo, as something new.. uh... http://www.albinoblacksheep.com is funny.", "Solution_6": "[url=http://www.gnu.org]www.gnu.org[/url] - Lots of free (open source) software\r\n\"Software is like sex, it's better when it's free.\" - Linus Torvalds" } { "Tag": [ "trigonometry", "ratio", "geometry", "circumcircle", "incenter", "conics", "parabola" ], "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14764[/img]\r\n\r\n :(", "Solution_1": "[color=darkblue]I'm not sure about this, but...\n\nWe have $ \\frac {b \\minus{} c}{a} \\equal{} \\frac {\\sin{\\frac {B \\minus{} C}{2}}}{\\cos{\\frac {A}{2}}}$, so we can find $ \\cos{\\frac {A}{2}}$ and $ \\sin{\\frac {A}{2}}$.\n\nThen $ \\frac {b \\plus{} c}{a} \\equal{} \\frac {\\cos{\\frac {B \\minus{} C}{2}}}{\\sin{\\frac {A}{2}}}$ and we can find $ b \\plus{} c$. But we know $ b \\minus{} c$, so we can find lenghts of the sides $ b,c$. Then construction is trivial.\n\nI've got only one question: Can we say that we can calculate the $ \\sin$ in consctruction problems.[/color]", "Solution_2": "Ahiles construction is absolutely correct, even though it looks ugly when attempted. Sine or cosine can be used /calculated in construction problems. $ \\sin \\phi$ or $ \\cos \\phi$ mean ratio of the proper leg and hypotenuse of some right triangle with one angle $ \\phi.$ The ratio can be constructed given the angle, and the angle can be constructed given the ratio.\r\n______________________________________________\r\n\r\nLet $ O$ be the circumcenter and $ H$ the orthocenter, $ \\angle OAH \\equal{} \\angle B \\minus{} \\angle C.$ $ M$ is midpoint of the segment $ a \\equal{} BC.$ Place the segment $ b \\minus{} c \\equal{} DE$ parallel to $ BC,$ so that $ M$ is also midpoint of $ DE.$ Then $ D, E$ are tangency points of the incircle $ (I)$ and the A-excircle $ (J)$ with $ BC.$ The perpendicular bisector $ OM$ of $ BC$ cuts the circumcircle at $ P, Q,$ such that $ P$ is on the opposite side of $ BC$ then $ A.$ The internal bisector $ AI$ of the $ \\angle A$ cuts $ (O)$ again at $ P.$\r\n \r\n$ R$ is the circumradius. The incenter $ I$ is on circle $ (P)$ with radius $ PB \\equal{} PC \\equal{} PI$ and $ MD \\equal{} PI \\sin \\frac {B \\minus{} C}{2}.$ From similar right $ \\triangle PQB \\sim PBM,$ $ MP \\equal{} \\frac {PB^2}{PQ} \\equal{} \\frac {PI^2}{2R}.$ Then $ \\frac {MD^2}{MP} \\equal{} 2R \\sin^2 \\frac {B \\minus{} C}{2} \\equal{} \\text{const}$ (disregarding $ b \\minus{} c$), which means that $ D$ is on a parabola $ \\mathcal P$ with vertex $ P,$ axis $ PQ,$ and passing through $ A.$ Since the ratio $ \\frac {DE}{BC} \\equal{} \\frac {b \\minus{} c}{a}$ is fixed and the 2 segments have a common midpoint $ M$, $ D$ is also on ellipse $ \\mathcal Q$ with center $ O,$ the major axis $ PQ,$ and minor axis length $ PQ \\cdot \\frac {b \\minus{} c}{a}.$ To construct intersections of the parabola and ellipse, project the ellipse $ \\mathcal Q$ to the circumcircle $ (O)$ and the parabola $ \\mathcal P$ to a larger parabola $ \\mathcal P^*.$ Let $ t \\perp PQ$ be the common tangent $ t$ at $ P$ of the parabola $ \\mathcal P$ and ellipse $ \\mathcal Q,$ same as the common tangent of the projected parabola $ \\mathcal P^*$ and circle $ (O).$ Put the coordinate origin at $ P,$ the x-axis along $ t$ and y-axis along $ PQ.$ Let $ F \\in PQ$ be the focus of $ \\mathcal P^*.$ Equations of the circle $ (O)$ and parabola $ \\mathcal P^*$ with focus $ F$ are\r\n\r\n$ x^2 \\plus{} (y \\minus{} OP)^2 \\equal{} OP^2,\\ \\ y \\equal{} \\frac {x^2}{4\\ PF}.$\r\n\r\nSubtracting and reducing by $ x^2,$\r\n\r\n$ x^2 \\equal{} 4\\ PF (2\\ OP \\minus{} 4\\ PF),\\ \\ y \\equal{} 2\\ OP \\minus{} 4\\ PF \\equal{} PQ \\minus{} 4\\ PF.$\r\n\r\n$ y$ is the distance of the parabola $ \\mathcal P^*$ / circle $ (O)$ intersections from their common tangent $ t.$\r\n_____________________________________________\r\n\r\n[b]Construction:[/b] Construct a right $ \\triangle A'KL$ with the side $ A'L \\equal{} a,$ right angle $ \\angle A'KL,$ and $ \\angle LA'K \\equal{} \\angle B \\minus{} \\angle C.$ Draw its circumcircle $ (O)$ centered at the midpoint $ O$ of $ A'L.$ Parallel to $ KL$ through $ O$ cuts the circle $ (O)$ at $ B, C$ and $ BC \\equal{} a.$ Construct points $ D, E \\in BC$ such that $ DE, BC$ have the common midpoint $ O$ and $ DE \\equal{} b \\minus{} c.$ Perpendicular bisector of $ BC$ cuts $ (O)$ at $ P, Q,$ such that $ P$ is on the opposite side of $ BC$ than $ A'.$\r\n\r\nProject the ellipse $ \\mathcal Q$ (not drawn) with major/minor axes $ PQ, DE$ to the circle $ (O).$ That is, do nothing, we already have the circle. To project the parabola $ \\mathcal P,$ draw perpendiculars $ n_B, n_D$ to $ BC$ at $ B, D.$ Perpendicular $ n_D$ cuts $ A'P$ at $ S.$ Draw parallels $ p_A, p_S$ to $ BC$ at $ A', S.$ Perpendicular $ n_B$ cuts the parallel $ p_S$ at $ S^*.$ Parallel $ p_A$ cuts the line $ PS^*$ at the projection $ A^*$ of $ A'.$ The projected parabola $ \\mathcal P^*$ (not drawn) has the same vertex $ P,$ the same axis $ PQ,$ and it passes through $ A^*.$\r\n\r\nLet $ t \\perp PQ$ at $ P$ be the parabola vertex tangent, the locus of perpendiculars from the parabola focus to the parabola tangents. Parallel to the parabola axis through $ A^*$ cuts $ t$ at $ U.$ Let $ V$ be the midpoint of $ PU.$ $ A^*V$ is the parabola tangent at $ A^*.$ Perpendicular to $ A^*V$ at $ V$ cuts the parabola axis $ PQ$ at the focus $ F.$\r\n\r\nQuadruple the segment $ PF$ into $ PG,$ so that $ PG \\equal{} 4\\ PF$ and reflect $ G$ in $ BC$ into $ M'.$ Perpendicular to $ PQ$ at $ M'$ cuts the circle $ (O)$ at $ B'C'.$ The $ \\triangle A'B'C'$ is just similar to the desired $ \\triangle ABC,$ because, not knowing the circumradius, we started with a circle $ (O)$ with diameter $ BC \\equal{} a.$ To finish, draw parallels to $ A'B', A'C'$ through $ B, C,$ respectively, intersecting at $ A.$", "Solution_3": "cuts the circle (O) at B,C \r\n\r\nwhere's B and where\u00b4s C ?\r\n\r\nConstruct points D,E\\in BC such that DE, BC have the common midpoint O and DE=b-c\r\n\r\nwhere's D and where\u00b4s E ? \r\n\r\n :(", "Solution_4": "$ A'L \\equal{} BC \\equal{} PQ \\equal{} a$ are all diameters of the circle $ (O).$ (This will not be circumcircle of the resulting $ \\triangle ABC$, but of the similar $ \\triangle A'B'C'.$) The diameter $ BC$ is parallel to $ KL$ and the diameter $ PQ$ perpendicular to $ KL.$ Small concentric circle with center $ O$ and radius $ \\frac {b \\minus{} c}{2}$ cuts $ BC$ at $ D, E.$ See the figure." } { "Tag": [], "Problem": "Salut! Are cineva posibilitatea sa posteze unele din problemele mai interesante de la Unirea 2005? Multumesc.", "Solution_1": "intrebarea este daca stie cineva care a fost la concurs despre forum? :)", "Solution_2": "Subiectele sunt in fisierul atasat (din pacate, .doc, deocamdata)", "Solution_3": "Multumesc. Va trebui sa mai astept putin pana sa le pot citi. Separe ca francezilor nu le plac prea mult documentele word.", "Solution_4": "Vezi c\u0103 \u0163i-am ata\u015fat o variant\u0103 .pdf a fi\u015fierului (creat\u0103 din .doc-ul ala) :)", "Solution_5": "Multumesc, Valentin, acum il pot citi. Mersi." } { "Tag": [ "calculus", "integration", "function", "limit", "calculus computations" ], "Problem": "Can anyone please explain the following problems? I don't know how to show them. Thanks. \r\n\r\n1. a. Show that if $f$ is an even function and the necessary integrals exist, then $\\displaystyle\\int^\\infty_{-\\infty} f(x)\\ dx = 2 \\displaystyle\\int^\\infty_0 f(x)\\ dx $.\r\n\r\nb. Show that if $f$ is odd and the necessary integrals exist, then $\\displaystyle\\int^\\infty_{-\\infty} f(x)\\ dx = 0$.\r\n\r\nAny help is greatly appreciated.", "Solution_1": "The key step we need is the following lemma:\r\n\r\nLemma: If $\\int_{-\\infty}^{\\infty}f(x)\\, dx$ exists, $\\int_{-\\infty}^{\\infty}f(x)\\, dx=\\lim_{M\\to\\infty}\\int_{-M}^M f(x)\\, dx.$\r\nProof: The limit of a sum is equal to the sum of the limits (if those limits exist), so\r\n$\\lim_{M\\to\\infty}\\int_{-M}^M f(x)\\, dx=\\lim_{M\\to\\infty}\\left(\\int_{-M}^0 f(x)\\, dx+\\int_{0}^M f(x)\\, dx\\right)$\r\n$=\\lim_{M\\to\\infty}\\int_{-M}^0 f(x)\\, dx+\\lim_{M\\to\\infty}\\int_{0}^M f(x)\\, dx$\r\n$=\\int_{-\\infty}^0f(x)\\, dx+\\int_0^\\infty f(x)\\, dx=\\int_{-\\infty}^{\\infty}f(x)\\, dx.$\r\n\r\nNow you can apply what you know about integrals of odd and even functions on finite intervals." } { "Tag": [ "geometry", "ratio", "algebra proposed", "algebra" ], "Problem": "Let $ \\left(a_1,a_2,...,a_n\\right)$ and $ \\left(b_1,b_2,...,b_n\\right)$ be two sequences of positive reals. We are searching for a permutation $ \\pi$ of the set $ \\left\\{1,2,...,n\\right\\}$ that minimizes the sum\r\n\r\n$ \\sum_{k\\equal{}1}^n a_{\\pi\\left(k\\right)} \\sum_{i\\equal{}k}^n b_{\\pi\\left(i\\right)}$\r\n$ \\equal{}a_{\\pi\\left(1\\right)}\\left(b_{\\pi\\left(1\\right)}\\plus{}b_{\\pi\\left(2\\right)}\\plus{}...\\plus{}b_{\\pi\\left(n\\right)}\\right)$\r\n$ \\plus{}a_{\\pi\\left(2\\right)}\\left(b_{\\pi\\left(2\\right)}\\plus{}b_{\\pi\\left(3\\right)}\\plus{}...\\plus{}b_{\\pi\\left(n\\right)}\\right)$\r\n$ \\plus{}...\\plus{}a_{\\pi\\left(n\\right)}b_{\\pi\\left(n\\right)}$.\r\n\r\nProve that any permutation $ \\pi$ that satisfies\r\n\r\n$ \\frac{a_{\\pi\\left(1\\right)}}{b_{\\pi\\left(1\\right)}}\\leq\\frac{a_{\\pi\\left(2\\right)}}{b_{\\pi\\left(2\\right)}}\\leq ...\\leq\\frac{a_{\\pi\\left(n\\right)}}{b_{\\pi\\left(n\\right)}}$\r\n\r\nminimizes this sum.\r\n\r\n[i]Remark.[/i] As an application, this yields an optimal strategy for fighting a group of monsters in a simplified RPG (no area damage, no pulling of enemies, etc.): Assume that one has to face $ n$ enemies, numbered $ 1$, $ 2$, ..., $ n$. For every $ i$, let $ a_i$ be the time that one needs to kill enemy $ i$ if one ignores the other enemies, and $ b_i$ the damage that enemy $ i$ deals to the player in one second. Then, in order to lose as little health as possible during the battle, one should first take on the enemy with the lowest ratio $ \\frac{a_i}{b_i}$, then after killing him turn on the one with the second lowest $ \\frac{a_i}{b_i}$, and so on.", "Solution_1": "Assume the contrary. Let $ k$ be the largest index so that $ \\frac {a_{\\pi(k)}}{b_{\\pi(k)}}\\le\\frac {a_{\\pi(k \\plus{} 1)}}{b_{\\pi(k \\plus{} 1)}}\\le...\\le\\frac {a_{\\pi\\left(n\\right)}}{b_{\\pi\\left(n\\right)}}$.\r\nIf $ k \\equal{} 1$, we are done, so assume now $ k > 1$.\r\n\r\nLet $ S$ be the current sum in question, and $ S'$ be the sum when $ a_{\\pi(k)}$ and $ a_{\\pi(k \\minus{} 1)}$ are switched.\r\n\r\nThen, $ S' \\minus{} S \\equal{} a_{\\pi(k)}b_{\\pi(k \\minus{} 1)} \\minus{} a_{\\pi(k \\minus{} 1)}b_{\\pi(k)} \\equal{} b_{\\pi(k)}b_{\\pi(k \\minus{} 1)}(\\frac {a_{\\pi(k)}}{b_{\\pi(k)}} \\minus{} \\frac {a_{\\pi(k \\minus{} 1)}}{b_{\\pi(k \\minus{}1 )}}) < 0$, contradicting the that $ S$ has the smallest sum of any arrangement, contradiction! Hence, the result.\r\n\r\nCheers,\r\n\r\nRofler", "Solution_2": "Yep. I have taken the liberty to fix some typos in your post (for instance, we are searching for the smallest, not the largest sum). Your solution is the first solution given in American Mathematical Monthly 1973, pp. 437-438.\r\n\r\nI will post the second proposed solution (by R. J. Dickson, without his overcomplicated writeup) in a moment.", "Solution_3": "Here comes Dickson's solution. Note that the sequence $ \\left(a_1,a_2,...,a_n\\right)$ needs not consist of positive reals; arbitrary reals will do.\r\n\r\n darij" } { "Tag": [], "Problem": "[b][size=100][color=Darkblue]\u0394\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf $ \\bigtriangleup ABC$ \u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf \u03c3\u03b5 \u03ba\u03cd\u03ba\u03bb\u03bf $ (O)$ \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ M,\\ N,$ \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b5\u03c0\u03b9 \u03c4\u03c9\u03bd \u03b5\u03c5\u03b8\u03b5\u03b9\u03ce\u03bd \u03c4\u03c9\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ce\u03bd \u03c4\u03bf\u03c5 $ AB,\\ AC$ \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03af\u03c7\u03c9\u03c2, \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03c9\u03c3\u03c4\u03b5 $ BM \\equal{} BC \\equal{} CN$ $ ($ $ B,$ \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03c9\u03bd $ A,\\ M$ \u03ba\u03b1\u03b9 $ C,$ \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03c9\u03bd $ A,\\ N$ $ ).$ \u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ OI_{a}\\perp MN,$ \u03cc\u03c0\u03bf\u03c5 $ O$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 $ \\bigtriangleup ABC$ \u03ba\u03b1\u03b9 $ I_{a}$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03ba\u03ad\u03bd\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03c0\u03b1\u03c1\u03b5\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c3\u03c4\u03b7 \u03b3\u03c9\u03bd\u03af\u03b1 $ \\angle A.$[/color][/size][/b]\r\n\r\n\u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b1\u03c5\u03c4\u03cc \u03bc\u03bf\u03c5 \u03c4\u03bf \u03ad\u03c3\u03c4\u03b5\u03b9\u03bb\u03b5 \u03bc\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7, \u03ad\u03bd\u03b1\u03c2 \u03c6\u03af\u03bb\u03bf\u03c2 \u03bc\u03b1\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf \u0392\u03b9\u03b5\u03c4\u03bd\u03ac\u03bc \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03be\u03b5\u03bd\u03b5\u03cd\u03c4\u03b7\u03ba\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03cc\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c4\u03cc\u03c3\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b1 ( \u03cc\u03c0\u03c9\u03c2 \u03b5\u03b3\u03ce \u03bc\u03c0\u03cc\u03c1\u03b5\u03c3\u03b1 ), \u03ad\u03bd\u03b1 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c4\u03cc\u03c3\u03bf \u03bb\u03b9\u03c4\u03ae \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03b7.\r\n\r\n\u0388\u03c7\u03c9 \u03c4\u03b7\u03bd \u03b1\u03af\u03c3\u03b8\u03b7\u03c3\u03b7 \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b1\u03c0\u03bb\u03bf\u03cd\u03c3\u03c4\u03b5\u03c1\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b2\u03bb\u03ad\u03c0\u03c9 \u03c3\u03c4\u03b7 \u03b4\u03b9\u03ba\u03ae \u03c3\u03b1\u03c2 \u03c0\u03c1\u03bf\u03c3\u03ad\u03b3\u03b3\u03b9\u03c3\u03b7, \u03b3\u03b9\u03b1\u03c4\u03af \u03cc\u03c0\u03c9\u03c2 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03c0\u03b5\u03c1\u03bd\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b1 \u03c7\u03c1\u03cc\u03bd\u03b9\u03b1, \u03c4\u03bf \u03bc\u03c5\u03b1\u03bb\u03cc \u03b1\u03c1\u03ad\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03cc\u03bb\u03bf \u03ba\u03b1\u03b9 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03cd\u03c0\u03bb\u03bf\u03ba\u03b5\u03c2 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 ( = complicated solutions... ) :( \r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.", "Solution_1": "\u0391\u03bd \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03b1\u03c1\u03ad\u03c3\u03b5\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 (\u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae) \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03b8\u03b5\u03c9\u03c1\u03b7\u03b8\u03b5\u03af \u03b1\u03c0\u03bb\u03ae. \r\n\r\n\u03a3\u03c5\u03b3\u03c7\u03c9\u03c1\u03ad\u03c3\u03c4\u03b5 \u03bc\u03b5, \u03b1\u03bb\u03bb\u03ac \u03b8\u03b1 \u03c4\u03b7 \u03b3\u03c1\u03ac\u03c8\u03c9 \u03bb\u03af\u03b3\u03bf \u03c3\u03c5\u03bd\u03bf\u03c0\u03c4\u03b9\u03ba\u03ac. \u03a4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03bf \u03bc\u03b5:\r\n$ OM^2 \\minus{} ON^2 \\equal{} I_aM^2 \\minus{} I_aN^2$. \u039a\u03b1\u03b9 \u03c4\u03ce\u03c1\u03b1 \u03b4\u03bf\u03c5\u03bb\u03ad\u03c5\u03c9 \u03bc\u03b5 \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf\u03c5. \u0394\u03b7\u03bb\u03b1\u03b4\u03ae:\r\n\r\n$ OM^2 \\minus{} ON^2 \\equal{} MB\\cdot MA \\minus{} MC\\cdot NA \\equal{} BC(AB \\minus{} AC)$\r\n\r\n\u039f\u03bc\u03bf\u03b9\u03b1, \u03b8\u03b5\u03c9\u03c1\u03ce\u03bd\u03c4\u03b1\u03c2 $ K,L$ \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c4\u03bf\u03bc\u03ae\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b1\u03c1\u03b1\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03bc\u03b5 \u03c4\u03b9\u03c2 $ AB, AC$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1, \u03b8\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ I_aM^2 \\minus{} I_aN^2 \\equal{} MK^2 \\minus{} NL^2 \\equal{} (BM \\minus{} BK)^2 \\minus{} (NC \\minus{} CL)^2 \\equal{} ...\\equal{}$\r\n\r\n$ \\equal{}BC (AB \\minus{}AC)$\r\n\r\n\u0395\u03c0\u03b5\u03c4\u03b1\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf... :wink:", "Solution_2": "\u03a0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 . \u0391\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 . \u039d\u03b1 \u03ad\u03bd\u03b1 \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03c0\u03b1\u03c1\u03cc\u03bc\u03bf\u03b9\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03bf\u03c5", "Solution_3": "\u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7 \u03c3' \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03c0\u03bf\u03bb\u03cd \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b2\u03ac\u03c1\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03ad\u03b2\u03b3\u03b1\u03bb\u03b5\u03c2 \u03b1\u03c0\u03cc \u03c0\u03ac\u03bd\u03c9 \u03bc\u03bf\u03c5, \u03b3\u03b9\u03b1 \u03ad\u03bd\u03b1 \u03bc\u03b1\u03ba\u03c1\u03cd \u03b1\u03c0\u03b1\u03bd\u03c4\u03b7\u03c4\u03b9\u03ba\u03cc \u03bc\u03ae\u03bd\u03c5\u03bc\u03b1.\r\n\r\n\u0391\u03c5\u03c4\u03ae \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03ce \u03b3\u03b9' \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae. \u0391\u03bd \u03b1\u03be\u03af\u03b6\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03c1\u03b1\u03c4\u03ae\u03c3\u03c9 \u03ba\u03ac\u03c4\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 ( \u03ad\u03c6\u03c4\u03b1\u03c3\u03b1 \u03bc\u03ad\u03c7\u03c1\u03b9 \u03a0\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2 \u03ba\u03b1\u03b9 Desarques... :ewpu: ), \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5\u03c1\u03b9\u03ba\u03ac \u03b5\u03bd\u03b4\u03b9\u03ac\u03bc\u03b5\u03c3\u03b1 \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03ad\u03c3\u03bc\u03b1\u03c4\u03b1, \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03c4\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03b1\u03c1\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03c3\u03b1\u03bd \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03c4\u03ad\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2.\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Given an acute-angled triangle $ ABC$ with circumcenter $ O.$ The circle passing through two points $ A,\\ O$ intersects with the line $ AB$ and $ AC$ at $ P,\\ Q$ other than $ A$ respectively. If the lengths of the line segments $ PQ,\\ BC$ are equal, then find the angle $ \\leq 90^\\circ$ that the lines $ PQ$ and $ BC$ make.", "Solution_1": "Consider circumcircles of $ APQ$ and $ ABC$. Since $ PQ \\equal{} BC$ and $ \\angle PAQ \\equal{} \\angle BAC$, it follows that the circles have same radius. let $ O_{1}$ be circumcentre of $ APQ$. Since $ AO \\equal{} R \\equal{} AO_{1} \\equal{} OO_{1}$, $ AOO_{1}$ is equilatrel and $ \\angle AO_{1}O \\equal{} 60$. Then $ \\angle OPC \\equal{} \\frac {\\angle AO_{1}O}{2} \\equal{} 30$, and since $ \\angle OAC \\equal{} 90\\minus{}B$, then $ \\angle POA \\equal{} B \\minus{} 60$, so $ \\angle PQB \\equal{} \\angle B\\minus{}60$. Since $ \\angle ABC \\equal{} B$, it follows the angle between $ PQ$ and $ BC$ is $ 60$ degrees.", "Solution_2": "Let D be the second intersection of $ (O')$ and $ (O). PQ\\cap AC\\equal{}\\{L\\}$\r\n$ \\angle PLC\\equal{}\\angle AQP\\plus{}\\angle ACB\\equal{}\\angle ADP\\plus{}\\angle ADB\\equal{}\\angle PDB$\r\nBut $ (O)$ and $ (O')$ have same radius and O' lies on $ (O)$ then $ \\angle APD\\equal{}\\angle ABD\\equal{}60^o \\Rightarrow \\angle PDB\\equal{}60^o$\r\nSo $ \\angle PLC\\equal{}60^o$", "Solution_3": "Let $M,N$ be the midpoints of $AB,AC$ respectively, note $OM\\perp AB, ON\\perp AC$. Also, $A,O,M,N$ are cyclic $\\implies \\triangle OMN\\sim \\triangle OPQ$, since $PQ=BC=2MN$ it follows that $\\triangle OPQ$ and $\\triangle OMN$ are a $60^{\\circ}$ rotation of each other hence the angle between $PQ$ and $BC$ is $60^{\\circ}$." } { "Tag": [ "number theory", "least common multiple", "number theory open" ], "Problem": "first consider Fibbonaci's seq: F1=F2=1, Fn=F(n-1)+F(n-2)\r\nnow consider this seq: Pn=Fn mod B ( B is a fixed integer)\r\nthis sequence is periodical with the maximum period B(B-1)/2\r\nhow can i actually determine the period... ?", "Solution_1": "As far as I know, the general case is an open problem.\r\nLet m \\geq 2 be an integer.\r\nLucas (1878) has proved that the Fibonacci sequence is periodic modulo m, and that there exists n such that F_n = 0 mod[m].\r\nLet k(m) the length of the lesat period of (F_i) mod[m], anf f(m) be the least n such that F_n = 0 mod[m].\r\n\r\nWalls has found many conditions on k(m) and f(m). In particular, f(m) divides k(m) and if m = prod (p_i) a_i where the p_i's are distinct primes, then k(m) = lcm( k((p_i) a_i) ).\r\nMoreover, if p is a prime and t is the largest integer such that k(p t ) = k(p) then k(p a+t ) = p a k(p) for all integers a \\geq 0.\r\n\r\nWalls results may be used to determine some values of k(m), but as said above, the general determination of k(m) is difficult.\r\nThus, it may be useful to look at the two following theorems of Vinson. Let t(m) = k(m)/f(m) :\r\n\r\nTh. 1 :\r\nIf m > 2 then :\r\nt(m) = 4 if f(m) is odd.\r\nt(m) = 1 if m =/= 0 mop[8] and f(p) = 2 mod[4] for each odd prime p dividing m.\r\nt(m) = 2 in every other case.\r\n\r\nThus, it shows that the main problem is to determine f(m).\r\n\r\nTh. 2 :\r\nLet p be an odd prime. Then, for each integer e > 0 :\r\nt(p e ) = 1 if p = 11 or 19 mod[20]\r\nt(p e ) = 2 if p = 3 or 7 mod[20]\r\nt(p e ) = 4 if p = 13 or 17 mod[20]\r\nt(p e ) =/= 2 if p = 21 or 29 mod[40].\r\nIn the other cases 'everything' may happen...\r\n\r\nMoreover, it is known [3] that k(m) \\leq 6m for all m.\r\n\r\nReferences :\r\n[1] D.D.Walls, 'Fibonacci series modulo m', A.M.M. (1960), p.525-532.\r\n[2] J.Vinson, 'The relation of the period modulo to the rank of apparition of m in the Fibonacci sequence', Fibonacci Quaterly, 1963, p.37-45.\r\n[3] A.M.M., 1992, exercice n E3410, p.272-273.\r\n\r\nPierre.", "Solution_2": "This problems resembles that of finding the order of 2 mod p for a prime p as much as finding F_p such that F_p is prime resembles that of finding all the primes of the form 2 p+1.\r\nThese are all, I think, open problems." } { "Tag": [ "number theory", "greatest common divisor", "geometry", "USAMTS", "AMC" ], "Problem": "Given that P,Q are primes, with P>50.\r\n\r\nFind the largest integer that must divide (P+Q) given that PQ is one less than a perfect square", "Solution_1": "I hope it's right:\n\n\n\n[hide]Since PQ is 1 less than a perfect square, we could have a parametric form of P=x+1 and Q=x-1 so that PQ=x^2-1. Thus, we are basically finding the largest integer that must divide P+Q=2x. We know that x must also be even since all primes greater than two are odd (and so P is odd and Q is odd, which makes x even). Given this amount of information, we could conclude that 2*2=4 must divide P+Q.[/hide]\n\n\n\n-interesting_move", "Solution_2": "You can get more factors than just 2.", "Solution_3": "Can you? Two possible pairs are (3,5) and (5,7) (basically, any pair of twin primes), and their sums are 8 and 12 respectively, whose gcd is 4.\r\n\r\nI'm sorry; I thought it said P<50. Sigh. Allright; we know that all large primes are neighbors of 6, therefore any twin primes must surround a multiple of 6, making their sum a multiple of 12.\r\n59, 61 are twin primes; so are 71, 73 and 101, 103. (i think they are all primes). anyways, their sums will be 120, 144, and 204. The gcd of these three expressions is 12, so that is the largest integer possible.\r\nHow's that?", "Solution_4": "Yes, but it says that P>50...so you can get more.", "Solution_5": "Right. After someone posts the solution to this problem, remind me to tell an interesting anecdote about a similar problem.", "Solution_6": "So the answer is [hide]12[/hide]?\n\n\n\n-interesting_move", "Solution_7": "Yes.\r\n\r\nIn the final round of Math Bowl at Bay Area Math Meet 2000, one question that Paul Zeitz asked was (worded differently almost certainly) ``a and b are 12-digit numbers differing by 2, and they're both prime. What is the remainder when ab is divided by 9?'' No one in the finals had a clue how to do it, but it's actually not that hard. a and b (mod 9) must be (2,4), (5,7), or (8,1) if we assume (WLOG) that a\\frac{\\sigma(k)}{k}.$\nProve that there exists an infinity of superabundant numbers.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366619&#p366619]3. (BEL 3)[/url][b]IMO4[/b] Let $ABC$ be an equilateral triangle and $\\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$, and $CA$ (including $A$, $B$, and $C$). Determine whether, for every partition of $\\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013660#p2013660]4. (BEL 5)[/url] On the sides of the triangle $ABC$, three similar isosceles triangles $ABP \\ (AP = PB)$, $AQC \\ (AQ = QC)$, and $BRC \\ (BR = RC)$ are constructed. The first two are constructed externally to the triangle $ABC$, but the third is placed in the same half-plane determined by the line $BC$ as the triangle $ABC$. Prove that $APRQ$ is a parallelogram.\n\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013665#p2013665]5. (BRA 1)[/url] Consider the set of all strictly decreasing sequences of $n$ natural numbers having the property that in each sequence no term divides any other term of the sequence. Let $A = (a_j)$ and $B = (b_j)$ be any two such sequences. We say that $A$ precedes $B$ if for some $k$, $a_k < b_k$ and $a_i = b_i$ for $i < k$. Find the terms of the first sequence of the set under this ordering.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013667#p2013667]6. (CAN 2)[/url] Suppose that ${x_1, x_2, \\dots , x_n}$ are positive integers for which $x_1 + x_2 + \\cdots+ x_n = 2(n + 1)$. Show that there exists an integer $r$ with $0 \\leq r \\leq n - 1$ for which the following $n - 1$ inequalities hold:\n\\[x_{r+1} + \\cdots + x_{r+i} \\leq 2i+ 1, \\qquad \\qquad \\forall i, 1 \\leq i \\leq n - r; \\]\n\\[x_{r+1} + \\cdots + xn + x_1 + \\cdots+ x_i \\leq 2(n - r + i) + 1, \\qquad \\qquad \\forall i, 1 \\leq i \\leq r - 1.\\]\nProve that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013668#p2013668]7. (CAN 5)[/url] Let $a$ be a positive integer and let $\\{a_n\\}$ be defined by $a_0 = 0$ and\n\\[a_{n+1 }= (a_n + 1)a + (a + 1)a_n + 2 \\sqrt{a(a + 1)a_n(a_n + 1)} \\qquad (n = 1, 2 ,\\dots ).\\]\nShow that for each positive integer $n$, $a_n$ is a positive integer.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013675#p2013675]8. (SPA 2)[/url] In a test, $3n$ students participate, who are located in three rows of $n$ students in each. The students leave the test room one by one. If $N_1(t), N_2(t), N_3(t)$ denote the numbers of students in the first, second, and third row respectively at time $t$, find the probability that for each t during the test,\n\\[|N_i(t) - N_j(t)| < 2, i \\neq j, i, j = 1, 2, \\dots .\\]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343813&#p343813]9. (USA 3)[/url][b]IMO6[/b] Let $ a$, $ b$ and $ c$ be the lengths of the sides of a triangle. Prove that\n\\[ a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\\ge 0.\n\\]\nDetermine when equality occurs.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013677#p2013677]10. (FIN 1)[/url] Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1/q$ and a polynomial $P$ with integral coefficients such that\n\\[ |P(x)-\\frac pq | < \\frac{1}{q^2}\\]\nfor all $x \\in I.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013680#p2013680]11. (FIN 2\u2019)[/url] Let $f : [0, 1] \\to \\mathbb R$ be continuous and satisfy:\n\\[ \\begin{cases}bf(2x) = f(x), &\\mbox{ if } 0 \\leq x \\leq 1/2,\\\\ f(x) = b + (1 - b)f(2x - 1), &\\mbox{ if } 1/2 \\leq x \\leq 1,\\end{cases}\\]\nwhere $b = \\frac{1+c}{2+c}$, $c > 0$. Show that $0 < f(x)-x < c$ for every $x, 0 < x < 1.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366613&#p366613]12. (GBR 4)[/url][b]IMO1[/b] Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy: $f(xf(y))=yf(x)$ for all $x,y$; and $f(x)\\to0$ as $x\\to\\infty$.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013682#p2013682]13. (LUX 2)[/url] Let $E$ be the set of $1983^3$ points of the space $\\mathbb R^3$ all three of whose coordinates are integers between $0$ and $1982$ (including $0$ and $1982$). A coloring of $E$ is a map from $E$ to the set {red, blue}. How many colorings of $E$ are there satisfying the following property: The number of red vertices among the $8$ vertices of any right-angled parallelepiped is a multiple of $4$ ?\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366621&#p366621]14. (POL 2)[/url][b]IMO5[/b] Is it possible to choose $1983$ distinct positive integers, all less than or equal to $10^5$, no three of which are consecutive terms of an arithmetic progression?\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013684#p2013684]15. (POL 3)[/url] Decide whether there exists a set $M$ of natural numbers satisfying the following conditions:\n\n(i) For any natural numberm $>1$ there are $a, b \\in M$ such that $a+b = m.$\n\n(ii) If $a, b, c, d \\in M , a, b, c, d > 10$ and $a + b = c + d$, then $a = c$ or $a = d.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013686#p2013686]16. (GDR 1)[/url] Let $F(n)$ be the set of polynomials $P(x) = a_0+a_1x+\\cdots+a_nx^n$, with $a_0, a_1, . . . , a_n \\in \\mathbb R$ and $0 \\leq a_0 = a_n \\leq a_1 = a_{n-1 } \\leq \\cdots \\leq a_{[n/2] }= a_{[(n+1)/2]}.$ Prove that if $f \\in F(m)$ and $g \\in F(n)$, then $fg \\in F(m + n).$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013689#p2013689]17. (GDR 3)[/url] Let $P_1, P_2, \\dots , P_n$ be distinct points of the plane, $n \\geq 2$. Prove that\n\\[ \\max_{1\\leq i \\frac{\\sqrt 3}{2}(\\sqrt n -1) \\min_{1\\leq i\\frac{\\sigma(k)}{k}.$\nProve that there exists an infinity of superabundant numbers.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2037701#p2037701]5. (BEL 2)[/url] Consider the set $\\mathbb Q^2$ of points in $\\mathbb R^2$, both of whose coordinates are rational.\n\n(a)\tProve that the union of segments with vertices from $\\mathbb Q^2$ is the entire set $\\mathbb R^2$.\n(b)\tIs the convex hull of $\\mathbb Q^2$ (i.e., the smallest convex set in $\\mathbb R^2$ that contains $\\mathbb Q^2$) equal to $\\mathbb R^2$ ?\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366619&#p366619]6. (BEL 3)[/url][b]IMO4[/b] Let $ABC$ be an equilateral triangle and $\\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$, and $CA$ (including $A$, $B$, and $C$). Determine whether, for every partition of $\\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2037703#p2037703]7. (BEL 4)[/url] Find all numbers $x \\in \\mathbb Z$ for which the number\n\\[x^4 + x^3 + x^2 + x + 1\\]\nis a perfect square.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013660#p2013660]8. (BEL 5)[/url] [b](SL83-4).[/b]On the sides of the triangle $ABC$, three similar isosceles triangles $ABP \\ (AP = PB)$, $AQC \\ (AQ = QC)$, and $BRC \\ (BR = RC)$ are constructed. The first two are constructed externally to the triangle $ABC$, but the third is placed in the same half-plane determined by the line $BC$ as the triangle $ABC$. Prove that $APRQ$ is a parallelogram.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013665#p2013665]9. (BRA 1)[/url] [b](SL83-5).[/b] Consider the set of all strictly decreasing sequences of $n$ natural numbers having the property that in each sequence no term divides any other term of the sequence. Let $A = (a_j)$ and $B = (b_j)$ be any two such sequences. We say that $A$ precedes $B$ if for some $k$, $a_k < b_k$ and $a_i = b_i$ for $i < k$. Find the terms of the first sequence of the set under this ordering.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2037704#p2037704]10. (BRA 2)[/url] Which of the numbers 1, 2, . . . , 1983 has the largest number of divisors?\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2037707#p2037707]11. (BRA 3)[/url] A boy at point $A$ wants to get water at a circular lake and carry it to point $B$. Find the point $C$ on the lake such that the distance walked by the boy is the shortest possible given that the line $AB$ and the lake are exterior to each other.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2037716#p2037716]12. (BRA 4)[/url] The number $0$ or $1$ is to be assigned to each of the $n$ vertices of a regular polygon. In how many different ways can this be done (if we consider two assignments that can be obtained one from the other through rotation in the plane of the polygon to be identical)?\n\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039309#p2039309]13. (BUL 1)[/url] Let $p$ be a prime number and $a_1, a_2, \\ldots, a_{(p+1)/2}$ different natural numbers less than or equal to $p.$ Prove that for each natural number $r$ less than or equal to $p$, there exist two numbers (perhaps equal) $a_i$ and $a_j$ such that\n\\[p \\equiv a_i a_j \\pmod r.\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039312#p2039312]14. (BUL 2)[/url] Let $\\ell$ be tangent to the circle $k$ at $B$. Let $A$ be a point on $k$ and $P$ the foot of perpendicular from $A$ to $\\ell$. Let $M$ be symmetric to $P$ with respect to $AB$. Find the set of all such points $M.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039315#p2039315]15. (CAN 1)[/url] Find all possible finite sequences $\\{n_0, n_1, n_2, \\ldots, n_k \\}$ of integers such that for each $i, i$ appears in the sequence $n_i$ times $(0 \\leq i \\leq k).$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013667#p2013667]16. (CAN 2)[/url] (SL83-6). Suppose that ${x_1, x_2, \\dots , x_n}$ are positive integers for which $x_1 + x_2 + \\cdots+ x_n = 2(n + 1)$. Show that there exists an integer $r$ with $0 \\leq r \\leq n - 1$ for which the following $n - 1$ inequalities hold:\n\\[x_{r+1} + \\cdots + x_{r+i} \\leq 2i+ 1, \\qquad \\qquad \\forall i, 1 \\leq i \\leq n - r; \\]\n\\[x_{r+1} + \\cdots + xn + x_1 + \\cdots+ x_i \\leq 2(n - r + i) + 1, \\qquad \\qquad \\forall i, 1 \\leq i \\leq r - 1.\\]\nProve that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039320#p2039320]17. (CAN 3)[/url] In how many ways can $1, 2,\\ldots, 2n$ be arranged in a $2 \\times n$ rectangular array $\\left(\\begin{array}{cccc}a_1& a_2 & \\cdots & a_n\\\\b_1& b_2 & \\cdots & b_n\\end{array}\\right)$ for which:\n(i) $a_1 < a_2 < \\cdots < a_n,$\n\n(ii) $b_1 < b_2 <\\cdots < b_n,$\n\n(iii) $a_1 < b_1, a_2 < b_2, \\ldots, a_n < b_n \\ ?$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039324#p2039324]18. (CAN 4)[/url] Let $b \\geq 2$ be a positive integer.\n\n(a) Show that for an integer $N$, written in base $b$, to be equal to the sum of the squares of its digits, it is necessary either that $N = 1$ or that $N$ have only two digits.\n\n(b) Give a complete list of all integers not exceeding $50$ that, relative to some base $b$, are equal to the sum of the squares of their digits.\n\n(c) Show that for any base b the number of two-digit integers that are equal to the sum of the squares of their digits is even.\n\n(d) Show that for any odd base $b$ there is an integer other than $1$ that is equal to the sum of the squares of its digits.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013668#p2013668]19. (CAN 5)[/url] [b](SL83-7).[/b] Let $a$ be a positive integer and let $\\{a_n\\}$ be defined by $a_0 = 0$ and\n\\[a_{n+1 }= (a_n + 1)a + (a + 1)a_n + 2 \\sqrt{a(a + 1)a_n(a_n + 1)} \\qquad (n = 1, 2 ,\\dots ).\\]\nShow that for each positive integer $n$, $a_n$ is a positive integer.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039347#p2039347]20. (COL 1)[/url] Let $f$ and $g$ be functions from the set $A$ to the same set $A$. We define $f$ to be a functional $n$-th root of $g$ ($n$ is a positive integer) if $f^n(x) = g(x)$, where $f^n(x) = f^{n-1}(f(x)).$\n\n[b](a)[/b] Prove that the function $g : \\mathbb R \\to \\mathbb R, g(x) = 1/x$ has an infinite number of $n$-th functional roots for each positive integer $n.$\n\n[b](b)[/b] Prove that there is a bijection from $\\mathbb R$ onto $\\mathbb R$ that has no nth functional root for each positive integer $n.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039348#p2039348]21. (COL 2)[/url] Prove that there are infinitely many positive integers $n$ for which it is possible for a knight, starting at one of the squares of an $n \\times n$ chessboard, to go through each of the squares exactly once.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039349#p2039349]22. (CUB 1)[/url] Does there exist an infinite number of sets $C$ consisting of $1983$ consecutive natural numbers such that each of the numbers is divisible by some number of the form $a^{1983}$, with $a \\in \\mathbb N, a \\neq 1 \\ ?$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013677#p2013677]23. (FIN 1)[/url] [b](SL83-10).[/b] Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1/q$ and a polynomial $P$ with integral coefficients such that\n\\[ \\left|P(x)-\\frac pq \\right| < \\frac{1}{q^2}\\]\nfor all $x \\in I.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039359#p2039359]24. (FIN 2)[/url] Every $x, 0 \\leq x \\leq 1$, admits a unique representation $x = \\sum_{j=0}^{\\infty} a_j 2^{-j}$, where all the $a_j$ belong to $\\{0, 1\\}$ and infinitely many of them are $0$. If $b(0) = \\frac{1+c}{2+c}, b(1) =\\frac{1}{2+c},c > 0$, and\n\\[f(x)=a_0 + \\sum_{j=0}^{\\infty}b(a_0) \\cdots b(a_j) a_{j+1}\\]\nshow that $0 < f(x) -x < c$ for every $x, 0 < x < 1.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013680#p2013680](FIN 2\u2019)[/url] [b](SL83-11)[/b] Let $f : [0, 1] \\to \\mathbb R$ be continuous and satisfy:\n\\[ \\begin{array}{cc}bf(2x) = f(x),& 0 \\leq x \\leq 1/2;\\\\ f(x) = b + (1 - b)f(2x - 1), & 1/2 \\leq x \\leq 1,\\end{array}\\]\nwhere $b = \\frac{1+c}{2+c}$, $c > 0$. Show that $0 < f(x)-x < c$ for every $x, 0 < x < 1.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039365#p2039365]25. (FRG 1)[/url] How many permutations $a_1, a_2, \\ldots, a_n$ of $\\{1, 2, . . ., n \\}$ are sorted into increasing order by at most three repetitions of the following operation: Move from left to right and interchange $a_i$ and $a_{i+1}$ whenever $a_i > a_{i+1}$ for $i$ running from $1$ up to $n - 1 \\ ?$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039366#p2039366]26. (FRG 2)[/url] Let $a, b, c$ be positive integers satisfying $\\gcd (a, b) = \\gcd (b, c) = \\gcd (c, a) = 1$. Show that $2abc-ab-bc-ca$ cannot be represented as $bcx+cay +abz$ with nonnegative integers $x, y, z.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366618&#p366618]27. (FRG 3)[/url] [b]IMO3[/b] Let $a,b$ and $c$ be positive integers, no two of which have a common divisor greater than $1$. Show that $2abc-ab-bc-ca$ is the largest integer which cannot be expressed in the form $xbc+yca+zab$, where $x,y,z$ are non-negative integers.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039367#p2039367]28. (GBR 1)[/url] Show that if the sides $a, b, c$ of a triangle satisfy the equation\n\\[2(ab^2 + bc^2 + ca^2) = a^2b + b^2c + c^2a + 3abc,\\]\nthen the triangle is equilateral. Show also that the equation can be satisfied by positive real numbers that are not the sides of a triangle.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039373#p2039373]29. (GBR 2)[/url] Let $O$ be a point outside a given circle. Two lines $OAB, OCD$ through $O$ meet the circle at $A,B,C,D$, where $A,C$ are the midpoints of $OB,OD$, respectively. Additionally, the acute angle $\\theta$ between the lines is equal to the acute angle at which each line cuts the circle. Find $\\cos \\theta$ and show that the tangents at $A,D$ to the circle meet on the line $BC.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039376#p2039376]30. (GBR 3)[/url] Prove the existence of a unique sequence $\\{u_n\\} \\ (n = 0, 1, 2 \\ldots )$ of positive integers such that\n\\[u_n^2 = \\sum_{r=0}^n \\binom{n+r}{r} u_{n-r} \\qquad \\text{for all } n \\geq 0\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366613&#p366613]31. (GBR 4)[/url] [b]IMO1[/b] Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy: $f(xf(y))=yf(x)$ for all $x,y$; and $f(x)\\to0$ as $x\\to\\infty$.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2039383#p2039383]32. (GBR 5)[/url]Let $a, b, c$ be positive real numbers and let $[x]$ denote the greatest integer that does not exceed the real number $x$. Suppose that $f$ is a function defined on the set of non-negative integers $n$ and taking real values such that $f(0) = 0$ and\n\\[f(n) \\leq an + f([bn]) + f([cn]), \\qquad \\text{ for all } n \\geq 1.\\]\nProve that if $b + c < 1$, there is a real number $k$ such that\n\\[f(n) \\leq kn \\qquad \\text{ for all } n \\qquad (1)\\]\nwhile if $b + c = 1$, there is a real number $K$ such that $f(n) \\leq K n \\log_2 n$ for all $n \\geq 2$. Show that if $b + c = 1$, there may not be a real number $k$ that satisfies $(1).$\n\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013686#p2013686]33. (GDR 1)[/url] [b](SL83-16).[/b] Let $F(n)$ be the set of polynomials $P(x) = a_0+a_1x+\\cdots+a_nx^n$, with $a_0, a_1, . . . , a_n \\in \\mathbb R$ and $0 \\leq a_0 = a_n \\leq a_1 = a_{n-1 } \\leq \\cdots \\leq a_{[n/2] }= a_{[(n+1)/2]}.$ Prove that if $f \\in F(m)$ and $g \\in F(n)$, then $fg \\in F(m + n).$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041514#p2041514]34. (GDR 2)[/url] In a plane are given n points $P_i \\ (i = 1, 2, \\ldots , n)$ and two angles $\\alpha$ and $\\beta$. Over each of the segments $P_iP_{i+1} \\ (P_{n+1} = P_1)$ a point $Q_i$ is constructed such that for all $i$:\n(i) upon moving from $P_i$ to $P_{i+1}, Q_i$ is seen on the same side of $P_iP_{i+1}$,\n(ii) $\\angle P_{i+1}P_iQ_i = \\alpha,$\n(iii) $\\angle P_iP_{i+1}Q_i = \\beta.$\nFurthermore, let $g$ be a line in the same plane with the property that all the points $P_i,Q_i$ lie on the same side of $g$. Prove that\n\\[\\sum_{i=1}^n d(P_i, g)= \\sum_{i=1}^n d(Q_i, g).\\]\nwhere $d(M,g)$ denotes the distance from point $M$ to line $g.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013689#p2013689]35. (GDR 3)[/url] [b](SL83-17).[/b] Let $P_1, P_2, \\dots , P_n$ be distinct points of the plane, $n \\geq 2$. Prove that\n\\[ \\max_{1\\leq i \\frac{\\sqrt 3}{2}(\\sqrt n -1) \\min_{1\\leq i1$ there are $a, b \\in M$ such that $a+b = m.$\n\n(ii) If $a, b, c, d \\in M , a, b, c, d > 10$ and $a + b = c + d$, then $a = c$ or $a = d.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013692#p2013692]52. (ROM 1)[/url] [b](SL83-19).[/b] Let $(F_n)_{n\\geq 1} $ be the Fibonacci sequence $F_1 = F_2 = 1, F_{n+2} = F_{n+1} + F_n (n \\geq 1),$ and $P(x)$ the polynomial of degree $990$ satisfying\n\\[ P(k) = F_k, \\qquad \\text{ for } k = 992,\\ldots , 1982.\\]\nProve that $P(1983) = F_{1983} - 1.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041734#p2041734]53. (ROM 2)[/url] Let $a \\in \\mathbb R$ and let $z_1, z_2, \\ldots, z_n$ be complex numbers of modulus $1$ satisfying the relation\n\\[\\sum_{k=1}^n z_k^3=4(a+(a-n)i)- 3 \\sum_{k=1}^n \\bar{z_k}\\]\nProve that $a \\in \\{0, 1,\\ldots, n \\}$ and $z_k \\in \\{1, i \\}$ for all $k.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2012303&#p2012303]54. (ROM 3)[/url] [b]ISL 1984 P1[/b] Find all solutions of the following system of $n$ equations in $n$ variables:\n\\[\\begin{array}{c}\\ x_1|x_1| - (x_1 - a)|x_1 - a| = x_2|x_2|,x_2|x_2| - (x_2 - a)|x_2 - a| = x_3|x_3|,\\ \\vdots \\ x_n|x_n| - (x_n - a)|x_n - a| = x_1|x_1|\\end{array}\\]\nwhere $a$ is a given number.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041738#p2041738]55. (ROM 4)[/url] For every $a \\in \\mathbb N$ denote by $M(a)$ the number of elements of the set\n\\[ \\{ b \\in \\mathbb N | a + b \\text{ is a divisor of } ab \\}.\\]\nFind $\\max_{a\\leq 1983} M(a).$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041747#p2041747]56. (ROM 5)[/url] Consider the expansion\n\\[(1 + x + x^2 + x^3 + x^4)^{496} = a_0 + a_1x + \\cdots + a_{1984}x^{1984}.\\]\n[b](a)[/b] Determine the greatest common divisor of the coefficients $a_3, a_8, a_{13}, \\ldots , a_{1983}.$\n\n[b](b)[/b] Prove that $10^{340 }< a_{992} < 10^{347}.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041752#p2041752]57. (SPA 1)[/url] In the system of base $n^2 + 1$ find a number $N$ with $n$ different digits such that:\n\n[b](i)[/b] $N$ is a multiple of $n$. Let $N = nN'.$\n\n[b](ii)[/b] The number $N$ and $N'$ have the same number $n$ of different digits in base $n^2 + 1$, none of them being zero.\n[b]\n(iii)[/b] If $s(C)$ denotes the number in base $n^2 + 1$ obtained by applying the permutation $s$ to the $n$ digits of the number $C$, then for each permutation $s, s(N) = ns(N').$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013675#p2013675]58. (SPA 2)[/url] [b](SL83-8). [/b] In a test, $3n$ students participate, who are located in three rows of $n$ students in each. The students leave the test room one by one. If $N_1(t), N_2(t), N_3(t)$ denote the numbers of students in the first, second, and third row respectively at time $t$, find the probability that for each t during the test,\n\\[|N_i(t) - N_j(t)| < 2, i \\neq j, i, j = 1, 2, \\dots .\\]\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041754#p2041754]59. (SPA 3)[/url] Solve the equation\n\\[\\tan^2(2x) + 2 \\tan(2x) \\cdot \\tan(3x) -1 = 0.\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013700#p2013700]60. (SWE 1)[/url] [b](SL83-21).[/b] Find the greatest integer less than or equal to $\\sum_{k=1}^{2^{1983}} k^{\\frac{1}{1983} -1}.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041758#p2041758]61. (SWE 2)[/url] Let $a$ and $b$ be integers. Is it possible to find integers $p$ and $q$ such that the integers $p+na$ and $q +nb$ have no common prime factor no matter how the integer $n$ is chosen.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041844#p2041844]62. (SWE 3)[/url] $A$ circle $\\gamma$ is drawn and let $AB$ be a diameter. The point $C$ on $\\gamma$ is the midpoint of the line segment $BD$. The line segments $AC$ and $DO$, where $O$ is the center of $\\gamma$, intersect at $P$. Prove that there is a point $E$ on $AB$ such that $P$ is on the circle with diameter $AE.$\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013701#p2013701]63. (SWE 4)[/url] [b](SL83-22).[/b] Let $n$ be a positive integer having at least two different prime factors. Show that there exists a permutation $a_1, a_2, \\dots , a_n$ of the integers $1, 2, \\dots , n$ such that\n\\[\\sum_{k=1}^{n} k \\cdot \\cos \\frac{2 \\pi a_k}{n}=0.\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041848#p2041848]64. (USA 1)[/url] The sum of all the face angles about all of the vertices except one of a given polyhedron is $5160$. Find the sum of all of the face angles of the polyhedron.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041861#p2041861]65. (USA 2)[/url] Let $ABCD$ be a convex quadrilateral whose diagonals $AC$ and $BD$ intersect in a point $P$. Prove that\n\\[\\frac{AP}{PC}=\\frac{\\cot \\angle BAC + \\cot \\angle DAC}{\\cot \\angle BCA + \\cot \\angle DCA}\\]\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343813&#p343813]66. (USA 3)[/url][b]IMO6[/b] Let $ a$, $ b$ and $ c$ be the lengths of the sides of a triangle. Prove that\n\\[ a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\\ge 0.\n\\]\nDetermine when equality occurs.\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041862#p2041862]67. (USA 4)[/url]The altitude from a vertex of a given tetrahedron intersects the opposite face in its orthocenter. Prove that all four altitudes of the tetrahedron are concurrent.\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041866#p2041866]68. (USA 5)[/url] Three of the roots of the equation $x^4 -px^3 +qx^2 -rx+s = 0$ are $\\tan A, \\tan B$, and $\\tan C$, where $A, B$, and $C$ are angles of a triangle. Determine the fourth root as a function only of $p, q, r$, and $s.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366615&#p366615]69. (USS 1)[/url][b]IMO2[/b] Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\\angle O_1AO_2=\\angle M_1AM_2$.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013703#p2013703]70. (USS 2)[/url] [b](SL83-24). [/b] Let $d_n$ be the last nonzero digit of the decimal representation of $n!$. Prove that $d_n$ is aperiodic; that is, there do not exist $T$ and $n_0$ such that for all $n \\geq n_0, d_{n+T} = d_n.$\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2013705#p2013705]71. (USS 3)[/url] [b](SL83-25).[/b] Prove that every partition of $3$-dimensional space into three disjoint subsets has the following property: One of these subsets contains all possible distances; i.e., for every $a \\in \\mathbb R^+$, there are points $M$ and $N$ inside that subset such that distance between $M$ and $N$ is exactly $a.$\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041872#p2041872]72. (USS 4)[/url] Prove that for all $x_1, x_2,\\ldots , x_n \\in \\mathbb R$ the following inequality holds:\n\\[\\sum_{n \\geq i >j \\geq 1} \\cos^2(x_i - x_j ) \\geq \\frac{n(n-2)}{4}.\\]\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041939#p2041939]73. (VIE 1)[/url] Let $ABC$ be a nonequilateral triangle. Prove that there exist two points $P$ and $Q$ in the plane of the triangle, one in the interior and one in the exterior of the circumcircle of $ABC$, such that the orthogonal projections of any of these two points on the sides of the triangle are vertices of an equilateral triangle.\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041947#p2041947]74. (VIE 2)[/url] In a plane we are given two distinct points $A,B$ and two lines $a, b$ passing through $B$ and $A$ respectively $(a \\ni B, b \\ni A)$ such that the line $AB$ is equally inclined to a and b. Find the locus of points $M$ in the plane such that the product of distances from $M$ to $A$ and a equals the product of distances from $M$ to $B$ and $b$ (i.e., $MA \\cdot MA' = MB \\cdot MB'$, where $A'$ and $B'$ are the feet of the perpendiculars from $M$ to $a$ and $b$ respectively).\n\n\n\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2041952#p2041952]75. (VIE 3)[/url] Find the sum of the fiftieth powers of all sides and diagonals of a regular $100$-gon inscribed in a circle of radius $R.$" } { "Tag": [ "probability", "integration", "limit", "calculus", "function", "parameterization", "algebra" ], "Problem": "Hi, I read some demonstrations of the central limit theorem and wrote some of my own, but I always end up requiring a certain hypothesis.\r\nLet f(x) b\u00e8 a probability distribution:\r\n\\[ \\begin{array}{l} f(x) \\ge 0 \\\\ \\int\\limits_{ - \\infty }^\\infty {f(x)dx} = 1 \\\\ E[x] = \\int\\limits_{ - \\infty }^\\infty {\\left[ {f(x)x} \\right]dx} = \\mu \\\\ x_c = x - \\mu \\\\ E[x_c ] = 0 \\\\ E[x_c^2 ] = \\int\\limits_{ - \\infty }^\\infty {\\left[ {f(x_c )x_c^2 } \\right]dx_c } = \\sigma ^2 \\\\ \\end{array} \\]\r\n\r\nLet \\[ x_1 ,x_2 ,...x_n \\] be n different samples of the variable x. Let z be \\[ z = \\sum\\limits_{i = 1}^k {x_i } \\]\r\nThen the probability distribution of z as the sum of n variables, called \\[ f_n (z) \\], can be defined recursively as:\\[ f_n (z) = \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)f_{n - 1} (z - x)dx} \\]\r\n\r\nWhere \\[ f_1 (z) = f(z) \\]\r\n\r\nNow we can use this relationship to prove that for every integer n:\r\n\\[ \\begin{array}{l} \\int\\limits_{ - \\infty }^{ + \\infty } {f_n (z)dz = } 1 \\\\ E_n [z] = \\int\\limits_{ - \\infty }^{ + \\infty } {f_n (z)zdz = } n\\mu \\\\ E_n [(z - n\\mu )^2 ] = n\\sigma ^2 \\\\ \\end{array} \\]\r\n\r\nNow, the central limit theorem states that:\r\n\\[ \\mathop {\\lim }\\limits_{n \\to \\infty } f_n \\left( {\\frac{{z - n\\mu }}{{\\sigma \\sqrt n }}} \\right) = \\frac{1}{{\\sqrt{2\\pi } }}\\exp \\left[ {\\frac{{ - w^2 }}{2}} \\right] \\]\r\n\r\nIn all the central limit theorem demonstrations I have seen it is necessary that for every integer k \\[ \\left| {E[x_c^k ]} \\right| = \\left| {\\int\\limits_{ - \\infty }^\\infty {\\left[ {f(x_c )x_c^k } \\right]dx_c } } \\right| < L < + \\infty \\]\r\n.\r\n\r\nIs it necessary for f(x) to be a probability distribution to satisfy this condition. Because otherwise, I can define a f(x) such that has overall integral 1, finite mean and variance, but for certain integers k E[x^k] diverges.\r\nFor instance, consider this function:\r\n\\[ \\begin{array}{l} \\left| x \\right| < = L \\to f(x) = a \\\\ \\left| x \\right| > L \\to f(x) = \\frac{{aL^4 }}{{x^4 }} \\\\ \\end{array} \\]\r\n\r\nWe can see that E[x]=0, E[x^2] is finite, E[x^(2k+1)]=0, but E[x^2k] diverges for k>1.\r\n\r\n\\[ \\begin{array}{l} f(x) = f( - x) \\to \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)x^{2k + 1} } dx = 0 \\\\ \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)dx = 2\\left[ {\\int\\limits_0^{ + \\infty } {f(x)dx} } \\right]} = 2\\left[ {\\int\\limits_0^L {adx + \\int\\limits_L^{ + \\infty } {\\frac{{aL^4 }}{{x^4 }}dx} } } \\right] = \\frac{8}{3}aL \\\\ aL = \\frac{3}{8} \\\\ \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)x^2 dx = \\frac{2}{3}} aL^3 + 2aL^3 = \\left( {\\frac{8}{3}aL} \\right)L^2 = L^2 \\\\ L = \\sigma \\\\ a = \\frac{3}{8\\sigma} \\end{array} \\]\r\n\r\nBut as we said:\r\n\\[ k > 1 \\to \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)x^{2k} dx = + \\infty } \\]\r\n\r\nNow my question is: is the central limit theorem valid for this particular f(x) ?", "Solution_1": "[quote=\"andrea_roxbury\"]Now, the central limit theorem states that:\n\n\\[ \\mathop {\\lim }\\limits_{n \\to \\infty } f_n \\left( {\\frac{{z - n\\mu }}{{\\sigma \\sqrt n }}} \\right) = \\frac{1}{{\\sqrt{2\\pi } }}\\exp \\left[ {\\frac{{ - w^2 }}{2}} \\right] \\] [/quote]\r\nThat statement seems a little technically garbled (what's $w$?), but that's not the point I'm going to make. What you have there is pointwise convergence of the density functions to the density of the standard normal distribution. That's far too strong a notion of convergence, and with it leaves you with a far too narrow and restricted a theorem. For instance: wouldn't you like it to be true that the binomial $(n,p)$ distribution, properly scaled, tends to the standard normal as $n\\to\\infty$? But for each $n,$ the scaled binomial distribution doesn't even have a density, since it is a discrete distribution.\r\n\r\nThe appropriate standard for a useful version of the Central Limit Theorem is [b]convergence in distribution.[/b] Here's a sample version:\r\n\r\nSuppose $X$ is a random variable with a finite second moment. $(E(X^2)<\\infty.)$ Let $\\mu=E(X)$ and $\\sigma=\\sqrt{\\text{Var}(X)}.$ Suppose $(X_k)$ is a sequence if independent identically distributed random variables with the same distribution as $X.$\r\n\r\nLet $Z_n=\\frac1{\\sqrt{n}\\,\\sigma}\\left(\\sum_{k=1}^nX_k-n\\mu\\right).$\r\n\r\nLet $F_n(z)=P(Z_n\\le z).$ This is the cumulative distribution function.\r\n\r\nThen, $\\forall\\,z,\\,\\lim_{n\\to\\infty}F_n(z)= \\int_{-\\infty}^z\\frac1{\\sqrt{2\\pi}}e^{-x^2/2}\\,dx.$\r\n\r\nThat's convergence in distribution. It's certainly possible for a sequence of discrete random variables to converge in distribution to a continuous random variable.\r\n\r\nThe proofs I know involve Fourier-analytic methods (or as the probabilists would say, the \"characteristic function\" $\\chi_X(t)=E(e^{itX}).)$ But in answer to your question: the proofs I know do not require that moments higher than the second moment be finite. Your example (after we tweak the parameters $a$ and $L$ to make sure it's a random variable) would be included in this theory; its second moment is finite and that's all we need.", "Solution_2": "First of all, the case of discrete distributions can be easily treated using Dirac's delta functions.\r\nThe operation of calculating the probability ${\\rm{P(z) = P(Z}}_{\\rm{n}} \\le z)$ where $Z_n=\\frac1{\\sqrt{n}\\,\\sigma}\\left(\\sum_{k=1}^nX_k-n\\mu\\right)$, is \"regularizing\": this means that even if we start with an irregular probability distribution (such as with the discrete case), we end up at the limit with a very regular function: in fact, the probability of $Z_n$ being a precise value is 0, since a single point has zero measure.\r\nAnyway, I wondered how you could use the characteristic function (which is an inverse fourier transform) without requiring that moments of greater than second order must be finite: for instance, the page [url]http://mathworld.wolfram.com/CentralLimitTheorem.html[/url]. It is used the fact that the distribution of the variable $s = \\sum\\limits_{i = 1}^N {x_i }$, $f_N (s)$, has as its characteristic function $E_N \\left[ {e^{2\\pi ifx} } \\right] = \\left[ {E\\left[ {e^{2\\pi ifx} } \\right]} \\right]^N$. Then a substitution of variables, taylor series etc are used to reach the limit of the characteristic of a normal distribution. However, we suppose that ${E\\left[ {e^{2\\pi ifx} } \\right]}$ exists (at least for certain values of f). But we know that:\r\n$E\\left[ {e^{2\\pi ifx} } \\right] = \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)e^{2\\pi ifx} dx = } \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)\\left( {\\sum\\limits_{k = 0}^\\infty {\\frac{{\\left( {2\\pi if} \\right)^k }}{{k!}}x^k } } \\right)dx = \\sum\\limits_{k = 0}^\\infty {\\frac{{\\left( {2\\pi if} \\right)^k }}{{k!}}} } \\int\\limits_{ - \\infty }^{ + \\infty } {f(x)x^k dx = } \\sum\\limits_{k = 0}^\\infty {\\frac{{\\left( {2\\pi if} \\right)^k }}{{k!}}} E\\left[ {x^k } \\right]$\r\n\r\nNow I can't see how it can be possible that the characteristic of f exists when f has moments of finite order that are not finite. I suspect I might be making a basic calculus mistake in my question, then could you please show me where I'm wrong?\r\n\r\nI can see anyway that:\r\n\r\n$\\left\\| {\\int\\limits_{ - \\infty }^{ + \\infty } {f(x)e^{2\\pi ifx} dx} } \\right\\| \\le \\int\\limits_{ - \\infty }^{ + \\infty } {\\left\\| {f(x)e^{2\\pi ifx} } \\right\\|dx} = \\int\\limits_{ - \\infty }^{ + \\infty } {\\left| {f(x)} \\right|dx} = 1$\r\n\r\nAnyway it is possible that, although finite in modulus, the limit might not exist because of the oscillating properties of the imaginary exponential.\r\n\r\nI tried to see if we could find a characteristic function of the cumulative probability function, but it turned useless, since erf(z) has no characteristic function. In fact, we can see that given a probability distribution function f(x) and the cumulative probability function $g(x) = \\int\\limits_{ - \\infty }^x {f(\\tau )d\\tau }$, if f has a characteristic function then g hasn't, and if g has a characteristic function then f hasn't. I'll prove it in the following lines, but notice the consequence that since we know that a normal distribution has a well known characteristic function, then erf(z) hasn't.\r\n$\\int {g(x)e^{sx} dx = \\frac{1}{s}} g(x)e^{sx} - \\frac{1}{s}\\int {\\frac{{dg}}{{dx}}e^{sx} dx} = \\frac{1}{s}g(x)e^{sx} - \\frac{1}{s}\\int {f(x)e^{sx} dx}$\r\n\r\n$\\int\\limits_{ - \\infty }^{ + \\infty } {g(x)e^{sx} } dx = \\frac{1}{s}\\mathop {\\lim }\\limits_{x \\to + \\infty } \\left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \\right] - \\frac{1}{s}\\int\\limits_{ - \\infty }^{ + \\infty } {f(x)e^{sx} dx} = \\frac{1}{s}\\mathop {\\lim }\\limits_{x \\to + \\infty } \\left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \\right] - \\frac{1}{s}E\\left[ {e^{sx} } \\right]$\r\n\r\nNow, if we know that $E\\left[ {e^{sx} } \\right]$ exists, in order for the characteristic of g(x) to exist, it is necessary that the limit $\\mathop {\\lim }\\limits_{x \\to + \\infty } \\left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \\right]$ exists.\r\n\r\n$s = 2\\pi if$\r\n\r\n$\\mathop {\\lim }\\limits_{x \\to + \\infty } \\left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \\right] = \\mathop {\\lim }\\limits_{x \\to + \\infty } e^{ - sx} \\left[ {g(x)e^{2sx} - g( - x)} \\right] = \\mathop {\\lim }\\limits_{x \\to + \\infty } e^{ - sx} \\mathop {\\lim }\\limits_{x \\to + \\infty } \\left[ {g(x)e^{2sx} - g( - x)} \\right]$\r\n\r\nTherefore, since \r\n$\\begin{array}{l} \\mathop {\\lim }\\limits_{x \\to + \\infty } e^{ - 2\\pi ifx} = ? \\\\ \\mathop {\\lim }\\limits_{x \\to + \\infty } \\left\\| {e^{ - 2\\pi ifx} } \\right\\| = 1 \\\\ \\end{array}$\r\n\r\nIt is necessary that $\\mathop {\\lim }\\limits_{x \\to + \\infty } \\left[ {g(x)e^{2sx} - g( - x)} \\right] = 0$\r\n\r\nBut we can see that this limit doesn't exist:\r\n$\\mathop {\\lim }\\limits_{x \\to + \\infty } g(x)e^{2sx} - \\mathop {\\lim g( - x) = }\\limits_{x \\to + \\infty } \\mathop {\\lim }\\limits_{x \\to + \\infty } g(x)e^{2sx} = \\left[ {\\mathop {\\lim }\\limits_{x \\to + \\infty } g(x)} \\right]\\mathop {\\lim }\\limits_{x \\to + \\infty } e^{2sx} = \\mathop {\\lim }\\limits_{x \\to + \\infty } e^{2sx} = ?$\r\n\r\nsince $s = 2\\pi if$", "Solution_3": "I don't have time right now to reply to all of this, and it may be a few days until I get back to it. There are some other users who could comment on this, and I hope one of them does.\r\n\r\nA couple of quick points:\r\n\r\n[quote]First of all, the case of discrete distributions can be easily treated using Dirac's delta functions.[/quote]\nBut the point wasn't about being able to express discrete distributions (which is easy enough); the point was about the mode of convergence. When you say this sequence of distributions converges to the standard normal, exactly what sense of convergence do you mean?\n\n\n[quote]Now I can't see how it can be possible that the characteristic function of f exists when f has moments of finite order that are not finite.[/quote]\r\nThe characteristic function does not have to be infinitely differentiable and does not have to be represented by a power series. For instance, we could take the distribution whose density is $\\frac1{\\pi(x^2+1)}.$ This distribution doesn't even have a first moment - its expectation is undefined. But it has a perfectly well-defined characteristic function of $e^{-|t|}.$ (Or, putting $2\\pi$ in the exponent as you seem to be doing, $e^{-2\\pi |t|}.)$ Of course that characteristic function isn't even differentiable at zero.\r\n\r\nThe manipulations you refer to in the proof of the CLT do not require the characteristic function to have a Taylor series - just a 2nd order Taylor polynomial with some kind of control of the error. And if the distribution has a finite 2nd moment, its characteristic function will be $C^2.$ That's the standard Fourier transform tradoff between smoothness on one side and rate of decay on the other.", "Solution_4": "I spent some time on the problem, I tried to put it in a different way, but ended up using Laplace/Fourier transforms again.\r\nGiven a probability distribution $f(v)$ of finite mean $\\mu$ and variance $\\sigma ^2$, the problem of determining the distribution of the variable $x = \\sum\\limits_{i = 1}^N {\\frac{{\\left( {v_i - \\mu } \\right)}}{{\\sqrt n \\sigma }}}$, $f_n (x)$, is similar of determining the probability distribution of the position of a particle at a certain time, when it changes velocity at regular intervals $\\Delta t_n = \\frac{1}{{\\sqrt n \\sigma }}$, and where the velocity is a random variable of distribution $f(v)$, and finally given the fact that the particle is in x=0 for t=0.\r\nTo simplify everything, let us suppose that $\\mu=0$.\r\n\r\nThe relationship we're interested in is:\r\n$g\\left( {(n + 1)\\Delta t_n ,x} \\right) = \\int\\limits_{ - \\infty }^{ + \\infty } {g(n\\Delta t_n ,x - v\\Delta t_n )} f(v)dv$\r\n\r\nNow it is quite delicate to treat this problem. Let $g\\left( {n\\Delta t_n ,x} \\right)$ be the solution to this equation, with a certain dependance from $f(v)$. We have to show that for n that approaches infinity, the solution depends only from the first and second moment of f(v), and is a normal distribution. Maybe we could take out a differential equation that has as its solution the normal distribution. However it would be easy to make mistakes in calculating the limits, therefore transforms are a very practical solution.\r\n\\[ \\begin{array}{l} g\\left( {(n + 1)\\Delta t_n ,x} \\right) = \\int\\limits_{ - \\infty }^{ + \\infty } {g(n\\Delta t_n ,x - v\\Delta t_n )} f(v)dv \\\\ g\\left( {(n + 1)\\Delta t_n ,x} \\right) = \\frac{1}{{\\Delta t_n }}\\left[ {\\int\\limits_{ - \\infty }^{ + \\infty } {g(n\\Delta t_n ,x - z)} f\\left( {\\frac{z}{{\\Delta t_n }}} \\right)dz} \\right] \\\\ \\\\ T[g\\left( {n\\Delta t_n ,x} \\right)] = G(n\\Delta t_n ,s) \\\\ T[f(z)] = F(s) \\\\ T[f(\\lambda z)] = \\frac{1}{\\lambda }F\\left( {\\frac{s}{\\lambda }} \\right) \\\\ G\\left( {(n + 1)\\Delta t_n ,s} \\right) = G\\left( {n\\Delta t_n ,s} \\right)\\left( {\\frac{1}{{\\Delta t_n }}T\\left[ {f\\left( {\\frac{z}{{\\Delta t_n }}} \\right)} \\right]} \\right) \\\\ G\\left( {(n + 1)\\Delta t_n ,s} \\right) = G\\left( {n\\Delta t_n ,s} \\right)F\\left( {\\Delta t_n s} \\right) \\\\ \\end{array} \\]\r\n\r\nAnd finally we obtain: $G(n\\Delta t_n ,s) = G(0,s)\\left[ {F\\left( {\\Delta t_n s} \\right)} \\right]^n$\r\n\r\nNow we can find the transform of our final solution if we calculate the limit:\r\n$\\mathop {\\lim }\\limits_{n \\to \\infty } \\left[ {F\\left( {\\Delta t_n s} \\right)} \\right]^n$\r\n\r\n$\\begin{array}{l} \\mathop {\\lim }\\limits_{n \\to \\infty } s\\Delta t_n = 0 \\\\ \\bar s = s\\Delta t_n \\\\ \\mathop {\\lim }\\limits_{\\bar s \\to 0} F(\\bar s) = \\int\\limits_{ - \\infty }^{ + \\infty } {f(v)dv = 1} \\\\ \\\\ \\frac{{d^k F(\\bar s)}}{{d\\bar s^k }} = \\int\\limits_{ - \\infty }^{ + \\infty } {f(v)v^k e^{\\bar sv} dv} \\\\ \\\\ \\mathop {\\lim }\\limits_{\\bar s \\to 0} \\frac{{d^k F(\\bar s)}}{{d\\bar s^k }} = \\int\\limits_{ - \\infty }^{ + \\infty } {f(v)v^k dv} = M_k \\\\ \\end{array}$\r\n\r\nYou probably know how we end the demonstration, developing F in taylor series etc, so my real question is: until now we assumed that f(v) had a transform F(s). Now when we calculate the limit for n that approaches infinity, we are using the transform in a region where s approaches 0. If we know that f(v) has overall integral 1, we know that the transform is defined for s=0, and if we know that has finite first and second moment, we know that the first and second derivatives of F(s) in 0 would be finite. Is this enough to know that F(s) is defined for at least a small region around s=0 ?", "Solution_5": "Another thing: there's sort of a weak point, about the convergence as you said. It is quite difficult to speak of convergence of a irregular function, since we could have dirac's deltas in the discrete probability case. Anyway we have that the transforms of such irregular function are much more regular, so another question of mine is: once we have a very good convergence of the transforms, what kind of convergence we have for the antitransforms?\r\nI'm sorry I couldn't manage to put the problem in terms of the cumulative probability functions (that are in fact much more regular than the probabilty distributions)." } { "Tag": [], "Problem": "What is the number of distinct triangles in the diagram?\n\n[asy]size(3cm,3cm);\n\ndraw((-1,1)--(1,1)--(1,-1)--(-1,-1)--(-1,1));\n\ndraw((0,1)--(1,0)--(0,-1)--(-1,0)--(0,1));\n\ndraw((-1,1)--(1,-1));\ndraw((-1,-1)--(1,1));[/asy]", "Solution_1": "I counted $ 20$. Did anyone count more?", "Solution_2": "I got $ \\boxed{20}$ as well.", "Solution_3": "Solution:\n\nThe smallest triangles: 8\nGroup two of the smallest triangles together: 4\nGroup two of the smallest triangles together combined with a square: 4\nGroup four of the smallest triangles together with two squares: 4\n\n$ 8 + 4 + 4 + 4 = \\boxed{20}$", "Solution_4": "Let the smallest possible triangle in the picture have area/size $t$, and the smallest possible square have area/size $s$. Now, we count:\n\nArea $t$ $\\implies$ $8\\triangle$s\nArea $2t$ $\\implies$ $4\\triangle$s\nArea $2t+s$ $\\implies$ $4\\triangle$s\nArea $4t+s$ $\\implies$ $4\\triangle$s\n\n$$8+4\\cdot3 = \\boxed{20}$$" } { "Tag": [ "geometry" ], "Problem": "Find the area of triangle ABC if a=40ft b=25ft and $ x \\equal{} y \\equal{} z \\equal{} 0$\r\nIf $ x,y,z \\neq0$, we multiplicates ever all together thigh is three equation \r\n$ (1 \\plus{} x^2)(1 \\plus{} y^2)(1 \\plus{} z^2) \\equal{} 8xyz$\r\nTake Cauchy inequality We has equal bar happen\r\n.............\r\n\r\n$ (x;y;z) \\equal{} (0;0;0); (1;1;1)$", "Solution_6": "Thank you.", "Solution_7": "It is a very well known problem similar problem is given at Canadian Math Olympiad - 1996.\r\nThe same problem is given at Bulgarian Math Olympiad 1977 or 1978 as well as Moskow Math Olympiad.", "Solution_8": "yes! Thank." } { "Tag": [ "probability" ], "Problem": "The table below represents the male population of students attending\nWalker Middle School. One boy is selected at random from the male\nstudents. What is the probability that he is in seventh grade? Express\nyour answer as a common fraction.\n\\begin{tabular}{|l|c|c|c|} \\hline\n& 7th Grade & 8th Grade & 9th Grade \\\\ \\hline\nMales & 25 & 22 & 23 \\\\ \\hline\n\\end{tabular}", "Solution_1": "The total number of boys is 25+22+23=70. Only 25 of these are 7th grade boys, so the answer is 25/70 or 5/14.", "Solution_2": "$ \\begin{tabular}{|l|c|c|c|} \\hline\r\n& 7th Grade & 8th Grade & 9th Grade \\\\ \\hline\r\nMales & 25 & 22 & 23 \\\\ \\hline\r\n\\end{tabular}$" } { "Tag": [ "function", "inequalities", "inequalities proposed" ], "Problem": "Let $x_1,x_2,\\dots , x_n$ be nonnegative reals such that\r\n\\[x_1x_2\\cdots x_n+(n-1) = x_1+x_2+\\cdots + x_n.\\]\r\nProve that $\\text{max}\\{x_1,x_2,\\dots , x_n\\} \\geq 1$.", "Solution_1": "The function $x_1+x_2+\\ldots+x_n-(n-1)-x_1x_2\\ldots x_n$ is linear in $x_1,\\ldots,x_n$.\r\n\r\nHence it takes its maximum over $[0,1]^n$ at the vertices of $[0,1]^n$ (include $0$ as an admissible value for now)\r\n\r\nIf $x_1=1$, the function becomes $x_2+\\ldots+x_n-(n-2)-x_2\\ldots x_n$, and we can apply induction.\r\n\r\nIf $x_1=0$, the function becomes $x_2+\\ldots+x_n-(n-1)$, which means that all of the remaining $x_i$s must be $1$. In either case, the inequality has a maximum value of $0$, and since $(0,1]^n\\subseteq [0,1]^n$, we're done.\r\n\r\nIt remains to show that equality is only achieved over $(0,1]^n$ if all of the variables are $1$. If one of the variables $x_j$ is not $1$, we may increase the value of the function by changing $x_j$ to one, since\r\n\r\n$x_j\\left(1-\\prod_{i\\not=j}x_i\\right)\\geq 0$", "Solution_2": "Cool... I was thinking more along the lines of... say $y_i = x_1x_2\\cdots x_i$ so assume $x_i < 1$ and hence $y_i <1$. We have\r\n\r\n$x_1x_2\\cdots x_n + (n-1) - (x_1+x_2+\\cdots +x_n) = \\sum_{i=1}^{n-1} (1+y_{i+1}-y_i-x_{i+1})$\r\n$ = \\sum_{i=1}^{n-1}(1-y_i)(1-x_{i+1}) > 0$,\r\n\r\ncontradiction!" } { "Tag": [ "SFFT", "special factorizations" ], "Problem": "2. You are given a piece of paper. You can cut the paper into 8 or 12 pieces. Then you can do so for any of the new pieces or let them uncut and so on. \r\nCan you get exactly 60 pieces? Show that you can get every number of pieces greater than 60.", "Solution_1": "[hide]Whenever you cut up a piece of paper, you add either $12$ or $8$ new pieces of paper, but lose the one you originally had (the one that's now cut up). So, you can only add $11$ or $7$ pieces of paper every time. Since you start with $1$ piece, all numbers that can be reached are of the form $1+7x+11y$. By the chicken mcnugget theorem, the number $1+7\\cdot11-7-11=60$ is the largest number that cannot be reached. The answer follows.[/hide]", "Solution_2": "[quote=\"cosinator\"][hide]Whenever you cut up a piece of paper, you add either $12$ or $8$ new pieces of paper, but lose the one you originally had (the one that's now cut up). So, you can only add $11$ or $7$ pieces of paper every time. Since you start with $1$ piece, all numbers that can be reached are of the form $1+7x+11y$. By the chicken mcnugget theorem, the number $1+7\\cdot11-7-11=60$ is the largest number that cannot be reached. The answer follows.[/hide][/quote]\r\n\r\nI don't think you should directly cite the chicken mcnugget theorem since it trivializes the problem.", "Solution_3": "it was pretty trivial to begin with\r\n\r\nand you can be like 1 - 7 - 11 + 7*11 = (7 - 1)(11 - 1) = 60 and cite simon's favorite factoring trick\r\n\r\nfun fun fun" } { "Tag": [ "function", "linear algebra" ], "Problem": "How would I proof the following:\r\nsuppose f: X--->Y is a function and A, B are subsets of Y\r\na) show that f inverse(A U B) = f inverse (A) U f inverse (B)\r\nb) show that f inverse (A \u2229 B)= f inverse (A) \u2229 f inverse (B).\r\n\r\n\r\n\r\nanother one:\r\n suppose f: X--->Y , A C X, and B C Y\r\na) show that A C f inverse(f(A)). give an example in which the containment is proper.\r\nb) show that f(f inverse(B)) C B. give an example in which the conatainment is proper.", "Solution_1": "These aren't linear algebra questions, of course - they're part of what I sometimes informally call \"functionspeak,\" which is the elementary background of many different parts of mathematics. I'd move this if I could think of a better place to put it - but i don't know where that would be.\r\n\r\nYou just have to consider the definitions, and keep asking what it would mean for a particular point to belong to a particular set.\r\n\r\nFor instance, $x\\in f^{-1}(A)$ if and only if $f(x)\\in A.$\r\n\r\nThus $x\\in f^{-1}(A\\cup B)$ iff $f(x)\\in A\\cup B$ iff $f(x)\\in A$ or $f(x)\\in B.$ \r\n\r\nCan you finish that thought?\r\n\r\nAs for the examples for your second pair of questions:\r\n\r\n[hide=\"Hint for (a):\"]Consider an example of a function that is not one-to-one.[/hide]\n[hide=\"Hint for (b):\"]Consider an example of a function that is not onto (does not map $X$ onto $Y$).[/hide]" } { "Tag": [ "function", "floor function" ], "Problem": "Find all non-decreasing functions $ f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that $ f(x\\plus{}f(y))\\equal{}f(f(x))\\plus{}f(y)$ for all reals $ x$ and $ y$.", "Solution_1": "Nobody? :D", "Solution_2": "Let x,y=0. Then, f(f(0)) = f(f(0)) + f(0) or f(0) = 0. \r\n\r\nNow let y = 0. We have f(x + f(0)) = f(f(x)) + f(0) or f(x) = f(f(x)).\r\n\r\nf(x) = x is the only function that satisfies this. [/code]", "Solution_3": "or f(x) = 0", "Solution_4": "... or $ f(x) \\equal{} \\lfloor x \\rfloor$, or many other functions. This solution is not complete by any means.", "Solution_5": "Is there like a way of proving that functions with ceilings or floors work\r\n\r\nLike that seems kind of hard.", "Solution_6": "I think beezees has established that f(f(x)) = f(x). Here is a demonstration of how varied the solutions to that equation can be:\r\n\r\n[hide]\n\nLet S be the finite set of points in the range of f. f(f(x))=f(x) means that we have fixed points at (z,z) for every z in S.... these points lie along the line y = x.\n\nConsider two consecutive elements of S, say z1 and z2. If there lies an x between these two points, then f(x) must be either z1 or z2 to preserve the non-decreasing criterion. Hence, we may have one of four schemes for each such interval:\n\n$ \\forall x : \\; z_1 < x < z_2, \\; f(x) \\equal{} z_1$\n$ \\forall x : \\; z_1 < x < z_2, \\; f(x) \\equal{} z_2$\n$ \\exists m : \\; z_1 < m < z_2 \\; : \\; \\forall x : \\; z_1 < x < m, \\; f(x) \\equal{} z_1, \\; \\forall x : \\; m < x \\leq z_2, \\; f(x) \\equal{} z_2$\n$ \\exists m : \\; z_1 < m < z_2 \\; : \\; \\forall x : \\; z_1 < x \\leq m, \\; f(x) \\equal{} z_1, \\; \\forall x : \\; m < x < z_2, \\; f(x) \\equal{} z_2$\n\nLet's label these schemes 1, 2, 3, 4, and apply them to each finite-length interval between fixed-points (there are a countable number).\n\nWe can thus consider the set S of fixed points, an ordered list of m values, and an ordered list of scheme numbers, and it will uniquely define a function that satisfies f(f(x)) = f(x).\n\n[/hide]\r\n\r\nNow, you can look at this set of functions and determine which solve the given equation" } { "Tag": [ "geometry", "rectangle", "Pythagorean Theorem" ], "Problem": "In a circle of 20 cm of diameter, the distance from the center to a chord AB is 6 cm. Find the length of the chord AB.\r\n\r\nThis problem appears in the test to enter in the university here in Chile (equivalent to SAT in USA)\r\nSo, by the Chord theorem we have : 16*4 = X*Y\r\nHence, we have one equation and we need another.\r\nSo, the answer is trivial if we \u201cassume\u201d that \u201cthe line from the center of the circle to the chord is \u2018perpendicular\u2019 to the chord\u201c.\r\nIn this case that segment also bisects the chord. So X = Y ==> 16*4 = X*X\r\nThen, X= 8 => chord AB = 16\r\n\r\nHowever, my opinion is that this problem is wrong because there isn\u2019t enough information.\r\nWhy ? in which part of the \u201cannouncement\u201d of the problem said \u201cexplicitly\u201d that the chord is \u2018perpendicular\u2019 to the line (in this case the radius) ?\r\nSo, this is my question, I means: Who, where? in the Math community defined (as a theorem) that each time that we read : \u201cThe distance of a point (p) to a line, by definition is always the length of the perpendicular segment from the point \u2018p\u2019 to that line.\u201c Who defined that? \r\nCould some one please give me a reference book or valid link in the internet that accept this definition as the \u201conly convention accepted in the math world\u201d. So, this is the only true assumption that we always need to do in this type of problem.\r\nThanks,", "Solution_1": "I don't have a good link, but here's the basic idea: The distance between a point and a line is defined as the length of the [i]shortest[/i] segment that connects the point and the line, which is perpendicular to the line. Otherwise, there would be infinitely many possible values for the distance. That would be very inconvenient. I hope this makes sense...", "Solution_2": "You mention: \u201c\u2026The distance between a point and a line is defined as the length of the shortest segment that connects the point and the line, which is perpendicular to the line.\u201d \r\n\r\nWell, I already know that, so you repeat the same. My question is : where is a valid reference, link etc that stablish that \u201cDEFINITION\u201d as a valid theorem or the only valid assumption in the math community. \r\n\r\nAlso you mention : \u201c.. Otherwise, there would be infinitely many possible values for the distance. That would be very inconvenient\u2026.\u201d\r\n\r\nIf I say : The area of a rectangle is 80. Find the \u201clength\u201d of the rectangle.\r\nUsing your logic or resoning, you will conclude the same \u201c\u2026. would be infinitely many possible values for the Length. That would be very inconvenient\u2026.\u201d\r\nObviously, The problem is NOT full determinate ( have another degree of freedom ) so you need to establish another equation. (see the attached document)", "Solution_3": "[quote=\"gva\"]You mention: \u201c\u2026The distance between a point and a line is defined as the length of the shortest segment that connects the point and the line, which is perpendicular to the line.\u201d \n\nWell, I already know that, so you repeat the same. My question is : where is a valid reference, link etc that stablish that \u201cDEFINITION\u201d as a valid theorem or the only valid assumption in the math community. \n\nAlso you mention : \u201c.. Otherwise, there would be infinitely many possible values for the distance. That would be very inconvenient\u2026.\u201d\n\nIf I say : The area of a rectangle is 80. Find the \u201clength\u201d of the rectangle.\nUsing your logic or resoning, you will conclude the same \u201c\u2026. would be infinitely many possible values for the Length. That would be very inconvenient\u2026.\u201d\nObviously, The problem is NOT full determinate ( have another degree of freedom ) so you need to establish another equation. (see the attached document)[/quote]\r\n\r\nwow, you wrote a lot in the pdf...but as frt said, the distance from a point to a line is defined as the shortest distance...which happens to be the perpendicular distance\r\n\r\nthe proof is pretty simple, it is a consequence of the pythagorean theorem...here is a proof anyway\r\n\r\nsay we have line $m$, point $P$ not on that line, $P'$ on line $m$ such that $P'P$ is perpendicular to $m$, and point $Q$ on $m$, distinct from $P'$,\r\nthen we have $P'M^2+P'P^2=MP^2$, but $P'M>0$ since those are distinct points, so we get\r\n$PM>PP'$\r\nso basically, the minimum distance occurs when you do the perpendicular distance...\r\n\r\nas for a direct reference, look in a geometry textbook...this is kind of a standard knowledge thing...here is a link...\r\n\r\n[url]http://softsurfer.com/Archive/algorithm_0102/algorithm_0102.htm#Distance%20to%20an%20Infinite%20Line[/url]\r\n\r\nas for your length of a rectangle, that is a different problem...this problem is fully determined since we have that the perpendicular distance is 6, and a perpendicular line from the center of a circle bisects the chord, the rest is just pythagorean theorem", "Solution_4": "[quote=\"gva\"]If I say : The area of a rectangle is 80. Find the \u201clength\u201d of the rectangle.[/quote]\r\nWe would never see this problem, unless they're asking for $0 4$\r\n$ 1>\\{x\\}\\geq\\frac{3}{[x]}$\r\nHave many $ x$ satisfy.", "Solution_2": "[quote=\"TTsphn\"]Easy to check $ x > 4$\n$ 1 >\\{x\\}\\geq\\frac{3}{[x]}$\nHave many $ x$ satisfy.[/quote]\r\n\r\nYour answer is : $ \\forall x>4$, TTsphn ? :( . Try $ x\\equal{}4,0001$", "Solution_3": "All $ x>4$ satisfy\r\n$ \\{x\\}\\geq\\frac{3}{[x]}$\r\nI don't understand what is problem.\r\nWe can find out many number as above.", "Solution_4": "I agree with you TTsphn :wink:", "Solution_5": "x>=n+3+1/n+3\r\nn-natural number" } { "Tag": [ "inequalities", "SFFT", "special factorizations" ], "Problem": "These were on a NEML last week:\r\n\r\n1. 20 people are lined up in a row. Every minute, the people in positions 10 and 20 move up to positions 1 and 2, respectively, and the whole line shifts to make room for them. After how many minutes will the two people originally in positions 1 and 2 return to this position?\r\n\r\n2. Solve for (a, b, c, d, e):\r\n\r\nabcde = a + b + c + d + e", "Solution_1": "Umm...the first one, I got 20 minutes. And on the second one, I got (0,0,0,0,0) Correct? Wait, are answers allowed to be trivial?", "Solution_2": "For the second one(2, 2, 2, 1, 1) would work also (I'm guessing abcde is a*b*c*d*e rather than just one five-digit number because then there'll be 10 answers or more)", "Solution_3": "Yeah, 20 minutes for 1. For 2, there is more than one solution.", "Solution_4": "for 2, I got 8, 9 , and 10 as possible solutions. I broke it up according to the number of ones, i think. There can't be 4 ones, if there are 3 ones, we use Simon's favorite factoring trick, and if there are 2 ones, there is only one possible solution (which I think tare gave). This is a close to my original thinking as I can remember.", "Solution_5": "For the second one, there are definitely infinitely many solutions, unless you give me a restriction on a, b, c, d, and e.", "Solution_6": "i think you could solve the second one with AM-GM, knowing that the product is greater than 5 and less than whatever. by the way, the second one wanted all values for the product if a,b,c,d,e are pos integers JBL.", "Solution_7": "Eh, yeah, don't have the test in front of me, #2 might be for positive integers only.", "Solution_8": "Well, I don't think AM-GM really tells you very much in that case. If it's restricted to positive integers, try setting first 4 of them equal 1, then 3, then 2, then 1, then none, and see what happens in each of the cases.", "Solution_9": "yah... that's what i did, except it is along the lines of guess and check... i got them both... but missed two easy ones... taught me to always read the questions carefully...", "Solution_10": "3,3,1,1,1 would also work for # 2", "Solution_11": "[quote]If it's restricted to positive integers, try setting first 4 of them equal 1, then 3, then 2, then 1, then none, and see what happens in each of the cases.[/quote]\r\n\r\nBut why would you do that?\r\n\r\nAnd for #1, does anyone have a nice solution that does not involve listing out the positions for each of the first 20 min (which is more or less what I did not the actual test)?", "Solution_12": "well just find the pattern of the changing of the positions. i beleive it changes by 1 until like the 11th time then by 2 or something. i worked it out some time back, so i dont really remember the exact changes", "Solution_13": "Ok... at first one clang moves the person back 2 spots... so after the first clang, he is in the 3rd spot... and he goes like that until after the 5th clang, he is in the 11th position. Now after 1 clang, he only moves back one, because the 10th is already ahead of him, and only 20th moves in front, so after the 14th clang, he is in the 20th spot. On the 15th clang, he moves back to the 2nd spot. Again, after each clang, he moves back 2 spots like at the beginning. so after the 19th clang, he'll be in spot 10. So the 20th clang will bring him back to 1st position.", "Solution_14": "[quote=\"Fierytycoon\"][quote]If it's restricted to positive integers, try setting first 4 of them equal 1, then 3, then 2, then 1, then none, and see what happens in each of the cases.[/quote]\n\nBut why would you do that?[/quote]\r\n\r\nBecause 1 is clearly a special number for this problem -- it only changes one of the two sides.", "Solution_15": "For #1, ok, I guess I sorta did the same thing you all did.\r\n\r\nFor #2, \r\n\r\n[quote]Because 1 is clearly a special number for this problem -- it only changes one of the two sides.[/quote]\r\n\r\nSo how can you prove that there can't be a solution w/o at least one of a,b,c,d,e being equal to 1?", "Solution_16": "try 2*2*2*2*2... its the smallest product that doesn't contain a 1 in it, but its sum isn't big enough... so there's not a combination w/0 1.", "Solution_17": "yeah Chinaboi got it. For each case (according to number of ones) try the rest all equal to 2 (that's the smallest product) and see whether there are any solutions.", "Solution_18": "Ah, ok, that was simple.", "Solution_19": "did anyone read the solution that they gave for #2? I didnt get to read it fully, but i saw they had some proof, im not exactly sure what it was though.", "Solution_20": "no, i just glanced at the answer page, and never saw it again...", "Solution_21": "for number 2, I think it meant all the numbers must be different or else it would probably be written like this aaaaa=a+a+a+a+a", "Solution_22": "[quote]2. Solve for (a, b, c, d, e): \n\nabcde = a + b + c + d + e[/quote]\r\n I think a,b,c,d,e are integers. We can assume that e >= a,b,c,d. Then abcde=a+b+c+d+e <= 5e.\r\nSo abcd <= 5. From the last inequality we will find solutions. :)", "Solution_23": "e[quote]2. Solve for (a, b, c, d, e): \n\nabcde = a + b + c + d + e[/quote]\r\nOoops, it is not given that a,b,c,d,e are positive integers. So my first solution it is not full.\r\n 1) if a,b,c,d,e are positive it is easy to find solutions.\r\n 2) if a,b,c,d,e are negative there are no any problem.\r\n 3) If some of a,b,c,d,e are negative and some are positive.\r\n In the third case : We can assume that a >= b,c,d >= e. So 5a >= a+b+c+d+e=abcde >= 5e.\r\nThen bcde <= 5 <= abcd. If a is negative then b,c,d,e are negative. So a is nonnegative and bcd is nonnegative , too.\r\n Since bcd >= 0. If e >= 0 then a,b,c,d >= e >= 0. So e <= 0. Let's denote e=-k then k >= 0.\r\nWe have a+b+c+d-k+abcdk=a+b+c+d+k(abcd-1)=0. If a,b,c,d >= 0 there are no problem. Since bcd >= 0\r\nonly two of them can be negative, say b and c. Let b=-m, c=-n. Then a+d-m-n-k+abmnk=0\r\nFor a >= 3 a+d+abmnk >= 3+d+ 3bnmk >= 3+d+ bnmk+bnmk+bnmk>m+n+k. We have to check a=2,a=1,a=0\r\nI think this solution is true, but I don't like it myself :blush: . Is there another solution?", "Solution_24": "Is my solution true?" } { "Tag": [ "Alcumus", "probability", "videos", "Support" ], "Problem": "Maybe Alcumus can provide hints to problems? I currently do not own Intro. to Counting and Probability (although I ordered it over a week ago) so this would be very helpful with most of the problems.", "Solution_1": "me agrees. This way the woreser ppl( like me) don't have to give up as much.", "Solution_2": "I volunteer to help write hints, if such is decided a good idea.", "Solution_3": "I volunteer too, to hint on the C&P problems. And watch the videos currently available if you want help. Or if you are in the diagnostic stage, then you'll have to wait." } { "Tag": [ "trigonometry" ], "Problem": "FInd the relationship between arcsin (cos (arcsin x)) and arccos (sin (arccos x))", "Solution_1": "in other words we have to link them in some way....ne ideas?", "Solution_2": "Let $\\theta=\\arcsin x$. So then\r\n\r\n\\begin{eqnarray*} \\arcsin(\\cos(\\arcsin x)) &=& \\arcsin(\\cos\\theta)\\\\ &=& \\arcsin(\\sin(90^\\circ-\\theta))\\\\ &=& 90^\\circ-\\theta \\end{eqnarray*}\r\n\r\nand \r\n\r\n\\begin{eqnarray*} \\arccos(\\sin(\\arccos x)) &=& \\arccos(\\sin(90-\\theta))\\\\ &=& \\arccos(\\cos \\theta)\\\\ &=& \\theta \\end{eqnarray*}\r\n\r\nSo they add to $90^\\circ.$", "Solution_3": "Try with complementar angles: i suppose the sum of the two terms is $\\frac{\\pi}{2}$" } { "Tag": [ "trigonometry", "function", "Support", "geometry", "geometric transformation", "homothety", "geometry proposed" ], "Problem": "Let $ABCD$ be a convex quadrilateral with its sides tangent to a circle $(O)$. Let $(O_1, r_1)$ and $(O_3, r_3)$ be circles that are different from $(O)$ and respectively tangent to segment $AB$, ray $CB$, ray $DA$ and sement $CD$, ray $BC$, ray $AD$. Similarly, we define $(O_2, r_2)$ and $(O_4, r_4)$ be circles that are different from $(O)$ and respectively tangent to segment $DA$, ray $BA$, ray $CD$ and sement $BC$, ray $AB$, ray $CD$. Prove that $r_1.r_3 = r_2.r_4$.", "Solution_1": "i just did copy/paste so try inderestandig serbian. It was in some journal and it's well known Sangaku problem\r\n\r\n\r\n\r\n4. Ako su poluprecnici spolja pripisanih kruznica tangentnog\r\ncetvorougla $ABCD$ nad strancima $AB,BC,CD,DA$ redom\r\n$r_1,r_2,r_3,r_4$, dokazati da je $r_1 \\cdot r_3=r_2 \\cdot r_4$\\\\\r\n\r\n\r\nResenje:\\\\\r\n\r\nNeka je tacka $O_1$ centar pripisane kruznice nad stranicom $AB$ i\r\nneka je podnozje normale iz $O_1$ na $AB$ tacka $M$. Onda je\r\n$r_1=O_1M$. Neka su unutrasnji uglovi cetvorougla $ABCD$ redom\r\n$\\alpha, \\beta, \\gamma, \\delta$, a duzine stranica redom\r\n$a,b,c,d$. Onda iz trougla $AO_1M$ imamo ${r_1 \\over \\cos{{\\alpha \\over 2}}}=AO_1$. Slicno je i $O_1B={r_1 \\over \\cos{{\\beta \\over 2}}}$. Izjednacavanjem povrsine trougla $ABO_1$ dobijamo\r\n$ar_1={AO_1\\cdot BO_1 \\cdot \\sin {{\\alpha + \\beta \\over 2}} \\over 2}$. Iz ovoga je $r_1={a \\cos{{\\alpha \\over 2 }}\\cdot \\cos{{\\beta \\over 2 }} \\over \\sin{{\\alpha+\\beta \\over 2}}}$.\\\\\r\n\r\nSada se slicno dobijaju i formule $r_2={b\\cdot \\sin{{\\beta+\\gamma \\over 2}} \\over \\cos{{\\beta \\over 2}}\\cdot \\cos{{\\gamma \\over 2}}}$, $r_3={c\\cdot \\sin{{\\gamma+\\delta \\over 2}} \\over \\cos{{\\gamma \\over 2}}\\cdot \\cos{{\\delta \\over 2}}}$ i\r\n$r_4={d\\cdot \\sin{{\\delta+\\alpha \\over 2}} \\over \\cos{{\\delta \\over 2}}\\cdot \\cos{{\\alpha \\over 2}}}$, pa se zadatak svodi na to\r\nda se dokaze\r\n\\begin{center}\r\n$ac \\cdot \\sin{{\\alpha+\\beta \\over 2}}\\cdot \\sin{{\\gamma+\\delta \\over 2}}=bd \\cdot \\sin{{\\beta+\\gamma \\over 2}} \\cdot \\sin{{\\alpha+\\delta \\over 2}}$ ($\\ast$)\r\n\\end{center}\\\\\r\n\r\n\r\n\r\nNeka je centar upisane kruznice u cetvorougao $ABCD$ tacka $O$ i\r\nneka upisana kruznica dodiruje stranice $AB,BC,CD,AD$ u tackama\r\n$M,N,P,Q$ redom. Neka je i poluprecnik upisane kruznice $r$. Onda\r\nje $a=AM+BM=r \\cot{{\\alpha \\over 2}}+r \\cot{{\\beta \\over 2}}=r{\\sin{\\alpha \\over 2}\\cdot \\cos{{\\beta \\over 2}}+\\sin{{\\beta \\over 2}}\\cdot \\cos{{\\alpha \\over 2}} \\over \\sin{{\\alpha \\over 2}}\\cdot \\sin{{\\beta \\over 2}}}=r{\\sin{{\\alpha+\\beta \\over 2}}\\over \\sin{{\\alpha \\over 2}}\\cdot \\sin{{\\beta \\over 2}}}$.\r\nSlicno se dobija i $b=r{\\sin{{\\beta+\\gamma \\over 2}}\\over \\sin{{\\beta \\over 2}}\\cdot \\sin{{\\gamma \\over 2}}}$,\r\n$c=r{\\sin{{\\gamma+\\delta \\over 2}}\\over \\sin{{\\gamma \\over 2}}\\cdot \\sin{{\\delta \\over 2}}}$ i $d=r{\\sin{{\\delta+\\alpha \\over 2}}\\over \\sin{{\\delta \\over 2}}\\cdot \\sin{{\\alpha \\over 2}}}$.\r\n\r\nSada jednostavnom proverom ( zamenjujuci ove izraze u ($\\ast$) )\r\ndobijamo da je ($\\ast$) tacno pa je tacna i njegova ekvivalencija,\r\nodnosno polazna pretpostacka $r_1\\cdot r_3=r_2 \\cdot r_4$, sto je\r\ni trebalo dokazati.", "Solution_2": "Dule-00, \r\n\r\n treegoner gave us a hint that it was a silly problem, \r\n\r\n so there was no need to get so serious and post solution in serbo-horvathian.\r\n\r\n So, next time please try to post in some silly language, like ancient babylonian.\r\n\r\n\r\n I've just found out that trigonometric functions were not available in 17th century\r\n \r\n Japan, the time of Sangaku(s). They struggled with geometric problems using \r\n\r\n medieval methods.\r\n \r\n\r\n Thank you.\r\n\r\n M.T.", "Solution_3": "Dule_00, although Armpist uses a callous language (with good reason), he is right. You must understand here can use only the English language and/or the U.M.L. with the its support LaTeX.\r\nI will use the universal mathematical language ([b]U.M.L.[/b]). Denote:\r\n$\\blacktriangleright AB=a,\\ BC=b,\\ CD=c,\\ DA=d;\\ A=2x,\\ B=2y,\\ C=2z,\\ D=2t$.\r\n$\\blacktriangleright$ the incircle $C(I,r)$ is tangent to the sides $AB$, $BC$, $CD$, $DA$ in the points $M$, $N$, $P$, $R$ respectively;\r\n$\\blacktriangleright$ the exincircles $C(I_1,r_1)$, $C(I_2,r_2)$, $C(I_3,r_3)$, $C(I_4,r_4)$ are tangent to the sides $AB$, $BC$, $CD$, $DA$ in the points $M'$, $N'$, $P'$, $R'$ respectively.\r\nFrom the well-known property result: $AM'=BM$, $BN'=CN$, $CP'=DP$, $DR'=AR$.\r\nThus, $AM'=r_1\\tan x,\\ BM=r\\cot y\\Longrightarrow r_1=r\\cot x\\cot y$.\r\nSimilarly, $r_2=r\\cot y\\cot z$, $r_3=r\\cot z\\cot t$ and $r_4=r\\cot t\\cot x$.\r\nTherefore, $r_1r_3=r_2r_4=r^2\\cot x\\cot y\\cot z\\cot t$.", "Solution_4": "The only thing armpist is right about is that there is no need to post a solution in Serbian, but that's only because there is no need to post a solution of any problem in any language whatsoever. I remember him quite well posting in Russian on an English forum ([url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=205&forum=DCForumID3#4]here[/url], message #6, Golland = armpist). As for the rest of his rattle, you would think that carving the solution in wood would make him content.\r\n\r\n \r\nLet $P \\equiv AB \\cap CD$, $Q \\equiv BC \\cap DA$. WLOG, let P lie on the opposite side of BC than A and let Q lie on the opposite side of CD than A. Let $T_1, T_2, T_3, T_4$ be the tangency points of $(O_1), (O_2), (O_3), (O_4)$ with AB, BC, CD, DA. The vertex A is the internal homothety center of $(O_4), (O_1)$ and the vertex C is the internal homothety center of $(O_2), (O_3)$. Hence, $\\frac{r_4}{r_1} = \\frac{AT_4}{AT_1},\\ \\frac{r_2}{r_3} = \\frac{CT_2}{CT_3}$. The convex quadrilateral ABCD and the concave quadrilateral APCQ are both tangential with an incircle, hence,\r\n\r\n$AB + CD = BC + DA,\\ \\ AP + CQ = QA + PC$\r\n\r\nStarting from $BC + CQ = BQ$, add $AP = AB + BP$ to both sides:\r\n\r\n$AP + BC + CQ = AB + BP + BQ$\r\n\r\nSubstitute $AP + CQ = QA + PC$ on the left:\r\n\r\n$QA + PC + BC = AB + BP + BQ$\r\n\r\nRearrange:\r\n\r\n$BC + CP - PB = AB + BQ - QA$\r\n\r\n$s_2 - PB = s_1 - QA$\r\n\r\nwhere $s_2, s_1$ are semiperimeters of the triangles $\\triangle BCP, \\triangle ABQ$. Since $(O_2)$ is an incircle of the $\\triangle BCP$ and $(O_1)$ is an excircle of the $\\triangle ABQ$,\r\n\r\n$CT_2 = AT_1$\r\n\r\nSimilarly, we obtain $CT_3 = AT_4$. Consequently,\r\n\r\n$\\frac{r_4}{r_1} = \\frac{AT_4}{AT_1} = \\frac{CT_3}{CT_2} = \\frac{r_3}{r_2}$.", "Solution_5": "Dear treegoner,\r\n\r\n\r\n What was the reason to call this problem 'very silly' ?\r\n\r\n What is your proof?\r\n\r\n\r\n\r\n T.Y.\r\n\r\n M.T.", "Solution_6": "I have no idea why I put it \"very silly\" :ninja: My proof is similar and based on the following lemma:\r\n\r\n[i]Lemma[/i] Suppose $ABC$ is a triangle. Then $\\frac{r_a}{r} = tg{\\frac{B}{2}}tg{\\frac{C}{2}}$." } { "Tag": [ "LaTeX", "induction", "strong induction", "number theory unsolved", "number theory" ], "Problem": "Please! prove quickly.... \r\nEvery integer y>=b can be expressed in the form a + x_i_1 +x_i_2+...+x_i_a, \r\nwhere 0<=a= for which x_i<= y < x_i + 1 . Using strong induction on i...if y=xi=1 such that x_i<= y-x_i < x_(j+1), and j=b can be expressed in the form a + x_i_1 +x_i_2+...+x_i_a, \nwhere 0<=a=0, equation x^3 + tx - 8 has the only positive real solution x(t). Calculate \\int_{0..7}[x(t)]^2dt.\r\n\r\nSource: VUMC 1997 ", "Solution_1": "Of course, the equation has at least one real root and it is obvious that it is positive. Of course, we cannot have two positive roots for a certain t. For x>0 define f(x)=-x^2+8/x. Then x(t)=f^(-1)(t) for all t>0 and the integral is integral from 0 to 7 from (f^(-1)(t))^2. Make the change of variable t=f(u) and it becomes integral from 2 to 1 from (u^2*f'(u))du and this one is 15,5." } { "Tag": [], "Problem": "This time of year always reminds me of Tolkien's The Lord of the Rings. It was november when I read the book for a first time and couple years later I was waiting for the movie. \r\nRight now I can't decide what to read:Hiperion by Den Simons Shards of Honor by Lois McMaster Bujold.\r\nWhat books you like to read? Do you have any recomendation for reading?", "Solution_1": "Have you ever read Enders Game? If not, then you are a crazy person!! It's by Oson Scott Card, and is extremely riviting. 10 out of 10! :10:", "Solution_2": "I agree. Then read Enders Shadow, which is just as good or getter in my opinion!", "Solution_3": "There's a little book I recommend called The Art and Craft of Problem Solving, by Paul Zeitz. Good choice.\r\n\r\nA good one: Fyodor Dostoevsky's Crime and Punishment, if you're into somewhat dense books which provoke thought about psychology. Dostoevsky's books are generally good for that purpose.\r\n\r\nAnd then there's that classic, How to Eat Fried Worms (the author's name escapes me at this point). I have to remember to cite that book in an SAT essay sometime...", "Solution_4": "[quote=\"K81o7\"]There's a little book I recommend called The Art and Craft of Problem Solving, by Paul Zeitz. Good choice.\n\nA good one: Fyodor Dostoevsky's Crime and Punishment, if you're into somewhat dense books which provoke thought about psychology. Dostoevsky's books are generally good for that purpose.\n\nAnd then there's that classic, How to Eat Fried Worms (the author's name escapes me at this point). I have to remember to cite that book in an SAT essay sometime...[/quote]\r\n\r\nlol :rotfl: \r\n\r\nthe sorrows of young werther by j. w. goethe\r\nthe stranger by a. camus", "Solution_5": "LORD OF THE RINGS ALL THE WAY!!!\r\n\r\nI reccomend reading Gulliver's Travels if you have not yet.", "Solution_6": "Lord of the Flies- William Golding, pretty good book.\r\n\r\nThe Alex Rider Series (Book 1,2,3,4,5,6)- Anthony Horowitz.", "Solution_7": "[quote=\"neelnal\"]LORD OF THE RINGS ALL THE WAY!!!\n\nI reccomend reading Gulliver's Travels if you have not yet.[/quote]\r\n\r\nLord of the Rings is a good book to read. Fun and Entertaining!", "Solution_8": "The Books of Genesis, Exodus, Deuteronomy, Judges, Ruth, I Samuel, and II Samuel are all very good.", "Solution_9": "Thanks for suggestions. :) \r\n\r\nI read LOTR 4 times and planning to read it again. :jump: \r\n\r\nIgnite168-who is the writer?\r\n\r\nK81o7-I read Dostoevsky and he's really fantastic. Please remember who is the writer of How to Eat Fried Worms .\r\n\r\ndeimos-I managed to skip The sorrows of young werther when I was in high scool, but maybe I should try it now. :maybe:", "Solution_10": "[quote=\"Ignite168\"]The Books of Genesis, Exodus, Deuteronomy, Judges, Ruth, I Samuel, and II Samuel are all very good.[/quote]\n\n\n[quote=\"Nienna\"]Ignite168-who is the writer?[/quote]\n\naccording to the bible and church, the real author is god.\n\n[quote=\"Nienna\"]deimos-I managed to skip The sorrows of young werther when I was in high scool, but maybe I should try it now. :maybe: [/quote]\r\n\r\nwell, it's a good book. you should read it. have you read the stranger by camus? any of camus's works? his philosphy is quite interesting.\r\nschiller is ... oh well, never mind.", "Solution_11": "[quote=\"Nienna\"]Please remember who is the writer of How to Eat Fried Worms.[/quote]\r\n\r\nYou do realize that's a book for..maybe 7-8 year olds?\r\n\r\nNevertheless, it is a classic.", "Solution_12": "http://en.wikipedia.org/wiki/How_to_Eat_Fried_Worms\r\n\"How to Eat Fried Worms is the title of a children's novel written by Thomas Rockwell [...]\"\r\n :wink:" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial" ], "Problem": "How can I compute the eigenvalues for a four by four matrix, all of whose entries are ones? Also, how can I compute the eigenvalues of a four by four matrix with (0 1 0 1) in row 1, (1 0 1 0) in row 2, (0 1 0 1) in row 3, (1 0 1 0) in row 4? \r\n\r\nIs there a shortcut? \r\n\r\nThanks", "Solution_1": "Yes, there's a shortcut. If $ J$ is the $ n\\times n$ matrix with all entries equal to $ 1$, $ J^2\\equal{}nJ$. It has rank 1, so you get the multiplicities as well: $ n$ with multiplicity $ 1$ and $ 0$ with multiplicity $ n\\minus{}1$.\r\nTry something similar with your other matrix.", "Solution_2": "[quote=\"bythecliff\"]How can I compute the eigenvalues for a four by four matrix, all of whose entries are ones?[/quote]\r\nLet's make this even more general: the $ n\\times n$ matrix of all $ 1$s, conventionally known as $ J.$ Multiply that by itself, and you find that $ J^2\\equal{}nJ.$\r\n\r\n[Side note: I am assuming that we are working over a field of characteristic zero. bythecliff is probably assuming the field is $ \\mathbb{R}.$]\r\n\r\nThe matrix satisfies the polynomial equation $ x^2\\minus{}nx\\equal{}0.$ If a matrix satisfies a polynomial equation, then the only possible eigenvalues are the roots of that equation. In this case, that means that the only possible eigenvalues are $ 0$ and $ n.$\r\n\r\nWe further note that $ J$ has rank $ 1.$ By the rank-nullity theorem, that means that its null space has dimension $ n\\minus{}1.$ That means that $ 0$ is an eigenvalue of geometric multiplicity $ n\\minus{}1,$ hence algebraic multiplicity at least $ n\\minus{}1.$\r\n\r\nOn the other hand, $ \\text{trace }J\\equal{}n,$ and the trace is the sum of the eigenvalues. Hence the eigenvalues can't all be zero. We must in fact have $ n\\minus{}1$ copy of the eigenvalue $ 0$ and one copy of the eigenvalue $ n.$\r\n\r\nIt's also not hard to see that $ [1\\;1\\;1\\;\\dots\\;1]^T$ is an eigenvector. The structure of $ J$ is just begging for us to try this out. It turns out to be the eigenvector associated with the eigenvalue $ n.$", "Solution_3": "Ah, jmerry beat me to that one. We both had this way of thinking drilled into us by the same individual, a fellow professor of mine at CSULB named Robert Mena.\r\n\r\nLet's look at your second question: $ A\\equal{}\\begin{bmatrix}0&1&0&1\\\\ 1&0&1&0\\\\ 0&1&0&1\\\\ 1&0&1&0\\end{bmatrix}.$\r\n\r\nThis is a symmetric matrix, hence diagonalizable (with real eigenvalues), hence the algebraic multiplicity of each eigenvalue equals its geometric multiplicity.\r\n\r\n$ A$ fairly clearly has rank $ 2,$ hence nullity $ 2,$ hence $ 0$ as an eigenvalue of multiplicity $ 2.$ But the trace is zero, so the other two eigenvalues must be a balanced $ \\pm$ pair.\r\n\r\nIt would be nice to find a polynomial that $ A$ satisfies. Let me use some block structuring to get at that. Write $ B\\equal{}\\begin{bmatrix}0&1\\\\1&0\\end{bmatrix}.$ Note that $ B^2\\equal{}I.$\r\n\r\nThen, $ A\\equal{}\\begin{pmatrix}B&B\\\\B&B\\end{pmatrix}.$ \r\n\r\nSquare that, using that $ B^2\\equal{}I,$ to get;\r\n\r\n$ A^2\\equal{}\\begin{pmatrix}2I&2I\\\\2I&2I\\end{pmatrix}.$\r\n\r\nMultiply that through again:\r\n\r\n$ A^3\\equal{}A^2A\\equal{}\\begin{pmatrix}4B&4B\\\\4B&4B\\end{pmatrix}\\equal{}4A.$\r\n\r\nHence $ A$ satisfies $ x^3\\minus{}4x\\equal{}0,$ which has roots $ 0$ and $ \\pm2.$\r\n\r\nWe can put together what we've found to say that the eigenvalues of $ A$ are $ 2,\\minus{}2,0,0.$" } { "Tag": [ "function", "algebra", "polynomial", "blogs", "algebra unsolved" ], "Problem": "Define $ f_n(x)$ as $ \\underbrace{f(f(...(f(x))...))}_{n\\ times}$ and let $ f: \\mathbb{Z}^{\\plus{}}\\leftarrow\\mathbb{Z}^{\\plus{}}$.\r\nDoes this sentence is true?\r\nThere exist a function $ f$ such that $ f_n(x)\\equal{}kx$ iff $ \\sqrt[n]{k}\\in\\mathbb{Z}^{\\plus{}}$ and $ f$ can be represented in one case (i.e. \"$ f(x)\\equal{}x$ when $ x$ is odd and $ \\equal{}2x\\plus{}3$ when $ x$ is even\" is prohibited.).\r\nAnd how do you call a function that can be represented in one case?", "Solution_1": "You have to give a precise definition of \"representation in one case\", for example, $ f$ is a polynomial.", "Solution_2": "What happens if $ f$ is not a polynomial?", "Solution_3": "Let's say this \"representation in one case\" means a polynomial. Then it is fairly easy to prove that $ f_n(x) \\equal{} kx$ exists iff $ k$ is an $ n$th power. Just show that $ 1$ is an upper bound for $ f$'s degree.\r\n\r\nLet's omit this representation condition entirely. Then there always exists a function $ f$ such that $ f_n(x) \\equal{} kx$ if $ n$ and $ k$ are positive integers. See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=185272]my blog[/url] for a generalization.", "Solution_4": "[quote=\"MellowMelon\"]Let's say this \"representation in one case\" means a polynomial. Then it is fairly easy to prove that $ f_n(x) \\equal{} kx$ exists iff $ k$ is an $ n$th power. Just show that $ 1$ is an upper bound for $ f$'s degree.\n\nLet's omit this representation condition entirely. Then there always exists a function $ f$ such that $ f_n(x) \\equal{} kx$ if $ n$ and $ k$ are positive integers. See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=185272]my blog[/url] for a generalization.[/quote]\r\ni have see you blog $ MellowMelon$.\r\ni will searsh for $ c)$ and for the cas where $ k\\not|S|$\r\n\r\n$ a)$ we suppose that $ S\\equal{}\\{b_i: \\ i\\in\\mathbb{N}\\}$ where $ b_{i\\plus{}1}>b_i$\r\nwe have $ \\forall n\\in\\mathbb{N}:$ exists exacly one $ (m,p,h)\\in\\mathbb{N}\\times \\mathbb{N}\\times \\{0,1,2,..,k\\minus{}1\\}$ such that $ n\\equal{}f^m(b_{pk\\plus{}h})$\r\nif $ h\\in\\{0,1,2,...,k\\minus{}2\\}: \\ g(n)\\equal{}f^{m}(b_{pk\\plus{}h\\plus{}1})$ and if $ h\\equal{}k\\minus{}1: \\ g(n)\\equal{}f^{m\\plus{}1}(b_{pk})$\r\nso if easy to chek that $ \\forall n\\in\\mathbb{N}: \\ g^k(n)\\equal{}f(n)$\r\n\r\n$ b)$ we suppose that $ |S|\\equal{}dk$ and $ S\\equal{}\\{b_0,b_1,b_2...,b_{dk\\minus{}1}\\}$ where $ b_{i\\plus{}1}>b_i$\r\nwe have $ \\forall n\\in\\mathbb{N}:$ exists exacly one $ (m,p,h)\\in\\mathbb{N}\\times \\{0,1,2,..,d\\minus{}1\\}\\times \\{0,1,2,..,k\\minus{}1\\}$ such that $ n\\equal{}f^m(b_{pk\\plus{}h})$\r\nif $ h\\in\\{0,1,2,...,k\\minus{}2\\}: \\ g(n)\\equal{}f^{m}(b_{pk\\plus{}h\\plus{}1})$ and if $ h\\equal{}k\\minus{}1: \\ g(n)\\equal{}f^{m\\plus{}1}(b_{pk})$\r\nso if easy to chek that $ \\forall n\\in\\mathbb{N}: \\ g^k(n)\\equal{}f(n)$", "Solution_5": "$ f: \\mathbb{N}\\to\\mathbb{N},n\\mapsto\\{\\begin{array}{cl}2n &\\text{ for }n \\text{ odd}\\\\\r\n4 &\\text{ for }n=2\\\\\r\n2 &\\text{ for }n=4 \\\\\r\nn &\\text{ for }n>4 \\text{ even}\\end{array}$\r\nThen $ \\mathbb{N}\\setminus f(\\mathbb{N})$ is infinite and $ f\\not =g^2$ for any $ g: \\mathbb{N}\\to\\mathbb{N}$.", "Solution_6": "[quote=\"olorin\"]$ f: \\mathbb{N}\\to\\mathbb{N},n\\mapsto\\{\\begin{array}{cl}2n & \\text{ for }n \\text{ odd} \\\\\n4 & \\text{ for }n = 2 \\\\\n2 & \\text{ for }n = 4 \\\\\nn & \\text{ for }n > 4 \\text{ even}\\end{array}$\nThen $ \\mathbb{N}\\setminus f(\\mathbb{N})$ is infinite and $ f\\not = g^2$ for any $ g: \\mathbb{N}\\to\\mathbb{N}$.[/quote]\r\nbut i have prove that $ g$ exists forall $ k\\in \\mathbb{N}^*$ if $ \\mathbb{N}\\setminus f(\\mathbb{N})$ is infinite and $ f$ injective.\r\nin your exemple $ f$ is not injective $ f(4) = f(1)$.\r\n\r\n :arrow: however that in your exemple that $ f\\neq g^2$", "Solution_7": "Sorry, forgot about $ f$ injective.\r\n\r\n$ f: \\mathbb{N}\\to\\mathbb{N},n\\mapsto\\{\\begin{array}{cl}\r\n2 & \\text{ for }n = 1 \\\\\r\n1 & \\text{ for }n = 2 \\\\\r\nn^2 & \\text{ for }n > 2\\end{array}$\r\nThen $ f: \\mathbb{N}\\to\\mathbb{N}$ injective and $ \\mathbb{N}\\setminus f(\\mathbb{N})$ is infinite \r\nand $ f\\not = g^2$ for any $ g: \\mathbb{N}\\to\\mathbb{N}$.\r\n\r\nIf $ g(1)=1\\Rightarrow 2=f(1)=g^2(1)=1$.\r\nIf $ g(1)=2\\Rightarrow 2=f(1)=g^2(1)=g(2)\\Rightarrow 1=f(2)=g^2(2)=2$.\r\nIf $ g(1)=n>2\\Rightarrow n^4=f^2(g(1))=g(f^2(1))=g(1)=n$.", "Solution_8": "[quote=\"olorin\"]Sorry, forgot about $ f$ injective.\n\n$ f: \\mathbb{N}\\to\\mathbb{N},n\\mapsto\\{\\begin{array}{cl} 2 & \\text{ for }n = 1 \\\\\n1 & \\text{ for }n = 2 \\\\\nn^2 & \\text{ for }n > 2\\end{array}$\nThen $ f: \\mathbb{N}\\to\\mathbb{N}$ injective and $ \\mathbb{N}\\setminus f(\\mathbb{N})$ is infinite \nand $ f\\not = g^2$ for any $ g: \\mathbb{N}\\to\\mathbb{N}$.\n\nIf $ g(1) = 1\\Rightarrow 2 = f(1) = g^2(1) = 1$.\nIf $ g(1) = 2\\Rightarrow 2 = f(1) = g^2(1) = g(2)\\Rightarrow 1 = f(2) = g^2(2) = 2$.\nIf $ g(1) = n > 2\\Rightarrow n^4 = f^2(g(1)) = g(f^2(1)) = g(1) = n$.[/quote]\r\nyes $ good$ :D ,\r\nmy fault is that $ Imf=\\bigcup_{m\\in\\mathbb{N}}f^m(S)$ where $ S=\\mathbb{N}-Im(f)$\r\nin your exemple we have $ 1,2\\in Imf$ but $ 1,2\\not\\in \\bigcup_{m\\in\\mathbb{N}}f^m(S)$ :wink:" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Find the lenght of each edge of a cube that has a volume twise that of a cube with and adge of 10cm . give your answer to the nearest tenth of a cuntimeter", "Solution_1": "The volume of the original cube is 1000 cm^3. Thus, the new cube has a volume of 2000 cm^3. To find the side length of the new cube, compute the cube root of its volume. The cube root of 2000 is 2^(1/3) x 10. Thus, the side length of the new cube is 2^(1/3) x 10 cm." } { "Tag": [ "symmetry", "number theory open", "number theory" ], "Problem": "Find all pairs $(x,y)$ of natural numbers $x>1$, $y>1$, such that $3x+1$ is divisible by $y$ and simultaneously $3y+1$ is divisible by $x$.", "Solution_1": "Because of the symmetry, if $(x_0,y_0)$ is a solution, then $(y_0,x_0)$ is also a solution.\r\n\r\nPut\r\n\r\n$(S): =\\begin{cases}3x+1=ay & (1) \\\\3y+1=bx & (2)\\end{cases}$\r\n\r\nwhere $a,b\\in\\mathbb{Z}^{+}$. We have\r\n\r\n\\begin{eqnarray*}xy|(3x+1)(3y+1)\\implies xy|3x+3y+1\\implies xy\\leqslant 3x+3y+1\\implies (x-3)(y-3)\\leqslant 10\\end{eqnarray*}\r\n\r\nTherefore, $2\\leqslant x\\leqslant 13$ and $3\\nmid x$ because of $(2)$, giving $x\\in\\{2,4,5,7,8,10,11,13\\}.$\r\n\r\n(Note: for $x>13$ we get either $y=3$, which can't satisfy $(1)$, or $y=2$, which is covered below)\r\n\r\nCasework:\r\n\r\nFor $x=2$ $(S)$ becomes $7=ay\\land 3y+1=2b$. The only possibility is $y=7$, which gives an integer solution for $b$. Hence, one pair is $(2,7)$, which also covers $(7,2)$.\r\n\r\nFor $x=4$ $(S)$ becomes $13=ay\\land 3y+1=4b$. The only possibility is $y=13$, which gives an integer solution for $b$. Hence, another pair is $(4,13)$, which also covers $(13,4)$.\r\n\r\nFor $x=5$ $(S)$ becomes $16=ay\\land 3y+1=5b.$ Since $y-3\\leq\\frac{10}{x-3}=5$, the possibilities are $y=8,4,2$, but only $y=8$ gives an integer $b$. Hence, another pair is $(5,8)$, which also covers $(8,5)$\r\n\r\nFor $x=10$ $(S)$ becomes $31=ay\\land 3y+1=10b.$ The only possibility is $y=31$, but it contradicts $(x-3)(y-3)\\leqslant 10.$\r\n\r\nFor $x=11$ $(S)$ becomes $34=ay\\land 3y+1=11b$. The possibilities are $y=2,17,34$, but the first one gives no integer $b$, and the other two contradict $(x-3)(y-3)\\leqslant 10.$\r\n\r\nIt's easy to check that $x=7,8,13$ give no new solutions.\r\n\r\nSummary: $\\{x,y\\}\\in\\{\\{2,7\\},\\{4,13\\},\\{5,8\\}\\}$", "Solution_2": "Let $x>y \\ 3x+1=my, \\ (m>3) \\ 3y+1=nx \\ (n<3) \\Longrightarrow 9x+3+m=mnx.$ It give\r\n$x=\\frac{3+m}{mn-9},y=\\frac{mn+3m-6}{m(mn-9)}$ It give 2 solutions:\r\n$n=1,m=10,x=13,y=4, \\ n=2.m=5,x=8,y=5,$ and symmetric (xy \\ 3x+1=my, \\ (m>3) \\ 3y+1=nx \\ (n<3) \\Longrightarrow 9x+3+m=mnx.$ It give\n$x=\\frac{3+m}{mn-9},y=\\frac{mn+3m-6}{m(mn-9)}$ It give 2 solutions:\n$n=1,m=10,x=13,y=4, \\ n=2.m=5,x=8,y=5,$ and symmetric (x1$, prove that $ p_n'(x)\\equal{}nq_{n\\minus{}1}(x),\\ q_n'(x)\\equal{}\\minus{}np_{n\\minus{}1}(x).$", "Solution_4": "Thanks for your help guys, this problem was driving me insane last night. I've never used complex numbers in trig before, but after I looked up the definitions of trig functions using $ r \\cdot cis(\\theta)$ I understood your proof." } { "Tag": [ "limit", "calculus", "integration", "function", "real analysis", "calculus computations" ], "Problem": "find this limit: :D \r\n\r\n\r\n$\\lim_{k\\to\\infty}\\;\\sum_{j=1}^{k}\\;\\;\\left( \\;\\;\\sqrt{\\;\\frac{1}{k^{2}}\\;+\\;\\left( \\;\\sqrt{1-\\frac{j^{2}}{k^{2}}\\;}\\;-\\;\\sqrt{1-\\frac{(j-1)^{2}}{k^{2}}\\;}\\;\\;\\right)^{2}\\;\\;}\\;\\;\\right) \\;\\;$\r\n\r\n\r\nanswer below:\r\n\r\n[hide]\n$answer: \\;\\;\\;\\quad \\boxed{\\frac{\\pi}{2}}$\n[/hide]", "Solution_1": "Let $x_{j}= \\frac{j}{k}$ and $\\Delta x = \\frac{1}{k}$. By MVT,\r\n\r\n$\\sqrt{1-x_{j}^{2}}-\\sqrt{1-x_{j-1}^{2}}=-\\frac{c_{j}\\Delta x}{\\sqrt{1-c_{j}^{2}}}$\r\n\r\nfor some $x_{j-1}< c_{j}< x_{j}$. Then this Riemann sum [u]converges[/u](really?) to the integral\r\n\r\n$\\int_{0}^{1}\\sqrt{1+\\left(\\frac{x}{\\sqrt{1-x^{2}}}\\right)^{2}}\\; dx = \\int_{0}^{1}\\frac{1}{\\sqrt{1-x^{2}}}\\; dx = \\frac{\\pi}{2}$\r\n\r\n\r\nI'm not sure if my solution is correct...", "Solution_2": "You're right to be cautious. Riemann sums don't automatically converge when the function is unbounded.\r\nYou'll need another argument to show the part for $x$ near 1 doesn't blow up.\r\n\r\nA simple version of that argument:\r\nFor each $j |b|$ then $ (1)$ never holds, so $ f'$ has fixed sign, so $ f$ is either increasing or decreasing\r\n\r\n$ \\bullet$ Even if $ |a| \\equal{} |b|$, then $ f'$ has a discrete set of roots, so $ f$ is still increasing or decreasing\r\n\r\n$ \\bullet$ Only if $ |a| < |b|$ we have intervals at which $ f'$ is positive and other intervals at which $ f'$ is negative\r\n\r\nFor example, $ f'\\left(\\frac {\\pi}{2}\\right)\\cdot f'\\left( \\minus{} \\frac {\\pi}{2}\\right) \\equal{} (a \\plus{} b)(a \\minus{} b) \\equal{} a^2 \\minus{} b^2 < 0$\r\n\r\nSo, at this final case there is no inverse" } { "Tag": [ "LaTeX", "AoPSwiki" ], "Problem": "I am new to this forum. How to copy mathequaitons from math-equation editor to the thread? \r\nRegards\r\nSridhar[/code]", "Solution_1": "Do you mean like this? $ \\frac{\\pi}{\\sqrt{2}}$\r\n\r\nIf so, this would probably be the wrong forum. In the forums, you place dollar signs (\\$) around your LaTeX equation. Here are a few basic commands:\r\n\r\n\\frac{}{} : the two curly braces take in two arguments for the numerator and denominator of the fraction\r\n\\sqrt{} : the braces puts your argument under a square root sign\r\n\\pi : the symbol pi\r\n\r\nThere are more in the AoPSWiki Guide.", "Solution_2": "By math-equation editor, do you mean the one on word? If so, copying is not possible. You must learn $ \\text{\\LaTeX}$ and rewrite them (see post above).", "Solution_3": "A more thorough FAQ can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=700733#700733]here[/url]." } { "Tag": [ "number theory", "relatively prime", "algebra unsolved", "algebra" ], "Problem": "Determine all integer m for which all solutions of the equation \r\n\r\n$ 3x^{3}\\minus{}3x^{2}\\plus{}m\\equal{}0$ \r\n\r\nare rational", "Solution_1": "For root's to be rational, the discriminant must be a perfect square. In other words:\r\n\r\n$ 9 \\minus{} 12m\\equal{}k^2$\r\n\r\nwhere $ k \\in \\mathbb{N}$. And since $ m \\in \\mathbb{N}$ we can say that there is no solution since $ k^2$ must be positive. But, if we take $ 0$ to be an element of $ \\mathbb{N}$ then the solution $ m\\equal{}0$ would give the appropriate result.", "Solution_2": "But its not quadratic. Not necessarily $ m\\in \\mathbb{N}$ I think there's something wrong.\r\nConverting rational to relatively prime integers $ (a,b)$:\r\n$ 3a^3\\minus{}3a^2b\\plus{}mb^3\\equal{}0$. Thus $ b|3$. For $ b\\equal{}3$, $ a^3\\minus{}3a^2\\plus{}9m\\equal{}0$, meaning $ 3|a$, impossible. Thus $ b\\equal{}1$ and $ m\\equal{}3t^2\\minus{}3t^3$ for all integers $ t$.\r\n\r\nP.S.I really have the feeling I did something wrong.", "Solution_3": "I apologize.. I misread the post. My solution is WRONG!!" } { "Tag": [ "calculus", "derivative", "analytic geometry", "graphing lines", "slope", "limit" ], "Problem": "what the heck is it? I mean i know that its the natural base, but why can't ten be the natural base or i don't know 42 be the natural base? :huh: :huh:", "Solution_1": "E is the limit of $ (1 \\plus{} (1/x))^x$ as x approaches infinity. This number has many special properties though. For example, think of the formula $ (1 \\plus{} (1/x))^x$ as a small amount of interest (1/x) compounded a large (x) amount of times. Thus, it is useful in computing things that compound continuously, such as money or bacteria. \r\nAlso, e is important in calculus. The derivative of $ e^x$ is itself and the derivative of $ ln x$ is $ 1/x$. This means if you look at the graph of $ e^x$, the slope of the tangent line at any point will be the y coordinate! And if you look at the graph of $ ln x$, the slope of the tangent line at any point will be the recipriocal of the x coordinate.", "Solution_2": "EDIT:(cleaner)\r\n\r\n$ e\\equal{}\\lim_{x\\to\\infty}(1\\plus{}\\frac{1}{x})^x$\r\n\r\n$ e\\equal{}\\lim_{x\\to0}(1\\plus{}x)^{\\frac{1}{x}}$", "Solution_3": "i suppose you'll not find $ e$ as important till you start calculus.\r\n\r\nOnce you'll start calculus,you'll tend to forget base $ 10$ or anything,then only $ e$ remains.It is equally important for calculus as $ \\pi$ is for geometry(circles)\r\n\r\nFor more details check [url]http://mathworld.wolfram.com/e.html[/url]", "Solution_4": "well, i am doing derivative calculus and i just read that $ \\frac{\\text{d}e}{dx} \\equal{} e$. That's pretty cool!", "Solution_5": "Well, in fact, $ \\frac {de}{dx} \\equal{} 0$, as $ e$ is just a real constant. What you probably wanted to write is $ \\frac {de^x}{dx} \\equal{} e^x$.", "Solution_6": "yes i did :D :D :blush: i like emotes" } { "Tag": [ "LaTeX" ], "Problem": "I am using the itemize environment to list some stuff but the spacing between the items of the list is too great for my liking. How can I decrease the this spacing between the items to something like the normal line spacing in a paragraph of text?", "Solution_1": "Inside the itemize environment you can change two vertical distances, \\parsep and \\itemsep. \\parsep sets the vertical spacing between paragraphs and itemsep adds extra spacing between items. Try using \\setlength{\\itemsep}{0pt}to remove the extra spacing (if that's what you are thinking of)\r\n[code]\\begin{itemize}\n\\setlength{\\itemsep}{0pt}\n\\item Item one\n\\item Item two\n\\end{itemize}[/code]", "Solution_2": "That worked a treat. You are a legend." } { "Tag": [ "analytic geometry", "Ramsey Theory", "combinatorics open", "combinatorics" ], "Problem": "Color all points in plane with RED or GREEN .\r\nLet $ABC$ is a triangle .\r\nProve:\r\nWe can choose $M,N,P$ are the same color ,such that :\r\ntriangle$MNP$ ~$ABC$.\r\n[i](PS: It 's from a easy question:ABC is a regular-triangle)[/i]", "Solution_1": "Use Ramsey theory.", "Solution_2": "HOW???\r\nCan you do the details?", "Solution_3": "In fact the result follows more directly from Gallai's theorem than from Ramsey's one.\r\nGallai's theorem states that :\r\nFor any finite subset $E$ of $\\mathbb{R}^{n}$ and any finite coloring of $\\mathbb{R}^{n}$ then we may find a monochromatic set $F$ which is homothetic to $E$.\r\n\r\nPierre.", "Solution_4": "For triangles, you can even work only on a suitably chosen lattice, and imitate [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=waerden&t=31161]this[/url] proof. More precisely, we can perform an affine transformation on the plane that turns $ABC$ into a right isosceles triangle, say, and then apply the cited problem directly.\r\n\r\nOf course, that's not such a big deal, since it uses Van der Waerden's Theorem, and once you've proved that, Gallai's Theorem shouldn't be too hard to prove either; the idea is the same :).", "Solution_5": "But why do the affine transformation would preserve coloration?\r\n\r\nPierre.", "Solution_6": "I don't really know what that means :?. \r\n\r\nThe affine transformation $T$ sends each point $x$ to $Tx$. In this new setting, color the point $y$ with the same color that $T^{-1}y$ used to have in the old setting. You know that you can find a monochromatic triangle $\\Delta$ homothetic to $T(ABC)$ (by that problem that I posted a link to), so in the old coloring, $T^{-1}(\\Delta)$ must have been a monochromatic triangle homothetic to $ABC$.", "Solution_7": "Yes, sorry, I understood that you transport the coloration too a little late.\r\n\r\nPierre.", "Solution_8": "Here's a proof which doesn't need the results mentioned. We will first give the proof assuming the following (a very very simple case of Van der Waerden's theorem):\r\n [b](1)[/b] For any coloring of a line with two colors there exist three points of the same color on the line with one of them being a midpoint of the other two.\r\n\r\n Let $P, Q,R$ be such points all colored, let's say, red with $Q$ as midpoint of $PR$. Let $PMR$ be a triangle similar to $ABC$. Let $R_{1}, P_{1}$ be the midpoints of $PM, RM$ respectively. Note that triangles $PR_{1}Q$, $QP_{1}R$, and $PMR$ are all similar to triangle $ABC$. Therefore if one of $R_{1}, P_{1}, M$ is red we are done and if not we have $R_{1}, P_{1}, M$ to be of the same color and $R_{1}, P_{1}, M$ is similar to $ABC$. Thus proved.\r\n\r\n[b]Proof of (1):[/b] As I can't draw a diagram I'm going to use a coordinate system. Pick two points $P,Q$ of the same color, say red, and set up a coordinate system with $P$ as origin and $PQ$ as x-axis. Let $Q=(o,a)$. Let $Q_{1}=(0,-a), Q_{2}=(0,a/2), Q_{3}=(0,2a)$. Now, if $Q_{1}$ is red we are done ($Q_{1}, P, Q$ are the desired triple). Similarly if $Q_{2}, Q_{3}$ are red we are done as well. And if not $Q_{1}, Q_{2}, Q_{3}$ are all of the same color and $Q_{2}$ is midpoint of the other two.\r\n\r\nThis would been simpler to describe if I could draw pictures." } { "Tag": [ "calculus", "derivative", "function", "integration", "Support", "limit", "real analysis" ], "Problem": "Suppose that $f: \\mathbb{R}\\to \\mathbb{R}$ is a continuous function such that $\\int_{\\mathbb{R}}f(x) \\phi^{''}(x) dx \\geq 0$ for every smooth function $\\phi$ positive with compact support. Prove that $f$ is convex.", "Solution_1": "Let $\\psi$ be a smooth \"bump\" function: nonnegative, supported on $[-1,1],$ and $\\int\\psi=1.$ Define $\\psi_{\\epsilon}(x)= \\frac1{\\epsilon}\\psi\\left(\\frac{x}{\\epsilon}\\right).$\r\n\r\nFix $a tag. I was able to use that to fix three things that just refused to render.\r\n\r\nThe only downside is that they all rendered forum-style. Is this discouraged?", "Solution_1": "They are meant to be rendered forumstyle... eum... wait:\r\n- do they work with 2 different caches? (what a waste?)\r\n- are they letting the cache being polluted again by non-displaystyled posts? :fool:" } { "Tag": [ "vector", "group theory", "abstract algebra", "induction", "superior algebra", "superior algebra solved" ], "Problem": "This a famous problem, i just want to see different elementary solutions for it :\r\n$G$ is a groupe such that $\\forall x \\in G{\\text{ }}: {\\text{ }}x^2 = e$\r\nshow that $\\left| G \\right|$ is a power of 2.", "Solution_1": "because it can be seen as a vector space of finite dimension over $\\frac{Z}{2Z}$.\r\nBut I bet it is your solution :)", "Solution_2": "that's why i talked about \"elementary solution\" , more precisely no linear algebra and without cauchy theorem :)", "Solution_3": "First, since $x \\mapsto x^2$ is a homomorphism, we see that it is abelian.\r\nNow take any nontrivial element, and let $H$ be the subgroup generated by it, so $|H|=2$.\r\nWhen we have already proved the theorem for smaller groups, we get that $|G/H|=2^s$ for some $s$, so $|G| = |G/H| |H| = 2^{s+1}$.\r\nThus, it's simple induction" } { "Tag": [ "number theory", "number theory unsolved" ], "Problem": "Prove the following Theorem:\r\n\r\nLet W be the set of all positive integers whose decimal expansion only contains the digit 9 (i.e. W = {9, 99, 999, 9999, ...}). For and odd prime p>5, p divides infinitely many distinct elements of the set W.", "Solution_1": "$ W \\equal{} \\{10^n\\minus{}1\\: : n\\in N\\}$\r\n\r\nIf $ p>5$, then $ p \\nmid 10$ and by Fermat's little theorem\r\n\r\n$ 10^{p\\minus{}1}\\minus{}1 \\equiv 0 \\mod p$\r\n$ 10^{2(p\\minus{}1)}\\minus{}1 \\equiv 0 \\mod p$\r\n$ 10^{3(p\\minus{}1)}\\minus{}1 \\equiv 0 \\mod p$\r\n..." } { "Tag": [ "geometry", "geometric transformation", "rotation" ], "Problem": "How many noncongruent triangles have vertices at the vertices of a regular 101-gon?", "Solution_1": "Note that there is a 1-1 Correspondence to this and the number of solutions to the equation $ a + b + c = 101$ such that $ a\\geq b\\geq c\\geq1$ (do you see why?).\r\n\r\nSince $ a,b,c\\geq1$, there are $ \\dbinom{(101 - 3) + 3 - 1}{3 - 1} = \\dbinom{100}2$ choices with only the restriction $ a,b,c\\geq1$.\r\n\r\nFor every $ (a,b,c)$ such that $ a,b,c$ are all different, there are $ 6$ permutations of that, and only one which satisfies $ a > b > c\\geq1$.\r\n\r\nIt is not possible to have an equilateral triangle, since $ 3\\not|101$. However, it is possible to have isosceles triangles, which there are only $ 3$ permutations of (and only one that satisfies $ a\\geq b > c\\geq1$).\r\n\r\nThus, we must break up $ \\dbinom{100}2$ into two cases: isosceles and scalene triangles.\r\n\r\n[hide=\"Isosceles Triangles\"]\nLet the sides be $ a,a,c$. Then we want $ 2a + c = 101$, which has a unique solution for each of $ 1\\leq a\\leq50$, or $ 50$ solutions (not counting order).\n\nWith order, there are $ 3\\cdot50 = 150$ ways to arrange this, but $ 50$ isosceles triangles.\n[/hide]\r\n\r\nBecause there are $ \\dbinom{100}2 = 4950$ triangles total (counting order of $ a,b,c$), $ 4950 - 150 = 4800$ scalene triangles (counting the rotations of $ a,b,c$).\r\n\r\nFinally, we see that there are $ \\frac {4800}6 + 50 = \\boxed{850}$ triangles that satisfy the given condition.", "Solution_2": "Isn't it just (101 C 3)/101/2 for rotations and flips of the same triangle. That would give 825, not 850, right?", "Solution_3": "[quote=\"math154\"]Note that there is a 1-1 Correspondence to this and the number of solutions to the equation $ a \\plus{} b \\plus{} c \\equal{} 101$ such that $ a\\geq b\\geq c\\geq1$ (do you see why?).\n[/quote]\nI don't see why this is so. Could someone explain?", "Solution_4": "The 101-gon has 101 sides and 101 vertices, of which you pick three for the triangle. Let's call these vertices $x$, $y$, and $z$. A certain number $a$ of the 101-gon's sides lie between vertices $x$ and $y$, $b$ between $y$ and $z$, and $c$ between $x$ and $z$.\n\nThe values of $a$, $b$, and $c$ determine the side-lengths of the triangles, and whenever one of them changes you get a different triangle. Therefore to solve the problem you count the different values for $a$, $b$, and $c$. Because the 101-gon has 101 sides, $a+b+c=101$." } { "Tag": [ "Harvard", "college", "calculus" ], "Problem": "Another question:\r\n\r\nI also need to do get these mini-recommendations from various teachers so that my counselor has something to write about me when he does his recommendation for me. He suggests that I choose five teachers who I will NOT be asking for my teacher recommendation for college. I have narrowed it down to exactly five teachers who I might ask for my college recommendation, and I am not sure who to pick.\r\n\r\n\r\nTeacher A\r\n\r\nsubject: English\r\n\r\n- taught me for several years (9th, 10th, 11th) for Honors English\r\n- advisor of a club for which I hold a leadership position\r\n- might exaggerate how smart I am more so than other teachers\r\n- on friendly terms with me\r\n \r\nTeacher B\r\n\r\nsubject: Social Studies\r\n\r\n- taught me for AP US History and AP US Gov & Politics\r\n- went to Harvard (which is where I plan to apply)\r\n- on friendly terms with me\r\n\r\nTeacher C\r\n\r\nsubject: Science\r\n\r\n- taught me for Honors Biology\r\n- might exaggerate how smart I am more so than other teachers\r\n- on friendly terms with me\r\n\r\nTeacher D\r\n\r\nsubject: Math\r\n\r\n- taught me for AP Calculus AB\r\n- might teach me again for AP Statistics next year\r\n- does not know me as well, but I might get to know him better near the end of the year (when all the seniors leave, and I am the only junior in the class)\r\n- i'm doing better than all the seniors in my class (A+)\r\n\r\nTeacher E\r\n\r\nsubject: Science\r\n\r\n- taught me for Accelerated Physics, AP Physics\r\n- thinks that I am: hardworking, intelligent\r\n- if he says something negative it would be that I am a bit too grade conscious\r\n- on friendly terms with me\r\n\r\nI know I should have one math/science teacher recommendation and one humanities recommendation, but I cannot decide which two to choose for my teacher recommendation and which three to leave for the \"mini\" recommendation.\r\n\r\nCan anyone give me suggestions?", "Solution_1": "Also, would it be better to choose an AP teacher rather than an honors teacher?", "Solution_2": "My vote is to ask the English teacher for a college recommendation and the social studies teacher for a mini-rec.", "Solution_3": "[quote=\"JBL\"]My vote is to ask the English teacher for a college recommendation and the social studies teacher for a mini-rec.[/quote]\r\n\r\nI agree; the English teacher seems to know you better and probably has a pretty high opinion of you.\r\n\r\nGood luck =)", "Solution_4": "What about my math/science recommendation? Who should I go with?", "Solution_5": "I would suggest using Teacher B for your college app. I'm pretty sure that Harvard weighs advice from its previous students pretty well.", "Solution_6": "[quote=\"Mr. T\"]I would suggest using Teacher B for your college app. I'm pretty sure that Harvard weighs advice from its previous students pretty well.[/quote]\r\n\r\nThey won't know it's their own student. This is not a good reason to use him for applying to Harvard.", "Solution_7": "I'd say E for your math/science, just b/c the other two options don't look so hot.", "Solution_8": "I return! Anyways, it might be too late but I would recommend the Harvard teacher because a good rec letter for Harvard coming from a teacher who went to Harvard would definitely state that he is a Harvard alumnus. This would look VERY good for you..." } { "Tag": [ "algebra", "polynomial", "modular arithmetic", "complex numbers", "superior algebra", "superior algebra solved" ], "Problem": "Prove that the quotient ring $C(x,y)/P(x,y)$ isn't a field, where $C(x,y)$-the ring of the polynomials in two variables, $P(x,y)$-some elemnt from $C(x,y)$. By the way,the elements of the factorazitaion are said to be equal if their difference is divisible by $P(x,y)$ (I don't know whether the word \"quotient ring\" is correct, because in my books it called \"factor ring\")", "Solution_1": "That just means that the ideal generated by $P$ is not maximal. In other words, we have to show that we can find another polynomial $Q$ with $P\\not |\\ Q$, s.t. the ideal generated by $P,Q$ is not the entire ring. \r\n\r\nIn particular, it suffices find $Q$ s.t. the identity $P(x,1)F(x,1)+Q(x,1)R(x,1)=1$ doesn't hold identically for any $F,R\\in\\mathbb C[x,y]$ (I supose $\\mathbb C$ is the set of complex numbers? :?). If $P$ has degree more than $1$ in $x$, take $Q$ s.t. $Q(x,1)$ is a divisor of $P$. If $P$ ahs degree $\\le 1$ in $x$ but $>1$ in $y$, do the same for $y$. If its degree is $\\le 1$ in both variables, then it's easy to work out manually :). I haven't looked at this case properly.\r\n\r\nP.S. This doesn't work if by $C(x,y)$ you didn't mean $\\mathbb C[x,y]$, the ring of polynomials in two variables with [b]complex[/b] coefficients. Is this what you meant, or is your $C$ something else?", "Solution_2": "By the $C(x,y)$ i meant $\\mathbb{C}[x,y]$- so wuth you solution there are no problems\r\nI just want you to find the mistake in my solition, because it seems to me that it's almost trivial and it must be wrong \r\nSupposing contary for every ${(f+P)\\in{\\mathbb{C}[x,y]}/P[x,y]}$ there must be inverse $(g+P)$ s.t.\r\n$(f+P)(g+P)=1$, taking there $f=1-P$(which is not divisible by P apart from the trivial cases) we get\r\n$g=1-P$, so $(1-P)^2=1$, so $P^2-2P=1$ which is possible only when $P$-constant", "Solution_3": "But the fact that the quotient ring is a field means simply this: given any $f$ which is not divisible by $P$, we can find another polynomiual $g$ s.t. $fg-1$ is divisible by $P$. Maybe I am misunderstanding what you wrote, but it seems to me like you think you have to find $g$ s.t. $fg=1$ not just $fg\\equiv 1\\pmod P$. Your counterewxample does not work: you simply have $f\\equiv g\\equiv 1\\pmod P$.", "Solution_4": "thanks,grobber, for showing me my stupidness-yes, of course it needs to be $fg\\equiv1(mod P)$ i forgot about equivalences\r\nthanks" } { "Tag": [ "Mafia" ], "Problem": "um.......this might be a weird request, but can someone start a game, like beat the mod, mafia, etc? I dont have time every day, and plus, my friends say i suck at story lines, :lol: o well........ but, can someone plz start one? o, and i join the newest game...........", "Solution_1": "There are ongoing and new games all around you in this forum. Take a look at them.", "Solution_2": "any NEW games?\r\n\r\ni dnt relly see n e......", "Solution_3": "Try the AoPS apprentice. I THINK that's a parody of \"The Apprentice\" on television, but I could be wrong...", "Solution_4": "parody? It's the real deal.", "Solution_5": "Also Empire* is still open for signups, although the next time anyone can get in is around turn 35.", "Solution_6": "*cough* 2 or 3 weeks *cough*" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "(i)Each point on the side of a square is coloured one of n colours. The square has side lengths of magnitude 1, and two points that are apart by distance one are of differing colors. Find the smallest such n. \r\n(ii) Solve the problem in (i), with the change that every single point inside the square and on the boundary are to be coloured one of n colors.\r\n- - - \r\nThe following problem is incredibly simple and a routine counting problem and perhaps shouldn't be on this forum, but just checking that I've got the correct answer: \r\n\r\nThere are 7 lanes in a shop where one can check out, labeled from 1 to 7; but only lanes 1 to 4 are able to accept credit cards. Three shoppers A,B and C are shopping, and while A intends to use his credit card, B and C intend to pay with cash instead. How many different ways are there of the three shoppers to stand in line (Note two people can be in the same column, and if this is the case order does matter.)", "Solution_1": "For the first part of the first question, the answer is $3$:\r\n\r\nLet $ABCD$ be the square and take $X_a\\in AB,X_b\\in BC,X_c\\in CD,X_d\\in DA$ s.t. $X_aX_b=X_cX_d=1$, and both $X_aX_b,X_cX_d$ make angles of $45^{\\circ}$ with the sides of the square. Color $[AX_a],[AX_d),[CX_c],[CX_b)$ with color $1,\\ [BX_b],[BX_a)$ with color $2$, and $[DX_d],[DX_c)$ with color $3$.\r\n\r\nThat shows that $3$ colors are enough. On the other hand, we need at least $3$ colors, since it's easy to inscribe an equilateral triangle of side $1$ in a unit square, and the vertices of such a triangle must have different colors." } { "Tag": [], "Problem": "Ann starts counting the letters of the alphabet beginning with A. When she gets to Z,\nshe goes backwards from Y to A and then reverses again going from B to Z. If she\ncontinues this process, what is the $ 2005^{\\text{th}}$ letter that she will count?\n\n[asy]label(\"21\",(0,20),S);\nlabel(\"22\",(7,20),S);\nlabel(\"23\",(14,20),S);\nlabel(\"24\",(21,20),S);\nlabel(\"25\",(28,10),S);\nlabel(\"U\",(0,15),S);\nlabel(\"V\",(7,15),S);\nlabel(\"W\",(14,15),S);\nlabel(\"X\",(21,15),S);\nlabel(\"Y\",(25,10),S);\nlabel(\"Z\",(28,2),S);\nlabel(\"26\",(32,2),S);\nlabel(\"30\",(7,-20),N);\nlabel(\"29\",(14,-20),N);\nlabel(\"28\",(21,-20),N);\nlabel(\"27\",(28,-10),N);\nlabel(\"V\",(7,-15),N);\nlabel(\"W\",(14,-15),N);\nlabel(\"X\",(21,-15),N);\nlabel(\"Y\",(25,-10),N);\nlabel(\"...\",(-2,13),W);\nlabel(\"...\",(-2,-13),W);[/asy]", "Solution_1": "Make one single group go from A to Z and back to B. This is 50 letters long. So, now we divide 2005 by 50 and get 5 as a remainder. Look at the sequence- A,B,C,D,E. E is 5th in a group so [b]E[/b] is the letter, I think.", "Solution_2": "When Ann first counts from A to Z, she counts 26 letters. After that, from Y to A there are 25 letters and from B to Z there are also 25 letters. So $ 25x\\equal{}2005\\minus{}26\\equal{}1979\\implies{x\\equal{}79R4}.$ Since 79 is odd, Ann will be counting from B to Z. Thus, E is the 2005th letter." } { "Tag": [ "videos" ], "Problem": "watch this [url]http://www.youtube.com/watch?v=1VJwn-At244[/url] and the related videos\r\n\"Just wait\", said President Bush.", "Solution_1": "Um, well, Saddam killed thousands of innocent people in cold blood, and I don't think Putin is capable of doing that.... Speaking of President Bush, on a different topic, anyone seen the videos of him grabbing the German ambassador, or cursing with Tony Blair (all at the G8)?", "Solution_2": "I think Mr. Putin said that: Moscow doesn't want the kind of violence-plagued democracy the United States has fostered in Iraq.\r\n\r\nSo why do you think that he shall be the next Saddam? Well, anyway, I believe that he will not be the next Bush who put the kind of violence-plagued democracy in Iraq.", "Solution_3": "WHEN PUTIN STRIDES FOR THE UNCONSTITUTIONAL THIRD TERM, SUPER KASPAROV IS GOING TO FLY DOWN AND UNLEASH DISCOVERD SLAP!", "Solution_4": "[quote=\"shobber\"]I think Mr. Putin said that: Moscow doesn't want the kind of violence-plagued democracy the United States has fostered in Iraq.[/quote]\nOf course, Putin doesn't want any kind of real democracy in Russia.\n\n[quote=\"Buffalo\"]I never know that US invaded Iraq because Saddam killed thousands of innocent people in cold blood, what I do know is, USA troops killed tens of thousands or more, in boiled blood, maybe. [/quote]\nAre you familiar with the word, \"non-sequitur?\"\n\n[quote=\"mysmartmouth\"]Um, well, Saddam killed thousands of innocent people in cold blood, and I don't think Putin is capable of doing that.[/quote]\r\nWhy on earth not?", "Solution_5": "Bush is indeed a really interesting president! \r\n\r\nhttp://www.youtube.com/watch?v=FOomGp8AnH8&search=bush%20G8", "Solution_6": "Well the title doesn't really describe your question.\r\n\r\nDoes he act like him, or will he end up like him?\r\n\r\nRussia is no perfect democracy, I recently saw a documentary where they managed to interview one of his spin doctors, and I frowned at their practices (especially if you know that this was what the people were still allowed to see).\r\n\r\nDid he kill innocents? Well some say his war in Chechnya (the second war) was meant to be a short war to gain popularity, and some even dare to say the bombing of buildings in moscow in 1999 was connected to him somehow.\r\n\r\n\r\n\r\nBut the truth is : USA does not attack countries that can really fight back. If Saddam really had WMD that could be deployed in 45 minutes, they wouldn't have dared. They will never attack Russia.\r\nDon't forget Reagan once said (unintentionally) on the radio that they were gonna bomb Moscow in ten minutes.\r\n[url]http://politicalhumor.about.com/cs/quotethis/a/reaganquotes.htm[/url]" } { "Tag": [], "Problem": "Hm...no real biology \"problems\" yet. I'll take the honors of becoming the first.\r\n\r\nQuestion:\r\n\r\nWhat does D.N.A stand for?", "Solution_1": "At least this I know: DNA = Desoxyribo Nucleic Acid :D", "Solution_2": "Hm... deoxyribonucleic acid?", "Solution_3": "[quote=\"Valentin Vornicu\"]At least this I know: DNA = Desoxyribo Nucleic Acid :D[/quote]Hehe, at least you were close. :D", "Solution_4": "yeah deoxyribonucleic acid...some say it had an even longer formulaic name originally...", "Solution_5": "nucleic acids carry genetic information, which is present in every cell. They have a high molecular weight and are composed of C, H, O, N.\r\njust a question, does deoxyribonucleic acids have $O$?", "Solution_6": "of course theres oxygen in DNA, it has lots of phosphates $\\text{PO}^{3-}_{4}$ which also makes it negatively charged.", "Solution_7": "*laughs* my friends and I have a CHON dance... =D", "Solution_8": "[quote=\"rippledance\"]*laughs* my friends and I have a CHON dance... =D[/quote]\r\nWhat?! lol How do you do it? I want to learn!", "Solution_9": "DNA stands for the word deoxyribonucleic acid.", "Solution_10": "DNA stands for the word deoxyribonucleic acid.", "Solution_11": "DNA...... :mad: don t ask about htat non-sense thread who thinks he knows everythin and poke shis nose in all the lifeprocesses :ninja: if i find who find somebody who gave all the control to that DNA i will....beg him to exlplain me how these four nitrogen bases with that five headed sugar and phosphate are ruling everything :rotfl:", "Solution_12": "D.N.A. stands for deoxyribonucleic acid.", "Solution_13": "Come. You gotta ask harder questions." } { "Tag": [ "inequalities", "algebra solved", "algebra" ], "Problem": "Solve in N the equation:\r\n\r\n(1+x!)(1+y!)=(x+y)!.\r\n\r\n\r\n\r\nIf a,b,c,d are pozitive real numbers then we have:\r\n\r\nsqrt((a+b)/(c+d))+sqrt((b+c)/(a+d))+sqrt((a+c)/(b+d))>=4(a+b+c)/(a+b+c+d).\r\n\r\n\r\n\r\nSolve 1/x*5^x+x*5^(1/x)=10.", "Solution_1": "Solve 1/x*5^x+x*5^(1/x)=10.\r\n\r\n[b]Solution:[/b]\r\nit's a classical! :D\r\n\r\n1/x*5^x+x*5^(1/x)>=2sqrt(1/x*5^x*x*5^(1/x))=2*5^(x+1/x)/2>=2*5=10 (x+1/x>=2).\r\nso you have AM=GM==> x=1 .", "Solution_2": "Solve in N the equation:\r\n\r\n(1+x!)(1+y!)=(x+y)!.\r\n\r\n[b]Solution:[/b]\r\n\r\nwe see that if x>=2 and y>=2 then both (1+x!) and (1+y!) are odd, so their product is odd so we can never have(1+x!)(1+y!)=(x+y)! in this case! :D\r\nWE consider that x=1. So we get that 2(1+y!)=(y+1)!. SO for y>=3 we get that 3 | (y+1)! but 3 not| 2(1+y!). So we get that y<3. \r\nwe get now the only solutions: S={(1,2);(2,1)}.", "Solution_3": "You should try numbering the problems when you post more than one in a post! :D :D \r\n\r\ncheers! :D", "Solution_4": "If a,b,c,d are pozitive real numbers then we have:\r\n\r\nsqrt((a+b)/(c+d))+sqrt((b+c)/(a+d))+sqrt((a+c)/(b+d))>=4(a+b+c)/(a+b+c+d).\r\n\r\n[b]Solution:[/b]\r\napply the mean inequality for \r\n\r\nsqrt((a+b)/(c+d)*1)>=2(a+b)/(a+b+c+d) and you do the same for \r\nsqrt((b+c)/(a+d)*1) and sqrt((a+c)/(b+d)*1) and you get the ineq! :D :D\r\ncheers! :D :D" } { "Tag": [ "IMO Shortlist", "combinatorics solved", "combinatorics" ], "Problem": "There are $2005$ students at a university. Some students join together to form several clubs( a student may belong to several clubs). Some clubs join together to form several societies (a club may belong to several societies). There are a total of $k$ societies. Suppose the following hols: \r\n\r\n(a) Each pair of students is exactly in one club;\r\n\r\n(b) For each stusent and for each society, the student is in exactly one club of the society;\r\n\r\n(c) Each club has an odd number of students. In addition, a club with $2m+1$ students is in exactly $m$ societies.\r\n\r\nFind all possible values of $k$.\r\n[hide=\"Remark\"]This problem is basically the same with [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=258304#p258304]IMO Shortlist 2004, Combinatorics, Problem 1[/url][/hide]", "Solution_1": "Have a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=41116 :)", "Solution_2": "Thanks!!!", "Solution_3": "By the way, what is IMOTC?", "Solution_4": "[quote=\"warut_suk\"]By the way, what is IMOTC?[/quote]Read [url=http://www.mathlinks.ro/Forum/resources.php?c=78&cid=49]here[/url] about it. Continue discussion about the problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=41116 ;)" } { "Tag": [], "Problem": "what is dimension 0? its not empty space, cause that's 3d\r\n\r\nwhat is dimension 4? \"space\" or time?", "Solution_1": "The dimension 0 is just a point.", "Solution_2": "is there such thing as 4D? if so explain it to me", "Solution_3": "4D, which comprises the known universe (I think) constitutes of a fabric called \"space-time,\" which includes the three dimensions of space (height, width, depth) and the dimension of time.", "Solution_4": "But it's with a funny metric. I'd rather think of space in $n$ dimensions, for $n\\ge 0$.", "Solution_5": "so the 4th dimension is about math?", "Solution_6": "[quote=\"RobinHood3000\"]4D, which comprises the known universe (I think) constitutes of a fabric called \"space-time,\" which includes the three dimensions of space (height, width, depth) and the dimension of time.[/quote]\r\ni doubt we have any access whatsoever to any part of the fourth dimension and we have access to time (not dimensions of it but stil...)", "Solution_7": "if we move freely in the 3d dimension and only one way in the 4th are we standing still in the 5th?", "Solution_8": "[quote=\"peter\"]if we move freely in the 3d dimension and only one way in the 4th are we standing still in the 5th?[/quote]\r\n\r\nmy opinion:\r\nif you only move one way in 4th dimension ur moving in max 4 ways or min. 1 way. That means you are not still in the 5th you are moving in1-4 directions.", "Solution_9": "1st dimension, 2nd dimenstion, 3rd dimension, 4th dimension, 5th dimesntion, 6th dimention, you guys are reallyy confusing me know." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Consider the interval [0,1] in R with the inherited metric. Consider the open cover consisting of [0, 1/10), (1/2,1], and, for each natural number n, the interval (1/(n+2), 1/n). Find the Lebesgue number of this cover.\r\n\r\nI took the obvious finite subcovering consisting of [0, 1/10), (1/2,1] and (1/(i+2), 1/i) for i = 1, 2, ..., 9. I then considered the intersection of [0, 1/10) and (1/11, 1/9), or (1/11, 1/10) and divided the diameter of this interval by 2 to get 1/220. Does this work?", "Solution_1": "Probably, yes, if you are also able to show that this interval around any point fits into one of the covering intervals. \"Probably\", because I'm not sure whether the question really was about [b]the[/b] (as you wrote) or about [b]a[/b] (which would be much more agreeing with the general spirit of analysis) Lebesgue number, i.e., whether you need also to check that no larger number works.", "Solution_2": "Hmm I was wondering about that too. I don't think it's too hard to show that 1/220 is a lebesgue number, but finding the largest would seem annoying. I have seen one or two proofs of the corresponding lemma, and it's usually done by contradiction. The proof on wikipedia ([url]http://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma[/url]) takes a direct approach, but it's not all that clear how one would go about finding the lebesgue number prescribed in the proof in practice. Thanks for confirming and clarifying the problem." } { "Tag": [ "geometry", "MATHCOUNTS" ], "Problem": "My state emailed everybody the scores of the 181 individuals and 28 teams who participated at states (no names, just their rank and score) You can correspond the top lists released along with their score so I was wondering why we aren't allowed to post this yet the state released scores for everybody?", "Solution_1": "I think you can post your scores, but not the problems or any hint of the problems' topics.", "Solution_2": "That would give an indication of the difficulty of the test. If I got a low score and still placed high, the test would be difficult. If I got a high score and still placed low, the test would be easy.", "Solution_3": "I dont think your allowed to post scores.", "Solution_4": "I know your not allowed to post scores, that's why I'm asking why my state just gave a list of everybody's score and the team scores.\r\n\r\nAlso, the difficulty of the test wouldn't really help you or hurt you. Saying \"It was harder than last years\" doesn't tell you what areas the test is mainly on, what kind of problems, what you need to study, etc. at all.", "Solution_5": "[quote=\"rcv\"]Do not discuss any problems from the MATHCOUNTS State Competition until further notice. Do not discuss your score. You may state your rank and whether or not you qualified for Nationals, but you may not solicit others to discuss any aspect of the State Competition. \n\n[quote=\"Last year, MCrawford wrote:\"] \nStudents who discuss any information including scores and difficulty level of the test will be put in the AoPS penalty box and not be allowed to post for a considerable period of time. [/quote]\n...\n[/quote]\r\n\r\nFrom the Announcement.", "Solution_6": "The topic is only a couple of pages long, is it that hard to read it to find out what I'm asking?", "Solution_7": "Because states take tests at different times, another person from a state that has not taken the test yet could use your scores to find out the difficulty of the test. You have taken the test, so it doesn't matter that you know your score. But somebody who hasn't taken the test would have access this forum as well and see your scores.", "Solution_8": "[quote=\"ehehheehee\"]Because states take tests at different times, another person from a state that has not taken the test yet could use your scores to find out the difficulty of the test. You have taken the test, so it doesn't matter that you know your score. But somebody who hasn't taken the test would have access this forum as well and see your scores.[/quote]\r\n\r\nExactly.", "Solution_9": "[size=200][color=blue]DO NOT, I REPEAT, DO NOT DISCUSS SCORES, PROBLEMS, OR DIFFICULTY OF THE 2006 STATE COMPETITION UNTILL AFTER THE COMPETITION WINDOW (should be by the end of March).[/color][/size]", "Solution_10": "[quote=\"Ignite168\"]I know your not allowed to post scores, that's why I'm asking why my state just gave a list of everybody's score and the team scores.\n\nAlso, the difficulty of the test wouldn't really help you or hurt you. Saying \"It was harder than last years\" doesn't tell you what areas the test is mainly on, what kind of problems, what you need to study, etc. at all.[/quote]\r\n\r\nWell it doesn't really matter than everybody in your state knows everybody else's score. The problem is if the information spreads to other states that haven't had their competition yet. I think that knowing whether or not the test will be easy is advantageous.", "Solution_11": "Yeah, you can talk about anything in team meetings, in your state, etc. \r\n\r\nBut:\r\n\r\n1. Do not post it in the forums, even the communities (and private ones too!).\r\n\r\n2. Do not talk about it in some mathcounts gathering with other states.", "Solution_12": "In my state, they wouldn't even tell anyone what the top score was, only our team's scores. When they announced the countdown, they only talked about the tiebreakers, but nothing about scores. I only know the scores because my mom figured them out somehow.", "Solution_13": "Meh...\r\nThe emcee told the whole crowd the score of the first place guy, basically. (He missed only X questions today!)\r\nAnd this dude told me my score when I was filling out forms for nats.\r\nBut we didn't get ANY other scores, in fact the only other thing we DID get was team rankings.\r\nWe didn't even get indiv rankings...", "Solution_14": "I haven't had my state yet, and you posting your scores will help me. We don't usually get our scores until a month after. No one told me at the previous chapters either. Hmmm. It's a conspiracy!", "Solution_15": "SO DON'T POST SCORES", "Solution_16": "[quote=\"Ignite168\"]I know your not allowed to post scores, that's why I'm asking why my state just gave a list of everybody's score and the team scores.\n\nAlso, the difficulty of the test wouldn't really help you or hurt you. Saying \"It was harder than last years\" doesn't tell you what areas the test is mainly on, what kind of problems, what you need to study, etc. at all.[/quote]\r\nAs the Announcements state, and as others have repeated, [b][size=150]DO NOT DISCUSS ANY ASPECT OF THE STATE COMPETITION[/size][/b], except your own rank.\r\n\r\nThere is no point discussing this matter further. Thread locked." } { "Tag": [ "geometry", "rectangle" ], "Problem": "A barn with faces which are all rectangles or isosceles triangles has \ndimensions as shown. Given that one gallon of paint covers 450 square \nfeet, how many full gallons of paint must you buy to paint the entire \nexterior of the barn, excluding its roof (shaded) and floor?\n\n[asy]draw((0,0)--(20,0));\ndraw((20,0)--(20,10));\ndraw((20,10)--(0,10));\ndraw((0,0)--(0,10));\ndraw((20,0)--(35,5));\ndraw((35,5)--(35,15));\ndraw((35,15)--(20,10));\ndraw((0,10)--(10,15));\ndraw((20,10)--(10,15));\ndraw((10,15)--(25,20));\ndraw((35,15)--(25,20));\ndraw((10,10)--(10,15),dashed);\ndraw((2,15)--(10,12.5),EndArrow);\nlabel(\"40 ft\",(10,0),S);\nlabel(\"50 ft\",(27.5,2.5),SE);\nlabel(\"10 ft\",(35,10),E);\nlabel(\"8 ft\",(2,15),W);\nfill((20,10)--(35,15)--(25,20)--(10,15)--cycle,gray(.6));[/asy]", "Solution_1": "Front: 180(10)+8(40)=1800+320=2120", "Solution_2": "The question asked for the amount of gallons of paint, so:\n\n$2120\\div{450}=4.7\\overline{1}$\n\nSo we need 5 gallons of paint." } { "Tag": [ "algebra", "polynomial", "AMC", "AIME" ], "Problem": "Let $x,y \\in R$ sastisfying that $x+y,x^{2}+y^{2},x^{4}+y^{4}\\in Z$.Prove that $x^{3}+y^{3}\\in Z$ :)", "Solution_1": "[hide=\"solution\"]\n$(x+y) \\in \\mathbb{Z}\\Rightarrow (x+y)^{2}=x^{2}+y^{2}+2xy \\in \\mathbb{Z}$.\nSince $x^{2}+y^{2}\\in \\mathbb{Z}$ also, this implies $2xy \\in \\mathbb{Z}$\n\n$(x^{2}+y^{2}) \\in \\mathbb{Z}\\Rightarrow (x^{2}+y^{2})^{2}=x^{4}+2x^{2}y^{2}+y^{4}\\in \\mathbb{Z}$\nSince $x^{4}+y^{4}\\in \\mathbb{Z}$, also, this implies $2x^{2}y^{2}\\in \\mathbb{Z}$ also. Along with $2xy \\in \\mathbb{Z}$ we get that $xy \\in \\mathbb{Q}$.\nSuppose $xy$ is not an integer. Then it can be expressed as $\\frac{a}{2}$ with $a$ an odd integer. Then $2x^{2}y^{2}=\\frac{a^{2}}{2}$- which is a contradiction since $2x^{2}y^{2}\\in \\mathbb{Z}$. Hence $xy \\in \\mathbb{Z}$.\n\nNow $(x+y)^{3}=x^{3}+y^{3}+xy(x+y) \\in \\mathbb{Z}$ and $xy$, $x+y$ and hence $xy(x+y) \\in \\mathbb{Z}$. Hence $x^{3}+y^{3}\\in \\mathbb{Z}$.\n\n[/hide]", "Solution_2": "I have a shorter idea.\r\n\r\n2xy ; 2(x^2)(y^2) :arrow: Z\r\n~> 2xy(xy+1) :arrow: Z\r\n~> xy :arrow: Z\r\n\r\n \r\nBtw can we prove that it's the same with n.\r\n\r\nAnd if we have x+y ; x^2 + y^2 ; x^3 + y^3 ... Z. Can we prove that x^4 + y^4 ... Z?", "Solution_3": "[quote=\"bibobeo\"]I have a shorter idea.\n\n2xy ; 2(x^2)(y^2) :arrow: Z\n~> 2xy(xy+1) :arrow: Z\n~> xy :arrow: Z\n\n \nBtw can we prove that it's the same with n.\n\nAnd if we have x+y ; x^2 + y^2 ; x^3 + y^3 ... Z. Can we prove that x^4 + y^4 ... Z?[/quote]\r\n\r\n[hide]\nYes. $x^{3}+y^{3}\\in \\mathbb{Z}\\ \\text{ and }\\ x+y \\in \\mathbb{Z}\\implies x^{2}-xy+y^{2}\\in \\mathbb{Z}\\ \\stackrel{x^{2}+y^{2}\\in \\mathbb{Z}}{\\Longrightarrow}\\ xy \\in \\mathbb{Z}$. Then, $x^{4}+y^{4}-2(xy)^{2}= (x+y)(x^{3}+y^{3}-(x+y)(xy)) \\in \\mathbb{Z}\\Longrightarrow \\boxed{ x^{4}+y^{4}\\in \\mathbb{Z}}$. [/hide]", "Solution_4": "[hide]$x+y\\in\\mathbb{Z}$, so $(x+y)^{2}\\in\\mathbb{Z}$.\nWe get $x^{2}+y^{2}+2xy\\in\\mathbb{Z}$. Since $x^{2}+y^{2}\\in\\mathbb{Z}$, we see that $2xy\\in\\mathbb{Z}$, so $xy\\in\\mathbb{Z}$. So $-xy\\in\\mathbb{Z}$, as well. So we can say that $x^{2}-xy+y^{2}\\in\\mathbb{Z}$, too.\nSo $(x+y)(x^{2}-xy+y^{2})\\in\\mathbb{Z}$, meaning $x^{3}+y^{3}\\in\\mathbb{Z}$.\nYou don't even need to know that $x^{4}+y^{4}\\in\\mathbb{Z}$.[/hide]", "Solution_5": "[b]Follow-up:[/b]\r\n\r\nDo $x+y \\in \\mathbb{Z}$ and $x^{2}+y^{2}\\in \\mathbb{Z}$ imply $x^{n}+y^{n}\\in \\mathbb{Z}$? Do they imply $x, y \\in \\mathbb{Z}$?", "Solution_6": "[quote=\"cincodemayo5590\"][b]Follow-up:[/b]\n\nDo $x+y \\in \\mathbb{Z}$ and $x^{2}+y^{2}\\in \\mathbb{Z}$ imply $x^{n}+y^{n}\\in \\mathbb{Z}$? Do they imply $x, y \\in \\mathbb{Z}$?[/quote]\r\n\r\n1. Yes. Since $x^{n}+y^{n}$ can be expressed as a polynomial with integer coefficients in $x+y$ and $xy$, the conclusion follows.\r\n[size=59]For those who wonder: Take the binomial expansion of (x+y)^n, rearrange it and then induct over n.[/size]\r\n\r\n2. No. Take $x=1-\\sqrt{2}, y=1+\\sqrt{2}$. Then $x+y=2\\in\\mathbb{Z}$ and $x^{2}+y^{2}=6\\in\\mathbb{Z}$, but $x,y\\notin\\mathbb{Z}$", "Solution_7": "[quote=\"lingomaniac88\"][hide]$x+y\\in\\mathbb{Z}$, so $(x+y)^{2}\\in\\mathbb{Z}$.\nWe get $x^{2}+y^{2}+2xy\\in\\mathbb{Z}$. Since $x^{2}+y^{2}\\in\\mathbb{Z}$, we see that $2xy\\in\\mathbb{Z}$, so $xy\\in\\mathbb{Z}$. So $-xy\\in\\mathbb{Z}$, as well. So we can say that $x^{2}-xy+y^{2}\\in\\mathbb{Z}$, too.\nSo $(x+y)(x^{2}-xy+y^{2})\\in\\mathbb{Z}$, meaning $x^{3}+y^{3}\\in\\mathbb{Z}$.\nYou don't even need to know that $x^{4}+y^{4}\\in\\mathbb{Z}$.[/hide][/quote]\n\num, that's not quite correct. You actually need $x^{4}+y^{4}$ to be an integer, otherwise you could end up with\n[hide=\"this\"]\n\\[x=\\frac{3+\\sqrt{7}}{2}, y=\\frac{3-\\sqrt{7}}{2}\\]\n\\[x+y=3, x^{2}+y^{2}=8, x^{3}+y^{3}=\\frac{45}{2}\\]\n[/hide]", "Solution_8": "[quote=\"lingomaniac88\"] we see that $2xy\\in\\mathbb{Z}$, so $xy\\in\\mathbb{Z}$.[/hide][/quote]\r\n :P \r\nI nominate this for a \"what's wrong with this proof\" puzzle", "Solution_9": "[b]Another Follow-Up (Why not :P ):[/b]\r\n\r\nAfter what point (if any) will $x+y, \\ x^{2}+y^{2}, \\ \\ldots, \\ x^{n}+y^{n}\\in \\mathbb{Z}\\Longrightarrow x,y \\in \\mathbb{Z}$?", "Solution_10": "[quote=\"cincodemayo5590\"][b]Another Follow-Up (Why not :P ):[/b]\n\nAfter what point (if any) will $x+y, \\ x^{2}+y^{2}, \\ \\ldots, \\ x^{n}+y^{n}\\in \\mathbb{Z}\\Longrightarrow x,y \\in \\mathbb{Z}$?[/quote]\n[hide]Never.\n\n[quote=\"Farenhajt\"][quote=\"cincodemayo5590\"][b]Follow-up:[/b]\n\nDo $x+y \\in \\mathbb{Z}$ and $x^{2}+y^{2}\\in \\mathbb{Z}$ imply $x^{n}+y^{n}\\in \\mathbb{Z}$? Do they imply $x, y \\in \\mathbb{Z}$?[/quote]\n\n1. Yes. Since $x^{n}+y^{n}$ can be expressed as a polynomial with integer coefficients in $x+y$ and $xy$, the conclusion follows.\n[size=59]For those who wonder: Take the binomial expansion of (x+y)^n, rearrange it and then induct over n.[/size]\n\n2. No. Take $x=1-\\sqrt{2}, y=1+\\sqrt{2}$. Then $x+y=2\\in\\mathbb{Z}$ and $x^{2}+y^{2}=6\\in\\mathbb{Z}$, but $x,y\\notin\\mathbb{Z}$[/quote]\n\nWe can use the counterexample from part 2 again: from part 1 we know that $x^{n}+y^{n}$ is an integer for all $n$, satisfying the conditions for any $n$, but $x$ and $y$ are not integers.[/hide]", "Solution_11": "[quote=\"randomdragoon\"][quote=\"lingomaniac88\"] we see that $2xy\\in\\mathbb{Z}$, so $xy\\in\\mathbb{Z}$.[/hide][/quote]\n :P \nI nominate this for a \"what's wrong with this proof\" puzzle[/quote]\r\n\r\nSorry. I must've thought I was working in $\\mathbb{Q}$ for a second there... that's what happens when you think too fast.", "Solution_12": "[quote=\"cincodemayo5590\"]After what point (if any) will $x+y, \\ x^{2}+y^{2}, \\ \\ldots, \\ x^{n}+y^{n}\\in \\mathbb{Z}\\Longrightarrow x,y \\in \\mathbb{Z}$?[/quote]\r\n\r\n[hide=\"Remark\"] This is the general term of a second-order homogeneous linear recursion\n\n$a_{n+1}= A a_{n}+B a_{n-1}$\n\nWith $A = x+y, B =-xy$. The first two conditions imply both that the first two terms are integer and that the coefficients of the recursion are integer, so all subsequent terms are integer. :) \n\nA generalized version of this was the subject of an interesting AIME problem, and noting this recursion made for a very easy solution. [/hide]", "Solution_13": "[quote=\"cincodemayo5590\"][quote=\"bibobeo\"]I have a shorter idea.\n\n2xy ; 2(x^2)(y^2) :arrow: Z\n~> 2xy(xy+1) :arrow: Z\n~> xy :arrow: Z\n\n \nBtw can we prove that it's the same with n.\n\nAnd if we have x+y ; x^2 + y^2 ; x^3 + y^3 ... Z. Can we prove that x^4 + y^4 ... Z?[/quote]\n\n[hide]\nYes. $x^{3}+y^{3}\\in \\mathbb{Z}\\ \\text{ and }\\ x+y \\in \\mathbb{Z}\\implies x^{2}-xy+y^{2}\\in \\mathbb{Z}\\ \\stackrel{x^{2}+y^{2}\\in \\mathbb{Z}}{\\Longrightarrow}\\ xy \\in \\mathbb{Z}$. Then, $x^{4}+y^{4}-2(xy)^{2}= (x+y)(x^{3}+y^{3}-(x+y)(xy)) \\in \\mathbb{Z}\\Longrightarrow \\boxed{ x^{4}+y^{4}\\in \\mathbb{Z}}$. [/hide][/quote]\r\n\r\n\r\nHey, are you sure?\r\nJust have a look at this.\r\nx = 1/sqrt(2) y = -1/sqrt(2)\r\nthen x + y = 0 ... Z\r\n x^2 + y^2 = 1 ... Z\r\n x^3 + y^3 = 0 ... Z\r\nbut x^4 + y^4 = 1/2 :roll:", "Solution_14": "[quote=\"cincodemayo5590\"][b]Another Follow-Up (Why not :P ):[/b]\n\nAfter what point (if any) will $x+y, \\ x^{2}+y^{2}, \\ \\ldots, \\ x^{n}+y^{n}\\in \\mathbb{Z}\\Longrightarrow x,y \\in \\mathbb{Z}$?[/quote]\r\n\r\nthe sequence can be recursively defined (see newton's sums), so if the first few are integers, then the rest of them will be integers (that is, if the first 4 sums are integers, then the rest of the sequence will be rational)\r\n\r\nanyway, a quick example: $a_{n}=(\\sqrt{2})^{n}+(-\\sqrt{2})^{n}$ shows this to be false", "Solution_15": "[quote=\"cincodemayo5590\"] $x^{3}+y^{3}\\in \\mathbb{Z}\\ \\text{ and }\\ x+y \\in \\mathbb{Z}\\implies x^{2}-xy+y^{2}\\in \\mathbb{Z}\\$[/quote]\nOf course, this is the mistaken line. $x^{2}-xy+y^{2}= \\frac{x^{3}+y^{3}}{x+y}$ need not be an integer.", "Solution_16": "[quote=\"t0rajir0u\"][quote=\"cincodemayo5590\"] $x^{3}+y^{3}\\in \\mathbb{Z}\\ \\text{ and }\\ x+y \\in \\mathbb{Z}\\implies x^{2}-xy+y^{2}\\in \\mathbb{Z}$[/quote]\n\nOf course, this is the mistaken line.$x^{2}-xy+y^{2}= \\frac{x^{3}+y^{3}}{x+y}$ need not be an integer.[/quote]\r\n\r\nOh. :oops: I'll delete what I've posted.\r\n\r\n(edit: The forum says i can't edit it! :o Darn. )" } { "Tag": [ "analytic geometry", "conics", "parameterization" ], "Problem": "Good morning\r\n\r\nI need to do a piece like as barrel in a CNC (Computer Numeric Control) lathe machine.\r\n\r\nOn the piece I need to do several grooves with 15mm of depth. The barrel's radius is 6260mm. The piece length is 1000mm. The tool (disc)radius is 150mm. The grooves have to finish with 150mm from the top of each side.\r\n\r\nSo, the question is: How can I found the exact center coordinates of tool to respect the job coordinates?\r\n\r\nI will appreciate your prompt help\r\nBest regards\r\n\r\nPedro Ferrer", "Solution_1": "You're new, so you probably don't yet understand the forums, but in the Middle School Classroom Math forum, you're not going to get the appropriate feedback you're looking for. Try to look around and get a grip on the levels of the forums so that you know best where to post about problems. I'll move this thread.", "Solution_2": "Good morning \r\n\r\nOk. But can you help me?\r\nYou can't transfer my topic to the correct forum? Please, tell me where you puted my topic.\r\n\r\nThank you.\r\n\r\nBest regards\r\n\r\nPedro Ferrer", "Solution_3": "Hi Andre \r\n\r\nThank for your attention \r\n\r\nThe drawing was designed in AutoCad only to show my problem. The following attachment shows what I want to do. I need to do parametric CNC programs to make the jobs an easier way. So, I already have parametric CNC programs to do grooves in a cylindrical and conic columns. CNC machines works with G code. \r\n\r\nSo, the operator that works with the machine usually knows the following parameters: \r\n\r\n- the column length (1000mm) \r\n- the distance of the tops to start and finish the groove (150mm both sides) \r\n- the groove depth (15mm) \r\n- the tool radius (150mm) \r\n- the barrel radius \r\n\r\nThe ideia is the operator insert the values that want in the text box, then press \"Compile\" button and automatically the computer generates the CNC program. For that, I need to calculate the tool center coodinates. \r\n\r\nImagine that the point (0,0) is on the upper left side of the column, and all the coordinates are in absolute mode and not in incremental mode. \r\n\r\nSo, I need this kind of information to create myself a canned cycle, to this kind of job in an easier way many times as I want. \r\n\r\nI will appreciate your prompt help \r\nBest regards \r\n\r\nPedro Ferrer" } { "Tag": [ "logarithms", "geometry", "rectangle", "SFFT", "calculus", "calculus computations" ], "Problem": "I need help on this problem.\r\n\r\nFind the constant k such that the equation\r\n\r\ndy/dx = (ln x)(ln y) + ln( kxy) becomes seperable.\r\n\r\nI tried using the identity ln( kxy) = ln(k) + ln(x) + ln(y) but I can't seem to figure out where to go from there.", "Solution_1": "For what value of $\\ln k$ does the right hand side factor as a product of two \"linear\" terms? Complete the rectangle.", "Solution_2": "\"Complete the rectangle\" also goes by the AoPS nickname of \"Simon's favorite factoring trick.\" Just remember to think of $\\ln x$ and $\\ln y$ as your variables for the purpose of the factorization.", "Solution_3": "Sorry but I am not sure how to \"complete the rectangle\" or even what you mean by that.", "Solution_4": "SFFT is the factorization $ab+a+b+1 = (a+1)(b+1)$ and other variants. Then follow Mr. Merryfield's hint and you should be set.", "Solution_5": "Ah! Thanks. That helps out a lot." } { "Tag": [ "calculus", "integration", "algebra proposed", "algebra" ], "Problem": "Show that for any nonzero integer $ a$ one can find a nonzero integer $ b$ such that the equation $ ax^2\\minus{}(a^2 \\plus{} b)x \\plus{} b \\equal{} 0$ has integral roots.", "Solution_1": "[quote=\"tdl\"]Show that for any nonzero integer $ a$ one can find a nonzero integer $ b$ such that the equation $ ax^2 \\minus{} (a^2 \\plus{} b)x \\plus{} b \\equal{} 0$ has integral roots.[/quote]\r\n\r\nThat's wrong : $ a\\equal{}2$ implies equation $ 2x^2 \\minus{}(b\\plus{}4)x\\plus{}b\\equal{}0$, so $ b$ even and $ b\\equal{}2c$ and so $ x^2\\minus{}(c\\plus{}2)x\\plus{}c\\equal{}0$ which can only have integers roots if $ c\\equal{}0$, which would imply $ b\\equal{}0$, which is forbidden since you are looking for nonzero $ b$.", "Solution_2": "if $ a\\neq 2$\r\n$ b \\equal{} 2a(2 \\minus{} a)$ gives $ ax^2 \\minus{} (b \\plus{} a^2)x \\plus{} b \\equal{} a(x \\minus{} 2)(x \\minus{} 2 \\plus{} a)$\r\n\r\nif $ a \\equal{} 2$ if $ 2x^2 \\minus{} (4 \\plus{} b)x \\plus{} b \\equal{} 2(x \\minus{} i)(x \\minus{} j)$ for some $ i,j\\in N$ then $ 1 \\equal{} (1 \\minus{} i)(j \\minus{} 1)$ so $ b \\equal{} 2ij \\equal{} 0$" } { "Tag": [ "complex analysis", "complex analysis unsolved" ], "Problem": "Show that X(t, 0, 1 , infinity) = t and deduce that four distinct points z[size=75]i[/size] lie on the circle if and only if X(z1,z2,z3,z4) is real.", "Solution_1": "Please use different titles for different topics, so your posts can be distinguished (You can still edit a new title/description in for the next day). In particular, \"Complex Analysis\" as a title is so generic as to be useless, especially in this forum. I thought I was going to have to delete some threads for multiposting there.", "Solution_2": "What is X(...)?" } { "Tag": [ "combinatorial geometry", "superior algebra", "superior algebra unsolved" ], "Problem": "Recall that a [i]convex polytope[/i] $\\mathcal{P}$ in $\\mathbb{R}^n$ is defined as the convex hull of a finite set of points in $\\mathbb{R}^n$ (or, equivalently, as a bounded intersection of finitely many half-hyperplanes). A [i]face[/i] of the polytope $\\mathcal{P}$ is the set of all points in $\\mathcal{P}$ that maximize some linear function. \r\n\r\nFor any sets $A,B \\subset \\mathbb{R}^n$, define the [i]Minkowski sum[/i] $A + B := \\{a+b \\mid a \\in A, b \\in B \\}$.\r\nIt is obvious that if $A$ and $B$ are convex polytopes, then so is $A + B$.\r\n\r\n[b]Problem. [/b] (a) Let $A$ and $B$ be convex polytopes in $\\mathbb{R}^n$. Show that every face of $A+B$ is the Minkowski sum of unique faces of $A$ and $B$.\r\n(b) Strengthen (a) as follows: show that if $A_1 \\subset A$ and $B_1 \\subset B$ are convex subsets such that $A_1 + B_1$ is a face of $A+B$, then $A_1$ and $B_1$ must be faces of $A$ and $B$, respectively.", "Solution_1": "First of all, it's clear that every vertex of $C=A+B$ is the unique sum of a vertex from each of $A$ and $B$. \r\n\r\nLet $A_1+B_1=C_1$ be a representation of a face $C_1$ of $C$ as a sum of convex subsets of $A,B$ (the notation is the same as in (b) above). If $c_1,\\ldots,c_n$ are the vertices of $C_1$, then let $a_i+b_i=c_i$ be the unique representations of the $c_i$'s as sums of vertices of $A,B$ (note that the $a_i$'s need not be distinct, and the same holds for the $b_i$'s). $A_1$ must contain $\\{a_i\\}$, and $B_1$ must contain $\\{b_i\\}$. Let $A_1'\\supseteq A_1$ be the face of $A$ spanned by $\\{a_i\\}$, and define $B_1'$ similarly. \r\n\r\nTake any $i,j$. We have $a_i+b_i=c_i,a_j+b_j=c_j$. Let $f$ be a linear functional which reaches its minimum $m$ on $C$ in $C_1$. We have $\\frac{f(a_i+b_j)+f(a_j+b_i)}2=f\\left(\\frac{a_i+b_i+a_j+b_j}2\\right)=\\frac{f(c_i)+f(c_j)}2=m$, so $f(a_i+b_j)=m$, meaning that $a_i+b_j\\in C_1$. This shows that $\\mbox{conv}(a_i)+\\mbox{conv}(b_i)\\subseteq C_1$, and since the reverse inclusion is obvious, we have $\\mbox{conv}(a_i)+\\mbox{conv}(b_i)=C_1\\ (*)$. \r\n\r\nNotice that $f$ takes the same value in all the $a_i$'s, and the same value in all the $b_i$'s, so it takes a constant value on $A_1'$, and also a constant value (which may be different from the other one, of course) on $B_1'$. This means that $A_1'+B_1'\\subseteq C_1$, so, since $A_1'\\supseteq\\mbox{conv}(a_i),B_1'\\supseteq\\mbox{conv}(b_i),\\ (*)$ gives us $A_1'+B_1'=C_1$.\r\n\r\nSuppose we can find a point $u\\in A_1'\\setminus\\mbox{conv}(a_i)$. We can then find a linear functional $g$ for which $g(u)<\\min_{\\mbox{conv}(a_i)}g$, and if $g$ reaches its minimum on $\\mbox{conv}(b_i)$ in $v$, then $u+v$ cannot belong to $\\mbox{conv}(a_i)+\\mbox{conv}(b_i)=C_1=A_1'+B_1'$, which is a contradiction. This shows that $A_1'=\\mbox{conv}(a_i)$, and the same type of argument gives $B_1'=\\mbox{conv}(b_i)$. Now, since $A_1\\supseteq\\mbox{conv}(a_i)=A_1',B_1\\supseteq\\mbox{conv}(b_i)=B_1'$, and the same type of argument as a few lines above shows that $A_1,B_1$ cannot contain points outside $A_1',B_1'$ respectively, we get $A_1=A_1',B_1=B_1'$, and we're done.\r\n\r\nI hope it's correct." } { "Tag": [ "function", "calculus", "derivative", "quadratics", "trigonometry", "algebra", "algebra unsolved" ], "Problem": "[i]I have put the problem in the College section too.In this section the problem can be solved ,as well more elementary.MLinkres in Romany have similar problems in their school marerial.[/i]\r\n\r\nFind the functions $f , g$ with $f'(x) = f(x) +g(x) , g'(x) = g(x) - f(x), f(0) = 0 , g(0) = 1$ \r\n\r\n for all $x\\in IR$ and $f , g$ are not constant functions.\r\n\r\n [u]Babis[/u]\r\n\r\n I suspect $f(x) =sinx e^x , g(x) = cos x e^x$. Please try with elementry way, not with the general form of a linear differential equation.", "Solution_1": "Finally I solved it. The function\r\n\r\n $H(x) = (f(x)e^{-x} - sinx)^2 + (g(x)e^{-x} - cosx)^2$\r\n\r\n has derivative 0 ect , so H(x) = 0.We used the fact that $f' - f = g , g' - g = -f$\r\n I still try to find them , without to guess them.\r\n\r\n [u]Babis[/u]", "Solution_2": "From $f'=g+f$ $\\implies f''=g'+f'=2g=2f'-2f$ So $f''-2f'+2f=0$ .The roots for the quadratic equation $m^2-2m+2=0$ are $m=1\\pm i$ , which mean the general solution for $f$ is\r\n\r\n$f(x)=e^x(A\\cos x+B\\sin x)$ . Then using $f(0)=0$ and $f'(0)=g(0)+f(0)=1$ to get $A=0,B=1$ . \r\n\r\nHence $f(x)=e^x\\sin x$\r\n\r\nWe do the same thing for $g$ to get $g''-2g'+2g=0$ $\\implies g=e^x(C\\cos x+D\\sin x)$. Since $g(0)=0 and g'(0)=1$ , we get $C=1,D=0$ so $g(x)=e^x\\cos x$\r\n\r\nEdit : Opps , i didnt see that you want an elementary way .... :roll:", "Solution_3": "Thank you.\r\n\r\n Really this was the simlest way - I mean your solution - based on homogenous linear differential equations theory. You helped me to refresh my poor steps in Dif. Equations( I was taught them in 1978 !!!). \r\n If you find something more elementary, please remember me !\r\n\r\n [u]Babis[/u]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "1) Determine all $n\\in\\mathbb{N}$, such that $\\{(a,b)\\in[2,\\infty)^2\\cap\\mathbb{N}^2|(a^a)^n=b^b\\}$ is non-empty.\r\n2) Solve $(a^a)^5=b^b$ in $\\mathbb{N}^2$.", "Solution_1": "This problem is from Belarus 2000 (or 2002), if I remember correctly.", "Solution_2": "What is N^2...\r\n\r\nBomb, I think prime factorization would help.", "Solution_3": "[quote=\"bomb\"]What is N^2...[/quote]\r\n$\\mathbb{N}^2=\\{(x,y)|x,y\\in\\mathbb{N}\\}$.", "Solution_4": "*Something wrong*", "Solution_5": "There is a non-trivial solution for the second part: $a = 2^8$ and $b = 2^{10}$." } { "Tag": [ "algorithm", "quadratics" ], "Problem": "How does one determine the sorting algorithm used (insertion, selection, merge, quick, heap) based on a chart of Size of Data and Time to Sort? Specifically:\r\n\r\nData in sorted order:\r\n[u]Size of Data Set and Time to Sort (microsec)[/u]\r\n100,000 and 10\r\n200,000 and 40\r\n\r\nData in random order:\r\n[u]Size of Data Set and Time to Sort (microsec)[/u]\r\n300,000 and 84\r\n600,000 and 336\r\n\r\n\r\nI looked around a lot for an explanation, but I couldn't find one... thanks in advance!", "Solution_1": "You have an algorithm that takes time quadratic in the size of the data set and performs better on sorted data than unsorted data. Which of those algorithms have both of these properties?", "Solution_2": "I know quadratic is probably insertion or selection... And I [i]think[/i] insertion works better for sorted data?", "Solution_3": "Whoops; what I wanted to say is, it takes time quadratic in the size of the data set regardless of whether it was already sorted or not. Insertion, selection, and bubble sort are all quadratic in the average case; which one is also quadratic in the best case?", "Solution_4": "Well insertion is O(n) in the best case. Selection is O(n^2) in both cases. So the answer would be selection?\r\n(I'm not sure what a bubble sort is...)", "Solution_5": "Well bubble sort is also quadratic in the worst case :maybe:.", "Solution_6": "Isn't bubble sort O(n) in the best case?\r\n\r\nhttp://en.wikipedia.org/wiki/Bubble_sort", "Solution_7": "I don't think anyone uses bubble sort anymore anyways...", "Solution_8": "That's irrelevant to this question. Of the five algorithms you listed, one satisfies both requirements.", "Solution_9": "True, I'm just saying that bubble sort isn't the answer.", "Solution_10": "Well bubble sort was the first sorting algorithm I learned [i]ever[/i]. But apparently they thought us a different variation... sorry for confusing you :).\r\n\r\n(It went like this:[code] for (int i = 1; i < n; i ++)\n for (int j = 0; j < n - i; j ++)\n if (a[j] > a[j + 1]) swap(a[j], a[j + 1]);\n[/code])" } { "Tag": [ "geometry", "geometry solved" ], "Problem": "triangle ABC is isosceles triangle with AB=AC, AD is perpendicular to BC, DE is perpendicular to AC . F is midpoint of DE. AF and BE are drawn they intersect in point G. then prove that AG is perpendicular to BE. (assume triangle ABC is an acute angled triangle and don't use co-ordinate geometry for this.)", "Solution_1": "Here if you use co-ordinate geometry problem becomes very simple.", "Solution_2": "Let $H$ be the midpoint of $CE$. Then $DH$ is a midline in triangle $BCE$, so $DH||BE$.\r\nNow, observe that $F$ is the orthocenter of triangle $ADH$. Indeed, $DF\\perp AC$ and $HF\\perp AD$ since $HF||CD$. It follows that $AF\\perp DH$, hence $AF\\perp BE$, as desired.", "Solution_3": "This problem is a particular case of the problem proposed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18084 . In order to make the relation clear, I rewrite the initial problem with shifted notations.\r\n\r\n[color=blue]Let ABC be an isosceles triangle with CA = CB, and let D be the orthogonal projection of its apex C on its base AB. Let E be the orthogonal projection of the point D on the side BC. Let F be the midpoint of the segment DE. Then prove that $CF\\perp AE$.[/color]\r\n\r\nNow, since the triangle ABC is isosceles with CA = CB, the orthogonal projection D of its apex C on its base AB is simply the midpoint of this base AB. Thus, $\\frac{AD}{DB}=1$. On the other hand, since F is the midpoint of the segment DE, we have $\\frac{EF}{FD}=1$. Thus, $\\frac{EF}{FD}=\\frac{AD}{DB}$, and hence this problem is a particular case of the one discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18084 .\r\n\r\n darij", "Solution_4": "Precisely the same construction ($F' \\in \\overline{BE}$ such that $\\overline{BD} \\parallel \\overline{FF'}$) manages in other problem as well." } { "Tag": [ "topology", "function", "real analysis", "real analysis unsolved" ], "Problem": "A map $f: \\mathbb{R}^n \\rightarrow \\mathbb{R}^n$ is called a $C^r$ diffeomorphism, if $f$ is a homeomorphism and $f$ and its inverse $f^{-1}$ are r-times continuously differentiable, i.e. they are both $C^r$ maps.\r\n\r\nHow can one prove the following:\r\n\r\nIf $f: \\mathbb{R}^n \\rightarrow \\mathbb{R}^n$ is a $C^1$ diffeomorphism and a $C^r$ map, then $f$ is a $C^r$ diffeomorphism.\r\n\r\nCan you please help me?", "Solution_1": "I am no specialist in this, but i think when you have a $C^r$ function that has a jacobian determinant nonzero in some point p, there is a neighbourhood of p on which it works as a $C^r$ diffeomorphism (this is a theorem)\r\n\r\nwhen you know that it is a $C^1$ diffeomorphism, it gives you jacobian determinant nonzero everywhere and I think this should help", "Solution_2": "Putting $f(u)=v$ we know that $matr(df(u))=matr(df^{-1}(v))$ so if the functions in the rows of\r\n $matr(df(u))$ are \r\n$C^{r-1}$ then the functions in the rows of $matr(df^{-1}(v))$ are $C^{r-1}$ and vice versa." } { "Tag": [ "inequalities", "inequalities proposed", "algebra" ], "Problem": "Show that in acute triangle ABC we have:\r\n$\\frac{a^5+b^5+c^5}{a^4+b^4+c^4} \\ge \\sqrt{3}R$", "Solution_1": "Any solution? :(", "Solution_2": "Can this be useful?\r\n\r\n$\\displaystyle (a+b+c)^5-a^5-b^5-c^5=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$", "Solution_3": "Maybe,It's very straight.\r\nBut in my solution I didn't use it.\r\n :)", "Solution_4": "So what 's yours ? give some hints about your solution .....", "Solution_5": "A tighter inequality is this one $\\frac{a^2+b^2+c^2}{a+b+c} \\geq \\sqrt{3}R$, which is the Garfulkel 's inequality.", "Solution_6": "I 'm sorry for my bad memory. It should be $\\frac{a^2+b^2+c^2}{a+b+c}\\leq \\sqrt{3}R$.", "Solution_7": "Oh,along time and no one post the solution for this problem\r\nHere is then hint ,very neccessary:\r\n$sinAsin3A+sinBsin3B+sinCsin3C \\le 0$\r\nPlease check it!", "Solution_8": "Then, the following better inequality holds\r\n\r\n$\\frac{a^4+b^4+c^4}{a^3+b^3+c^3} \\ge \\sqrt{3}R$.", "Solution_9": "I start with the ineq:\r\n\r\n$\\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$\r\n\r\nWhich is equal with:\r\n\r\n$sinAsin3A+sinBsin3B+sinCsin3C \\le 0$\r\n\r\nAnd easy to prove $(\\frac{a^5+b^5+c^5}{a^4+b^4+c^4})^2 \\ge \\frac{a^4+b^4+c^4}{a^2+b^2+c^2}$\r\nBecause $(a^5+b^5+c^5)(a^5+b^5+c^5)(a^2+b^2+c^2) \\ge (a^4+b^4+c^4)^3$ (Holder)\r\n\r\nVasc,Please proof $\\frac{a^4+b^4+c^4}{a^3+b^3+c^3} \\ge \\sqrt{3}R$ \r\nI think it's tronger.", "Solution_10": "It is not very hard.\r\nWe get the desired result by multiplying the inequalities\r\n\r\n$(a^4+b^4+c^4)(a^2+b^2+c^2) \\ge (a^3+b^3+c^3)^2$\r\nand\r\n$a^4+b^4+c^4 \\ge 3R^2(a^2+b^2+c^2)$. ;)", "Solution_11": "Oh yes.\r\nI am wrong.Thank you.", "Solution_12": "Can anybody please give a hint to that Jack Garfunkel's inequality? And also does anybody knows in what magazine (issue-year) was published?\r\n\r\nThanks", "Solution_13": "[quote=\"hungkhtn\"]I start with the ineq:\n\n$\\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$\n\nWhich is equal with:\n\n$sinAsin3A+sinBsin3B+sinCsin3C \\le 0$[/quote]\r\nYesterday I found in my collection the inequality\r\n$\\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$.\r\n\r\nThe proof is the following. Since\r\n$R^2=\\frac{a^2b^2c^2}{16S^2}=\\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$,\r\nusing the notations $a^2=\\frac{y+z}{2}$ etc., the inequality becomes as follows\r\n$xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \\ge 0.$ ;)", "Solution_14": "GREAT PROOF, VASC ;)", "Solution_15": "Also, I nearby found the following complementary inequalities for any triangle: \r\n\r\n(a) $(a+b+c)R \\ge \\sqrt{ab^3+bc^3+ca^3}$;\r\n\r\n(b) $(a+b+c)R \\ge \\sqrt{a^2b^2+b^2c^2+c^2a^2}$.", "Solution_16": "Why do you use this substitution?\r\n\r\n$\\displaystyle a^2=\\frac{y+z}{2} $\r\n\r\nIn my opinion $a^2,b^2,c^2$ are not sides of a triangle.\r\n\r\n\r\n$a^2+b^2 \\geq \\frac{(a+b)^2}{2} \\geq \\frac{c^2}{2}$\r\n\r\ninstead of\r\n\r\n$a^2+b^2 \\geq c^2$\r\n\r\nWhere am I wrong, Vasc?", "Solution_17": "manlio,\r\n\\[ a^2+b^2-c^2 = \\frac{y+z+z+x-x-y}{2} = z \\]\r\nand so on, so they are sidelengths of a triangle.", "Solution_18": "Manlio, the triangle is acute. ;)\r\n\r\n[quote=\"hungkhtn\"]Show that in acute triangle ABC we have:\n$\\frac{a^5+b^5+c^5}{a^4+b^4+c^4} \\ge \\sqrt{3}R$[/quote]", "Solution_19": "Show that in acute triangle $ABC$ , prove or disprove:\n\n$\\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \\ge \\sqrt{3}R$ $(n\\in R^+,n\\ge 2)$\n\n$\\frac{a^{n+2}+b^{n+2}+c^{n+2}}{a^n+b^n+c^n} \\ge 3R^2$ $(n\\in R^+,n\\ge 2)$", "Solution_20": "We :\nuse $ 2s\\geq 3\\sqrt{3}R, (*) $ \n $ \\frac{\\sum a^{n+1}}{\\sum a^{n}}(CBS)\\geq \\frac{\\sum a}{3}=\\frac{2s}{3}\\geq (*)=\\sqrt{3}R $", "Solution_21": "We :\nuse $ 2s\\geq 3\\sqrt{3}R, (*) $ \n $ \\frac{\\sum a^{n+2}}{\\sum a^{n}}(CBS)\\geq (\\frac{\\sum a}{3})^{2}\\geq (*)=3R^{2} $", "Solution_22": "generalization :\nLet $ k,p\\in N,k-p\\geq 1 $ , then we : $ \\frac{\\sum a^{k}}{\\sum a^{p}}\\geq (\\frac{a+b+c}{3})^{k-p}\\geq (\\sqrt{3}R)^{k-p} $", "Solution_23": "sinA+sinB+sinC\u2264\n@nicusorz is right\uff1f", "Solution_24": "@Vasc\nyour solving isn't right\n\n\n\n", "Solution_25": " wrong\uff1a $ 2s\\geq 3\\sqrt{3}R, (*) $", "Solution_26": "boi, this thread is 13 years old", "Solution_27": "[quote=Vasc]Also, I nearby found the following complementary inequalities for any triangle: \n(b) $(a+b+c)R \\ge \\sqrt{a^2b^2+b^2c^2+c^2a^2}$.[/quote]\nWe have\n\\[LHS^2 - RHS^2 = \\left[12Rr+(p^2+5r^2-16Rr)\\right] (4R^2+4Rr+3r^2-p^2)+4r(4R^2+3Rr+2r^2)(R-2r) \\geqslant 0.\\]\n", "Solution_28": "[quote=sqing]Show that in acute triangle $ABC$ , prove or disprove:\n\n$\\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \\ge \\sqrt{3}R$ $(n\\in R^+,n\\ge 2)$\n\n$\\frac{a^{n+2}+b^{n+2}+c^{n+2}}{a^n+b^n+c^n} \\ge 3R^2$ $(n\\in R^+,n\\ge 2)$[/quote]\n\nThe first one holds for $\\boxed{n\\ge 3}$ :!: . And they are both trivial because of the very nice inequality in [b]non-obtuse-angled[/b] triangles which has already been mentioned, that is:\n\\[\\boxed{\\frac {a^4 + b^4 + c^4}{a^2 + b^2 + c^2} \\ge 3R^2}\\ .\\]\nThis was also proven by other means [url=https://artofproblemsolving.com/community/c6h361604p1980265][b]here[/b][/url].\n\nNow, $n\\mapsto \\frac {a^{n+1} + b^{n+1} + c^{n+1}}{a^n + b^n + c^n}$ is decreasing and this is an immediate consequence of CS. And the base case for $n = 3$ follows again from CS together with the above-mentioned inequality. \n\nSimilarly, $n\\mapsto \\frac {a^{n+2} + b^{n+2} + c^{n+2}}{a^n + b^n + c^n}$ is also a decreasing function on $n$ because $f(n)\\ge f(n - 1)$ reduces to:\n\\[\n\\begin{array}{lc}\n & \\displaystyle \\sum \\left(b^{n-1}c^{n+2} + c^{n-1}b^{n+2}\\right) \\ge \\sum \\left(b^nc^{n+1} + c^nb^{n+1}\\right) \\\\\\\\\n\\iff & \\displaystyle \\sum b^nc^n\\left(\\frac {c^2}b + \\frac {b^2}c\\right) \\ge \\sum b^nc^n\\left(b + c\\right),\n\\end{array}\\]\nwhich is obvious. As the base case coincides with the previous \"boxed\" inequality, there is nothing else to be done ;).", "Solution_29": "Nice.....\n\n\n", "Solution_30": "\t\n#14Feb 14, 2005, 9:05 pm\nhungkhtn wrote:\nI start with the ineq:\n\n$\\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$\n\nWhich is equal with:\n\n$sinAsin3A+sinBsin3B+sinCsin3C \\le 0$\nYesterday I found in my collection the inequality\n$\\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$.\n\nThe proof is the following. Since\n$R^2=\\frac{a^2b^2c^2}{16S^2}=\\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$,\nusing the notations $a^2=\\frac{y+z}{2}$ etc., the inequality becomes as follows\n$xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \\ge 0.$ ;)\nright??????", "Solution_31": "111111111dddddddd" } { "Tag": [ "arithmetic sequence" ], "Problem": "The interior angles of a convex polygon form an arithmetic progression with a common difference of $4^\\circ$. Determine the number of sides of the polygon if its largest interior angle is $172^\\circ.$", "Solution_1": "[hide]The external angles forms an arithmetic progression where $d = 4^\\circ$ and $a_1 = 8^\\circ$. Since the sum of the external angles of a convex polygon is equal to $360^\\circ$, we can determine $a_n$:\n$S = \\frac{(a_1 + a_n)d}{2}$\n$360 = \\frac{(8 + a_n)4}{2}$\n$a_n = 172^\\circ$\n\n$a_n = a_1 + (n - 1)d$\n$172 = 8 + (n - 1)4$\n$n = 42 \\Rightarrow \\boxed{\\mbox{42 sides}}$[/hide]" } { "Tag": [ "geometry", "circumcircle", "geometry solved" ], "Problem": "In triangle ABC, L,M,N is the midpoint of side BC, CA, AB, respectively. P is another point that isn't on the circumcircle of ABC. X, Y, Z are the foot of perpendicular from P to BC, CA, AB, respectively. Line MN and YZ, NL and ZX, LM and XY meet at D, E, F, respectively. Prove that:\r\n(1) The three lines XD, YE, ZF meet have a common point S.\r\n(2) S lies on the circumcircle of $ \\triangle LMN$ and $ \\triangle XYZ$.\r\n\r\nActually, the first part of this one is not very difficult. Applying Ceva of Menulaus, we can solve it (and it is not nessery for X,Y,Z to be the foot). However, the second part is rather difficult that till now it remains unsolved without lots of caculation.\r\nThanks!", "Solution_1": "Combine Theorems 1.1, 1.6, 1.7, in [url=http://www.cip.ifi.lmu.de/~grinberg/GenFeuerPDF.zip]\"Generalization of the Feuerbach point\"[/url] on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url]. (Your point S is called L in this note.)\r\n\r\n Darij" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a;b;c;d$ be real numbers satisfying\r\n\\[ \\Longleftrightarrow\\begin{cases}\r\n0-1\\wedge \\alpha>1$.\r\n\r\n\\[ (1+x)^{\\alpha} > 1+\\alpha x, \\]\r\n\\[ \\alpha\\ln(1+x)>\\ln(1+\\alpha x). \\]\r\n\r\nIt's Jensen for $f(x)=-\\ln x$.\r\n\r\n($0=\\left(1-\\frac{1}{\\alpha}\\right)\\ln 1$)." } { "Tag": [ "algebra", "polynomial", "geometry", "3D geometry", "ratio", "trigonometry", "number theory" ], "Problem": "[hide=\"Results\"]Big Schools\n\nLevel 1\n1st Place Individual - Sean\n2nd Place Individual - ?\n3rd Place Individual - Sage\n1st Place Team - AC Flora\n2nd Place Team - ?\n3rd Place Team - ?\n\nLevel 2\n1st Place Individual - John B.\n2nd Place Individual - Peter, Harry, Jim, Brian (tied, in order of \"tiebreakers\")\n1st Place Team - Spring Valley\n2nd Place Team - AC Flora\n3rd Place Team - Hoggard\n\nOverall\n1st Place Team - AC Flora\n2nd Place Team - ?\n3rd Place Team - ?\n\nI have a really bad memory, so someone fill in the blanks.\n[/hide]\r\n\r\nIt was a much better contest than last year, but I thought level 1 was too hard (a 12 won it in small schools) and level 2 was way too easy. But I liked the problems, and it was cool.\r\n\r\nSo I'm gonna post some level 1 problems some time soon. Can someone post the hard level 2 problems?", "Solution_1": "Well, here are the hardest level 2 problems. As was said, it was pretty easy. They should have known last year that they needed to make it harder, after getting a 23, 24, and 25. At least this year they specifically said they needed to make it harder.\r\n\r\n[hide]13. The equation $\\sqrt{x+5}-\\sqrt{x-2}+1=0$ has how many real roots and how many complex roots?\n\n14. When the polynomial $5x^{4}+4x^{3}+3x^{2}+Mx+N$ is divided by $x^{2}+1$, the remainder is $0$. If $M$ and $N$ are real numbers, what is $M-N$?\n\n16. Suppose that $sin{x}+cos{x}=a$. Write $sin^{4}{x}+cos^{4}{x}$ in terms of $a$. [i]These were multiple choice, so the only way I found to do this is plugging in values to the possibilities to see which works...[/i]\n\n20. If a 4x4x4 cube is made from 32 white unit cubes and 32 black unit cubes, what is the largest possible percentage of black surface area?\n\n21. Circle $C_{1}$ with center $P_{1}$ and circle $C_{2}$ with center $P_{2}$ intersect in points $Q$ and $R$ such that angle $QP_{1}R=30^{o}$ and angle $QP_{2}R=60^{o}$. What is the ratio of the area of circle $C_{1}$ to the area of circle $C_{2}$?\n\n24. In triangle ABC, the point D lies on BC, and AD is the bisector of angle $BAC$. If $|AB|=c$, $|AC|=b$, and angle $CAD=w$, then what is $|AD|$, in terms of $b$, $c$, and $w$?[/hide]\r\n\r\nHmm... complaints about the test: Problems 12-19, with the exception of 15, were all blah classroom problems. The last page was exclusively geometry (with not a single geometry problem occurring before that last page). There was a total of one number theory problem, and that was \"what is the last digit of 517^{1003}?\" Also, I tend to think poorly of any test that contains more than one messy trig problem (I mean, basic law of cosines with 120 degree angles for example is fine, but nothing with 75 and 15 degree angles).", "Solution_2": "[quote=\"jb05\"]16. Suppose that $sin{x}+cos{x}=a$. Write $sin^{4}{x}+cos^{4}{x}$ in terms of $a$.[/quote]\r\n[hide]Note that $\\sin^{4}x+2(\\sin x \\cos x)^{2}+\\cos^{4}x = (\\sin^{2}x+\\cos^{2}x)^{2}= 1$. Also, $a^{2}= (\\cos x+\\sin x)^{2}= 1+2 \\sin x \\cos x$. It follows that $\\sin x \\cos x = \\frac{a^{2}-1}{2}$, so $\\sin^{4}x+cos^{4}x = 1-\\frac{(a^{2}-1)^{2}}{2}$.[/hide]", "Solution_3": "i tried getting my school to go to ccu, but unsuccessfully. \r\nand assuming i may be able to find more than one other person in my school to go next year, does anyone happen to want to, say, post all the level II problems? thanks^_^" } { "Tag": [ "puzzles" ], "Problem": "I am the ruler of shovels\r\nI have a twin\r\nI am thin as a knife\r\nI have a wife\r\n\r\nWho am I?", "Solution_1": "Mary-Kate Olson :rotfl:", "Solution_2": "[quote=\"kstan013\"]Mary-Kate Olson :rotfl:[/quote]\r\nLol. Could be, but not quite the answer i was looking for", "Solution_3": "[hide=\"guess\"]\nking of spades?[/hide]", "Solution_4": "[quote=\"anyone\"][hide=\"guess\"]\nking of spades?[/hide][/quote]\r\nGood guess", "Solution_5": "[hide=\"Answer, I think.\"]The king of spades.[/hide]", "Solution_6": "[hide]The King Of Spades[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "Sorry, but I haven't been able to find this on the wiki...\r\n\r\nHow do you create a large brace for systems of equations? Thanks!", "Solution_1": "Well actually, I dunno if you want something like\r\n\r\n[code]\\begin{tabular}{rcl|}\n$$ & = & $$ \\\\\n$$ & = & $$ \\\\ \\hline\n\\end{tabular}[/code]", "Solution_2": "Something like this:", "Solution_3": "Ohhh, that's easy\r\n\r\n$ x_{k}\\equal{}\\left\\{\\begin{array}{cr}0 &\\text{if }x_{k\\minus{}1}\\equal{} 0,\\\\ \\left\\{\\frac{p_{k}}{x_{k\\minus{}1}}\\right\\}&\\text{if }x_{k\\minus{}1}\\ne0,\\end{array}\\right.$\r\n\r\n[color=red]x_k=\\left\\{\\begin{array}{cr}0&\\text{if }x_{k-1}=0,\\\\\n\\left\\{\\frac{p_k}{x_{k-1}}\\right\\}&\\text{if }x_{k-1}\\ne0,\\end{array}\\right.[/color]\r\n\r\nDoes that make sense?", "Solution_4": "The cases environment is simpler\r\n$ x_{k}\\equal{}\\begin{cases}0 &\\text{if }x_{k\\minus{}1}\\equal{}0,\\\\ \\left\\{\\frac{p_{k}}{x_{k\\minus{}1}}\\right\\}&\\text{if }x_{k\\minus{}1}\\ne 0\\end{cases}$\r\n\r\n[code]x_k=\n\\begin{cases}\n0 & \\text{if } x_{k-1}=0,\\\\ \n\\left\\{\\frac{p_k}{x_{k-1}}\\right\\} & \\text{if } x_{k-1} \\ne 0\n\\end{cases}[/code]", "Solution_5": "Thanks for your help! And what package is the cases environment included in?", "Solution_6": "It's in amsmath which you should use in every document along with amssymb (which means you can write $ x\\in\\mathbb{R}$ for example)." } { "Tag": [ "symmetry" ], "Problem": "$ \\sum_{k\\equal{}1}^{2010}k(2011\\minus{}k)\\equal{}335\\cdot2011n$\r\n\r\nFind $ n$.", "Solution_1": "Hi, my calculator gives the answer as 2012, which may serve as a hint.", "Solution_2": "[hide=\"my solution\"]\n$ \\sum_{k \\equal{} 1}^{2010}k(2011 \\minus{} k) \\equal{} \\sum_{k\\equal{}1}^{2010}2011k\\minus{}\\sum_{k\\equal{}1}^{2010}k^2$\n\n$ \\equal{}\\frac{2011.2010.2011}{2}\\minus{}\\frac{2010.2011.4021}{6}$\n\n$ \\equal{}2011(\\frac{2011.2010}{2}\\minus{}\\frac{2010.4021}{6})\\equal{}335.2011.2012$\n\n\nSo $ n\\equal{}2012$\n\n[/hide]" } { "Tag": [ "logarithms" ], "Problem": "Suppose $A$ and $B$ are positive real numbers for which $\\log_{A}B = \\log_{B}A$. If neither $A$ nor $B$ is $1$ and $A\\not=B$, find the value of $AB$.", "Solution_1": "[hide=\"hint\"]$A^{x}=B$ and $B^{x}=A$\n \nmultiplying them seems like a good idea....[/hide]", "Solution_2": "[quote=\"now a ranger\"]Suppose $A$ and $B$ are positive real numbers for which $\\log_{A}B = \\log_{B}A$. If neither $A$ nor $B$ is $1$ and $A\\not=B$, find the value of $AB$.[/quote]\r\n[hide]Let $\\log_{a}b=x$\n$x=\\frac{1}{x}$\n$x^{2}=1$\n$x= \\pm 1$\nx clearly can't be 1 (they told us that $a \\neq b$)\nSo x=-1\nmeaning that $A^{-1}=B$\nso AB=1.[/hide]", "Solution_3": "But what he doesn't get is why $\\log_{b}a=\\frac{1}{\\log_{a}b}$.\r\n\r\n[hide=\"Proof.\"]$\\log_{b}a=x\\\\b^{x}=a\\\\b=\\sqrt[x]{a}\\\\b=(a)^{\\frac{1}{x}}\\\\\\rightarrow\\boxed{\\log_{a}b=\\frac{1}{x}}$. Q.E.D.[/hide] :D" } { "Tag": [ "articles" ], "Problem": "Have any of you ever played [url=http://www.sudoku.com/]this game[/url]? It's pretty fun and simple: You have a 9 X 9 grid separated into 9 3 X 3 smaller squares. There are already integers between 1 and 9 in some cells. Your objective is to fill in all the cells w/ integers between 1 and 9 such that every row, column, and 3 X 3 box has each integer between 1 and 9 in it exactly once. It's pretty fun and involves lots of logic. Try it!", "Solution_1": "ah, yes, i found that article on MSN and read about it. downloaded and solved my first sudoku from DailySudoku. It's incredibly fun, but a warning- I had a \"hard\" one and I finished it during an errand trip. So the \"easy\" and \"mediums\" might be boring.", "Solution_2": "I love sudoku.\r\n\r\nI love PBN and cross-sum as well.", "Solution_3": "Sudoku and other similar games can also be found through [url=http://www.puzzle.jp/index-e.html]Puzzle Japan[/url]. I am personally a fan of Edel.", "Solution_4": "Edel is PBN!\r\nI knew in Korean it's \ub124\ubaa8\ub124\ubaa8\ub85c\uc9c1, but I did not know it was also called Edel.", "Solution_5": "Yeah. I've seen it. In fact, the math teacher I work for has gotten her eighth grade students practically addicted to those puzzles.\r\n\r\nAs for me, I love them, but I have done them in another incarnation--in the bimonthly issues of \"Dell Math Puzzles and Logic Problems.\" I can say I am pretty fast at these things.", "Solution_6": "I LOVE SUDOKU!!!!!!!!!!!!", "Solution_7": "[quote=\"bleumoose\"]ah, yes, i found that article on MSN and read about it. downloaded and solved my first sudoku from DailySudoku. It's incredibly fun, but a warning- I had a \"hard\" one and I finished it during an errand trip. So the \"easy\" and \"mediums\" might be boring.[/quote]\r\n\r\nI like to solve ones published in [b]The Times[/b] online newspaper. My another favorite among Japanese puzzles is PBN for more than a year already.", "Solution_8": "[quote=\"themonster\"]Have any of you ever played [url=http://www.sudoku.com/]this game[/url]? It's pretty fun and simple: You have a 9 X 9 grid separated into 9 3 X 3 smaller squares. There are already integers between 1 and 9 in some cells. Your objective is to fill in all the cells w/ integers between 1 and 9 such that every row, column, and 3 X 3 box has each integer between 1 and 9 in it exactly once. It's pretty fun and involves lots of logic. Try it![/quote]We've been playing this on the board already in the NEW [url=http://www.mathlinks.ro/Forum/index.php?f=360]Math & Strategy Games[/url] forum :) It's a ton of fun!", "Solution_9": "[quote=\"math92\"]I LOVE SUDOKU!!!!!!!!!!!![/quote] i love them too in a matter afact i just did one!!!!!", "Solution_10": "i'm doing one right now!!!", "Solution_11": "[quote=\"math92\"]i'm doing one right now!!![/quote]post it here :P", "Solution_12": "*8*5*14**\r\n1**3***89\r\n**7****25\r\n2*****6**\r\n*7*****5*\r\n***1****6\r\n73****8**\r\n85***4**3\r\n**48*3*7*", "Solution_13": "[quote=\"math92\"]6**\n*5*\n**6[/quote]\r\n\r\n\r\nwait... 6th box has two 6s...", "Solution_14": "The Los Angeles Times will add them starting Monday- they're definitely spreading.", "Solution_15": "[quote=\"Geehoon\"][quote=\"math92\"]6**\n*5*\n**6[/quote]\n\n\nwait... 6th box has two 6s...[/quote]\r\ni must have typed it wrong :( \r\ni solved it correctly though(without the six in the wrong place)\r\nI don't know where the paper with the Sudoku on it is :( :(" } { "Tag": [], "Problem": "A prjectile of intial speed V m/s is fired at an angle of elevation alpha from the origin O towards a target T, which is moving away from O along the x-axis. \r\n\r\nAt the instant the projectile is fired, the target T is d meteres from O and it is moving away at a constant speed of u m/s. It can be easily derived that if the projectile hits the target when fired at a n angle of elevation alpha, that\r\n\r\nu = V cos(alpha) - gd/(2Vsin(alpha))\r\n\r\ni) Show that if u > V/(root 3), the target cannot be hit by the projectile no matter what angle of elevation the projectile is fired.\r\n\r\nii) Suppsoe u < V (root 3), Show that the target can be hit when it is at precisely two distances from O.\r\n\r\nDifferent approaches for doing part i) would be appreciated, but part 2 is givine me problems. Thanks for any help.", "Solution_1": "ahh sorry guys, for part ii) it is actually V/root (3), not V root (3). anyone got any ideas on how to do it? any solutions??\r\n\r\n=D" } { "Tag": [ "calculus", "integration", "abstract algebra", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ X$ be a scheme of finite type over a field $ k$(not necessarily algebraically closed).\r\n(a)Show that the following three conditions are equivalent(in which case we say that $ X$ is geometrically irreducible).\r\n\r\n(i)$ X \\times _k \\bar{k}$is irreducible.where $ \\bar{k}$ denotes the algebraic closure of $ k$(By abuse of notation,we write $ X \\times _k \\bar{k}$ to denote $ X \\times _{Spec k} Spec{ \\bar{k}}$).\r\n(ii)$ X \\times _k k_s$is irreducible.where $ k_s$ denotes the separable closure of $ k$.\r\n(iii)$ X \\times _k K$ is irreducible for every extension field $ K$ of $ k$.\r\n\r\n(b)Show that the following three conditions are equivalent(in which case we say that $ X$ is geometrically reduced).\r\n\r\n(i)$ X \\times _k \\bar{k}$is reduced. \r\n(ii)$ X \\times _k k_p$is reduced,where $ k_p$ denotes the perfect closure of $ k$.\r\n(iii)$ X \\times _k K$ is reduced for every extension field $ K$ of $ k$.\r\n\r\nI can't translate the problem locally,and I can't tackle the problem when $ X$ is affine.Thank you for help.", "Solution_1": "Some hints for the nontrivial direction(s):\r\n\r\n1. If $ X,Y$ are reduced schemes of finite type over $ k$ algebraically closed, then $ X \\times_k Y$ is reduced: Reduce to $ X,Y$ affine, write a global section of $ \\mathrm{Spec} (A \\otimes_k B)$ as $ \\sum a_{ij} f_i \\otimes g_j$, $ (f_i),(g_i)$ $ k$-bases of $ A,B$. Restricting this to every closed point of $ X$ and using $ \\bigcap_{\\mathfrak{m} \\in \\mathrm{Spec}_\\mathrm{max} A} \\mathfrak{m} \\equal{} (0)$ (reduced) gives $ \\sum a_{ij}f_i(x) \\equal{} 0$ for all $ j$, so $ a_{ij} \\equal{} 0$. [Similar for integral]\r\n2. If $ f \\in A \\otimes_k K$ is nilpotent, this holds already in a finitely generated extension of $ k$. Then apply 1. and that every $ k$-module is flat. [Similar for integral]\r\n3. See http://en.wikipedia.org/wiki/Radicial_morphism", "Solution_2": "Sorry,Can somebody explain it,I can't understand it, :(", "Solution_3": "if you are still interested in this, I could elaborate it. but what is exactly unclear? have you followed myan's solution (sketch)?", "Solution_4": "bump...\r\n\r\nIf you could elaborate a bit more that would be great." } { "Tag": [ "search", "number theory proposed", "number theory" ], "Problem": "Find $ a;b;c \\in N$ such that $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3abc$", "Solution_1": "Hint: [hide=\"click text\"] Do a Google search on \"Markov numbers\"[/hide]", "Solution_2": "take a look at:\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159235[/url]\r\npost #8" } { "Tag": [], "Problem": "Which of the following numbers is a perfect square?\n\\\\\nA. $ 98! \\cdot 99!$\n\\\\\nB. $ 98! \\cdot 100!$\n\\\\\nC. $ 99! \\cdot 100!$\n\\\\\nD. $ 99! \\cdot 101!$\n\\\\\nE $ 100! \\cdot 101!$", "Solution_1": "let's first look at option a\r\n\r\n$ 98!\\cdot99!$\r\n\r\nwe know that every number up to $ 98$ is repeated, and therefore makes a perfect square\r\nhowever, we would still have a factor of $ 99$ left, and since $ 99$ is not a perfect square, this expression is not a perfect square\r\n\r\nfor option b, we use the same principle to see that the factors remaining are $ 99$ and $ 100$\r\nsince the product of these numbers is not a perfect square, the expression is not a perfect square\r\n\r\nfor option c, we can see that every factor up to $ 99$ is multiplied twice\r\nthe factor left is $ 100$, which is a perfect square, making the whole expression a perfect square\r\n\r\nfor options d and e, we would have a factor of 101 left over\r\nsince 101 is prime and it doesn't have a repeat, both d and e are not perfect squares\r\n\r\n\r\nso our answer is $ \\boxed{c}$" } { "Tag": [ "LaTeX" ], "Problem": "I'm trying to install LaTeX on my new computer, a vista. I have installed MikTex just fine and am having trouble downloading the TeXnic Center installer. I go to the SourceForge download page, the download options screen comes up and I click on \"save\", but then the options screen just disappears and won't let me choose a place to save the file to. Clicking on \"run\" doesn't seem to put the file anywhere either. Is this a problem with the site, my computer, or just the operator? Any help is appreciated. Thanks.", "Solution_1": "Is Internet Explorer blocking the download? Click on the yellow bar at the top to allow the download to proceed.", "Solution_2": "I do that, the screen refreshes, and then the menu comes up asking if I want to run, save, or cancel. I click on save, but it doesn't go into \"my computer\" and ask where I want to save it to. When I click on \"run\" it downloads as normal but then it's not saved anywhere, which doesn't do me any good.", "Solution_3": "It sounds like a permissions issue. Are you able to download other files from other places? 'Run' does save it to a temporary directory and you may be able to find it using Vista's search.\r\n\r\nWhat happens if you click on this link to TexnicCenter http://ovh.dl.sourceforge.net/sourceforge/texniccenter/TXCSetup_1Beta7_01.exe\r\nDoes that allow you to download?\r\n\r\nDo you have or are you able to download Firefox? If so, can you download from Sourceforge using Firefox?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove the following inequality $ (3a\\plus{}2b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c})\\le \\frac{45}{2}$\r\nwhere $ a,b,c\\in [1;2]$.When does equality occur?", "Solution_1": "This problem is in an ongoing contest in Vietnam." } { "Tag": [], "Problem": "What do you use?", "Solution_1": "I find that it takes less effort to type \"www.mathlinks.ro\" than \"www.artofproblemsolving.com\". And I have less chance of making a mistake.", "Solution_2": "i use aops because it is :coool: er\r\n\r\nfirst i was introduced to this site as http://www.mathlinks.ro though", "Solution_3": "[quote=\"mathnerd314\"]I find that it takes less effort to type \"www.mathlinks.ro\" than \"www.artofproblemsolving.com\". And I have less chance of making a mistake.[/quote]\r\n\r\nJust put it as a favorite, like me. :P", "Solution_4": "[quote=\"robinhe\"][quote=\"mathnerd314\"]I find that it takes less effort to type \"www.mathlinks.ro\" than \"www.artofproblemsolving.com\". And I have less chance of making a mistake.[/quote]\n\nJust put it as a favorite, like me. :P[/quote]\r\n\r\nThats exactly what I did. :)" } { "Tag": [ "group theory", "abstract algebra", "linear algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "How many Sylow $ p$-subgroups are there in the group $ GL_n(F_p)$?", "Solution_1": "A typical $ p$-Sylow subgroup is the matrices of the form $ I\\plus{}N$, where $ N$ is a strictly upper triangular nilpotent matrix. All other choices are of course similar.\r\n\r\n\r\nNow the structure: we have an eigenspace $ V_1$ for which $ Nv\\equal{}0$ whenever $ N$ is any of these matrices and $ v\\in V_1$, a space $ V_2$ for which $ N^2v\\equal{}0$, and so on.\r\nThe $ p$-Sylow subgroups are in 1-1 correspondence with chains of subspaces $ V_0\\subset V_1\\subset V_2\\subset\\cdots \\subset V_n\\equal{}F_p^n$ where each $ V_k$ has dimension $ k$.", "Solution_2": "Those chains of subspaces go by the name of [url=http://en.wikipedia.org/wiki/Flag_(linear_algebra)]complete flags[/url], and it's not too bad to count them directly. One can also embed the Sylow $ p$-subgroup $ S$ given by jmerry into the [url=http://en.wikipedia.org/wiki/Borel_subgroup]Borel subgroup[/url] $ B$ of upper triangular matrices and argue that the normalizer of $ B$ is $ B$ (as well as arguing that $ S$ is normal in $ B$)." } { "Tag": [ "ratio", "geometry", "3D geometry", "calculus", "integration" ], "Problem": "It takes you 1 gallon of paint to paint the inside of a box. The next day you paint a box that has the same ratios of length to width and length to height, but it takes you 2 gallons to paint the box. You conclude that the total inside area of the larger box is twice that of the smaller box. You conclude that the volume of the larger box equals_______ times that of the smaller box.\r\n\r\n\r\n\r\n You have two boxes of the same shape that hold flour. The larger box holds twice as much flour as the smaller one. You conclude that it would take you______ times as much paint to paint the larger box than it would take you to paint the smaller box.", "Solution_1": "[quote=\"grizzland\"]It takes you 1 gallon of paint to paint the inside of a box. The next day you paint a box that has the same ratios of length to width and length to height, but it takes you 2 gallons to paint the box. You conclude that the total inside area of the larger box is twice that of the smaller box. You conclude that the volume of the larger box equals_______ times that of the smaller box.\n\n\n\n You have two boxes of the same shape that hold flour. The larger box holds twice as much flour as the smaller one. You conclude that it would take you______ times as much paint to paint the larger box than it would take you to paint the smaller box.[/quote]\r\n\r\nFor the first one:\r\n\r\n[hide]Ratio of Areas is $1: 2$ thus ratio of sides is $\\sqrt{1}: \\sqrt{2}$ $\\rightarrow$ $1: \\sqrt{2}$. The ratio of volumes is the cube of the sides which is $1^{3}: \\sqrt{2}^{3}$\n\nThus the second box is $\\sqrt{2}^{3}=2^{3/2}=\\boxed{2\\sqrt{2}}$ times as large as the first in terms of volume.[/hide]\n\nFor the second:\n\n[hide]Same as above with volumes. $\\text{Small}: \\text{Big}\\rightarrow 1: 2$ in terms of volume ratios. Thus the side ratios are $1^{1/3}: 2^{1/3}\\rightarrow 1: 2^{1/3}$ and the ratio of the areas would be $1^{2}: (2^{1/3})^{2}\\rightarrow 1: 2^{2/3}$.\n\nThus the second box would require $\\boxed{2^{2/3}}$ times as much paint as the first.[/hide]\r\n\r\nBasically you have to look at it as:\r\nSide to side is $a: b$\r\nArea to area is $a^{2}: b^{2}$\r\nVolume to volume is $a^{3}: b^{3}$", "Solution_2": "The first blank is $2\\sqrt{2}$, since the an amount of paint covers a certain surface area, two gallons implies twice as much surface area as the one gallon; that then implies the product of two sides is twice as big. So, the sides increase in proportion by $\\sqrt{2}$, $\\sqrt{2}\\sqrt{2}= 2$. Then the volume increases accordingly:\r\n$xyz=V \\Rightarrow x\\sqrt{2}\\left(y\\sqrt{2}\\right)\\left(z\\sqrt{2}\\right)= 2V\\sqrt{2}$ .\r\n\r\nThe second blank is $\\sqrt[3]{4}$.\r\nTo get the increase per side:\r\n$2^{\\frac{1}{3}}$ .\r\nThen per surface area:\r\n$\\left(2^{\\frac{1}{3}}\\right)^{2}= \\sqrt [3]{2^{2}}= \\sqrt [3]{4}$ .\r\n\r\nThis may be redundant, to what was just posted. :P", "Solution_3": "I got the same.", "Solution_4": "I guess to prove the concepts behind scaling that grizzland was confused about requires calculus but it is really easy to formalize then. (just consider integration over large dimensions)" } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "reflection", "LaTeX", "angle bisector" ], "Problem": "In triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find angle B.", "Solution_1": "Siuhochung, you forgot the additional condition $\\angle A\\ge\\angle B$.\r\nLet M be the mid-point on AB and $\\theta$ be the degree of the four equal angles,\r\nthen \\[1=\\frac{AM}{MB}=\\frac{[AMC]}{[MBC]}=\\frac{AC\\cdot CM \\sin3\\theta}{BC\\cdot CM \\sin\\theta}=\\frac{AC}{BC}\\frac{\\sin3\\theta}{\\sin\\theta}=\\frac{\\sin(90^\\circ -3\\theta)}{\\sin(90^\\circ -\\theta)}\\frac{\\sin3\\theta}{\\sin\\theta}=\\frac{\\sin6\\theta}{\\sin2\\theta}\\]\r\n$(\\frac{AC}{BC}=\\frac{\\sin(90^\\circ -3\\theta)}{\\sin(90^\\circ -\\theta)}$ is obtained by sine rule in $\\triangle ABC. )$\r\nSo $6\\theta=2\\theta$ or $6\\theta+2\\theta=180^\\circ$\r\nRejecting the first one, we have $\\theta=22.5^\\circ$", "Solution_2": "Is there a geometrical way to find the answer?", "Solution_3": "[quote=\"shobber\"]Is there a geometrical way to find the answer?[/quote]\r\nmecrazywong's solution:\r\nSince O and H are isogonal conjugates, O is on the median, thus O is the midpoint of AB, then angle C is 90 and the answer comes out readily.", "Solution_4": "I got mine now: \r\n\r\nLet $M$ be the midpoint, $CD$ is the bisector of $\\angle{C}$, $CH \\perp AB$. Let $BH=HD=y$ and $MD=x$, therefore $AM=x+2y$. Use the angle bisector theorem: \r\n\\[\\frac{2x+y}{x}=\\frac{AC}{DC}=\\frac{AC}{BC}=\\frac{x+y}{y}\\].\r\nHence: $\\frac{x}{y}=\\sqrt{2}$. \r\n\r\nThen use the angle bisector theorem again in $\\triangle{CMH}$: $\\frac{x}{y}=\\frac{CM}{CH}$. And because $\\angle{CHM}=90^o$, so we have: $2CH^2=CH^2+(x+y)^2$, which gives: $CH=x+y$. Then $CM=\\sqrt{2} CH=2y+\\sqrt{2}y=BM=AM$. So $\\angle{C}=90^o$.", "Solution_5": "Would it be too stupid for me to ask what are 'isogonal conjugates'?", "Solution_6": "[quote=\"leepakhin\"]Would it be too stupid for me to ask what are 'isogonal conjugates'?[/quote]\r\n\r\nYou may read darij's post on isogonal conjugates.\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=18472\r\n\r\nIn brief, if you reflect three concurrent cevians with respect to their angle bisectors, the reflected cevians still concur and the concurring point is called the isogonal conjugate of the origianl point.", "Solution_7": "I know that this does not help , but\r\n\r\n a) the problem is from Prasolov' s << Plane geometry >> , where there is a simple solution.\r\n b) this problem was given again in the Greek Math Competition 2004 - 2005. I have one more elementary , beautiful and simple solution.\r\n I' ll ask to a friend to write this solution in Latex for the ... mathlinkers. \r\n (excuse me if I have done something wrong. It ' is the first time I write here ...)", "Solution_8": "Sorry ,\r\n the comments of my last post concern the topic\r\n << http://www.mathlinks.ro/Forum/viewtopic.php?p=322058#p322058 >>\r\n\r\n I kindly ask from the moderators to cancel these two posts from here.\r\n\r\nThank you", "Solution_9": "It is kownthat $CH$ and $CO$ are isogonal conjugates. From this we get that $O$ lies on the median from $C$. Hence, the triangle is rightangled or issoscelles. If it is issoscelles the hipothesys connot hold. Hence it is rightangled. Siince, in the right-angled triangle the median is a half of the side it bisects, we get that each of the four angles from $C$ is equal to $B$. Hence $B=\\frac \\pi 8$" } { "Tag": [], "Problem": "Express the following as a mixed number: \n\n\\[ \\minus{}7\\minus{}(\\minus{}7)^2\\plus{}(\\minus{}7)(\\minus{}7)\\plus{}7(\\minus{}\\frac{1}{7})^2\\]", "Solution_1": "to solve this, we simply evaluate the expression\r\n\r\n$ \\minus{}7\\minus{}(\\minus{}7)^2\\plus{}(\\minus{}7)(\\minus{}7)\\plus{}7(\\minus{}\\frac{1}{7})^2$\r\n\r\n$ \\minus{}7\\minus{}49\\plus{}49\\plus{}7(\\frac{1}{49})$\r\n\r\n$ \\minus{}7\\plus{}\\frac{1}{7}$\r\n\r\n$ \\boxed{\\minus{}6\\frac{6}{7}}$" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Suppose that $a_n$ is a sequence of non-zero real numbers that goes to 0 at infinity. Prove that for any real number $x$ one can find integer sequences $k_n, m_n$ such that $x$ is the sum of the infinite series of general term $a_n k_n$ and also product of all $m_n a_n$.", "Solution_1": "It seems we should just use the approach of Riemann when he shows that a conditional convergent series can be arranged to do anything. \r\n\r\nWe just go along and pick the smallest integer $k_N$ such that the partial sum of $k_na_n$ up to $N$ is greater than $x$. Then for the $N+1$ we pick the largest integer so that the partial sum is less than $x$. It is not hard to see we must get within $a_n$ each time, so it converges. \r\n\r\nThen for the $m_n$ one it is similar; we want to pick the one that makes teh product just greater/just smaller each time. Now we are sure we're multiplying by a number $1 + a_n \\ge a_nm_n \\ge 1 - a_n$. It seems it converges as well but for the moment it is not so obvious to me the reason. Maybe we take logs? I shouldn't be so lazy, I bet it is not really too hard." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Given triangle $ ABC$. $ 2008$.", "Solution_2": "[quote=\"Rust\"]Minimal set M is $ M \\equal{} \\{m|exist \\ prime \\ p\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$. Obviosly $ p \\minus{} 1\\not \\in M$ if p prime $ p > 2008$.[/quote]\r\n\r\nCan you show this $ M$ satisfies the condition?", "Solution_3": "It is obviosly.", "Solution_4": "[quote=\"nguyentrang\"][quote=\"Rust\"]Minimal set M is $ M \\equal{} \\{m|exist \\ prime \\ p\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$. Obviosly $ p \\minus{} 1\\not \\in M$ if p prime $ p > 2008$.[/quote]\n\nCan you show this $ M$ satisfies the condition?[/quote]\r\n\r\n$ n\\in \\{m|\\exists p\\ prime\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$\r\n$ \\Rightarrow \\forall k\\in\\mathbb{N}: \\ p|[m\\plus{}k(m\\plus{}1)]\\plus{}1$\r\n$ \\Rightarrow m\\plus{}k(m\\plus{}1)\\in \\{m|\\exists p\\ prime\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$\r\nso $ \\{m|\\exists p\\ prime\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$ checks the property.\r\nand $ \\forall p\\ prime>2008,\\forall q\\ prime\\le 2008: \\ q\\not|(p\\minus{}1)\\plus{}1$ then $ p\\minus{}1\\not\\in \\{m|\\exists p\\ prime\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$.\r\n\r\n$ \\{m|\\exists p\\ prime\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}$ is the mimal set have this poperty because.\r\nif $ M$ cheks then proberty then $ \\forall p\\ prime\\le 2007: \\{m\\in\\mathbb{N}|\\ m\\equiv p[p\\plus{}1]\\}\\equal{}\\{m\\in\\mathbb{N}|\\ p\\plus{}1|m\\plus{}1\\}\\subset M$\r\nthen $ {\\{m|\\exists p\\ prime\\le 2008, \\ suth \\ that \\ p|m \\plus{} 1\\}\\equal{}\\bigcup_{p\\ prim\\le 2008}\\{m\\in\\mathbb{N}|\\ p\\plus{}1|m\\plus{}1\\}\\subset M}$" } { "Tag": [ "linear algebra" ], "Problem": "Let A,B in M_n(K)\r\nProve that $r(A)+r(B)\\leq n+r(AB)$", "Solution_1": "We have $corank(AB)\\le corank(A)+corank(B)$, thus substracting this from 2n=2n, we get $(n-corank(A))+(n-corank(B)) \\le (n)+(n-corank(AB))$ which is simply $rank(A)+rank(B)\\le n+rank(AB)$." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ C_1,C_2$ and $ C_3$ be three pairwise disjoint circles. For each pair of disjoint circles, we define their internal tangent lines as the two common tangents which intersect in a point between the two centres. For each $ i,j$, we define $ (r_{ij},s_{ij})$ as the two internal tangent lines of $ (C_i,C_j)$. Let $ r_{12},r_{23},r_{13},s_{12},s_{13},s_{23}$ be the sides of $ ABCA'B'C'$.\r\nProve that $ AA',BB'$ and $ CC'$ are concurrent.\r\n(see [url=http://img407.imageshack.us/my.php?image=aaaaafz3.png]figure[/url] )\r\n\r\n\r\nSource: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]Oliforum contest, round 2[/url]", "Solution_1": "Don't use Imageshack please - this forum should be readable in several years, while Imageshack have a tendency to remove old images with time. You can upload images as attachments to your posts.\r\n\r\nYour problem has been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=140096 . I know a beautiful solution by April. April, have you published it already, or are you going to?\r\n\r\n darij", "Solution_2": "In fact i know that the problem was already posted, but without solution..", "Solution_3": "Did someone solve it in the contest ? If yes please post the solution", "Solution_4": "[quote=\"bboypa\"]In fact i know that the problem was already posted, but without solution..[/quote]\r\nWhen will the official solution become available?\r\nI can't wait to see the solution to this one :blush: .", "Solution_5": "i think this evenig i'll post gradation. you can find solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140096]here[/url](identical to official one (by gabriel)) (another two can be found in relativ topics :wink: )" } { "Tag": [ "\\/closed" ], "Problem": "Some of the pages on this site have a yellow background, which makes it very uncomfortable to look at. Is there any way we could change it to a less vibrant color. I found that it really strains my eyes to look at it...\r\n\r\n\r\nAnyone else agree? Or is it just me? :D", "Solution_1": "I like the neutral gray of the forum. There is probably a way for the admin to create custom color schemes, and the users to choose which to use. But I agree that the yellow backgrounds can get annoying.", "Solution_2": "For those that have ever used QBasic, remember the bright white (15 in Screen 12) and just plain white (7 in Screen 12)? I think the second white would be nice if it were a choice.", "Solution_3": "How about a softer yellow? \r\n\r\nWe tried whites and greys but they were just horrid for a site with this page structure." } { "Tag": [ "algebra", "linear equation" ], "Problem": "Ok this is a worked out example in the book, but it does not give an explanation on how to do it it just gives the answer\r\n\r\n-x +3y=10\r\nx+y=2\r\nSolution: On a single set axes, graph each linear equation\r\n\r\n-x+3y=10 x+y=2\r\nX Y X Y\r\n0 10/3 0 2\r\n-4 2 2 0\r\n2 4 1 1\r\n\r\nThose are the points in the chart and that is where im stuck I NEED TO KNOW HOW THEY GOT THOSE POINTS. The book does not explain how they arrived at those points so can someone please help me out. Thanks in advance.", "Solution_1": "[quote=\"dvcalculus\"]Ok this is a worked out example in the book, but it does not give an explanation on how to do it it just gives the answer\n\n-x +3y=10\nx+y=2\nSolution: On a single set axes, graph each linear equation\n\n-x+3y=10 x+y=2\nX Y X Y\n0 10/3 0 2\n-4 2 2 0\n2 4 1 1\n\nThose are the points in the chart and that is where im stuck I NEED TO KNOW HOW THEY GOT THOSE POINTS. The book does not explain how they arrived at those points so can someone please help me out. Thanks in advance.[/quote]\r\nFor example, use x = -4, y = -2. -(-4) is 4. Next, 4 + 3y is 10. Thus, 3y is 6. y is 2.", "Solution_2": "Well, in most solutions by graphing, you just assign random values to x and find the corresponding values for y. Then graph the lines and you get your answer. It's a pretty painful method of solving equations though... I usually prefer solving them algebraically.", "Solution_3": "Well when you graph the two linear equations they should cross at one point. Now using the Cartesian graphing system the point should be in a $(x,y)$ format, where $x$ is the solution to the x in your equation and $y$ is the solution to the y in your equation." } { "Tag": [ "trigonometry", "geometric series", "algebra unsolved", "algebra" ], "Problem": "Simplify the following sum-\r\n$ \\sum_{i\\equal{}1}^{n} \\frac{\\sin(ix)}{\\sin(x)^i}$", "Solution_1": "Using\r\n\r\n$ \\sin(x) \\equal{} \\frac{e^{x\\,i}\\minus{}e^{\\minus{}x\\,i}}{2\\,i}$\r\n\r\nwe can write the sum as\r\n\r\n$ \\frac{1}{2\\,i}\\sum^n_{k\\equal{}1}\\left(\\frac{2\\,i}{1\\minus{}e^{\\minus{}2\\,x\\,i}}\\right)^k\\minus{}\\frac{1}{2i}\\sum^n_{k\\equal{}1}\\left(\\frac{2i}{e^{2\\,x\\,i}\\minus{}1}\\right)^k$\r\n\r\nNow just use the expression for the geometric series and that's it." } { "Tag": [ "linear algebra", "matrix", "vector", "function", "linear algebra unsolved" ], "Problem": "Let $ A\\in\\mathbb{C}^{n\\times n}$ and $ \\left\\Vert\\cdot\\right\\Vert$ be a matrix norm on $ \\mathbb{C}^{n\\times n}$ such that $ \\left\\Vert A\\right\\Vert\\equal{}\\rho(A)$, where $ \\rho(A)$ is the spectral radius of $ A$. How can I show that for each eigenvalue $ \\lambda$ of $ A$ such that $ |\\lambda|\\equal{}\\rho(A)$, the geometric multiplicity of $ \\lambda$ equals to its algebraic multiplicity?", "Solution_1": "I'm going to assume that this matrix norm is the norm induced by a vector norm.\r\n\r\nWLOG, assume $ \\rho(A)\\equal{}\\|A\\|>0.$ \r\n\r\nSuppose the conclusion does not hold. Then there exists a $ \\lambda\\equal{}\\|A\\|$ and vectors $ x,y$ such that $ \\{x,y\\}$ is independent and $ Ax\\equal{}\\lambda x\\plus{}y,Ay\\equal{}\\lambda y.$\r\n\r\nFor $ t\\in\\mathbb{R},$ define $ z(t)\\equal{}x\\plus{}\\frac{t}{\\lambda}\\,y.$\r\n\r\nLet $ g(t)\\equal{}\\|z(t)\\|.$ We have that $ g$ is a convex function on $ \\mathbb{R}$ and that it goes to $ \\infty$ as $ t\\to\\pm\\infty.$ Thus is assumes a minimum value, which we call $ m.$ Let $ [a,b]\\equal{}\\{t\\in\\mathbb{R}: g(t)\\equal{}m\\}.$\r\n\r\nFrom the properties of $ x$ and $ y,$ we can compute that $ Az(t)\\equal{}\\lambda z(t\\plus{}1).$ In particular, $ Az(b)\\equal{}\\lambda z(b\\plus{}1).$\r\n\r\nBut then $ \\frac{\\|Az(b)\\|}{\\|z(b)\\|}\\equal{}\\frac{|\\lambda|\\,\\|z(b\\plus{}1)\\|}{\\|z(b)\\|}$ is strictly greater than $ |\\lambda|,$ contradicting that the norm of the matrix is $ |\\lambda|.$", "Solution_2": "Thank you!" } { "Tag": [ "linear algebra", "matrix", "geometry", "geometric transformation", "homothety", "vector", "algebra" ], "Problem": "find all matrices $M \\in M_{n}(\\mathbb R)$ such that $Vect( PMP^{-1}: P \\in GL_{n}{R}) = M_{n}( \\mathbb{R})$", "Solution_1": "I don't know what the situation is in positive characteristic (and I don't want to worry about that right now :)), but the answer is the same for all fields $K$ of characteristic zero: the only proper, non-trivial linear subspaces $\\mathcal L$ of $\\mathcal M_{n}(K)$ which are stabilized by all conjugations (i.e. $P\\mathcal LP^{-1}=\\mathcal L,\\ \\forall P\\in\\mbox{Gl}_{n}(K)$) are (a) the space of scalar matrices, and (b) the space $\\mbox{sl}_{n}(K)$ of matrices with trace zero. This is a consequence of the following two results, which hold for fields $K$ of characteristic zero (again, I'm not sure what happens for fields of characteristic $p>0$):\r\n\r\n\r\n$(1)$ If $\\mathcal L\\le\\mathcal M_{n}(K)$ is stabilized by all conjugations, then it's an ideal in $\\mathcal M_{n}(K)$, the latter being considered a Lie algebra with the usual Poisson bracket $[A,B]=AB-BA$. \r\n\r\n$(2)$ $\\mbox{sl}_{n}(K),\\ n\\ge 2$ is a simple Lie algebra; it follows that the only ideals of $\\mathcal M_{n}(K)$ are (a) and (b) in the statement above. \r\n\r\n\r\n$(2)$ is well-known. I'm only going to prove $(1)$. It's easy to prove it algebraically, but I prefer this:\r\n\r\nLet $X$ be an arbitrary matrix, and $M$ an element of $\\mathcal L$. We want to prove that $[X,M]$ belongs to $\\mathcal L$. Since the bilinear form $f(A,B)=\\mbox{trace}(AB)$ is non-degenerate on $\\mathcal M_{n}(K)$, it suffices to pick an arbitrary matrix $U$ in the orthogonal complement of $\\mathcal L$ with respect to $f$ and prove that $f(U,[X,M])=0$. By working in the field generated by the entries of $U, X, M$, we can assume that $K$ is a subfield of $\\mathbb C$. \r\n\r\nNow let $\\overline K$ be the completion of $K$ with respect to an Archimedean norm (so $\\overline K$ is either $\\mathbb R$ or $\\mathbb C$, and $K$ is dense in $\\overline K$). We know that $\\mbox{trace}(UPMP^{-1})=0,\\ \\forall P\\in\\mbox{Gl}_{n}(K)\\ (*)$, so, since $\\mbox{Gl}_{n}(K)$ is dense in $\\mbox{Gl}_{n}\\left(\\overline K\\right)$, $(*)$ holds for $P\\in\\mbox{Gl}_{n}\\left(\\overline K\\right)$. Now apply it to $P=\\exp(tX),\\ t\\in\\mathbb R$, and differentiate at $t=0$. We get precisely the desired relation $\\mbox{trace}(U[X,M])=0$.", "Solution_2": "A counterexample to the Grobber's questioning about $carac(K)\\not=0$.\r\n$K=\\mathbb{F}_{2},n=2$ and \\[M=\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}\\]\r\n$GL_{2}(\\mathbb{F}_{2})$ admits 6 elements and $E=\\{P^{-1}MP\\}$ admits also 6 elements. $E$ generates $\\mathcal{M}_{2}(\\mathbb{F}_{2})$, $M$ isn't a homothety and $tr(M)\\not=0$.", "Solution_3": "I suggest you look at the question again, because that is certainly [i]not[/i] a counterexample :). \r\n\r\nThe result, as I formulated it above, is: if $E=\\{PMP^{-1}\\}$ doesn't generate $\\mathcal M_{n}(K)$, then $M$ is either scalar or has trace zero. The contrapositive is: if $M$ is non-scalar and has non-zero trace, then $E$ generates $\\mathcal M_{n}(K)$. Your example confirms this rather than disproving it.", "Solution_4": "Here's my attempt to prove Grobber's claim for fields with at least three elements. It's not complete, but I strongly believe it's not far from that. Suppose $M$ is such that the vector space of its conjugates is not the entire $M_{n}(K)$, thus this vector space is included in some hyperplane of $M_{n}(K)$, which has the form $\\{X|Tr(AX)=0\\}$ for some nonzero matrix $X$. Therefore $Tr(PAP^{-1}M)=0$ for all invertible $P$. Fix for instant two distinct indices $i,j$ and take the matrices $P=I+aE_{ij}$. Then $P^{-1}=I-aE_{ij}$ and the condition $Tr(P^{-1}APM)=0$ implies that the polynomial $tr(AM)+a(Tr(AE_{ij}M)-Tr(E_{ij}AM)-a^{2}tr(E_{ij}AE_{ij}M)=0$ for all $a\\in K$. If $K$ has at least 3 elements, this implies that $tr(E_{ij}AE_{ij}M)=0$. Develop this to get $a_{ji}m_{ji}=0$. BUT (and as an idiot I spent lots of time without noticing this) if $A$ is a matrix satisfying $Tr(PAP^{-1}M)=0$ for all $P$, the same property holds for any conjugate of $A$ (idiot, but essential!). Therefore, for all $i\\ne j$ and all matrices $X$ conjugated to $A$ we have $x_{ij}m_{ij}=0$. Suppose that $A$ is not scalar, otherwise $Tr(M)=0$ and it's ok. Then there exists some nonzero vector $v$ with $v,Av$ not collinear. Thus completing this in a basis and puting $v$ as the $i$-th vector of the basis and $Av$ as the j-th vector, we have a conjugate of $A$ which has a $1$ on the $j,i$ position. This implies that $m_{ji}=0$ for all $j\\ne i$ and thus $M$ is diagonal. Unfortunately I'm blind and I don't see how to conclude easily from this. I am convinced the argument should be straightforward.", "Solution_5": "Yes Grobber,\r\nI was not awaked :blush:" } { "Tag": [ "algebra", "system of equations" ], "Problem": "Jim paddles a canoe upstream at a rate of 5 mph but travels a distance of only 9 miles in 3 hours. What was the rate of the current in miles per hour?", "Solution_1": "Let $ x$ be the rate of the current, in miles per hour.\r\n\r\n$ 3(5\\minus{}x)\\equal{}9$\r\n$ 5\\minus{}x\\equal{}3$\r\n$ x\\equal{}\\boxed{2}$ miles per hour is the rate of the current.", "Solution_2": "I think this is a shorter way.\r\n\r\nWithout the current, he would have went 15 miles, but he only went 9. 15-9=6, and 6/3 for the mph of the current gives you the current is going at 2 mph.\r\n\r\nEdit: Dope!!!! I forgot about that. GJ isabella.", "Solution_3": "Alternatively, we see that Jim only traveled 3 miles per hour, even though he was going 5 mph. Thus, the current is $ 5 \\minus{} 3 \\equal{} \\boxed{2}$ mph.", "Solution_4": "I thought that you could do it with a system of equations.", "Solution_5": "The system of equations is for when you go against the current and with the current. Like say you went 10 miles per hour with the current, and 6 miles per hour against the current.\r\nThus x+y=10, and x-y=6." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Suppose that $ f: N \\rightarrow N$ satisfies $ f(1)=1$ and for all $ n$,\r\na) $ 3f(n)f(2n+1)=f(2n)(1+3f(n))$\r\nb) $ f(2n)<6f(n)$\r\nFind all solutions to the equation $ f(k)+f(m)=293$", "Solution_1": "[quote=\"santosguzella\"]Suppose that $ f: N \\rightarrow N$ satisfies $ f(1)=1$ and for all $ n$,\na) $ 3f(n)f(2n+1)=f(2n)(1+3f(n))$\nb) $ f(2n)<6f(n)$\nFind all solutions to the equation $ f(k)+f(m)=293$[/quote]\r\nFrom $ f(2n)=3f(n)\\frac{f(2n+1)}{1+3f(n)}, \\ f(2n)<6f(n)$ we get $ f(2n)=3f(n),f(2n+1)=1+3f(n)=f(2n)+1$.\r\nTherefore $ f(0)=0,f(1)=1,f(2)=3,f(3)=4,f(4)=9,f(5)=10,...$\r\nIt give $ 3|f(2n), f(2n+1)=1\\mod 3$, because $ 293=2\\mod 3$ k,m are odd $ k=2k_{1}+1,m=2m_{1}+1$ and $ f(k_{1})+f(m_{1})=97$.\r\nTherefore one of them odd, another even. Let $ k_{1}$ even. It give $ k_{1}=2k_{2},m_{1}=2m_{2}+1$ and $ f(k_{2})+f(m_{2})=32$.\r\n$ k_{2},m_{2}$ are odd $ k_{2}=2k_{3}+1,m_{2}=2m_{3}+1$ and $ f(k_{3})+f(m_{3})=10$, therefore $ k_{3}=1,m_{3}=4$, or $ k_{3}=4,m_{3}=1$ or $ k_{3}=0,m_{3}=5$ or $ k_{3}=5,m_{3}=0$.\r\nIt give $ (k,m)=(13,39),(37,15),(39,13),(15,37),(5,47),(45,7),(47,5),(7,45)$." } { "Tag": [ "geometry", "circumcircle", "trapezoid", "number theory", "least common multiple", "geometry proposed" ], "Problem": "Given quadrilateral $ ABCD$. The circumcircle of $ ABC$ is tangent to side $ CD$, and the circumcircle of $ ACD$ is tangent to side $ AB$. Prove that the length of diagonal $ AC$ is less than the distance between the midpoints of $ AB$ and $ CD$.", "Solution_1": "$ K, L$ are midpoints of $ AB, CD.$ From circumcircle tangencies, $ \\angle ABC \\equal{} \\angle ACD,$ $ \\angle CAB \\equal{} \\angle CDA$ $ \\Longrightarrow$ $ \\triangle ABC \\sim \\triangle DCA$ are oppositely similar $ \\Longrightarrow$ $ \\angle BCA \\equal{} \\angle CAD$ $ \\Longrightarrow$ $ BC \\parallel AD.$ Midline $ KL \\parallel (BC \\parallel AD)$ of trapezoid $ ABCD$ cuts its diagonal $ AC$ in half at $ M$ $ \\Longrightarrow$ $ \\triangle AKM \\sim \\triangle LCM$ are oppositely similar $ \\Longrightarrow$ $ AKCL$ is cyclic with diagonal intersection $ M.$ Power of $ M$ to its circumcircle is $ \\minus{} KM \\cdot ML \\equal{} \\minus{} AM \\cdot MC \\equal{} \\minus{} \\frac {_1}{^4}AC^2$ and $ KL \\equal{} KM \\plus{} ML \\ge 2 \\sqrt {KM \\cdot ML} \\equal{} AC.$ Nothing prevents $ KL \\equal{} AC$ when $ AKCL$ becomes a square." } { "Tag": [], "Problem": "Do there exist 100 positive integers not exceeding 25000 such that the sum of every pair is distinct?", "Solution_1": "[hide=\"Hint\"]\nEvery positive integer can be expressed as the sum of distinct fibonacci numbers.\n[b]Demonstration:[/b] Note that for j>2, $ F_{j}=0.\"", "Solution_1": "\"Suppose we use synthetic division to find where has a positive leading coefficient and c>=0.\"\r\n\r\nso lets create an equation for $f(x)=x^{3}-x^{2}+2x+6$\r\n\r\nSo lets synthetically divide it by for instance $x-3$ so that:\r\n\r\n[code]3| 1 -1 2 6\n 3 6 24\n 1 2 8 30[/code]\nOk, so you see then when divided by 3 then the coefficients are all positive which means that there is no root greater than 3. Whenever you synthetically divide by a number and the coefficients and remainder come out positive there cannot be a root that is larger than that number, which is called the upper bound. This is because the remainder is too large to ever be able to equal zero, and you cannot get it smaller if all your values are positive, and they all will be if the number is larger than the experimented one. It's basically just logic.\n\nTo get the lower bound we once again are going to use $f(x)=x^{3}-x^{2}+2x+6$ as an example and divide this time by $x+3$:\n\n[code]-3| 1 -1 2 6\n -3 12 -42\n 1 -4 14 -36[/code]\r\n\r\nOk, so this time the signs in the last one alternate. If the signs in the answer alternate then you have your lower bound, which in this case is -3, so nothing is smaller than -3. The reason this works is because, well, try any smaller number and you'll see that the remainder will get closer to -infinity. It's logic again as well, since obviously you'll end up repeating the same pattern (positive, negative, positive, negative) and cannot get the remainder as 0.\r\n\r\nSo basically, you can zero in on the answer with this. So you try c, then -c, each proving to be either too large (if all resulting coefficients are positive) or too small (if resulting coefficients have alternating signs). You then try d (which is smaller than c) and -d to get possibly a closer upper and lower bound, and do it until you find the number which can be factored from the equation.\r\n\r\nI hope all this makes sense... :D \r\n\r\n(thak you solafidefarms for helping me out with this, I actually did my first equations using LaTeX :lol: )" } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "I'm having trouble seeing how:\r\n\r\n$ -\\dfrac {1}{2}mv^2 = mgd -\\dfrac {1}{2}kd^2$\r\n\r\nleads to:\r\n\r\n${ d = \\dfrac {mg + \\sqrt { m^2g^2 + mkv^2}} {k}}$\r\n\r\nCan anyone help with the initial step/s please? Thanks!", "Solution_1": "move everything to the right to get\r\n\r\n$ \\dfrac {1}{2}kd^2\\minus{}mgd\\minus{}\\dfrac {1}{2}mv^2\\equal{} 0$\r\n\r\nThen solve as a quadratic of d using the [url=http://en.wikipedia.org/wiki/Quadratic_equation]Quadratic Formula[/url]\r\n\r\nTip: If you're doing physics, unless it is the actual question, it is best not to memorize these formulas. It is often much more convenient to plug in a few numbers first... But yeah, spring problems sometimes require solutions to quadratic equations.", "Solution_2": "are you having trouble with the other 3 variables?\r\n\r\nyou can rewrite the equation as \r\n$ (\\frac{1}{2}k)d^2\\minus{}(mg)d\\minus{}(\\frac{1}{2}mv^2)$\r\nthen set the coefficients equal to a, b, and c", "Solution_3": "Thanks very much guys! :) \r\nYeah I solved it by plugging in numbers,\r\nbut wanted to see how the algebra worked; it's a physics problem combining\r\nthree separate formulas (formulae?)." } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "when does $ \\frac {d(\\frac {\\partial L}{\\partial \\dot{q}})}{dt} \\equal{} \\frac {\\partial \\dot{L}}{\\partial \\dot{q}}$ hold", "Solution_1": "not sure what dL/dq is but...\r\n\r\nanything that follows $ \\frac{dy}{dt}\\equal{}y$ is of the form $ y\\equal{}Ae^t$", "Solution_2": "Note that there is a dot over the $ L$ in the RHS. Since $ \\dot{L}\\equal{}\\frac{dL}{dt}$, this asks when we can change the order of differentiating with respect to $ t$ and $ \\dot{q}$.\r\n\r\nThis should be when the mixed partials exist and are continuous. Due to the fact that this is probably arising from the Hamiltonian or Lagrangian, there may be less strict requirements due to physical considerations. I don't really remember." } { "Tag": [], "Problem": "synthesize BENZENE from CALCIUM CARBIDE", "Solution_1": "hint:\r\ncalcuim carbide in water gives acetylene", "Solution_2": "[hide=\"Next\"]Then we could use an appropriate catalyst.[/hide]\r\n\r\nJust a question: what is the point of this problem?" } { "Tag": [ "algebra", "polynomial", "abstract algebra", "vector", "calculus", "integration", "Ring Theory" ], "Problem": "Let $ I$ be an ideal of the ring $\\mathbb{Z}\\left[x\\right]$ of all polynomials with integer coefficients such that \n\na) the elements of $ I$ do not have a common divisor of degree greater than $ 0$, and \n\nb) $ I$ contains of a polynomial with constant term $ 1$. \n\nProve that $ I$ contains the polynomial $ 1 + x + x^2 + ... + x^{r-1}$ for some natural number $ r$.\n\n[i]Gy. Szekeres[/i]", "Solution_1": "Actually $ I$ contains $ x^r\\plus{}x^{r\\plus{}1}\\plus{}\\cdots\\plus{}x^s$ for some $ r\\le s$, and that's the best you can do.", "Solution_2": "So you're saying that the statement is wrong? :? Because I'm quite sure it's not. I'll just sketch the proof I had in mind:\r\n\r\nI want to show that $ A \\equal{} \\mathbb Z[x]/I$ is finite. Condition (b) implies that $ x$ is invertible in $ A$, so $ x^t \\equal{} 1$ in $ A$ for some positive integer $ t$. Then, if, say, the positive integer $ n$ belongs to $ I$ (there is such a positive integer, for example the order of $ 1$ in the additive quotient group $ (A, \\plus{} )$), we have $ 1 \\plus{} x \\plus{} \\ldots \\plus{} x^{nt \\minus{} 1} \\equal{} 0$ in $ A$, i.e. $ 1 \\plus{} x \\plus{} \\ldots \\plus{} x^{nt \\minus{} 1}\\in I$. Hence it suffices to show that $ A$ is finite. \r\n\r\n$ I$ has a primary decomposition as $ I \\equal{} \\bigcap_i\\frak q_i$. Since $ A$ injects into $ \\prod_i\\left(\\mathbb Z[x]/\\frak q_i\\right)$ and the $ \\frak q_i$ obviously satisfy conditions (a) and (b), it suffices to prove that a primary ideal $ \\frak q$ satisfying (a) and (b) has finite quotient $ \\mathbb Z[x]/\\frak q$. Let $ \\frak p$ be the radical of $ \\frak q$. Some power $ \\frak p^n$ is contained in $ \\frak q$, so it suffices to show that $ \\mathbb Z[x]/\\frak p^n$ is finite. Finally, since $ \\mathbb Z[x]/\\frak p^n$ can be written as successive extensions of $ \\frak p^k/\\frak p^{k \\plus{} 1},\\ k \\equal{} \\overline{0,n \\minus{} 1}$, and these are finitely-generated modules over $ A/\\frak p$, it suffices to show that given a [i]prime[/i] ideal $ \\frak p$ with properties (a) and (b), the quotient ring $ A \\equal{} \\mathbb Z[x]/\\frak p$ is finite (in other words, the initial problem has been reduced to the case when $ I$ is prime). This is easy: (a) implies that the ideal contains a positive integer, so, being a prime ideal, it must contain some prime number $ p$. But now (b) shows that $ x\\in A$ is algebraic over the field $ \\mathbb F_p$ with $ p$ elements, so $ A$ is a finite-dimensional vector space over $ \\mathbb F_p$ (since $ A \\equal{} \\mathbb F_p[x]$).\r\n\r\n\r\nP.S.\r\n\r\nSometimes I used $ x$ as an indeterminate, and sometimes I used it, by a slight abuse of notation, in order to indicate the image of the indeterminate $ x$ through some canonical map of $ \\mathbb Z[x]$ onto a quotient ring; I hope it's clear from the context which is which.", "Solution_3": "Yes grobber, I misread b) and thought of leading coefficient $ 1$. Sorry.\r\n\r\nEDIT \r\nHere is a another proof.\r\n\r\nThe ideal $ \\mathbb{Q}I$ of $ \\mathbb{Q}[x]$ must be principal, say $ \\mathbb{Q}I \\equal{} t(x)\\mathbb{Q}[x]$ with some $ t(x)\\in\\mathbb{Z}[x]$ [url=http://planetmath.org/encyclopedia/PrimitivePolynomial.html]primitive[/url].\r\nThen $ I\\subseteq t(x)\\mathbb{Z}[x]$ and a) gives $ t(x) \\equal{} \\pm 1$.\r\nThen $ 1\\in \\mathbb{Q}I \\equal{} \\bigcup_{n\\in\\mathbb{N}}n^{ \\minus{} 1}I$ gives some natural number $ n\\in I$.\r\nSo $ \\mathbb{Z}[x]/I$ has some finite characteristic $ n\\in\\mathbb{N}$ and prime ring $ R\\cong\\mathbb{Z}/n\\mathbb{Z}$.\r\n\r\nBy b) we find $ f(x)\\in I$ of degree $ d$ with $ f(0) \\equal{} 1$. Then $ g(x) \\equal{} x^df(x^{ \\minus{} 1})\\in\\mathbb{Z}[x]$ is monic.\r\nLet $ y\\in \\mathbb{Z}[x]/I$ be the inverse of $ x \\plus{} I$.\r\nThen $ f(x) \\equal{} x^dg(x^{ \\minus{} 1})\\in I$ gives $ g(y) \\equal{} 0$, so $ y$ is [url=http://en.wikipedia.org/wiki/Integral_extension]integral[/url] over $ R$ and $ R[y]$ is finite.\r\nBut then $ 1,1 \\plus{} y,1 \\plus{} y \\plus{} y^2,\\ldots$ can't be all different \r\nand we find some $ r,s\\in\\mathbb{N},r < s$ with $ 1 \\plus{} y \\plus{} \\cdots \\plus{} y^r \\equal{} 1 \\plus{} y \\plus{} \\cdots \\plus{} y^s$.\r\nSo $ 0 \\equal{} y^{r \\plus{} 1} \\plus{} \\cdots \\plus{} y^s \\equal{} y^{s}(1 \\plus{} y^{ \\minus{} 1} \\plus{} \\cdots \\plus{} y^{ \\minus{} (s \\minus{} r \\minus{} 1)})$ and $ y$ invertible gives $ 1 \\plus{} x \\plus{} \\cdots \\plus{} x^{s \\minus{} r \\minus{} 1}\\in I$.\r\n\r\nPS Nice proof grobber! I like yours better than mine. :wink:", "Solution_4": "This problem can be solved in very easy way.\n\nEasy to see from $(a)$ that $n\\in I$. Let $f=1+a_1x+...+a_kx^k\\in I$, then if we proved that $x^r(1+x+x^2+...+x^j)\\in (n,f)$, then $x^r(1+x+x^2+...+x^j) = ng+fh$ where all coefficients of $h$ not divisible by $n$, so easy to see that $h=x^rh'$, $g=x^rg'$ and $1+x+x^2+...+x^j=ng'+fh'\\in I$.\n\nIn case where $\\gcd(a_k,n)=1$, easy to see that we can take mod $Z_n$ and see that for some $r_1< r_2$ polynoms $1+x+x^2+...+x^{r_1}$, $1+x+x^2+...+x^{r_2}$ will have same remains by $f$ in $Z_n$.\n\nLet polynom $f^*=1+d_1x+...+d_ex^e$ in $I$ with constant term $1$ and minimal value $a=d_e$ of highest term, easy to see that $d_e$ divides every other highest term of any polynom in $I$. Let $a\\not=1$.\n\nFixed number $b$ and take polynom $u=u_1+...+u_{l}x^{l}+u_tx^t$ in $I$ with minimal $\\deg\\not= 0$ in $Z_n$ and where $t-l=b$ and $u_t=a$. We can see that $u_t=a|n$. Take $\\frac{n}{u_t}u=\\frac{n}{u_t}u_1+...+\\frac{n}{u_t}u_{l}x^{l}+nx^t$ in $Z_n$ his $\\deg$ is smaller, so it is $0$ and so $n|\\frac{n}{u_t}u_{l}$ and $u_t|u_{l}$, so take $x^{t-l}u-\\frac{u_l}{u_t}u\\in I$ with bigger value of $t-l$ and highest term $a$.\n\nSo we can take polynom $Q=q_1+...+q_gx^g+ax^w$, where $w-g>e$. And from $f^*$ construct polynom in $I$ with smaller highest term. So $a=1$. done" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ x,y$ be positive number such that $ x\\plus{}y\\equal{}\\sqrt{10}$. Find the minimum of expression: $ M\\equal{}(x^4\\plus{}1)(y^4\\plus{}1)$", "Solution_1": "[quote=\"phuong\"]Let $ x,y$ be positive number such that $ x \\plus{} y \\equal{} \\sqrt {10}$. Find the minimum of expression: $ M \\equal{} (x^4 \\plus{} 1)(y^4 \\plus{} 1)$[/quote]\r\n$ M \\equal{} (x^4 \\plus{} 1)(y^4 \\plus{} 1)\\equal{}(x^4 \\plus{} 1)(1\\plus{}y^4) \\ge (x^2\\plus{}y^2)^2\\ge [\\frac{(x\\plus{}y)^2}{2}]^2\\equal{}25$", "Solution_2": "I think your answer has a little problem. The equality doesn't hold.", "Solution_3": "Let $ xy\\equal{}t$, $ M\\equal{}t^4\\plus{}2t^2\\minus{}40t\\plus{}101\\ \\left(0 3 setting x1 = x2 = ... = x(n-1) = 1, xn = 1/2 yields a contradiction,\r\nif n = 3 one can set x1 = x3 = 2, x2 = 3 and if n = 2 or n = 1 the inequality becomes obvious." } { "Tag": [ "geometry", "3D geometry", "Putnam", "college contests" ], "Problem": "A solid has a cylindrical middle with a conical cap at each end. The height of each cap equals the length of the middle. For a given surface area, what shape maximizes the volume?", "Solution_1": "I don't get it...It seems like they describe the shape pretty much perfectly in the problem.", "Solution_2": "I agree with [b]Rushil[/b]. The problem statement is not very clear.", "Solution_3": "Let $r$ be the radius of the cylinder and $h$ the \"length of the middle portion.\" Then I interpret this as an object whose overall length is $3h.$ \r\n\r\n$V=\\pi r^2h+\\frac23\\pi r^2h$\r\n\r\n$A=2\\pi rh+2\\pi r\\sqrt{r^2+h^2}$\r\n\r\nNow: maximize $V$ for fixed $A.$ (Or minimize $A$ for fixed $V,$ which would be the same problem.)", "Solution_4": "[quote=\"Kent Merryfield\"]Let $r$ be the radius of the cylinder and $h$ the \"length of the middle portion.\" Then I interpret this as an object whose overall length is $3h.$ \n\n$V=\\pi r^2h+\\frac23\\pi r^2h$\n\n$A=2\\pi rh+2\\pi r\\sqrt{r^2+h^2}$\n\nNow: maximize $V$ for fixed $A.$ (Or minimize $A$ for fixed $V,$ which would be the same problem.)[/quote]\r\n\r\nThanks, I see it now! I made a silly error in thinking the surface area ,$A = 2\\pi rh$ (surface of the middle)+ $2\\pi rh$(surface of the cone caps)\r\n\r\nI guess I need to brush up on my secondary school formulas." } { "Tag": [ "logarithms", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "$ \\sum_{n\\equal{}1}^{\\infty}(1\\minus{}\\frac{ \\ln n}{n})^n$\r\n\r\nThe above series seems convergence....\r\n\r\nUse Cauchy's Test , I can't get anything\r\n\r\nIt seems difficult to work out..... :o", "Solution_1": "edit: my hint was misleading. Thanks Kent.", "Solution_2": "It diverges by comparison to the harmonic series.\r\n\r\n$ \\left(1 \\minus{} \\frac {\\ln n}n\\right)^n \\equal{} \\text{exp}\\left(n\\ln\\left(1 \\minus{} \\frac {\\ln n}n\\right)\\right)$\r\n\r\n$ \\equal{} \\text{exp}\\left( \\minus{} \\ln n \\minus{} \\frac {\\ln^2n}{2n} \\plus{} o(n^{ \\minus{} 1})\\right)$\r\n\r\n$ \\equal{} e^{ \\minus{} \\ln n}\\cdot e^{o(1)} \\equal{} \\frac1n\\cdot e^{o(1)}.$\r\n\r\nBut $ e^{o(1)}$ is the sequence of exponentials of a sequence that tends to zero. It is both bounded above and bounded below away from zero. We conclude that\r\n\r\n$ \\left(1 \\minus{} \\frac {\\ln n}n\\right)^n > \\frac cn$ for some positive constant $ c.$\r\n\r\nHence the claim of divergence for the series.\r\n\r\nEdit: WWW's inequality points in the wrong direction. $ 1 \\minus{} \\frac {\\ln n}n\\to 1$ as $ n\\to \\infty.$", "Solution_3": "Thanks very much, Wonderful !\r\n\r\nIn your reply:\r\n\r\n$ \\text{exp}\\left(n\\ln\\left(1 \\minus{} \\frac {\\ln n}{n}\\right)\\right) \\equal{} \\text{exp}\\left( \\minus{} \\ln n \\minus{} \\frac {\\ln^{2}n}{2n} \\plus{} o(n^{ \\minus{} 1})\\right)$\r\n\r\nBut I think the right hand of the above equation maybe $ \\text{exp}\\left( \\minus{} \\ln n \\minus{} \\frac {\\ln^{2}n}{2n} \\plus{} o(\\frac {\\ln^{2}n}{n})\\right)$\r\n\r\nbecause:\r\n\r\n$ \\ln\\left(1 \\minus{} \\frac {\\ln n}{n}\\right) \\equal{} \\minus{} \\frac {\\ln n}{n} \\minus{} \\frac {\\ln^{2}n}{2n^2} \\plus{} o(\\frac {\\ln^{2}n}{n^2})$\r\n\r\nso plus $ n$\r\n\r\nit equal:\r\n\r\n$ \\minus{} \\ln n \\minus{} \\frac {\\ln^{2}n}{2n} \\plus{} o(\\frac {\\ln^{2}n}{n})$\r\n\r\nand $ o(\\frac {\\ln^{2}n}{n})$ seems not have the same order as $ o(\\frac {1}{n})$\r\n\r\nconsider: $ \\frac {\\ln n}{n}$", "Solution_4": "Use a better series estimate. We have $ \\ln(1 \\minus{} u) \\equal{} \\minus{} u \\minus{} \\frac {u^2}{2} \\plus{} O(u^3)$, so that term is $ n\\cdot O(\\frac {\\ln^3 n}{n^3}) \\equal{} O(\\frac {\\ln^3 n}{n^2}) \\equal{} o(n^{ \\minus{} 1})$, using that logarithms are smaller than powers of $ n$ in the last step.", "Solution_5": "The whole point of using the big-O, little-o notation is to avoid getting bogged down in details that don't really matter.\r\n\r\nIn fact, the next term of that Taylor expansion is $ O\\left(\\frac{\\ln^3n}{n^2}\\right),$ And it is true that that is smaller than $ \\frac1n,$ hence in calling it $ o(n^{\\minus{}1})$ I was making a true statement.\r\n\r\nI realized a few lines later that even there I was giving more detail than I needed to, and the whole expression $ \\minus{}\\frac{\\ln^2n}{2n}\\plus{}o(n^{\\minus{}1})$ need only be written as $ o(1).$\r\n\r\nIn your criticism of my argument, you're trying to be more precise, to give more information. My defense to that is that I didn't need to give that information, and the the effort involved in being more precise is wasted effort.", "Solution_6": "Thanks very much :D \r\n\r\nI know :D" } { "Tag": [ "geometry", "rectangle", "quadratics", "LaTeX", "algebra", "quadratic formula" ], "Problem": "A 39 ounce can of coffee requires 188.5 square inches of aluminum. If its height is 7 inches, what is its radius? \r\n\r\nHINT GIVEN: Think about a right circular cylinder.", "Solution_1": "[hide]\n$V = r^{2}h \\pi$\n$V = 188.5$\n$h = 7$\n$r = \\sqrt{\\frac{188.5}{7\\pi}}$\n[/hide]", "Solution_2": "Actually, AstroPhys, I think you did the formula for volume when the question is asking for the surface area.\r\n\r\n[hide=\"hide hide hide hide hide hide hide hide...a message brought to you by jli\"]Since a cylinder's surface can be \"flattened\" to a rectangle and two circles, with the rectangle's width being the circles' circumferences, the equation for this problem can be written:\n\n$188.5=2{\\pi}r^{2}+7(2{\\pi}r)$ or $0=2{\\pi}r^{2}=14{\\pi}r-188.5$.\n\nPutting this into the quadratic formula, we get:\n\n${(-14{\\pi}+/-{\\sqrt}{196{\\pi}^{2}-8{\\pi}+754)}/{4{\\pi}}$. This simplifies to $r$ being very close to either $-10$ or $3$. Since the result must be a real answer, $r=3$.[/hide]\r\n\r\nEDIT: I'm sorry, I can't quite get the latex on the quadratic formula to work. I hope it's understandable to everyone.", "Solution_3": "they already explained it.\r\n\r\n$Surface area = 2\\pi(r^{2}+h)$", "Solution_4": "[quote=\"sharkman\"]A 39 ounce can of coffee requires 188.5 square inches of aluminum. If its height is 7 inches, what is its radius? \n\nHINT GIVEN: Think about a right circular cylinder.[/quote]\r\n[hide=\"soltuion\"]\n39 ounces is completely irrelevant. \n\nThe surface area of a can is $2\\pi rh+2\\pi r^{2}=14\\pi r+2\\pi r^{2}=188.5$\n\nSo $r\\approx-10$ and $3$ so its $\\boxed{3}$.\n[/hide]\r\nTo igiul, the height needs to be multiplied by r." } { "Tag": [ "logarithms", "floor function", "number theory proposed", "number theory" ], "Problem": "Prove that for every natural number like $n\\geq 2$,\r\n\r\n$\\sum_{i=1}^{n-1}\\lfloor{\\sqrt[i+1]n}\\rfloor=\\sum_{i=1}^{n-1}\\lfloor{\\log_{i+1}n}\\rfloor$", "Solution_1": "Are you sure this is correct ? Mupad suggests that the statement is false :?", "Solution_2": "Look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=91284" } { "Tag": [ "geometry", "ratio", "geometric series", "area of a triangle", "Heron\\u0027s formula" ], "Problem": "The sequence $a_{1}, a_{2}, a_{3}, \\ldots$ is a geometric sequence.\r\nThe sequence $b_{1}, b_{2}, b_{3}, \\ldots$ is a geometric sequence.\r\n$b_{1}= 1$\r\n$b_{2}= \\sqrt[4]{7}-\\sqrt[4]{28}+1$\r\n$a_{1}= \\sqrt[4]{28}$\r\n$ \\sum_{n=1}^{\\infty}{\\frac{1}{a_{n}}}= \\sum_{n=1}^{\\infty}{b_{n}}$\r\n\r\nCompute the area of the triangle with side lengths $a_{1}, a_{2},$ and $a_{3}$.", "Solution_1": "[hide]$\\sum_{n=1}^{\\infty}{\\frac{1}{a_{n}}}= \\sum_{n=1}^{\\infty}{b_{n}}\\Rightarrow \\frac{1/\\sqrt[4]{28}}{1-1/k_{a}}= \\frac{1}{\\sqrt[4]{28}-\\sqrt[4]{7}}\\Rightarrow k_{a}= \\sqrt{2}\\Rightarrow a_{1}= \\sqrt[4]{28}, a_{2}= 2\\sqrt[4]{7}, a_{3}= 2\\sqrt[4]{28}\\Rightarrow \\boxed{A = \\frac{7}{2}}$\n\n$k_{a}$ is the constant ratio of the $a$ geometric series. I used the formula for the sum of a geometric series as $\\frac{t_{1}}{1-r}$ and used Heron's formula to find the area.[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Using only the numbers 1, 2, 3, 4and 5 (none more than once) and the operations +, -, * and / and brackets, how many of the following numbers is it possible to make: 76, 79, 86, 92, 97 and 98? Is theree a website for this type of question?", "Solution_1": "Hm... Maybe I could make a computer program for that, but that isn't so easy. Anyway, I'll try.", "Solution_2": "This link is for 6 numbers.\r\n\r\nhttp://www.ee.unb.ca/cgi-bin/tervo/compte.html" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "In a community of more than six people each member exchanges letters\r\nwith exactly three other members of the community. Show that the com-\r\nmunity can be partitioned into two nonempty groups so that each member\r\nexchanges letters with at least two members of the group he belongs to.", "Solution_1": "EDIT: You are right RealZeta. That old argument could be modified to show we can partition the group into two groups where each person knows at most 1 person in each group, but for this problem, since the complete community satisfies the condition, I cannot guarantee that one of the groups is non-empty. I'm working on another approach that I think will work...", "Solution_2": "Perhaps I am not seeing it, but is it guranteed that both sets remain non-empty?", "Solution_3": "Ok, lets try this. It may have a kink or two, so let me know if it does.\r\n\r\nFirst, let $n$ be the number of people in the community. We have $n\\ge 7$ is best possible, since a regular hexagon with the outside edges and $3$ pairs of longest diagnols connected cannot be partitioned correctly. So $n=6$ does not always work.\r\n\r\nNow, let $G=(V,E)$ be the graph of the community, $V$ be the set of people, and $E$ be the edges, such that two vertices are connected if and only if they share a letter. It is obvious that for $n>6$ that $|E|>n>n-1=|V|-1$. This means that $G$ cannot be a tree, in other words, $G$ is not acyclic.\r\n\r\nNow let $C$ be the cycle in $G$ with the shortest length. This leaves three cases:\r\n\r\n[b]Case I:[/b] $|C|$ is at least $5$.\r\nNow, because the degree of each vertex is $3$, each of these points in the cycle connects to exactly one more point. However, they cannot connect to any of the other points in the cycle besides their neighbors, for then the cycle would not be the shortest. Secondly, no two different points in the cycle can connect to the same point outside the cycle because, again, $C$ would not be the shortest. So each the points in the cycle connect to unique points outside the cycle, and since for all vertices outside $C$ will have degree $2$ in the subgraph of $G$ \\ $C$, we have the desired partition.\r\n\r\n[b] Case II:[/b] $|C|=3$.\r\nThis is the simplest case. Let $C=C_{1}C_{2}C_{3}$ be the vertices in the cycle. Then each of the $C_{i}$ connect to one more point outside $C$, and they can make at most one vertex outside $C$ have degree less than $2$ in subgraph $G$ \\ $C$. In this case, include that point with $C$, and since there are at least $3$ more points, we are done. Otherwise, we are done anyways.\r\n\r\n[b] Case III:[/b] $|C|=4$. This is a bit awkward. Let $C=C_{1}C_{2}C_{3}C_{4}$ like above. Now, like in Case I, we cannot connect anymore of the $C_{i}$'s together, otherwise $C$ is not the shortest. However, we can connect $C_{1}$ and $C_{3}$ to the same point outside and the same goes for $C_{2}$ and $C_{4}$. Now $n$ cannot be odd, since $\\sum_{v\\in V}deg(v)=3n=2|E|$ which is impossible. And if $n=8$, then we cannot have both pairs in the cycle latch to the same points, because then the $deg(v)=3$ for all vertices cannot hold (there are only a finite number of cases to check). And finally, for $n\\ge 10$, there are no problems to partition the groups. So we are done.", "Solution_4": "We make a graph $G(V, E)$ with $|V|=6$, we connect two vertice if and only if they communicate with each other.\r\n\r\nWe call a partition $(A,B), A+B=V$ is good such that the number of edges between $A$ and $B$ reaches the minimum.\r\n\r\nIt is easy to see a good partition is an answer, because if there is a member doesn't exchanges letters with at least two members in his group, then we put him in the other group, the value I mentioned above will decrease, it is a contradiction with the definition of good partition!", "Solution_5": "[quote=\"libra_gold\"]\nIt is easy to see a good partition is an answer.\n[/quote]\r\n\r\nbut, In this situation, A or B can be empty! right?" } { "Tag": [ "geometry" ], "Problem": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b5\u03af\u03bc\u03b1\u03b9 \u03ba\u03b1\u03b9\u03bd\u03bf\u03cd\u03c1\u03b9\u03bf\u03c2 \u03c3\u03c4\u03bf forum \u03ba \u03bc'\u03b1\u03c1\u03ad\u03c3\u03b5\u03b9 \u03b7 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1.\u039c'\u03b1\u03c1\u03ad\u03c3\u03b5\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b7 \u03c0\u03b1\u03c1\u03ad\u03b1 \u03c3\u03b1\u03c2.\u0388\u03c7\u03c9 \u03ad\u03bd\u03b1 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03b2\u03b1\u03c3\u03b1\u03bd\u03af\u03b6\u03bf\u03bc\u03b1\u03b9 2 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03ba \u03b8\u03ad\u03bb\u03c9 \u03bd\u03b1 \u03c4\u03bf \u03b4\u03b5\u03af\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b5 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03b5\u03c4\u03b5.\u0388\u03c7\u03bf\u03c5\u03bc\u03b5 \u03ad\u03bd\u03b1 \u03b9\u03c3\u03cc\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03ba \u03ad\u03bd\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c0\u03bf\u03c5 \u03b1\u03c0\u03ad\u03c7\u03b5\u03b9 1, 3/8 \u03ba 5/8 \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b9\u03c2 \u03ba\u03bf\u03c1\u03c5\u03c6\u03ad\u03c2 \u03c4\u03bf\u03c5.\u03a0\u03cc\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5?", "Solution_1": "\u039a\u03b1\u03bb\u03c9\u03c3\u03ae\u03c1\u03b8\u03b5\u03c2 billakos!!! \u0395\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03c5\u03c7\u03ac\u03c1\u03b9\u03c3\u03c4\u03bf \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03c4\u03af\u03b8\u03b5\u03bd\u03c4\u03b1\u03b9 \u03bd\u03ad\u03bc\u03b1 \u03bc\u03ad\u03bb\u03b7 \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03ad\u03b1 \u03bc\u03b1\u03c2 :) \r\n\r\n\u038c\u03c3\u03bf\u03bd \u03b1\u03c6\u03bf\u03c1\u03ac \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03ac \u03c3\u03bf\u03c5 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b1\u03c5\u03c4\u03cc \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03bd\u03b1\u03bb\u03c5\u03c4\u03b9\u03ba\u03ae \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1 (\u03b1\u03bd \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03c9\u03c0\u03b9\u03ba\u03ac \u03b4\u03b5\u03bd \u03c4\u03b7\u03bd \u03be\u03ad\u03c1\u03c9 \u03ba\u03b1\u03bb\u03ac :oops: ).\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03b1\u03bd \u03b8\u03b5\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 $ ABC$ \u03c4\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf, $ M$ \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03ba\u03b1\u03b9 $ MA \\equal{} 1, MB \\equal{} 3/8, MC \\equal{} 5/8$, \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03c4\u03b1\u03b3\u03bc\u03ad\u03bd\u03b5\u03c2 \u03c4\u03bf\u03c5 \u039c \u03c3\u03b5 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03c9\u03c2 \u03c4\u03b7\u03bd \u03c4\u03bf\u03bc\u03ae \u03b4\u03cd\u03bf \u03ba\u03cd\u03ba\u03bb\u03c9\u03bd \u03bc\u03b5 \u03ba\u03ad\u03bd\u03c4\u03c1\u03b1 \u03c4\u03b1 $ A, B$ \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03bd\u03b1 \u03c0\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03ad\u03c7\u03b5\u03b9 \u03b1\u03c0' \u03c4\u03bf \u03c4\u03c1\u03af\u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03b7\u03bd \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b7 \u03b1\u03c0\u03cc\u03c3\u03c4\u03b1\u03c3\u03b7 (\u03c0.\u03c7. $ MC \\equal{} 5/8$) \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03bc\u03b5 \u03bc\u03cc\u03bd\u03bf \u03ac\u03b3\u03bd\u03c9\u03c3\u03c4\u03bf \u03c4\u03b7\u03bd \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5.\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03c7\u03c9\u03c1\u03af\u03c2 \u03bd\u03b1 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03ad\u03bd\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03bb\u03c5\u03b8\u03b5\u03af \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03bc\u03b5\u03c4\u03c1\u03af\u03b1... \u03bc\u03b5 \u03c7\u03c1\u03ae\u03c3\u03b7 \u03bd\u03cc\u03bc\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 :wink: \r\n\r\n\u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c9\u03c0\u03b9\u03ba\u03ac \u03b8\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ae\u03c3\u03c9 \u03bd\u03b1 \u03b2\u03c1\u03c9 \u03bc\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03ba\u03b1\u03b8\u03b1\u03c1\u03ac \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03c0\u03bf\u03c5 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03cc\u03c4\u03b9 \u03b8\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 :P", "Solution_2": "\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ae \u03bb\u03cd\u03c3\u03b7 . (\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03bb\u03b1\u03b9\u03cc\u03c4\u03b5\u03c1\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03bc\u03ae\u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03b4\u03b9\u03b1 \u03b1\u03c0\u03bb\u03ac \u03bc\u03b5 \u03ac\u03bb\u03bb\u03b1 \u03bd\u03bf\u03cd\u03bc\u03b5\u03c1\u03b1 )\r\n\u0397 \u03b2\u03b1\u03c3\u03b9\u03ba\u03ae \u03b9\u03b4\u03ad\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03c0\u03ac\u03c1\u03b5\u03b9\u03c2 \u03c4\u03b1 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac \u03c4\u03bf\u03c5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf\u03c5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b9\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 :wink: \r\n(\u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03ac \u03bd\u03b1 \u03b4\u03bf\u03c5\u03bb\u03ad\u03c8\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03b5\u03bc\u03b2\u03b1\u03b4\u03ac)", "Solution_3": "\u0391\u03ba\u03cc\u03bc\u03b1 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03bb\u03cd\u03c3\u03b7 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03b4\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b1\u03bd \u03c3\u03ba\u03b5\u03c6\u03c4\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf \u039c \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03c4\u03bf \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf \u03c4\u03bf\u03c5 $ ABC$ \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03cc\u03c4\u03b9 $ MA \\equal{} MB \\plus{} MC$ \u03ba\u03b1\u03b9 \u03b1\u03c0' \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 \u03a0\u03c4\u03bf\u03bb\u03b5\u03bc\u03b1\u03af\u03bf\u03c5.\r\n\r\n\u039c\u03b5\u03c4\u03ac \u03b1\u03c0\u03bf \u03b1\u03c5\u03c4\u03cc \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03bd\u03cc\u03bc\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd \u03c3\u03c4\u03bf $ MAB$ \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03b9\u03c2 \u03b1\u03bc\u03ad\u03c3\u03c9\u03c2 \u03c4\u03b7\u03bd \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03b7 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac (\u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 $ \\frac {49}{64}$ \u03b1\u03bd \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2)", "Solution_4": "\u038c\u03bb\u03b1 \u03c3\u03c7\u03b4\u03cc\u03bd \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b7\u03b3\u03bf\u03c1\u03af\u03b1 \u03c4\u03bf\u03c5 \u03b8\u03b5\u03c9\u03c1\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 Pompeiu.\u0391\u03bd \u03c4\u03b1 \u03bc\u03ae\u03ba\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03c3\u03c5\u03bd\u03ae\u03b8\u03c9\u03c2 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c3\u03c4\u03c1\u03bf\u03c6\u03ae \u03b3\u03cd\u03c1\u03c9 \u03c0\u03c7 \u03c3\u03c4\u03bf \u0392\u039c \u03ba\u03b1\u03c4\u03ac 60 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c3\u03cd\u03b3\u03ba\u03c1\u03b9\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03bd\u03cc\u03bc\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd. \u0391\u039d \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf \u03bc\u03b5 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03b4\u03cd\u03bf \u03ac\u03bb\u03bb\u03c9\u03bd , \u03c4\u03cc\u03c4\u03b5 \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b1\u03bd \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac.\r\n \r\n\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2", "Solution_5": "\u039c\u03ae\u03c0\u03c9\u03c2 \u03ad\u03b9\u03bd\u03b1\u03b9 \u03b5\u03cd\u03ba\u03bf\u03bb\u03bf \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03af\u03c4\u03b5 \u03c0\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b7 \u03c4\u03bf\u03c5 \u03c0\u03c1\u03bf\u03b7\u03b3\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf\u03c5 \u03bc\u03ae\u03bd\u03b1 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b1\u03c4\u03b5?", "Solution_6": "[quote=\"stergiu\"]\u038c\u03bb\u03b1 \u03c3\u03c7\u03b4\u03cc\u03bd \u03b1\u03c5\u03c4\u03ac \u03c4\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b7\u03b3\u03bf\u03c1\u03af\u03b1 \u03c4\u03bf\u03c5 \u03b8\u03b5\u03c9\u03c1\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 Pompeiu.\u0391\u03bd \u03c4\u03b1 \u03bc\u03ae\u03ba\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03c3\u03c5\u03bd\u03ae\u03b8\u03c9\u03c2 \u03ba\u03ac\u03bd\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c3\u03c4\u03c1\u03bf\u03c6\u03ae \u03b3\u03cd\u03c1\u03c9 \u03c0\u03c7 \u03c3\u03c4\u03bf \u0392\u039c \u03ba\u03b1\u03c4\u03ac 60 \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c3\u03cd\u03b3\u03ba\u03c1\u03b9\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03bd\u03cc\u03bc\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03c9\u03bd. \u0391\u039d \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03bf \u03bc\u03b5 \u03c4\u03bf \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c4\u03c9\u03bd \u03b4\u03cd\u03bf \u03ac\u03bb\u03bb\u03c9\u03bd , \u03c4\u03cc\u03c4\u03b5 \u03b4\u03bf\u03c5\u03bb\u03b5\u03cd\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03c0\u03b1\u03bd \u03c4\u03b1 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac.\n \n\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2[/quote]\r\n\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03bc\u03bf\u03c5 \u03c0\u03b5\u03b9\u03c2 \u03c6\u03af\u03bb\u03b5 \u03c0\u03bf\u03b9\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 \u03c4\u03bf\u03c5 Pompeiu?" } { "Tag": [], "Problem": "I don't know if this the right forum, but what is the Pearson product-moment correlation? The problem gives a table of x's and y's and asks me to use a graphing calculator to find the Pearson product-moment correlation.", "Solution_1": "http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient\r\n\r\nThis isn't really the right forum for this kind of question." } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Given a natural number $n \\geq 2$, consider all the fractions of the form $\\frac{1}{ab}$, where $a$ and $b$ are natural numbers, relative primes and such that:\n$a < b \\leq n$, \n$a+b>n$.\nShow that for each $n$, the sum of all this fractions are $\\frac12$.", "Solution_1": "It is easy to use induction on this one. If by $T_n$ you denote desired sum and by $H_n= \\sum_{i=1}^n \\frac{1}{i}$ then $T_{n+1}=T_n + \\frac{H_n}{n+1} - \\frac{H_{n-1}}{n}$", "Solution_2": "[quote=\"Megus\"]It is easy to use induction on this one. If by $T_n$ you denote desired sum and by $H_n= \\sum_{i=1}^n \\frac{1}{i}$ then $T_{n+1}=T_n + \\frac{H_n}{n+1} - \\frac{H_{n-1}}{n}$[/quote]\n\n\nBut according to the problem, $T_{n+1} = T_n = \\frac{1}{2}$, which implies that \n\n$n H_n = (n+1)H_{n-1}$,\n\nwhich is false for $n>2$.", "Solution_3": "Does anyone have a valid solution for this problem?", "Solution_4": "Lemma. $A = \\sum_{(a,n+1)=1} \\dfrac {1} {a(n+1)} = \\dfrac {1} {2} \\sum_{(a,n+1)=1} \\dfrac {1} {a(n+1-a)} = B$.\n\nIndeed, $A = \\dfrac {1} {n+1} \\sum_{(a,n+1)=1} \\dfrac {1} {a}$, while $B = \\dfrac {1} {2(n+1)} \\sum_{(a,n+1)=1} \\left (\\dfrac {1} {a} + \\dfrac {1} {n+1-a}\\right) = \\dfrac {2} {2(n+1)} \\sum_{(a,n+1)=1} \\dfrac {1} {a}$, since $(a,n+1)=1$ if and only if $(n+1-a,n+1)=1$ if and only if $(a,n+1-a)=1$. \nBut for $(a,b)=1$, $a+b=n+1$, $1\\leq a < b \\leq n$ one clearly has $\\sum \\dfrac {1} {ab} = B$.\n\nNow, when passing from $n$ to $n+1$, the fractions with $b=n+1$ are added, while those with $a+b=n+1$ should be subtracted, hence the value of the sum remains the same, since $A-B=0$. This value is thus that for $n=2$, i.e. $\\dfrac {1} {2}$.", "Solution_5": "[quote=\"mavropnevma\"]Lemma. $A = \\sum_{(a,n+1)=1} \\dfrac {1} {a(n+1)} = \\dfrac {1} {2} \\sum_{(a,n+1)=1} \\dfrac {1} {a(n+1-a)} = B$.\n\nIndeed, $A = \\dfrac {1} {n+1} \\sum_{(a,n+1)=1} \\dfrac {1} {a}$, while $B = \\dfrac {1} {2(n+1)} \\sum_{(a,n+1)=1} \\left (\\dfrac {1} {a} + \\dfrac {1} {n+1-a}\\right) = \\dfrac {2} {2(n+1)} \\sum_{(a,n+1)=1} \\dfrac {1} {a}$, since $(a,n+1)=1$ if and only if $(n+1-a,n+1)=1$ if and only if $(a,n+1-a)=1$. \nBut for $(a,b)=1$, $a+b=n+1$, $1\\leq a < b \\leq n$ one clearly has $\\sum \\dfrac {1} {ab} = B$.\n\nNow, when passing from $n$ to $n+1$, the fractions with $b=n+1$ are added, while those with $a+b=n+1$ should be subtracted, hence the value of the sum remains the same, since $A-B=0$. This value is thus that for $n=2$, i.e. $\\dfrac {1} {2}$.[/quote]\n\n\nGreat solution!! I knew that after proving that, it solved the problem, but I didn't know how to." } { "Tag": [], "Problem": "Why is NO3- a stronger oxidizing agent than H+?", "Solution_1": "edit~if you read what i said before, IGNORE IT!!! it did not understand your question correctly.\r\n\r\nthe reason $NO_3$ is a stronger oxidizing agent is that it's reduction half reaction has a higher reduction $E^0$ . the half reactions are as following:\r\n\r\n$NO3^- +4H^+ 3e^-\\rightarrow NO +2H_2O; E^0=+1.23$\r\n\r\n$2H^+ +2e^-\\rightarrow H_2; E^0=0.00$" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "what is the no of solution\r\n $ 2^x \\plus{} 3^x \\plus{} 4^x \\minus{} 5^x \\equal{} 0$ :(", "Solution_1": "Rewrite your equation as $ \\left (\\frac {2} {5}\\right )^x \\plus{} \\left (\\frac {3} {5}\\right )^x \\plus{} \\left (\\frac {4} {5}\\right )^x \\equal{} 1$. Left-Hand-Side is a strictly decreasing continuous function, \r\ntaking value $ 3$ at zero, and with limit $ 0$ at infinity, so it crosses value $ 1$ exactly once.\r\n\r\nThe approximate solution is 2.373294368404781." } { "Tag": [ "limit", "logarithms", "calculus", "function", "calculus computations" ], "Problem": "Prove that $ \\lim_{x\\to0}x\\ln x\\equal{}0.$\r\n\r\nDon't apply L'H\u00f4pital's Rule.", "Solution_1": "thats kind of clear but let $ x\\equal{}e^{\\minus{}t}$ then we have:\r\n\r\n$ L\\equal{}\\minus{}\\lim_{t\\rightarrow\\plus{}\\infty}\\frac{t}{e^{t}}\\equal{}0$", "Solution_2": "[quote=\"sinajackson\"]\n$ L \\equal{} \\minus{} \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}} \\equal{} 0$[/quote]\r\nYeah, but, can you explain why $ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}} \\equal{} 0$? I know it's 'cause $ e^t$ grows faster than $ t.$ I'd like another explanation.", "Solution_3": "At first because $ t>0$ and $ e^t>0$, $ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}} \\ge 0$. At second, for positive $ t$:$ e^t\\equal{}1\\plus{}t\\plus{}\\frac{1}{2}t^2\\plus{}...>\\frac{1}{2}t^2$, so \r\n\\[ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}}\\le\\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{\\frac{1}{2}t^2}\\equal{}\\lim_{t\\rightarrow \\plus{} \\infty}\\frac {2}{t}\\equal{}0\\]\r\n and we have that $ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}}\\equal{}0$", "Solution_4": "[quote=\"greenvert\"]At first because $ t > 0$ and $ e^t > 0$, $ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}} \\ge 0$. At second, for positive $ t$:$ e^t \\equal{} 1 \\plus{} t \\plus{} \\frac {1}{2}t^2 \\plus{} ... > \\frac {1}{2}t^2$, so\n\\[ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}}\\le\\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{\\frac {1}{2}t^2} \\equal{} \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {2}{t} \\equal{} 0\n\\]\nand we have that $ \\lim_{t\\rightarrow \\plus{} \\infty}\\frac {t}{e^{t}} \\equal{} 0$[/quote]\r\n\r\nAre you sure we can write: $ e^t \\equal{} 1 \\plus{} t \\plus{} \\frac {1}{2}t^2 \\plus{} ... > \\frac {1}{2}t^2$ ?\r\nremember $ t\\rightarrow\\infty$!!! and we can write that just in case $ t\\rightarrow0$ ...\r\n\r\nbut we can say that $ e^{t} > t^{2}$ for big $ t$ and continue!", "Solution_5": "[quote=\"sinajackson\"]remember $ t\\rightarrow\\infty$!!! and we can write that just in case $ t\\rightarrow0$ ...[/quote]\r\n\r\nIt is an easy exercise in calculus to show that $ e^x > 1 \\plus{} x \\plus{} \\frac{x^2}{2!} \\plus{} \\cdots \\plus{} \\frac{x^n}{n!}$ for all $ x > 0$. This even does not need knowledge about power series, and if you know calculus, this becomes much more simple. all terms of $ e^x \\equal{} \\sum_{n\\equal{}0}^{\\infty} \\frac{x^n}{n!}$ are positive if $ x > 0$, so greenvert's argument is valid.", "Solution_6": "[quote=\"sos440\"][quote=\"sinajackson\"]remember $ t\\rightarrow\\infty$!!! and we can write that just in case $ t\\rightarrow0$ ...[/quote]\n\nIt is an easy exercise in calculus to show that $ e^x > 1 \\plus{} x \\plus{} \\frac {x^2}{2!} \\plus{} \\cdots \\plus{} \\frac {x^n}{n!}$ for all $ x > 0$.[/quote]\r\nthats right sos440, i didnt deny this :D , i just said that we cant have the equality for $ x\\rightarrow\\infty$! but $ >$ doesnt make any problem! :)", "Solution_7": "Here's an argument. (I don't claim it's a [i]good[/i] argument but at least it's different.)\r\n\r\nLet $ s>0$ be arbitrary and consider $ f(x)\\equal{}x^s\\ln x$ for $ x>0.$\r\n\r\n$ f(x)<0$ for $ 00.$\r\n\r\nBut for any fixed $ s,$ consider $ r$ such that $ 0=20\r\n\r\nm^3 < (m-3)*((m+2)^2) < (m-3)*(m+2)*(m+3) < (m-3)*((m+3)^2)<(m+1)^3\r\n\r\ncheck by expanding..\r\n\r\nTrivial solution, although it required me some time to arrive at...", "Solution_9": "Alternative Solution:\nSet:\n$a=m^2(m+3)$\\\\\n$b=(m+3)(m-1)^2$\\\\\n$c=m(m+3)(m-1)$\\\\\nSo $abc=[(m-1)(m)(m+3)]^3$\\\\\nAnd we can check $m^3< a,b,c <(m+1)^3$ for all $m \\geq 2$" } { "Tag": [ "search", "function", "\\/closed" ], "Problem": "You know the stars in your personal info on the side of every post? How many posts do you need to get\r\n1/2 a star?\r\n1 star?\r\n1 1/2 stars?\r\netc...", "Solution_1": "To get 4 stars, you click the search button.\r\n\r\nTo get 5 stars, you type in what you want to search and hit enter. For example, your question. \r\n\r\nDo not pass GO. Do not collect $ \\$200$.", "Solution_2": "For me it's not working (search)", "Solution_3": "Well since you couldn't find it, I found it for you. :maybe: \r\n0 - 19 New Member (Zero Stars) \r\n20 - 49 P versus NP (Half Star) \r\n50 - 99 Hodge Conjecture (One Star) \r\n100 - 249 Poincare Conjecture (Two Stars) \r\n250 - 499 Riemann's Hypothesis (Two and Half stars) \r\n500 - 999 Yang Mills Theory (Three Stars) \r\n1000 - 2499 Navier-Stokes Equation (Four Stars) \r\n2500 - Birch & Swinnerton Dyer. (Five Stars)", "Solution_4": "Thanks. Also, I found it--the search function works again =]", "Solution_5": "[quote=\"Dojo\"]Well since you couldn't find it, I found it for you. :maybe: \n0 - 19 New Member (Zero Stars) \n20 - 49 P versus NP (Half Star) \n50 - 99 Hodge Conjecture (One Star) \n100 - 249 Poincare Conjecture (Two Stars) \n250 - 499 Riemann's Hypothesis (Two and Half stars) \n500 - 999 Yang Mills Theory (Three Stars) \n1000 - 2499 Navier-Stokes Equation (Four Stars) \n2500 - Birch & Swinnerton Dyer. (Five Stars)[/quote]\r\n\r\nI thought it was 0-24 New Member and 25-49 P versus NP...", "Solution_6": "No, Dojo is correct." } { "Tag": [], "Problem": "a) prove that there are no right triangles with all integer sides such that the lengths of both legs are powers of 2.\r\n\r\nb) find the smallest integer $k$ such that there exists a right triangle with integer hypotenuse and legs of length $k^{a}$, $k^{b}$ for integers $a,b$.\r\n\r\nedit: forgot part b", "Solution_1": "[quote=\"chess64\"]prove that there are no right triangles with all integer sides such that the lengths of both legs are powers of 2.[/quote]\r\n\r\n[hide=\"Uh?\"]$\\sqrt{2^{2a}+2^{2b}}$, WLOG $a\\ge b$, $\\sqrt{2^{2b}(2^{2(a-b)}+1)}=2^{b}\\sqrt{2^{2(a-b)}+1}$, $2^{2(a-b)}$ is a square, so $2^{2(a-b)}+1$ is not a square, so $2^{b}\\sqrt{2^{2(a-b)}+1}$ is not an integer.[/hide]", "Solution_2": "oops, i forgot part b... i just added it", "Solution_3": "It seems like kind of a silly question to me (both parts). [hide=\"Because, I mean...\"] Obviously if we're looking at primitive triples, one of the legs must be $1$, and $1$ is not part of any integer triples. [/hide]", "Solution_4": "Oh, darn..." } { "Tag": [ "geometry", "circumcircle", "Euler", "perimeter", "LaTeX", "national olympiad" ], "Problem": "1. http://www.mathlinks.ro/Forum/viewtopic.php?t=137067\r\nLet $ABC$ be a triangle and $M,N,P$ be th midpoints of sides $BC, CA, AB$. The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$. Show that the area of triangle $ABC$ is at most the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$.\r\n\r\n2. http://www.mathlinks.ro/Forum/viewtopic.php?t=137071\r\nConsider $p$ a prime number and $p$ consecutive positive integers $m_{1}, m_{2}, \\ldots, m_{p}$. Choose a permutation $\\sigma$ of $1, 2, \\ldots, p$. Show that there exist two different numbers $k,l \\in \\{1,2, \\ldots, p\\}$ such that $m_{k}m_{\\sigma(k)}-m_{l}m_{\\sigma(l)}$ is divisible by $p$.\r\n\r\n3. http://www.mathlinks.ro/Forum/viewtopic.php?t=6188\r\nLet $ABC$ be a triangle with all angles $\\leq 120^{\\circ}$. Let $F$ be the Fermat point of triangle $ABC$, that is, the interior point of $ABC$ such that $\\angle AFB = \\angle BFC = \\angle CFA = 120^\\circ$. For each one of the three triangles $BFC$, $CFA$ and $AFB$, draw its Euler line - that is, the line connecting its circumcenter and its centroid.\r\n\r\nProve that these three Euler lines pass through one common point.\r\n\r\n[i]Remark.[/i] The Fermat point $F$ is also known as the [b]first Fermat point[/b] or the [b]first Toricelli point[/b] of triangle $ABC$.\r\n\r\n4. http://www.mathlinks.ro/Forum/viewtopic.php?t=137070\r\nConsider a convex polygon $A_{1}A_{2}\\ldots A_{n}$ and a point $M$ inside it. The lines $A_{i}M$ intersect the perimeter of the polygon second time in the points $B_{i}$. The polygon is called balanced if all sides of the polygon contain exactly one of points $B_{i}$ (strictly inside). Find all balanced polygons.\r\n\r\n[Note: The problem originally asked for which $n$ all convex polygons of $n$ sides are balanced. A misunderstanding made this version of the problem appear at the contest]\r\nThanks to iura :)", "Solution_1": "Here's the PDF version. I'm not too familiar with LaTex (In fact I'm doing one of my first files now!), so could someone give it the OK?" } { "Tag": [ "ratio", "Vieta" ], "Problem": "The roots of an equation $ x^{3}\\minus{}148x\\plus{}672\\equal{}0$ are A,B, and C. If A and B are in the ratio of 3 to 4, find AB+C.\r\nA) 34\r\nB) 36\r\nC) 42\r\nD) 62\r\nE) None of the above", "Solution_1": "(x+14)(x-6)(x-8)=0\r\n\r\nx=-14;6;8\r\n\r\nA=6, B=8, C=-14\r\n\r\n6*8-14=34 ----> A)", "Solution_2": "did you brute force that or did you set something up using vieta's?", "Solution_3": "Brute force, nothing special :)", "Solution_4": "hello\r\n [hide=\"solution\"]\nLet the roots be 3a,4a and -7a\n84$ a^3$=672\na=2\n\nAB+C=34 (a)[/hide]\r\n\r\nBachukas sorry for the previous post.\r\nI have edited my post.", "Solution_5": "The roots are 3a,4a and -7a, not 3a,4a and 7a, so $ AB \\plus{} C \\equal{} 6\\cdot8 \\plus{} ( \\minus{} 14) \\equal{} 34$" } { "Tag": [], "Problem": "How to do qn 5,6 and 7\r\n\r\nhttp://unitest.acer.edu.au/sample-questions/uniTEST_SampleQuestions.pdf", "Solution_1": "[hide=\"Question 5\"]\nDale worked for 1 hr. 20 min. or 80 min. 80 min. minus 15 min. (for the initial period) is 65 min. 65 min. divided by 15 is 4.333333... billing sections which rounds up to 5 billing sections of 15 dollars each. 5 times 15 is 75 dollars plus the initial 60 dollars is 135 dollars or answer $ \\boxed{D}$.\n[/hide]\n\n[hide=\"Questions 6 and 7\"]\nThe code does not take into account the initial 15 min. work for which Dale charges 60 dollars when Dale works for more than 15 minutes. However, the code does include code to account for 15 dollars worth of the first 15 minutes. Thus, the program will charge 45 dollars less than it should for all jobs exceeding 15 minutes which is answer $ \\boxed{B}$. This makes the answer to #7 $ \\boxed{B}$ as well.\n[/hide]", "Solution_2": "hey thanks for the solutions." } { "Tag": [ "trigonometry" ], "Problem": "Let $S$ be the set of all real solutions of the equation\r\n\\[ \\sec \\left[ \\arctan \\frac{1}{1+e^x} - \\arctan (1-e^x) \\right] = \\frac{\\sqrt{5}}{2} \\]\r\nThen:\r\n\r\na) $S=\\emptyset$\r\nb) $S=\\mathbb{R}$\r\nc) $S \\subset [1, 2]$\r\nd) $S \\subset [-1, 1]$\r\ne) $S = [-1, 2]$", "Solution_1": "[hide]Let $\\arctan \\frac{1}{1+e^x}-\\arctan (1-e^x)=\\theta$.\n\n$\\sec \\theta=\\frac{\\sqrt{5}}{2} \\Rightarrow \\sec^2 \\theta=\\frac{5}{4}\\Rightarrow \\tan^2 \\theta=\\frac{1}{4} \\Rightarrow \\tan \\theta=\\pm \\frac{1}{2}$.\n\nUse the subtraction formula for tangent to get $\\tan \\theta=\\frac{\\frac{1}{1+e^x}-(1-e^x)}{1+\\frac{1-e^x}{1+e^x}}$.\n\nSimplify the fraction by multiplying top and bottom by $1+e^x$: $\\tan \\theta=\\frac{1-(1-e^x)(1+e^x)}{2}=\\frac{e^{2x}}{2}$.\n\nSince this must equal $\\pm \\frac{1}{2}$, $e^{2x}=\\pm 1$. When -1, no solution and when 1, $x=0\\Rightarrow \\boxed{D}$.[/hide]", "Solution_2": "[hide]Let $\\theta = \\arctan \\frac{1}{1+e^x}$ and $\\alpha =\\arctan (1-e^x)$.\n\nHence \n\n$\\sec (\\theta - \\alpha) = \\frac{\\sqrt{5}}{2} \\Leftrightarrow \\cos(\\theta - \\alpha)=\\frac{2}{\\sqrt{5}}\\Leftrightarrow \\cos \\theta \\cos \\alpha + \\sin \\theta \\sin \\alpha = \\frac{2}{\\sqrt{5}} \\Rightarrow$\n\n$\\frac{1+e^x}{\\sqrt{1+(1+e^x)^2}} \\cdot \\frac{1}{\\sqrt{1+(1-e^x)^2}} + \\frac{1}{\\sqrt{1+(1+e^x)^2}} \\cdot \\frac{1-e^x}{\\sqrt{1+(1-e^x)^2}}=\\frac{2}{\\sqrt{5}}\\Leftrightarrow$\n\n$\\frac{2}{\\sqrt{4+e^{4x}}}=\\frac{2}{\\sqrt{5}} \\Rightarrow 4+e^{4x} = 5 \\Rightarrow x=0 \\Rightarrow \\boxed{D}$.[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let x, y , z be positive real numbers such that:\r\n$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$\r\nShow that:\r\n$\\sqrt{x+yz}+\\sqrt{y+zx}+\\sqrt{z+yx} \\ge \\sqrt{xyz}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z}$", "Solution_1": "http://www.kalva.demon.co.uk/apmo/asoln/asol024.html", "Solution_2": "ahem\r\n\r\nal.M.V, the meaning is to have fun solving the problem, not to post a link to the solution :D", "Solution_3": "well, put a=1/x b=1/y c=1/z, then the inequality becomes: \r\n :Sigma: :sqrt: (a+bc) >= 1+ :Sigma: :sqrt: ab\r\n\r\nnow, :sqrt: (a 2 +ab+bc+ca)>=a+ :sqrt: bc because its equivalent to ab+ac>=2a :sqrt: bc. suming the analogous inequalities, and taking into acount that a+b+c=1 and that a 2 +ab+bc+ca=a(a+b+c)+bc=a+bc one gets the desired result.", "Solution_4": "I visited the link, but I don't really understand how to do it with a short proof.\r\nI tried a different (longer) method, could anyone comment on my proof? Maybe there is a flaw to it. I skipped a few basic steps because it took up too much space and I assume that they are obvious. This seems to be too long a proof. If there is a way to write the proof in a similar way and shorter, could anyone comment on it please?\r\n\r\nMultiply the first equation by $ \\sqrt{abc}$\r\nSubstitute the value of $ \\sqrt{abc}$ in the inequality with the value of the new equation we got from above.\r\nThen if we prove that\r\n$ \\sqrt{a+bc}\\ge\\frac{\\sqrt{bc}}{\\sqrt{a}}+\\sqrt{a}$ , $ \\sqrt{b+ca}\\ge\\frac{\\sqrt{ac}}{\\sqrt{b}}+\\sqrt{b}$ and $ \\sqrt{c+ab}\\ge\\frac{\\sqrt{ab}}{\\sqrt{c}}+\\sqrt{c}$ the statement holds true.\r\n\r\nFor $ \\sqrt{a+bc}\\ge\\frac{\\sqrt{bc}}{\\sqrt{a}}+\\sqrt{a}$\r\n$ a+bc\\ge\\frac{bc}{a}+2\\sqrt{bc}+a$\r\n$ bc\\ge\\frac{bc}{a}+2\\sqrt{bc}$\r\n$ bc-\\frac{bc}{a}\\ge2\\sqrt{bc}$\r\n$ bc(1-\\frac{1}{a})^{2}\\ge4$\r\n$ \\frac{c}{b}+\\frac{b}{c}\\ge\\2$ has to be true.\r\n\r\nLet $ X=\\frac{c}{b}+{b}{c}$\r\n$ \\frac{dX}{db}=-\\frac{c}{b^{2}}+\\frac{1}{c}$\r\nstationary pt when $ -\\frac{c}{b^{2}}+\\frac{1}{c}=0$\r\n$ \\frac{c}{b^{2}}=\\frac{1}{c}$\r\ntherefore $ b=c$\r\n\r\n$ \\frac{d^{2}X}{db^{2}}=2\\frac{c}{b^{3}}$\r\nWhen $ b=c$, $ =2\\frac{c}{c^{3}}$\r\nAs $ c$ is a positive real, $ 2\\frac{1}{c^{2}}>0$\r\nTherefore $ b=c$ is a minimum value.\r\nTherefore $ \\frac{c}{b}+\\frac{b}{c}=\\frac{c}{c}+\\frac{c}{c}=2$\r\nTherefore $ \\frac{c}{b}+\\frac{b}{c}\\ge\\2$ is true.\r\n\r\nProve the same way for the rest of the two inequalities." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Find the $n$ th term of the sequence $\\{a_{n}\\}$ such that $a_{1}=1,\\ a_{n+1}=2a_{n}-n^{2}+2n\\ (n=1,\\ 2, \\ 3,\\ \\cdots).$", "Solution_1": "Standard trick:\r\n1) Calculate a polynomial $b_{n}$ with $b_{n+1}-2b_{n}=-n^{2}+2n$. Identifying coefficients, $b_{n}=n^{2}+1$, so $b_{1}=2$.\r\n2) Calculate the homegeneous $c_{n}$ with $c_{n+1}-2c_{n}=0$: $c_{n}=2^{n}c_{1}$.\r\n3) $a_{n}=b_{n}+c_{n}$\r\nThe only remaining thing to calculate is $c_{1}=a_{1}-b_{1}=-1$. So, $a_{n}=-2^{n}+n^{2}+1$.", "Solution_2": "Or by rearranging:\r\n\r\n$a_{n+1}=2a_{n}-n^{2}+2n\\iff a_{n+1}-(n+1)^{2}-1=2a_{n}-n^{2}+2n-(n+1)^{2}-1$\r\n\r\nwhich gives\r\n\r\n$a_{n+1}-(n+1)^{2}-1=2(a_{n}-n^{2}-1)$\r\n\r\nhence $a_{n}-n^{2}-1=2^{n-1}(a_{1}-1^{2}-1)=-2^{n-1}\\iff a_{n}=n^{2}-2^{n-1}+1$\r\n\r\nNOTE: There's a typo in [b]lordWings[/b]'s answer", "Solution_3": "Your answer is correct, Farenhajt. :)" } { "Tag": [ "inequalities", "algebra unsolved", "algebra" ], "Problem": "Solve the equation\r\n$ \\sqrt{x\\plus{}\\sqrt{x^2\\minus{}1}}\\equal{}\\frac{9\\sqrt{2}}{4}(x\\minus{}1)\\sqrt{x\\minus{}1}$", "Solution_1": "[quote=\"Algadin\"]Solve the equation\n$ \\sqrt {x + \\sqrt {x^2 - 1}} = \\frac {9\\sqrt {2}}{4}(x - 1)\\sqrt {x - 1}$[/quote]\r\nLet $ a=\\sqrt{x-1}$, $ b=\\sqrt{x+1}$. Then $ b^2-a^2=2$.\r\nThere fore :\r\n$ \\{ \\begin{array}{l} b^2-a^2=2 \\\\2(a+b)=9a^3 \\end{array} $ $ \\leftrightarrow \\{ \\begin{array}{l} b^2=a^2+2 \\\\2b=9a^3-2a \\end{array} $ $ \\leftrightarrow \\{ \\begin{array}{l} b^2=a^2+2 \\\\81a^6-36a^4-8=0 \\end{array} $\r\n$ \\leftrightarrow \\{ \\begin{array}{l} b^2=a^2+2 \\\\a^2=\\frac{2}{3} \\end{array} $ $ \\leftrightarrow \\{ \\begin{array}{l} b^2=\\frac{8}{3} \\\\2b=9a^3-2a \\end{array} $ $ \\leftrightarrow x=\\frac{5}{3}$ :)", "Solution_2": "hello, for $ x \\in \\mathbb{R}$ we have the range of definition $ x\\geq1$. By squaring both sides of the equation we get: $ \\sqrt{x^2\\minus{}1}\\equal{}\\frac{81}{8}(x\\minus{}1)^3\\minus{}x$. The inequality $ \\frac{81}{8}(x\\minus{}1)^3\\minus{}x\\geq0$ ist fulfilled for all $ x$ with $ \\frac{1}{27}(972\\plus{}12\\sqrt{6465}^{\\frac{1}{3}}\\plus{}\\frac{8}{9(972\\plus{}12\\sqrt{6465})^{\\frac{1}{3}}}\\plus{}1\\le x <\\infty$.Squaring again and factoring we have\r\n$ \\minus{}\\frac{1}{64}(3x\\minus{}5)(27x^2\\minus{}48x\\plus{}25)(81x^3\\minus{}207x^2\\plus{}171x\\minus{}53)\\equal{}0$. From here we have the solution $ x\\equal{}\\frac{5}{3}$.\r\nremark: $ x\\equal{}\\frac{1}{27}(908\\plus{}36\\sqrt{633})^{\\frac{1}{3}}\\plus{}\\frac{16}{27(908\\plus{}36\\sqrt{633})^{\\frac{1}{3}}}\\plus{}\\frac{23}{27}$ is not in our range of definition.\r\nSonnhard." } { "Tag": [ "quadratics", "trigonometry" ], "Problem": "Let $ az^2 \\plus{} bz \\plus{} c \\equal{} 0$ be a quadratic with complex coefficients\r\nThen after some manipulation we get\r\n$ (2az \\plus{} b)^2 \\equal{} \\Delta$. where $ \\Delta \\equal{} b^2 \\minus{} 4ac$\r\nNow let $ y \\equal{} (2az \\plus{} b)$ and $ \\Delta \\equal{} u \\plus{} vi$ where $ u,v$ are real\r\n$ y^2 \\equal{} u \\plus{} vi$\r\nThen $ y \\equal{} \\pm(\\sqrt {\\frac {|\\Delta| \\plus{} u}{2}} \\plus{} ($sign[v]$ )\\sqrt {\\frac {|\\Delta| \\minus{} u}{2}})$ where sign[v] is the sign of the number v\r\nI don't understand how the above conlusion was reached :?: .\r\n\r\nThanks for your time :)\r\n\r\nEDIT: OOPS :oops:", "Solution_1": "How do we have $ (2az\\plus{}b)^2\\equal{}\\Delta$ and $ (2az\\plus{}b)^2\\equal{}y,$ then $ y^2\\equal{}\\Delta?$", "Solution_2": "[quote=\"RoFlLoLcOpT\"]How do we have $ (2az \\plus{} b)^2 \\equal{} \\Delta$ and $ (2az \\plus{} b)^2 \\equal{} y,$ then $ y^2 \\equal{} \\Delta?$[/quote]\r\nSrry, its edited now", "Solution_3": "Write in polar form: $ y^2\\equal{}u\\plus{}vi\\equal{}re^{\\theta i}$: we have\r\n\r\n$ r\\cos\\theta\\equal{}u$, $ r\\sin\\theta\\equal{}v$, $ r\\equal{}\\sqrt{u^2\\plus{}v^2}\\equal{}|\\Delta|$.\r\n\r\nThen, $ y\\equal{}r^\\frac12e^{(\\frac{\\theta}2\\plus{}\\pi k)i}$ for $ k\\equal{}0,1$, by DeMoivre's.\r\n\r\nThat should give you:\r\n\r\n$ y\\equal{}\\pm r^\\frac12\\left(\\cos\\frac{\\theta}2\\plus{}i\\sin\\frac{\\theta}2\\right)$\r\n\r\n$ \\cos^2\\frac{\\theta}2\\equal{}\\frac{1\\plus{}\\cos\\theta}2$ and $ \\sin^2\\frac{\\theta}2\\equal{}\\frac{1\\minus{}\\cos\\theta}2$ follow from the double-angle formulas.\r\n\r\n$ \\cos\\theta\\equal{}\\frac{u}{r}$, $ \\sin\\theta\\equal{}\\frac{v}{r}$ implies $ \\cos^2\\frac{\\theta}2\\equal{}\\frac{1\\plus{}\\frac{u}{r}}2$ and $ \\sin^2\\frac{\\theta}2\\equal{}\\frac{1\\minus{}\\frac{u}{r}}2$.\r\n\r\n$ y\\equal{}\\pm\\left(\\sqrt{\\frac{|\\Delta|\\plus{}u}2}\\plus{}i\\sqrt{\\frac{|\\Delta|\\minus{}u}2}\\right)$\r\n\r\nDiscussion: the 'y' you have written is a [b]real number[/b] as opposed to the 'y' I have found is a [b]complex number[/b]. And the 'y' form that I have written is [u]correct[/u] as it stands when applying DeMoivre's. I triple checked every single detail and researched [b]all[/b] the formulas to make sure I'm not insane, so yea...", "Solution_4": "Thank you, and my \"y\" is also a complex number", "Solution_5": "[quote=\"Complex_Ninja\"]$ y \\equal{} \\pm(\\sqrt {\\frac {|\\Delta| \\plus{} u}{2}} \\plus{} ($sign[v]$ )\\sqrt {\\frac {|\\Delta| \\minus{} u}{2}})$ where sign[v] is the sign of the number v[/quote]This 'y' is a real number.\r\n\r\nIf that's the book's answer, it's still wrong anyways, strictly 'speaking' :P" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "$ a_{1},...,a_{n}>0$ and $ a_{1}\\plus{}...\\plus{}a_{n}\\equal{}1$ $ m\\geq1$ find min: \r\n$ (a_{1}\\plus{}\\frac{1}{a_{1}})^{m}\\plus{}...\\plus{}(a_{n}\\plus{}\\frac{1}{a_{n}})^{m}$", "Solution_1": "I am not very interest in inequalities so I will post solution with some property of function.\r\nBecause $ f(x)\\equal{}x^m$ ($ m>1)$ is convect then use Jensen 's inequalities we have :\r\n$ \\sum_{i\\equal{}1}^n (a_i\\plus{}\\frac{1}{a_i})^m\\geq \\ n(\\frac{\\sum_{i\\equal{}1}^n a_i\\plus{}\\frac{1}{a_i}}{n})^m$ \r\nOther :\r\n$ \\sum_{i\\equal{}1}^n\\frac{1}{a_i}\\geq \\frac{n^2}{\\sum_{i\\equal{}1}^n a_i}$", "Solution_2": "If $ m\\equal{}1$ $ \\sum \\left(a_i\\plus{}\\frac{1}{a_i}\\right)\\ge n^2\\plus{}1$ is an immediate consequence of Cauchy-Schwarz.\r\n\r\nIf $ m>1$ then\r\n$ \\left(\\sum\\left(a_i\\plus{}\\frac{1}{a_i}\\right)^m\\right)^\\frac{1}{m}\\left(\\sum1^\\frac{m}{m\\minus{}1}\\right)^\\frac{m\\minus{}1}{m}\\ge\r\n\\sum\\left(a_i\\plus{}\\frac{1}{a_i}\\right)$ (Holder).\r\nRaising both sides to the m-th power we obtain\r\n$ \\sum\\left(a_i\\plus{}\\frac{1}{a_i}\\right)^m\\cdot n^{m\\minus{}1} \\ge \\sum\\left(a_i\\plus{}\\frac{1}{a_i}\\right)^m$.\r\nBecause $ f(x)\\equal{}x^m$ is monotonically increasing ($ m>1$) and $ \\sum\\frac{1}{a_i}\\ge \\frac{n^2}{\\sum a_i}\\equal{}n^2$ (CS), \r\n$ \\sum\\left(a_i\\plus{}\\frac{1}{a_i}\\right)^m\\ge \\frac{1}{n^{m\\minus{}1}}\\left(n^2\\plus{}1\\right)^m\\equal{}n\\left(n\\plus{}\\frac{1}{n}\\right)^m$\r\nwith equality for $ a_1\\equal{}a_2\\equal{}...\\equal{}a_n\\equal{}\\frac{1}{n}$." } { "Tag": [ "linear algebra", "matrix", "linear algebra solved" ], "Problem": "Let be the set $\\mathcal M$\r\n\\[\r\n\\mathcal M=\\left\\{\\left(\\begin{array}{lll}\r\nx&0&-x\\\\\r\n0&y&0\\\\\r\n-x&0&x\r\n\\end{array}\\right)\\mid x,y\\in\\mathbb C\\right\\}\r\n\\]\r\nProve that $(\\mathcal M,+,\\cdot)$ is a commutative ring and is isomorphic with the set $\\mathbb C$X$\\mathbb C$", "Solution_1": "Notice that \r\n$\\begin{pmatrix}\r\nx&0&-x\\\\\r\n0&y&0\\\\\r\n-x&0&x\r\n\\end{pmatrix}\\begin{pmatrix}\r\nz&0&-z\\\\\r\n0&w&0\\\\\r\n-z&0&z\r\n\\end{pmatrix}\r\n=\r\n\\begin{pmatrix}\r\nz&0&-z\\\\\r\n0&w&0\\\\\r\n-z&0&z\r\n\\end{pmatrix}\r\n\\begin{pmatrix}\r\nx&0&-x\\\\\r\n0&y&0\\\\\r\n-x&0&x\r\n\\end{pmatrix}\r\n=\r\n\\begin{pmatrix}\r\n2xz&0&-2xz\\\\\r\n0&yw&0\\\\\r\n-2xz&0&2xz\r\n\\end{pmatrix}\r\n$\r\nThen it is easy to verify that $\\mathcal M$ is a commutative ring.", "Solution_2": "Define $f:\\mathcal M\\to \\mathbb{C}\\oplus\\mathbb{C}$ as\r\n$f\\left(\r\n\\begin{pmatrix}\r\nx & 0 & -x\\\\\r\n0 & y & 0\\\\\r\n-x& 0 &x\r\n\\end{pmatrix}\r\n\\right)=(2x,y)$\r\nIt is easy to verify that $f$ is an isomorphism of rings." } { "Tag": [ "inequalities" ], "Problem": "If a triangle has one side that is 9 cm and a second side that is 3 cm less than twice the third side, what are the possible lengths for the second and thirds sides?", "Solution_1": "[hide=\"Answer\"]\n$ 4 < x < 12$, where x is the shorter side.\n\nIf x is the shortest side, 2x-3 is the other unknown side, they add up to more than 9, hence 3x-3>9, x>4.\nAlso, x+9>2x-3, hence x<12. The inequality (2x-3)+9>x is always true. \n[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "1. Let $ a,b,c$ be nonnegative real numbers. Prove that\r\n\\[ \\left( \\frac{a}{a\\plus{}b}\\right)^2 \\plus{}\\left( \\frac{b}{b\\plus{}c}\\right)^2\\plus{}\\left( \\frac{c}{c\\plus{}a}\\right)^2 \\le \\frac{3\\plus{}\\sqrt{5}}{2}\\frac{a^2\\plus{}b^2\\plus{}c^2}{(a\\plus{}b\\plus{}c)^2}\\]\r\nWith equality holds if and only if $ (a,b,c)\\equal{}\\left( \\frac{\\sqrt{5}\\plus{}1}{2},1,0\\right)$.\r\n2. Let $ a,b,c$ be nonnegative real numbers. Prove that\r\n\\[ \\frac{a^3}{a\\plus{}b}\\plus{}\\frac{b^3}{b\\plus{}c}\\plus{}\\frac{c^3}{c\\plus{}a} \\le \\left( \\frac{\\sqrt[3]{4}}{3}\\plus{}1\\right) \\frac{a^3\\plus{}b^3\\plus{}c^3}{a\\plus{}b\\plus{}c}\\]\r\nWith equality holds if and only if $ (a,b,c)\\equal{}\\left( 0,1,\\sqrt[3]{2}\\right)$.\r\n:)", "Solution_1": "[quote=\"can_hang2007\"]1. Let $ a,b,c$ be nonnegative real numbers. Prove that\n\\[ \\left( \\frac {a}{a \\plus{} b}\\right)^2 \\plus{} \\left( \\frac {b}{b \\plus{} c}\\right)^2 \\plus{} \\left( \\frac {c}{c \\plus{} a}\\right)^2 \\le \\frac {3 \\plus{} \\sqrt {5}}{2}\\frac {a^2 \\plus{} b^2 \\plus{} c^2}{(a \\plus{} b \\plus{} c)^2}\n\\]\nWith equality holds if and only if $ (a,b,c) \\equal{} \\left( \\frac {\\sqrt {5} \\plus{} 1}{2},1,0\\right)$.\n2. Let $ a,b,c$ be nonnegative real numbers. Prove that\n\\[ \\frac {a^3}{a \\plus{} b} \\plus{} \\frac {b^3}{b \\plus{} c} \\plus{} \\frac {c^3}{c \\plus{} a} \\le \\left( \\frac {\\sqrt [3]{4}}{3} \\plus{} 1\\right) \\frac {a^3 \\plus{} b^3 \\plus{} c^3}{a \\plus{} b \\plus{} c}\n\\]\nWith equality holds if and only if $ (a,b,c) \\equal{} \\left( 0,1,\\sqrt [3]{2}\\right)$.\n:)[/quote]\r\nThe general of 2) is\r\n\\[ \\frac{a^n}{a\\plus{}b}\\plus{}\\frac{b^n}{b\\plus{}c}\\plus{}\\frac{c^n}{c\\plus{}a} \\le \\left( \\frac{(n\\minus{}1)^{\\frac{n\\minus{}1}{n}}}{n}\\plus{}1\\right) \\frac{a^n\\plus{}b^n\\plus{}c^n}{a\\plus{}b\\plus{}c}\\]\r\nfor all $ a,b,c \\ge 0, k \\ge 1$.\r\nI have proved this one for all $ k \\ge 3$ and $ k \\equal{}2$. But I cannot prove it for all $ 1 \\le k \\le 3$, I think it is true, too. Could anyone help me prove it or disprove it? Thanks", "Solution_2": "i wanna see solutions", "Solution_3": "[quote=\"Hong Quy\"]i wanna see solutions[/quote]\r\nTake a look here: http://www.mathlinks.ro/viewtopic.php?t=208794 :)" } { "Tag": [ "calculus", "derivative", "calculus computations" ], "Problem": "Let $ f: [0,\\plus{}\\infty\\rangle \\rightarrow \\langle 0,1],f\\in C^{\\infty}([0,\\plus{}\\infty\\rangle)$ such that $ \\forall k\\in\\mathbb{N}_0,x\\in [0,\\plus{}\\infty \\rangle$\r\n$ (\\minus{}1)^k f^{(k)}(x)\\geq 0$. Define $ g: \\langle 0,\\plus{}\\infty\\rangle \\rightarrow \\mathbb{R}$, with $ g(x)\\equal{}\\frac{1\\minus{}f(x)}{x}$. Prove that for $ \\forall k\\in\\mathbb{N}_0,x\\in \\langle 0,\\plus{}\\infty\\rangle$\r\n $ (\\minus{}1)^k g^{(k)}(x)\\geq 0$.\r\n\r\nI got that $ \\frac{d}{dx}( (\\minus{}1)^{k\\minus{}1}g^{(k\\minus{}1)}(x) x^k)\\geq 0$, what is not exactly what we're looking for.", "Solution_1": "Use Leibniz rule:\r\n\\[ (f\\cdot g)^{(n)} \\equal{} \\sum_{k \\equal{} 0}^{n}\\binom{n}{k}f^{(k)}g^{(n \\minus{} k)}\r\n\\]\r\n\r\nHere we have $ f\\equal{}\\frac1{x}$ and $ g\\equal{}1\\minus{}f(x)$." } { "Tag": [ "function", "logarithms", "number theory", "totient function", "relatively prime", "real analysis", "real analysis solved" ], "Problem": "Let f be Euler's totient function. Compute the limit of the Cesaro sequence associated to the sequence $ x_n=f(n)/n$. Find a good asimptotic equivalent of this Cesaro-sequence.", "Solution_1": "What's the Cesaro sequence associated to another sequence?", "Solution_2": "And what is totient function? There are many Euler's functions and their names may differ from country to country.", "Solution_3": "The totient function is $\\phi(n)$, the number of numbers $\\leq n$ and coprime with $n$.", "Solution_4": "The totient function is $\\phi(n)$, the number of numbers $\\leq n$ and coprime with $n$.\r\n\r\nIs the Cesaro sequence associated to $\\frac{f(n)}n$ the sequence $\\frac{\\phi(n+1)-\\phi(n)}{n+1-n}=\\phi(n+1)-\\phi(n)$?", "Solution_5": "It is called Euler's function here in Russia.", "Solution_6": "In Romania too, but if you google \"totient function\" you'll see that this name is really popular in English.", "Solution_7": "If Cesaro sequence is $\\phi(n+1)-\\phi(n)$ then harazi's question is meaningless.", "Solution_8": "Yeah, I know, so I suppose it was a rather stupid question. Anyway, I think we should wait for his reply :).", "Solution_9": "What about the limit of $\\frac{x_1+x_2+...+x_n}{n}$ ?", "Solution_10": "[quote=\"Moubinool\"]What about the limit of $\\frac{x_1+x_2+...+x_n}{n}$ ?[/quote]\r\nWhat does it mean? Is it explanation of harazi's question?", "Solution_11": "Using identities\r\n1) $\\frac{\\varphi(n)}{n}=\\sum_{t|n}\\frac{\\mu(t)}{t}$\r\nand\r\n2) $\\sum_{t=1}^\\infty \\frac{\\mu(t)}{t^2}=\\frac{1}{\\zeta(2)}=\\frac{6}{\\pi^2}$\r\nwe can easily obtain\r\n\\[\\frac{1}{n}\\sum_{k=1}^{n}\\frac{\\varphi(k)}{k}=\\frac{6}{\\pi^2}+O(\\frac{\\ln n}{n}).\\]", "Solution_12": "Yes harazi very nice result\r\n\r\n$\\frac{3}{\\pi^2}+O(\\frac{lnn}{n})$ Walfisz 1963 \r\n\r\nhttp://mathworld.wolfram.com/TotientSummatoryFunction.html", "Solution_13": "Moubinool! What is actually we need to calculate: $\\frac{1}{n}\\sum_{k=1}^{n}\\varphi(k)$ or $\\frac{1}{n}\\sum_{k=1}^{n}\\frac{\\varphi(k)}{k}$?", "Solution_14": "When I saw your anwser myth, I undestood find the limit of $W(n)=\\frac{\\phi(1)+...+\\phi(n)}{n^2}$ it is known see formulae (3)\r\n\r\nhttp://mathworld.wolfram.com/TotientSummatoryFunction.html\r\n\r\nWe have to wait harazi to get precision on its question.", "Solution_15": "Sorry for the late answer. The question is to find an asimptotic behaviour of the sequence $ x_n=\\frac{1}{n}*(\\frac{f(1)}{1}+\\frac{f(2)}{2}+...+\\frac{f(n)}{n})$ where f(n) is the number of numbers smaller than n and relatively prime to n.", "Solution_16": "So I solved this problem. Or you know more precise estimate?", "Solution_17": "Yes, thanks a lot Myth. Unfortunately, I do not know more precise estimation." } { "Tag": [ "rotation", "geometry unsolved", "geometry" ], "Problem": "triangle ABC\r\n \r\n\r\nSquares ABDE and BCGH are constructed on the sides AB and BC of a triangle ABC. Prove the following:\r\n\r\n1. The median BM of the triangle formed externally between the squares is also the altitude of triangle ABC.\r\n2. BM = AC/2.", "Solution_1": "Let $S$ be the midpoint of $AC$ and $N$ be the center of $\\Box BCGH$.\r\n\r\nFrom [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116421[/url],\r\nwe know that $\\angle SNI=90^\\circ$ and $NS=NI$.\r\n\r\nSo we can rotate $\\triangle CNS$ $90^\\circ$ with pivot $N$ to be $\\triangle BNI$.\r\n\r\nFrom this, we get those 2 properties." } { "Tag": [ "trigonometry", "geometry", "geometric transformation", "ratio", "parallelogram", "circumcircle", "cyclic quadrilateral" ], "Problem": "A convex pentagon $ABCDE$ is given so that $B=E=90^{\\circ}$ and $\\widehat{BAC}\\equiv \\widehat{EAD}$. Let $O$ be the intersection point of $BD$ and $CE$. Prove that the lines $AO$ and $BE$ are perpendicularly.", "Solution_1": "The problem proposed also here and vittasko gave a beautiful solution.\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=78598", "Solution_2": "[color=darkred][b]An equivalent enunciation.[/b] In outside of the triangle $ABC$ construct the triangles $MBC\\ ,\\ CRA\\ ,\\ APB$ so that \n\n$MB=MC\\ ;\\ PA\\perp PB\\ ;\\ RA\\perp RC\\ ;\\ \\widehat{ABP}\\equiv\\widehat{ACR}\\equiv \\widehat{BCM}\\ .$ Denote : the middlepoint $D$ of the side $[BC]\\ ;$ \n\nthe intersection $I\\in CP\\cap BR\\ .$ Prove that $DP=DR\\ ,\\ I\\in AM$ and $AM\\perp PR\\ .$[/color]\r\n\r\n[color=darkblue][b]Proof[/b] (shortly, synthetically and easy metrically). Denote the middlepoints $E$, $F$ of the sides $AC$, $AB$ respectively and $m(\\widehat{BCM})=x\\ .$\n\n$\\boxed{DP=DR}\\blacktriangleright\\ ED=FP=\\frac{c}{2}\\ ,\\ ER=FD=$ $\\frac{b}{2}\\ ,\\ m(\\widehat{PFD})=m(\\widehat{DER})=$ $A+2\\cdot (90^{\\circ}-x)$ $\\Longrightarrow$ $\\triangle PFD\\equiv\\triangle DER\\ \\mathrm{(s.a.s)}$ $\\Longrightarrow$ $DP=DR\\ .$\n\n$\\boxed{\\ \\ I\\in AM\\ \\ }\\blacktriangleright$ The pairs of rays $\\{\\ (AP\\ ,\\ (AR\\ \\}$, $\\{\\ (BM\\ ,\\ (BP\\ \\}$, $\\{\\ (CM\\ ,\\ (CR\\ \\}$ are isogonally for the angles $\\widehat{BAC}$, $\\widehat{CBA}$, $\\widehat{ACB}$ respectively. \n\nFrom the well-known [u]Jakob's theorem[/u] obtain immediately that the lines $AM$, $BR$ and $CP$ are concurrently.\n\n$\\boxed{AM\\perp PR}\\blacktriangleright$ Prove easily that : $MB=MC=\\frac{a}{2\\cdot\\cos x}\\ ,\\ \\frac{AP}{c}=\\frac{AR}{b}=\\sin x\\ ,\\ \\frac{PB}{c}=\\frac{RC}{b}=\\cos x$ $\\mathrm{\\ ,\\ }$ $\\underline{AP^{2}-AR^{2}=\\left(c^{2}-b^{2}\\right)\\cdot\\sin^{2}x}\\ \\ (1)\\ .$\n\n$MP^{2}=BM^{2}+BP^{2}-2\\cdot BM\\cdot BP\\cdot \\cos (B+2x)\\mathrm{\\ ,\\ }$ $MR^{2}=CM^{2}+CR^{2}-2\\cdot CM\\cdot CR\\cdot\\cos (C+2x)\\mathrm{\\ ,\\ }$ $BM=CM$ $\\Longrightarrow$\n \n\\begin{eqnarray*}MP^{2}-MR^{2}& = & BP^{2}-CR^{2}+2\\cdot BM\\cdot \\left[CR\\cdot\\cos (C+2x)-BP\\cdot\\cos (B+2x)\\right]\\\\\\\\ & = & (c^{2}-b^{2})\\cdot\\cos^{2}x+2\\cdot\\frac{a}{2\\cdot\\cos x}\\cdot\\left[b\\cdot\\cos x\\cdot\\cos (C+2x)-c\\cdot cos x\\cdot\\cos (B+2x)\\right]\\\\\\\\ & = & (c^{2}-b^{2})\\cdot \\cos^{2}x+ab\\cdot\\cos (C+2x)-ac\\cdot\\cos (B+2x)\\\\\\\\ & = & (c^{2}-b^{2})\\cdot\\cos^{2}x+(ab\\cdot\\cos C-ac\\cdot\\cos B)\\cdot\\cos 2x-(ab\\cdot\\sin C-ac\\cdot\\sin B)\\cdot\\sin 2x\\\\\\\\ & = & (c^{2}-b^{2})\\cdot\\cos^{2}x-(c^{2}-b^{2})\\cdot\\cos 2x=(c^{2}-b^{2})\\cdot\\sin^{2}x\\Longrightarrow \\underline{MP^{2}-MR^{2}=(c^{2}-b^{2})\\cdot\\sin^{2}x}\\ \\ (2)\\ .\\end{eqnarray*}\n\nFrom the relations $(1)$ and $(2)$ obtain $AM\\perp PR\\ .$\n\n[b]Remark.[/b] I used the simple relations : $ab\\cdot\\sin C=ac\\cdot\\sin B=2S\\ \\ ;\\ \\ ab\\cdot\\cos C-ac\\cdot\\cos B=b^{2}-c^{2}\\ .$[/color]", "Solution_3": "Dear all my friends.\r\n\r\nI would like to present the translation of the solution, as it has already been posted in the link mentioned by Silouan.\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=78598 \r\n\r\n[b][size=117]PROPOSITION.[/size][/b] - A convex pentagon $ABCDE$ is given, such that $\\angle B = \\angle E = 90^{o}$ and $\\angle BAC = \\angle EAD.$ If $O,$ is the intersection point of diagonals $BD$ and $CE,$ prove that the lines $AO\\perp BE.$ \r\n\r\n[b][size=117]SOLUTION.[/size][/b] \u2013 $($ In my drawing, $AC =7.5,$ $AD = 7.8,$ $CD = 6.3,$ $AB = 4.5$ $).$\r\n\r\nThe right triangles $\\bigtriangleup ABC$ $($ $\\angle ABC = 90^{o}$ $)$ and $\\bigtriangleup AED$ $($ $\\angle AED = 90^{o}$ $),$ are similar and so, we have that $\\angle ACB = \\angle ADE$ $,(1)$\r\n\r\nThrough vertices $C,$ $D,$ of the triangle $\\bigtriangleup ACD,$ we draw two lines isogonal to $CB,$ $DE$ respectively, with respect to the angles $\\angle ACD,$ $\\angle ADC$ and we denote as $A',$ their intersection point. Because of $(1),$ we have that $\\angle CDA' = \\angle DCA'$ and so, the triangle $\\bigtriangleup A'CD,$ is isosceles $($ $A'C = A'D$ $).$\r\n\r\nWe have now, the configuration of the triangle $\\bigtriangleup ACD$ and based on it\u2019s sidelines $CD,$ $AC,$ $AD,$ erected outwardly, the triangles $\\bigtriangleup A'CD,$ $\\bigtriangleup BAC,$ $\\bigtriangleup EAD$ respectively, whose their sidelines through the vertices of $\\bigtriangleup ACD,$ are isogonal with respect to the corresponded anlge $($ The lines $AB,$ $AE,$ are isogonal with respect to the angle $\\angle CAD,$ etc $).$\r\n\r\nSo, by the Jacobi\u2019s theorem, we have that the lines $AA',$ $BD,$ $CE,$ are concurrent and hence, we conclude that the line $AA'$ passes through the point $O,$ as the intersection point of $BD,$ $CE.$\r\n\r\nWe denote as $F,$ $G,$ the midpoints of the sidelines $AC,$ $AD$ of $\\bigtriangleup ACD,$ respectively and as $B',$ $E',$ the intersection points of the midperpendiculars of $AC,$ $AD,$ from the lines $AB,$ $AE,$ respectively.\r\n\r\nWe can easy to prove that $B'E'\\parallel BE$ and so, it is enough to prove that the line $AA',$ is perpendicular to $B'E'.$\r\n\r\nWe denote as $K,$ $L,$ the intersection points of the segment line $FG,$ from the lines $A'C,$ $A'D,$ respectively. From $CD\\parallel KL,$ $\\Longrightarrow$ $\\angle KLA' = \\angle CDA' = \\angle DCA' = \\angle LKA'$ $,(2)$\r\n\r\nSo, the triangle $\\bigtriangleup A'KL$ is isosceles $($ $A'K = A'L$ $)$ and if $K',$ $L',$ are the midpoints of $A'C,$ $A'D$ respectively, we have that $KK' = LL'$ $,(3)$ $($ because of $A'C = A'D$ $).$\r\n\r\nFrom the isosceles triangle $\\angle up B'AC$ $($ $B'A = B'C$ $)$ and the cyclic quadrilateral $BB'CF,$ we have that $\\angle CB'F = \\angle BB'F = \\angle ACB$ $,(4)$\r\n\r\nBut $\\angle ACB = angle DCA'$ $,(5)$ as isogonal lines with respect to the angle $\\angle ACD.$\r\n\r\nFrom $(2),$ $(4),$ $(5),$ $\\Longrightarrow$ $\\angle CB'F = \\angle LKA'$ $\\Longrightarrow$ $\\angle CB'F = \\angle CKF$ $,(6)$\r\n\r\nFrom $(6),$ we conclude that the quadrilateral $B'KCF$ is cyclic and so, we have that $\\angle B'KC = \\angle B'FC = 90^{o}.$\r\n\r\nHence, the line $B'K$ is perpendicular to $A'K$ and similarly we can prove that the line $E'L,$ is perpendicular to $A'L.$ \r\n\r\nFrom the triangle $\\bigtriangleup A'B'C,$ by the second theorem of medians, we have that $(A'B')^{2}-(CB')^{2}= 2(A'C)\\cdot (KK')$ $,(7)$\r\n\r\nSimilarly from the triangle $\\bigtriangleup A'E'D,$ $\\Longrightarrow$ $(A'E')^{2}-(DE')^{2}= 2(A'D)\\cdot (LL')$ $,(8)$\r\n\r\nFrom $(7),$ $(8),$ $\\Longrightarrow$ $(A'B')^{2}-(AE')^{2}= (CB')^{2}-(DE')^{2}$ $,(9)$ $($ because of $A'C = A'D$ and $KK' = LL'$ $).$\r\n\r\nFrom $(9)$ $\\Longrightarrow$ $(A'B')^{2}-(AE')^{2}= (AB')^{2}-(AE')^{2}$ $,(10)$ $($ because of $CB' = AB'$ and $DE' = AE'$ $).$\r\n\r\nFrom $(10),$ we conclude that the line $AA',$ is perpendicular to $B'E'$ and the proof is completed.\r\n \r\n[b][size=117]NOTE.[/size][/b] - From all the above, it is clear that the proposition is also true and when the given pentagon $ABCDE,$ is a non-convex one.\r\n\r\nKostas Vittas.\r\n\r\nPS. \u2013 I am waiting for some other ( simpler ) proofs, to be posted here next time. The above proof, which ( I think ) is presented first time in Hellenic publication, is dedicated to all the members of the Hellenic Mathlinks Forum.", "Solution_4": "This is very nice problem! I have a solution by Ceva theorem as following\r\nCalled lines $CP\\cap AR=P',BR\\cap AP=R'$ and called $AH$ is altitulde of triangle $APR$ we will show $CP,BR,AH$ are concurrent by Ceva theorem!\r\nIndeed let $\\angle PAB=\\angle RAC= x$ we consider\r\n$\\frac{HP}{HR}\\cdot\\frac{R'A}{R'P}\\cdot\\frac{P'R}{P'A}$\r\nUsing notation as my fig\r\n$\\frac{HP}{HR}=\\frac{AP\\sin A1}{AR\\sin A2}=\\frac{AP\\cos P1}{AR\\cos R1}$\r\n$\\frac{R'A}{R'P}=\\frac{S_{RR'A}}{S_{RR'P}}=\\frac{S_{BR'A}}{S_{BR'P}}=\\frac{S_{BR'A}-S_{RR'A}}{S_{BR'P}-S_{RR'P}}=\\frac{S_{APC}}{S_{BPR}}=\\frac{APAC\\sin(x+A)}{PRPB\\sin(90-R1)}=\\frac{APAC\\sin(x+A)}{PRPB\\cos R1}$\r\nSimilarly \r\n$\\frac{P'R}{P'A}=\\frac{RPRC\\cos R2}{ABAR\\sin(x+A)}$\r\nAnd by $\\triangle APB\\sim \\triangle ARC$ we see easily $\\frac{HP}{HR}\\cdot\\frac{R'A}{R'P}\\cdot\\frac{P'R}{P'A}=1$ easily.\r\nThere is other way to look this problem I will post later !", "Solution_5": "The problem is very interesting.This is the solution\r\nLet $F,K$ be the points on $EC,BD$ that $AF \\bot EC$ $AK \\bot BD$ $(1)$\r\nFrom (1) and $AB \\bot BC$, $AE \\bot ED$ we have $A,B,C,F$ are cyclic and $A,E,D,K$ are cyclic\r\nSo $\\angle AFB=\\angle ACB$ and $\\angle ADE=\\angle AKE$ $(2)$\r\nOtherwise triangle $AED$ and $ABC$ are similar so $\\angle ACB= \\angle ADE$ $(3)$\r\nFrom (2) and (3) $\\angle AKE=\\angle AFB$\r\nThus $90+\\angle AKE=90+\\angle AFB$ or $\\angle EFB=\\angle EKB$\r\nSo $E,F,K,B$ are cyclic\r\nSo $\\angle EBO=\\angle KFO=\\angle KAO$ ($F,A,K,O$ are cyclic)\r\nthis means $A,B,K,I$ are cyclic\r\nSo $\\angle AIB=\\angle AKB=90$ or $AO,EB$ are perpendicularly $q.e.d$", "Solution_6": "[quote=\"Huy\u1ec1n V\u0169\"]The problem is very interesting.This is the solution\nLet $F,K$ be the points on $EC,BD$ that $AF \\bot EC$ $AK \\bot BD$ $(1)$\nFrom (1) and $AB \\bot BC$, $AE \\bot ED$ we have $A,B,C,F$ are cyclic and $A,E,D,K$ are cyclic\nSo $\\angle AFB=\\angle ACB$ and $\\angle ADE=\\angle AKE$ $(2)$\nOtherwise triangle $AED$ and $ABC$ are similar so $\\angle ACB= \\angle ADE$ $(3)$\nFrom (2) and (3) $\\angle AKE=\\angle AFB$\nThus $90+\\angle AKE=90+\\angle AFB$ or $\\angle EFB=\\angle EKB$\nSo $E,F,K,B$ are cyclic\nSo $\\angle EBO=\\angle KFO=\\angle KAO$ ($F,A,K,O$ are cyclic)\nthis means $A,B,K,I$ are cyclic\nSo $\\angle AIB=\\angle AKB=90$ or $AO,EB$ are perpendicularly $q.e.d$[/quote]\r\n\r\n [color=red]Yes ! \n\n This solution is delightfull and deserves a gold medal ! :lol: [/color]\r\n\r\n [u]Babis[/u]", "Solution_7": "Thank you dear Huyen Vu, for a very nice solution.\r\n\r\nI was sure that this problem has a synthetic proof, simpler and sorter than the complicated one, I have been presented.\r\n\r\nBest regards, Kostas vittas.", "Solution_8": "Some more things about this nice problem\r\nlet $P$ be the projection of $D$ on $AB$\r\nlet $Q$ be the projection of $C$ on $AE$\r\n$CQ$ meets $DP$ at $T$. $DQ$ meets $CP$ at $J$. We have some following result\r\n[b]1)[/b] $A,I, T,O$ are collinear\r\n[b]2)[/b]The line $AO$ are the isogonal of the line $AJ$ with respect to the line $AB$ and $AE$ \r\nI think they are beautiful results about isogonal lines.", "Solution_9": "Interesting and new results for me, dear Huyen Vu.\r\n\r\nAs a consequence of the result that the point $T\\equiv CQ\\cap DP,$ lies on the segment line $AO,$ I see that the segment line $AJ,$ passes through the point $S,$ as the intersection point of $BC,$ $DE.$\r\n\r\nAlso, the segment lines $CD,$ $BE,$ $PQ,$ are concurrent at one point, so be it, $R.$\r\n\r\nI have some ideas about the proofs, but I must check more carefully my notes, before a definite answer. \r\n\r\nKostas Vittas.", "Solution_10": "Some more results\r\n$BC$ meets $DE$ at $S$. $BQ$ meets $PE$ at $V$\r\n[b]3)[/b]$S,M,T,V$ are collinear where $M$ is the midpoint of $CD$\r\n[b]4)[/b]$AV$ and $CD$ are perpendicular", "Solution_11": "[quote=\"Huy\u1ec1n V\u0169\"]Some more things about this nice problem\nlet $P$ be the projection of $D$ on $AB$\nlet $Q$ be the projection of $C$ on $AE$\n$CQ$ meets $DP$ at $T$. $DQ$ meets $CP$ at $J$. We have some following result\n[b]1)[/b] $A,I, T,O$ are collinear\n[b]2)[/b]The line $AO$ are the isogonal of the line $AJ$ with respect to the line $AB$ and $AE$ \nI think they are beautiful results about isogonal lines.[/quote]\r\n\r\n[b][size=109]PROOF OF THE RESULT 1.[/size][/b] \r\n\r\n$($ In my drawing $AC = 9.5,$ $AD = 8.8,$ $CD = 5.5,$ $AB = 7.7$ $)$\r\n\r\nWe will prove that the point $T\\equiv CQ\\cap DP,$ lies on the segment line $AO.$\r\n\r\nWe denote as $K,$ the intersection point of $AC,$ $DE$ and as $L,$ the one of $AD,$ $BC.$\r\n\r\nWe consider the pencil $D.ACBP,$ which is intersected from the segment line $BL\\parallel DP$ and so, we have that $(D.ACBP) = (L,C,B)$ $,(1)$\r\n\r\n$($ We symbolize as $(D.ACBP),$ the double ratio of the pencil $D.ACBP$ and as $(L,C,B),$ the simple ratio $\\frac{BL}{BC}$ $).$\r\n\r\nSimilarly the pencil $C.ADEQ,$ is intersected from the segment line $EK\\parallel CQ$ and so, we have $(C.ADEQ) = (K,D,E)$ $,(2)$ \r\n\r\nIt is easy to show that the triangles $\\bigtriangleup ABL,$ $\\bigtriangleup AEK$ are similar, because of they have $\\angle BAL = \\angle EAK.$\r\n\r\nBecause of now, $\\angle BAC = \\angle EAD,$ we conclude that $\\frac{BL}{BC}= \\frac{EK}{ED}$ $\\Longrightarrow$ $(L,C,B) = (K,D,E)$ $,(3)$\r\n\r\nFrom $(1),$ $(2),$ $(3)$ $\\Longrightarrow$ $(D.ACBP) = (C.ADEQ)$ $,(4)$\r\n\r\nFrom $(4)$ and because of the pencils $D.ACBP,$ $C.ADEQ,$ have the segment line $CD,$ as their common ray, we conclude that the points $A\\equiv DA\\cap CA,$ $T\\equiv DP\\cap CQ$ and $O\\equiv DB\\cap CE,$ as the intersection points of the other their homologous rays, are collinear.\r\n\r\nHence, the point $T,$ lies on the segment line $AO$ and the proof of the [b][size=109]result 1,[/size][/b] is completed.\r\n\r\n[b][size=109]NOTE.[/size][/b] - Because of $T\\equiv AO\\cap CQ\\cap DP,$ by Desarques\u2019s theorem, we conclude that the triangles $\\bigtriangleup APQ$ and $\\bigtriangleup ODC,$ are perspective and so, we conclude that the points $B\\equiv AP\\cap OD,$ $E\\equiv AQ\\cap OC$ and $R\\equiv PQ\\cap CD,$ are collinear.\r\n\r\nBecause of now, the collinearity of the points $B,$ $E,$ $R,$ we conclude that the triangles $\\bigtriangleup APQ$ and $\\bigtriangleup SCD,$ where $S\\equiv BC\\cap DE,$ by Desarques\u2019s theorem are perspective and hence, the segment lines $AS,$ $PC,$ $QD,$ are concurrent at one point, so be it $J.$\r\n\r\nThat is, the point $J\\equiv PC\\cap DQ,$ lies on the segment line $AS.$ \r\n\r\n\r\n[b][size=109]PROOF OF THE RESULT 2.[/size][/b]\r\n\r\nWe will prove that the segment lines $AO,$ $AJ\\equiv AS,$ are conjugates with respect to the angle $\\angle BAE.$\r\n\r\nFrom all the discussion about the problem in this topic, it is clear that all the results, have not been correlated with some property of the pentagon and I think is better to continue by considering the correct alternative configuration of the triangle $\\bigtriangleup ACD,$ as a reference triangle and the similar right triangles $\\bigtriangleup ABC$ and $\\bigtriangleup AED$ $($ $\\angle ABC = \\angle AED = 90^{o}$ and $\\angle BAC = \\angle EAD$ $),$ erected on it\u2019s sidelines $AC,$ $AD$ respectively, outwardly to it.\r\n\r\nIt has already been proved $($ see the Huyen Vu\u2019s best solution $),$ that $AO\\perp BE.$ \r\n\r\nWe consider the triangle $ACD$ and it is easy to show that the right triangles $\\bigtriangleup AQC,$ $\\bigtriangleup APD,$ which they are erected on it\u2019s sidelines $AC,$ $AD$ respectively, inwardly to it, are similar because of $\\angle QAC = \\angle PAD.$\r\n\r\nSo, by the same way as before, we have the result that $AJ\\perp PQ,$ where $J\\equiv DQ\\cap CP.$\r\n\r\n$($ The problem, as I have been mentioned, is also true for the non-convex pentagon $AQCDP$ $).$ \r\n\r\n$\\bullet$ From the cyclic quadrilaterals $APDE,$ $AQCB,$ we have that $\\angle AEP = \\angle ADP$ $,(1)$ and $\\angle ABQ = \\angle ACQ$ $,(2)$\r\n\r\nBut $\\angle ADP = 90^{o}-\\angle PAD = 90^{o}-\\angle QAC = \\angle ACQ$ $,(3)$\r\n\r\nFrom $(1),$ $(2),$ $(3)$ $\\Longrightarrow$ $\\angle AEP = \\angle ABQ$ or $\\angle QEP = \\angle QBP$ and hence, the quadrilateral $BEQP,$ is cyclic.\r\n\r\nThat is we have that the segment lines $BE,$ $PQ,$ are antiparallel each other with respect to the angle $\\angle BAE.$\r\n\r\nHence, because of $AO\\perp BE$ and $AS\\perp PQ,$ we conclude that the segment lines $AO,$ $AS\\equiv AJ,$ are isogonal conjugates with respect to the angle $\\angle BAE$ and the proof of the [b][size=109]result 2,[/size][/b] is completed.\r\n\r\nKostas Vittas.", "Solution_12": "[quote=\"Huy\u1ec1n V\u0169\"]Some more results\n$BC$ meets $DE$ at $S$. $BQ$ meets $PE$ at $V$\n[b]3)[/b]$S,M,T,V$ are collinear where $M$ is the midpoint of $CD$\n[b]4)[/b]$AV$ and $CD$ are perpendicular[/quote]\r\n\r\n[b][size=109]PROOF OF THE RESULT 3.[/size][/b]\r\n\r\nWe will prove that the points $V\\equiv BQ\\cap PE,$ $T\\equiv CQ\\cap DP$ and $S\\equiv BC\\cap DE,$ are collinear.\r\n\r\nIt has already been prove that the segment lines $CD,$ $BE,$ $PQ,$ are concurrent at the point $R.$\r\n\r\nSo, because of the collinearity of the points $B,$ $E,$ $R,$ by Desarques\u2019s theorem, we conclude that the triangles $\\bigtriangleup VPQ$ and $\\bigtriangleup SDC,$ are perspective and hence, the segment lines $VS,$ $PD,$ $QC,$ are concurrent at one point.\r\n\r\nThat is the segment line $VS,$ passes through the point $T,$ as the intersection point of $PD,$ $QC.$ \r\n\r\nHence, the points $V,$ $T,$ $S,$ are collinear and the proof of the [b][size=109]result 3,[/size][/b] is completed.\r\n\r\n[b][size=109]NOTE.[/size][/b] \u2013 It is clear that the segment line $VTS,$ passes through the midpoint of the segment $CD,$ because of the parallelogram $CSDT.$\r\n\r\n\r\n[b][size=109]PROOF OF THE RESULT 4.[/size][/b]\r\n\r\nThe circucircles $(O_{1}),$ $(O_{2}),$ of the cyclic quadrilaterals $ABCQ,$ $AEDP$ respectively, are intersected at one point $($ other than $A$ $),$ so be it $A_{1},$ which is easy to prove that lies on the segment line $CD$ $($ $\\angle AA_{1}C = \\angle ABC = 90^{o}= \\angle AED = \\angle AA_{1}D$ $).$\r\n\r\nIf now, we denote as $(O_{3}),$ the circumcircle of the cyclic quadrilateral $BEQP,$ it is easy to show that the segment lines $AA_{1},$ $BQ,$ $EP,$ as the radical axes of $(O_{1}),$ $(O_{2}),$ $(O_{3}),$ taken per two of them, are concurrent at one point.\r\n\r\nHence, the segment line $AA_{1},$ passes through the point $V,$ as the intersection point of $BQ,$ $EP.$ \r\n\r\nWe conclude now, that $AV\\perp CD$ and the proof of the [b][size=109]result 4,[/size][/b] is completed.\r\n\r\nKostas Vittas.", "Solution_13": "You've done well, Kostas Vittas\r\nSome things about [b]result 1[/b]:\r\nWe have $\\angle QTA=\\angle QPA=\\angle QEB$ ($A,Q,T,P$ are cyclic and $B,E,Q,P$ are cyclic)\r\nbut $QT$ and $EA$ are perpendicular so $AT$ and $EB$ are perpendicular $(1)$\r\nSo we can use (1) and AO,BE are perpendicular to have [b]result 1[/b]\r\nAlso we can use Kostas Vittas's proof for [b]result 1[/b] and (1) to prove the original problem.This is another proof for original problem\r\n\r\nTo prove $PQ,BE,CD$ concurrent at $R$ we can also use [b]result 4[/b]" } { "Tag": [], "Problem": "evaluate:\r\n\r\n100[size=200]C[/size]0 + 100[size=200]C[/size]1 +100[size=200]C[/size]2 + 100[size=200]C[/size]3 +100[size=200]C[/size]4 + ... + 100[size=200]C[/size]100", "Solution_1": "Madali lang yan eh\r\n\r\njust look at the binomial expansion of (x+y)^100\r\n = then you will see all the 100Ck for k:1,2,..100\r\n\r\nso if you want to get the sum. set x=1, y =1 so you will make all the x's and y's to become 1 and you will get what you need\r\nwhich is 2^100" } { "Tag": [ "LaTeX", "trigonometry" ], "Problem": "[i]\u039e\u03b5\u03ba\u03b9\u03bd\u03ac\u03c9 \u03c4\u03b7 \u03c3\u03c4\u03ae\u03bb\u03b7 \u03c4\u03b7\u03c2 \u03ac\u03bb\u03b3\u03b5\u03b2\u03c1\u03b1\u03c2 \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c6\u03c4\u03b9\u03b1\u03be\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b5\u03c2 , \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u0398\u03b1\u03bb\u03ae \u03b3\u03b9\u03b1 \u03c4\u03b7 \u0393\u0384\u03c4\u03ac\u03be\u03b7 \u03ae \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u0395\u03c5\u03ba\u03bb\u03b5\u03af\u03b4\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u0392\u0384\u03c4\u03ac\u03be\u03b7.\u0391\u03bd \u03b4\u03b5 \u03b2\u03c1\u03b5\u03af\u03c4\u03b5 \u03c7\u03c1\u03cc\u03bd\u03bf \u03bd\u03b1 \u03c4\u03b7\u03bd \u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03b5\u03c4\u03b5 , \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03bc\u03ad\u03c1\u03b1 \u03b8\u03b1 \u03c3\u03c4\u03b5\u03af\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03c5\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03ae \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7.\u0397 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u0392\u03b1\u03bb\u03ba\u03b1\u03bd\u03b9\u03ac\u03b4\u03b1 junior.\u0391\u03c5\u03c4\u03cc \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03b1\u03c0\u03bf\u03c4\u03b5\u03bb\u03b5\u03af \u03ba\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c5\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 senior ,\u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9![/i]\r\n\r\n[color=red]\u0391\u03a3\u039a\u0397\u03a3\u0397 - 1[/color]\r\n\r\n \u0391\u03bd $a , b , c >0$ \u03ba\u03b1\u03b9 $a+b+c =1$, \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 :\r\n\r\n $\\frac{a^{2}+a}{2a+b^{2}+c^{2}}+\\frac{b^{2}+b}{2b+c^{2}+a^{2}}+\\frac{c^{2}+c}{2c+a^{2}+b^{2}}\\leq \\frac{3}{2}$\r\n\r\n [u] \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2[/u]", "Solution_1": "[color=red]\u0391\u03a3\u039a\u0397\u03a3\u0397 1 - \u039b\u03a5\u03a3\u0397[/color]\r\n\r\nKathws $a+b+c=1$ ara $2a=2-2b-2c, 2b=2-2a-2c, 2c=2-2a-2b$ ki etsi i pros apodeiksi anisotita (an antikatastisoume ta parapanw ston paranomasti) ginetai:\r\n\r\n${\\frac{a(a+1)}{(b-1)^{2}+(c-1)^{2}}+\\frac{b(b+1)}{(c-1)^{2}+(a-1)^{2}}}+\\frac{c(c+1)}{(a-1)^{2}+(b-1)^{2}}\\leq \\frac{3}{2}$\r\n\r\nApo tin anisotita C-S exoume $(b-1)^{2}+(c-1)^{2}\\geq \\frac{(a+1)^{2}}{2}$ kai douleuontas kiklika pairnoume gia tous ipoloipous $2$ paranomastes tis analoges sxeseis ki etsi i pros apodeiksi anisotita ginetai:\r\n\r\n$\\frac{2a}{a+1}+\\frac{2b}{b+1}+\\frac{2c}{c+1}\\leq \\frac{3}{2}$\r\n\r\nkai ektelwntas ton metasximatismo $a+1=x, b+1=y, c+1=z$ (etsi $x+y+z=4$) pairnoume tin \r\n\r\n$\\frac{2(x-1)}{x}+\\frac{2(y-1)}{y}+\\frac{2(z-1)}{z}\\leq \\frac{3}{2}$\r\n\r\nkai kanontas tis prakseis $6-2\\big(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\big)\\leq\\frac{3}{2}$\r\nap'opou arkei na deiksoume oti $2\\big(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\big)\\geq \\frac{9}{2}$\r\n\r\npou prokiptei apo tin anisotita arithmitikou-armonikou kathws\r\n\r\n$\\frac{x+y+z}{3}\\geq \\frac{3}{\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}$\r\n\r\nH isotita profanws isxiei otan $x=y=z=\\frac{4}{3}$ 'h oti $a=b=c=\\frac{1}{3}$\r\n\r\nAlexandros", "Solution_2": "\u039a\u03b9 \u03b5\u03b3\u03c9 \u03c4\u03bf \u03b9\u03b4\u03b9\u03bf \u03b5\u03ba\u03b1\u03bd\u03b1!\u0391\u03bb\u03bb\u03b1 \u03b4\u03c5\u03c3\u03c4\u03c5\u03c7\u03c9\u03c2 \u03bc\u03b5 \u03c0\u03c1\u03bf\u03bb\u03b1\u03b2\u03b5\u03c2 cretanman,\u03c0\u03b1\u03bd\u03c9 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03bf\u03c5\u03c3\u03b1 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03b3\u03c1\u03b1\u03c8\u03c9...\u03b5\u03c0\u03b5\u03b9\u03b4\u03b7 \u03b5\u03b9\u03bc\u03b1\u03b9 \u03bd\u03b5\u03bf \u03bc\u03b5\u03bb\u03bf\u03c2 \u03b5\u03ba\u03b1\u03bd\u03b1 100 \u03c9\u03c1\u03b5\u03c2... :oops:", "Solution_3": "[quote=\"cretanman\"]\n$\\frac{2a}{a+1}+\\frac{2b}{b+1}+\\frac{2c}{c+1}\\leq \\frac{3}{2}$[/quote]\r\n\r\n\u0391\u03c0\u03bb\u03ac \u03b3\u03c1\u03ac\u03c6\u03c9 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03ae \u03c7\u03c9\u03c1\u03af\u03c2 \u03bd\u03b1 \u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 .\r\n\u0391\u03c6\u03b1\u03b9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03bf 3 \u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03b4\u03cd\u03bf \u03bc\u03ad\u03bb\u03b7 \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03c1\u03bf\u03c2 \u03b1\u03c0\u03cc\u03b4\u03b5\u03b9\u03be\u03b7 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \r\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}\\geq\\frac{9}{4}$\r\n\u0391\u03c0\u03cc Andreescu \u03cc\u03bc\u03c9\u03c2 \r\n$\\frac{1}{a+1}+\\frac{1}{b+1}+\\frac{1}{c+1}\\geq\\frac{9}{a+b+c+3}=\\frac{9}{4}$ \r\n\r\n :wink:", "Solution_4": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03c9 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 \r\n1/2\u03b1+\u03b2^2+\u03b3^2=<1/\u03b1^2+\u03b3^2+\u03b3^2 \u03b1\u03c1\u03b1 \u03b7 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03ad\u03b1 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \r\n2(\u03b1^2+\u03b1+\u03b2^2+\u03b2+\u03b3^2+\u03b3)=<3(\u03b1^2+\u03b2^2+\u03b3^2+1)\r\n\u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03bc\u03b5\u03c4\u03ac \u03c4\u03b9\u03c2 \u03bb\u03af\u03b3\u03b5\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03cc\u03c0\u03c9\u03c2 \u03b8\u03b1 \u03b4\u03b9\u03b1\u03c0\u03b9\u03c3\u03c4\u03ce\u03c3\u03b5\u03c4\u03b5 :D", "Solution_5": "[quote=\"ambros\"]\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03c9 \u03ba\u03b9 \u03b5\u03b3\u03ce \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7 \n1/2\u03b1+\u03b2^2+\u03b3^2=<1/\u03b1^2+\u03b3^2+\u03b3^2 [/quote]\r\n\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b2\u03ac\u03bb\u03b5 \u03b3\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 $a=b=c=\\frac{1}{3}$", "Solution_6": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b5\u03ba\u03b1\u03c4\u03c3\u03b1 \u03ba\u03b9 \u03b5\u03bb\u03c5\u03c3\u03b1 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03a3\u03a9\u03a3\u03a4\u0391 \u03a3\u0399\u039b\u039f\u03a5\u0391\u039d\u0395 \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03b7 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd\r\n\u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \r\n\u03b1(\u03b1+1)/\u03b2^2+\u03b3^2+2\u03b1=<3\u03b1/2\r\n\u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03bf \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \r\n2\u03b2\u03b3=<2\u03b1^2+(\u03b2+\u03b3)^2\r\n :D :D :D :D :D", "Solution_7": "[quote=\"ambros\"]\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03b5\u03ba\u03b1\u03c4\u03c3\u03b1 \u03ba\u03b9 \u03b5\u03bb\u03c5\u03c3\u03b1 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03a3\u03a9\u03a3\u03a4\u0391 \u03a3\u0399\u039b\u039f\u03a5\u0391\u039d\u0395 \u03b3\u03b9\u03b1\u03c4\u03af \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03bb\u03ac\u03b8\u03b7 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd\n\u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03c9 \u03cc\u03c4\u03b9 \n\u03b1(\u03b1+1)/\u03b2^2+\u03b3^2+2\u03b1=<3\u03b1/2\n\u03b7 \u03bf\u03c0\u03bf\u03af\u03b1 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03bf \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \n2\u03b2\u03b3=<2\u03b1^2+(\u03b2+\u03b3)^2\n :D :D :D :D :D[/quote]\r\nE\u03af\u03c3\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7 \u03c6\u03bf\u03c1\u03ac?\u039a\u03b1\u03c4\u03b1\u03c1\u03c7\u03ae\u03bd \u03b4\u03b5 \u03b3\u03c1\u03ac\u03c6\u03b5\u03b9\u03c2 \u03bc\u03b5 latex \u03ba\u03b1\u03b9 \u03b4\u03b5 \u03b2\u03b3\u03ac\u03b6\u03c9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1!\u0395\u03c0\u03b9\u03c0\u03bb\u03ad\u03bf\u03bd \u03b4\u03b5 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bf\u03c1\u03b8\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c4\u03c9\u03bd \u03c0\u03c1\u03ac\u03be\u03b5\u03c9\u03bd...\r\n\u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03b5\u03b9\u03c2 \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce?", "Solution_8": "\u039c\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03b3\u03c1\u03ac\u03c8\u03b5\u03b9\u03c2 \u03cc\u03bb\u03b7 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03bf\u03c5 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03bf\u03c3\u03c4 ?\r\n\u0398\u03b1 \u03c4\u03bf \u03ba\u03ac\u03bd\u03c9 \u03b5\u03b3\u03ce \u03bc\u03b5\u03c4\u03ac \u03c3\u03b5 Latex . ok ?", "Solution_9": "[color=red]\u0391\u039b\u0393\u0395\u0392\u03a1\u0391 2 - \u0395\u039a\u03a6\u03a9\u039d\u0397\u03a3\u0397[/color]\r\n\r\nEstw poliwnimo $P_{n}(x)=x^{n}-a_{n-1}x^{n-1}-\\cdots-a_{1}x-a_{0}$ me $n\\geq 2$, opou $a_{0},a_{1},\\ldots,a_{n-1}>0$ kai $a_{0}+a_{1}+\\cdots+a_{n-1}=1$. Na prosdiorisete oles tis rizes $\\rho$ tou $P_{n}(x)$ me $|\\rho|=1$, kathws kai ton vathmo pollaplotitas kathe tetoias rizas.\r\n\r\nAlexandros", "Solution_10": "[color=red]\u0391\u039b\u0393\u0395\u0392\u03a1\u0391 2 - \u039b\u03a5\u03a3\u0397[/color]\r\n\r\n\u0388\u03c3\u03c4\u03c9 $r$ \u03bc\u03b9\u03b1 \u03c1\u03af\u03b6\u03b1 \u03c4\u03bf\u03c5 $P_{n}(x)$ \u03bc\u03b5 $|r|=1$, \u03c4\u03cc\u03c4\u03b5\r\n\r\n$|P_{n}(r)| = |r^{n}-a_{n-1}r^{n-1}-a_{n-2}r^{n-2}+...-a_{0}| \\geq$\r\n$\\geq |r^{n}|-|a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+...+a_{0}| \\geq$\r\n$\\geq |r^{n}|-|a_{n-1}r^{n-1}|-|a_{n-2}r^{n-2}|-...-|a_{0}|=$\r\n$=|r^{n}|-a_{n-1}|r^{n-1}|-a_{n-2}|r^{n-2}|-...-a_{0}=$\r\n$= 1-a_{n-1}-a_{n-2}-...-a_{0}= 0$\r\n\r\n\u0397 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b1\u03bd $r^{n}(a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+...+a_{0}) \\in \\mathbb R_{+}$ \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03cc\u03c4\u03b1\u03bd\r\n$a_{n-1}r^{n-1},a_{n-2}r^{n-2},...,a_{1}r \\in \\mathbb R_{+}$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03bf $r$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c1\u03af\u03b6\u03b1 \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03b4\u03c5\u03bf \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2.\r\n\u0386\u03c1\u03b1, $r \\in \\mathbb R_{+}$ \u03ba\u03b1\u03b9 \u03b5\u03c0\u03b5\u03b9\u03b4\u03ae $|r|=1$, \u03b8\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $r=1$.\r\n \r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b9 $P_{n}(x)=(x-1)[x^{n-1}+(1-a_{n-1})x^{n-2}+...+(1-a_{n-1}-a_{n-2}-...-a_{1})]$. \u03a4\u03bf \u03c0\u03bf\u03bb\u03c5\u03ce\u03bd\u03c5\u03bc\u03bf \u03cc\u03bc\u03c9\u03c2 $t(x)=x^{n-1}+(1-a_{n-1})x^{n-2}+...+(1-a_{n-1}-a_{n-2}-...-a_{1})$ \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03c1\u03af\u03b6\u03b1 \u03c4\u03bf $1$, \u03b1\u03c6\u03bf\u03cd $t(1)=1+(1-a_{n-1})+...+(1-a_{n-1}-a_{n-2}-...-a_{1}) > 0$.\r\n\r\n\u03a3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 \u03b7 $r=1$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03ae \u03c1\u03af\u03b6\u03b1 \u03c4\u03bf\u03c5 $P_{n}(x)$ \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $|r|=1$, \u03ba\u03b1\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 \u03b2\u03b1\u03b8\u03bc\u03cc \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03cc\u03c4\u03b7\u03c4\u03b1\u03c2.\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2...", "Solution_11": "Stelio poli swsti i lisi sou!! \r\n\r\nTo kako einai oti i askisi stin ekfwnisi den dieukrinizei ean thelei na lithei sto $\\mathbb{R}$ 'h sto $\\mathbb{C}$ dioti sto $\\mathbb{C}$ diskoleuoun ligo ta pragmata kathws psaxneis oles ekeines tis rizes pou vriskontai panw ston monadiaio kiklo pou ikanopoioun tin dosmeni poliwnimiki eksiswsi. Einai nomizw arketa eukoli an prokeitai gia to $\\mathbb{R}$ giati mporeis akoma kai me paragwgous na deikseis oti to $P_{n}'(1)\\neq 0$ opote to $1$ einai riza pollaplotitas $1$.\r\n\r\nPantws epeidi den exw tin lisi apo to diagwnismo, tha tin meletisw ligaki stous migadikous giati einai arketa endiaferousa (an fisika prokiptei apotelesma). An thelete koitakste to! (H ekfwnisi pou exw einai daxtilografimeni se H/Y kai einai i episimi kolla pou dwthike tote ston diagwnismo).\r\n\r\nAlexandros", "Solution_12": "\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03b5, \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b8\u03b5\u03c9\u03c1\u03ce \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $\\mathbb C$ \u03ba\u03b1\u03b9 \u03bc\u03b5 $|a|$ \u03b5\u03bd\u03bd\u03bf\u03ce \u03c4\u03bf \u03bc\u03ad\u03c4\u03c1\u03bf \u03c4\u03bf\u03c5 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd $a$ \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03b1\u03c0\u03b1\u03c1\u03b1\u03af\u03c4\u03b7\u03c4\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c0\u03cc\u03bb\u03c5\u03c4\u03b7 \u03c4\u03b9\u03bc\u03ae. \u0388\u03c4\u03c3\u03b9 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03c9 \u03cc\u03c4\u03b9 \u03b7 \u03bc\u03cc\u03bd\u03b7 [b]\u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ae[/b] \u03c1\u03af\u03b6\u03b1 $r$ \u03c4\u03bf\u03c5 $P_{n}(x)$ \u03bc\u03b5 $|r|=1$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 $r=1$.\r\n\r\nP.S. \u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ad\u03c2 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $\\mathbb C$, \u03b1\u03bb\u03bb\u03b9\u03ce\u03c2 \u03b8\u03b1 \u03b1\u03c1\u03ba\u03bf\u03cd\u03c3\u03b5 \u03bd\u03b1 \u03b5\u03be\u03b5\u03c4\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03cc\u03bd\u03bf \u03c4\u03b9\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03c4\u03ce\u03c3\u03b5\u03b9\u03c2 $r=1$ \u03ba\u03b1\u03b9 $r=-1$.", "Solution_13": "[quote]\u0397 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03cc\u03c4\u03b1\u03bd $r^{n}(a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+...+a_{0}) \\in \\mathbb R_{+}$[/quote]\r\n\r\nDiladi isxirizesai oti h isotita $|z_{1}-z_{2}|=|z_{1}|-|z_{2}|$ isxiei otan $z_{1}z_{2}\\in \\mathbb R_{+}$ ? Auto den ennoeis 'h kanw kapoio lathos? Exei noima na milame gia diataksi stous migadikous an den kseroume oti einai pragmatikoi? An p.x. $z_{1}=5i$ kai $z_{2}=3i$ tote isxiei h parapanw isotita.\r\n\r\nMporeis na to analiseis ligo perissotero an ennoeis kati diaforetiko apo auto pou katalava?\r\n\r\nAlexandros", "Solution_14": "\u0395\u03b3\u03ce \u03b1\u03c0\u03bb\u03ac \u03b5\u03af\u03c0\u03b1 (\u03ae \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03b5\u03bd\u03bd\u03bf\u03bf\u03cd\u03c3\u03b1...) \u03cc\u03c4\u03b9 $|z_{1}-z_{2}|=|z_{1}|-|z_{2}|$ \u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b1\u03bd $z_{1}\\cdot w \\in \\mathbb R_{+}$, \u03cc\u03c0\u03bf\u03c5 $w$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03c3\u03c5\u03b6\u03c5\u03b3\u03ae\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03cc\u03c2 \u03c4\u03bf\u03c5 $z_{2}$. \r\n\r\n\u03a4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ae\u03be\u03b5\u03c1\u03b1 \u03c0\u03c9\u03c2 \u03c3\u03c5\u03bc\u03b2\u03bf\u03bb\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03bf \u03c3\u03c5\u03b6\u03c5\u03b3\u03ae\u03c2 \u03b5\u03bd\u03cc\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd \u03c3\u03c4\u03bf $Latex$ (...\u03b7 \u03c0\u03b1\u03cd\u03bb\u03b1 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc) \u03ba\u03b1\u03b9 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1 \u03ba\u03b1\u03c4\u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03bd \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac. :roll:\r\n\u0397 \u03bf\u03c5\u03c3\u03af\u03b1 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c4\u03b7\u03c2 \u03bb\u03cd\u03c3\u03b7\u03c2 \u03b4\u03b5\u03bd \u03b5\u03c0\u03b7\u03c1\u03b5\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9. \u0391\u03c0\u03bb\u03ac \u03b2\u03ac\u03bb\u03b5 \u03cc\u03c0\u03bf\u03c5 $r^{n}$ \u03c4\u03bf\u03bd \u03c3\u03c5\u03b6\u03c5\u03b3\u03ae \u03c4\u03bf\u03c5. \u0391\u03c0\u03cc \u03c4\u03b7 \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03b7 \u03ba\u03c5\u03c1\u03af\u03c9\u03c2 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03cc\u03c4\u03b9 $r \\in R_{+}$ \u03ba\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 \u03bf \u03c3\u03c5\u03b6\u03c5\u03b3\u03ae\u03c2 \u03c4\u03bf\u03c5 $r^{n}$ \u03b9\u03c3\u03bf\u03cd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf $r^{n}$ \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03c0\u03bf\u03c5 \u03ad\u03b3\u03c1\u03b1\u03c8\u03b1.\r\n\r\n\u038c\u03c3\u03bf\u03bd \u03b1\u03c6\u03bf\u03c1\u03ac \u03c4\u03ce\u03c1\u03b1 \u03c4\u03b7 \u03b4\u03b9\u03ac\u03c4\u03b1\u03be\u03b7 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2, \u03c0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03bd\u03cc\u03b7\u03bc\u03b1. \u0395\u03b3\u03ce \u03cc\u03bc\u03c9\u03c2 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ce \u03c4\u03b1 \u03bc\u03ad\u03c4\u03c1\u03b1 \u03c4\u03c9\u03bd \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ce\u03bd \u03ba\u03b1\u03b9 \u03cc\u03c7\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b4\u03b5\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1.\r\n\r\nP.S. \u039a\u03b1\u03bb\u03ac \u03ad\u03ba\u03b1\u03bd\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03b4\u03b9\u03b5\u03c5\u03ba\u03c1\u03af\u03bd\u03b9\u03c3\u03b5\u03c2... . :)", "Solution_15": "Ki egw eixa tin aporia auti otan tin eida RAKAR, alla eutixws exei texnasmataki kai vgainei poli omorfi xwris polles prakseis (kai xwris eksiswseis 16ou vathmou...)! Asxolisou... einai arketa sinarpastiki kai aksizei...! Na anaferw oti to pedio orismou paizei idiaitero rolo sti sigekrimeni askisi gia to texnasma pou xreiazetai! Min pw perissotera kai sas tin xalasw! \r\n\r\nAlexandros", "Solution_16": "cretanman\r\n\r\n\u03c4\u03b7\u03bd \u03ad\u03bb\u03c5\u03c3\u03b1 \u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03cc\u03c0\u03bf\u03c5 \u03c7 = 2\u03c3\u03c5\u03bd8\u03c9 , \u03bc\u03b5 8\u03c9 \u03b5 [ -\u03c0/2 , \u03c0/2] .\r\n\u0388\u03c4\u03c3\u03b9 \u03c4\u03b7\u03bd \u03ad\u03ba\u03b1\u03bd\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03c3\u03cd ?\r\n\r\n\u039d\u03af\u03ba\u03bf\u03c2 \u03a1", "Solution_17": "Eisai swstos Niko! :)\r\n\r\nEtsi akrivws... An theleis grapse tin pliri lisi gia to pws to skeftikes na to kaneis auto giati exei simasia na to doun didaktika kapoioi oi opoioi den exoun ksanadei to texnasma auto (gia alles paromoies askiseis kathws episis gia na mi fanei teleiws ouranokatevati idea giati den einai...) kathws episis kai poso evgales to $x$. Nomizw oti kai apo ekei kai epeita exei endiaferon i askisi. Tin idea omws tin anakalipses. Etsi akrivws tin exw doulepsei ki egw kai vrika $2$ trigwnometrikes liseis gia to $x$.\r\n\r\nAlexandros\r\n\r\nY.G. Xronia polla kai gia xthes, opws episis kai ston Niko Skompri (nickolas)", "Solution_18": "\u03a0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03b9\u03b4\u03ad\u03b1 \u039d\u03af\u03ba\u03bf .\u0397 \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03c4\u03cc\u03c4\u03b5 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03b5\u03b9 \u03b1\u03bd \u03b4\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b7\u03bd \r\n$cos\\ m+\\sqrt{3}sin\\ m=2cos\\ 8m$ \u03cc\u03c0\u03bf\u03c5 m \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b8 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 .\u03a4\u03ce\u03c1\u03b1 \u03b1\u03c5\u03c4\u03ae \u03ad\u03c7\u03b5\u03b9 \u03bd\u03bf\u03bc\u03b9\u03b6\u03c9 \u03b4\u03cd\u03bf \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 (\u03b4\u03b5\u03bd \u03b5\u03af\u03bc\u03b1\u03b9 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03cc\u03bc\u03c9\u03c2 :maybe: )", "Solution_19": "[quote]Xronia polla kai gia xthes[/quote]\r\n\r\n\u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03b5 .\r\n\r\n\u039d\u03b1\u03b9 \u03b5\u03ba\u03b5\u03af \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03b5\u03b9 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad .\r\n\u03a3\u03c4\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03c7\u03b5\u03af\u03bf \u03c4\u03b7\u03bd \u03ad\u03c7\u03c9 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03c9\u03bc\u03ad\u03bd\u03b7 \u2026\u2026\u2026.", "Solution_20": "Niko swstotata,\r\n\r\nEpitrepse mou na kanw mia diorthwsi stin 9h grammi tis lisis sou oti pairnoume tin $|\\cos{\\omega}|+\\sqrt{3}|\\sin{\\omega}|=2\\cos{\\omega}$ (dioti mporei i gwnia $\\omega$ na einai arnitiki ($-\\pi /16 \\leq \\omega \\leq 0$) kai to imitono tote einai arnitiko).\r\n\r\nParathetw kai tin diki mou lisi gia to pws skeftika na tin antimetwpisw arxika. Nomizw oti etsi diafainetai kai o logos gia ton opoio kaneis ton metasximatismo pou anaferei o Nikos.\r\n\r\nAs paroume $A=\\sqrt{2+\\sqrt{2+x}}, \\ \\ A\\geq 0$\r\n\r\nTote $A^{2}-2=\\sqrt{2+x}$ ara $x=(A^{2}-2)^{2}-2$\r\n\r\nTote h arxiki eksiswsi dinei\r\n$\\sqrt{2+A}+\\sqrt{3}\\sqrt{2-A}=2[(A^{2}-2)^{2}-2]$, ap'opou vgazoume (ligo xontroeidws afou veltiwnetai poli auti i anisotita, alla apo edw fainetai o logos tou metasximatismou) $-2\\leq A\\leq 2$\r\n\r\nApo tin teleutaia auti sxesi pairnoume $0\\leq \\frac{2+A}{4}\\leq 1$ kai $0\\leq \\frac{2-A}{4}\\leq 1$\r\n\r\n(Kapoios tha mporouse na pei sto simeio auto oti epeidi $-2\\leq A\\leq 2$ ara $-1\\leq \\frac{A}{2}\\leq 1$ ara na thesoume $\\frac{A}{2}=cos{a}$ 'h $\\frac{A}{2}=cos^{2}{a}$. Kati tetoio omws den mas voleuei dioti emeis theloume na \"diwksoume apo ti mesi\" ta $2+A$ kai $2-A$ pou emfanizontai katw apo tin riza)\r\n\r\n\r\nTo gegonos loipon auto, (prepei sto $\\frac{A+2}{4}$ na thesoume mia timi megaliteri 'h isi apo $0$ kai mikroteri 'h isi tou $1$) mas wthei na thesoume\r\n\\[\\frac{A+2}{4}=\\cos^{2}{a}\\]\r\nTote $A+2=4\\cos^{2}{a}$ kai $2-A=4\\sin^{2}{a}$\r\n\r\nkai kanontas eukoles prakseis $x=(A^{2}-2)^{2}-2$ kai xrisimopoiwntas tin tautotita $\\cos{2a}=2cos^{2}{a}-1$ pairnoume to polipothito\r\n\\[x=2\\cos{8a}\\]\r\nOmws $x\\geq 0$, ara $\\cos{8a}\\geq 0$ kai telika $-\\frac{\\pi}{16}\\leq a \\leq\\frac{\\pi}{16}$\r\n\r\nAra i arxiki eksiswsi pairnei tin morfi\r\n\r\n$|\\cos{a}|+\\sqrt{3}|\\sin{a}|=2\\cos{8a}$, me $-\\frac{\\pi}{16}\\leq a \\leq\\frac{\\pi}{16}$\r\n\r\nEksetazoume tin periptwsi opou $0\\leq a\\leq \\frac{\\pi}{16}$ (H alli periptwsi, einai sxedon i idia kai pairnoume akrivws tis idies liseis opws kai apo auti).\r\n\r\nSinepws exoume na lisoume tin eksiswsi\r\n\\[\\cos{a}+\\sqrt{3}\\sin{a}=2\\cos{8a}\\]\r\n'h tin eksiswsi\r\n\\[\\sin{\\big(a+\\frac{\\pi}{6}\\big)}=\\cos{8a}\\]\r\n'h (Xrisimopoiwntas tin tautotita\r\n$\\sin{\\omega}=\\cos{\\big(\\frac{\\pi}{2}-\\omega)}$)\r\n\\[\\cos{8a}=\\cos{\\big(\\frac{\\pi}{3}-a\\big)}\\]\r\nkai pairnoume tin lisi gia to a:\r\n\\[a=\\frac{2k\\pi}{9}+\\frac{\\pi}{27}, \\ k\\in\\mathbb{Z}\\ \\ 'h \\ \\ a=\\frac{2k\\pi}{7}-\\frac{\\pi}{21}, \\ k\\in\\mathbb{Z}\\]\r\n$\\bullet$ Ean $a=\\frac{2k\\pi}{9}+\\frac{\\pi}{27}, \\ k\\in\\mathbb{Z}$ tote\r\nlogw tou periorismou $0\\leq a\\leq \\frac{\\pi}{16}$ kai to gegonos oti $k\\in\\mathbb{Z}$, exoume $k=0$ to opoio simainei oti\r\n$a=\\frac{\\pi}{27}$ kai\r\n\\[\\boxed{x=2\\cos{\\frac{8\\pi}{27}}}\\]\r\n$\\bullet$ Ean $a=\\frac{2k\\pi}{7}-\\frac{\\pi}{21}, \\ k\\in\\mathbb{Z}$ tote\r\ndouleuontas me ton idio akrivws tropo pairnoume kai pali $k=0$ ara $a=-\\frac{\\pi}{21}$ kai telika\r\n\\[\\boxed{x=2\\cos{\\frac{8\\pi}{21}}}\\]\r\nElpizw na pirate kati apo tin sigekrimeni askisi kai gia tetoiou eidous metasximatismous genika (pou exoun pesei kata kairous se diagwnismous, akomi kai tis EME (nomizw to 1995 ston Thali tis 3hs Likeiou me poli poli eukoli ekfwnisi kai lisi)).\r\n \r\nFilika,\r\nAlexandros\r\n\r\nY.G. Einai arketa endiaferousa akomi kai meta pou ftanoume sto simeio $\\cos{8a}=\\cos{\\big(\\frac{\\pi}{3}-a\\big)}$ oson afora to $x$ pou emeis psaxnoume na vroume (ara tis times tou $k\\in\\mathbb{Z}$).", "Solution_21": "[quote]Epitrepse mou na kanw mia diorthwsi stin 9h grammi tis lisis sou [/quote]\r\n\r\n\u0388\u03c7\u03b5\u03b9\u03c2 \u03b4\u03af\u03ba\u03b9\u03bf \u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03b5 !\r\n\r\n\u0395\u03af\u03c7\u03b1 \u03c4\u03bf \u03bd\u03bf\u03c5 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03bf \u03c0\u03c1\u03cc\u03c3\u03b7\u03bc\u03bf \u03c4\u03bf\u03c5 \u03c3\u03c5\u03bd\u03b7\u03bc\u03b9\u03c4\u03cc\u03bd\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03bc\u03bf\u03c5 \u03be\u03ad\u03c6\u03c5\u03b3\u03b5 \u03c4\u03bf \u03b7\u03bc\u03af\u03c4\u03bf\u03bd\u03bf !!", "Solution_22": "\u03c7\u03c1\u03bf\u03bd\u03b9\u03b1 \u03c0\u03bf\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf\u03bd \u0391\u03bc\u03b2\u03c1\u03bf\u03c3\u03b9\u03bf.\u03b5\u03c0\u03b5\u03c3\u03b1\u03bd \u03bc\u03b1\u03b6\u03b5\u03bc\u03b5\u03bd\u03b5\u03c2 \u03b7 \u03b3\u03b9\u03bf\u03c1\u03c4\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c6\u03bf\u03c1\u03bf\u03c5\u03bc....\u03b5\u03bb\u03c0\u03b9\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b4\u03b9\u03b1\u03b2\u03b1\u03b6\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03be\u03b5\u03ba\u03bf\u03c5\u03c1\u03b1\u03c3\u03c4\u03bf\u03c2 \u03c4\u03bf \u03c3\u03b1\u03b2\u03b2\u03b1\u03c4\u03bf :D", "Solution_23": "[color=red]ASKHSH 9 \n[/color]\r\nNa vrethoun oi akeraioi $x$ tetoioi wste \r\n\r\n$(4-x)^{4-x}+(5-x)^{5-x}+10=4^{x}+5^{x}$\r\n\r\nAlexandros", "Solution_24": "akereoi???.....mipos i8eles na pis pragmatiki gia na exi pio ....endiaferon....\r\ngia $x\\geq 5$ ine profanos $4^{x}+5^{x}>4^{5}+5^{5}>10+(4-x)^{4-x}+(5-x)^{5-x}$\r\ngia $x=4,3,2,1,0$ dokimazume me to xeri...kai vriskume oti to$x=2$ ine lisi..\r\ngia $x\\leq-1$ ,8etume$x=-k$ ,$k\\geq 1$ opote ginete\r\n$(k+4)^{k+4}+(k+5)^{k+5}+10=(1/4)^{k}+(1/5)^{k}$ pu ine profanos adinati me $k\\geq 1$....\r\n\r\n :D", "Solution_25": "Me akeraious tin eixa vrei tin askisi opote tin afisa opws itan. Ontws itan eukoli gia na antistathmisei ligo tin proigoumeni.\r\n\r\nSwstos RAKAR.\r\n\r\nKai ksekinw kai thn 10h kai teleutaia askisi gia na kleisei to prwto meros tis Algevras\r\n[color=red]\nASKHSH 10[/color]\r\n\r\nEstw $p$ thetikos pragmatikos arithmos kai estw $|x_{0}|\\leq 2p$. Gia $n\\geq 1$ orizoume \r\n\r\n$x_{n}=3x_{n-1}-\\frac{1}{p^{2}}x_{n-1}^{3}$\r\n\r\nNa ekfrasete tin $x_{n}$ ws sinartisi twn $n$ kai $x_{0}$.\r\n\r\nAlexandros", "Solution_26": "[color=red]\u0391\u03a3\u039a\u0397\u03a3\u0397 10 - \u039b\u03a5\u03a3\u0397[/color]\r\n\r\n\u0388\u03c3\u03c4\u03c9 \u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $f(x)=3x-\\frac{x^{3}}{p^{2}}=x\\left[3-\\left(\\frac{x}{p}\\right)^{2}\\right]$\r\n\r\n\u039f \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03bf\u03c2 \u03cc\u03c1\u03bf\u03c2 \u03c4\u03bf\u03c5 $x$ \u03c3\u03c4\u03b7\u03bd \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 $y=f(x)$.\r\n\r\n\u0388\u03c3\u03c4\u03c9 $x$ \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03cc\u03c1\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1\u03c2 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf\u03c2 \u03ce\u03c3\u03c4\u03b5 $|x|\\leq 2p$. \u0398\u03b1 \u03b4\u03bf\u03cd\u03bc\u03b5 \u03cc\u03c4\u03b9 \u03ba\u03b1\u03b9 \u03bf \u03b5\u03c0\u03cc\u03bc\u03b5\u03bd\u03bf\u03c2 \u03cc\u03c1\u03bf\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03af\u03b4\u03b9\u03b1 \u03b9\u03b4\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1.\r\n\r\n\u039f x \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9 $x=(2p)\\sin\\theta$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b3\u03c9\u03bd\u03af\u03b1 $\\theta$\r\n\r\n\r\n$y = f(x) = x\\left[3-\\left(\\frac{x}{p}\\right)^{2}\\right]\\Rightarrow $\r\n\r\n$y = p\\cdot\\left(\\frac{x}{p}\\right)\\left[3-\\left(\\frac{x}{p}\\right)^{2}\\right]$\r\n\r\n$= (2p)\\sin \\theta\\left[3-(2\\sin \\theta)^{2}\\right]$\r\n\r\n$= (2p)\\sin \\theta (3-4\\sin^{2}\\theta)$\r\n\r\n$= (2p)\\cdot (3\\sin \\theta-4\\sin^{3}\\theta) \\Rightarrow \\boxed{y = (2p)\\cdot \\sin 3 \\theta}$, \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 $\\leq 2p$\r\n\r\n\u0388\u03c4\u03c3\u03b9, \u03c4\u03bf $x=(2p)\\sin\\theta$ \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03b9\u03c7\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf $y=(2p)\\sin 3\\theta$\r\n\r\n\u0388\u03c3\u03c4\u03c9 $\\theta_{o}$ \u03b7 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03b3\u03c9\u03bd\u03af\u03b1. \u03a4\u03cc\u03c4\u03b5 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $x_{n}=(2p)\\sin(3^{n}\\theta_{o})$\r\n\r\n\r\n\u039b\u03bf\u03c5\u03ba\u03ac\u03c2", "Solution_27": "\u039c\u03b5 \u03c0\u03c1\u03cc\u03bb\u03b1\u03b2\u03b5\u03c2 \u039b\u03bf\u03c5\u03ba\u03ac !!", "Solution_28": "\u03a4\u03bf\u03c5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd, \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b5\u03af\u03bc\u03b1\u03c3\u03c4\u03b5 \u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03bf\u03b9 \u03cc\u03c4\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03ae :) :) :D", "Solution_29": "Eksairetika! Mpravo kai stous dio sas!\r\n\r\nSinexizoume sto epomeno topic me tis askiseis 11-20.\r\n\r\nAlexandros" } { "Tag": [ "\\/closed" ], "Problem": "Link http://www.mathlinks.ro/LaTeX/AoPS_L_About.php died , please fix it!", "Solution_1": "Hmm, well, It's working for me... (now?)", "Solution_2": "I fixed the problem. Should work fine now. This will go under supports." } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "There are infinite lines on a planar, then we can color it with 2 colors.\r\nHelp me prove it. :blush:", "Solution_1": "Could you rephrase the question?\r\nWhat are we coloring?", "Solution_2": "I would guess it refers to well known problem of coloring parts in black and white in such a way that neighbourhood parts are colored in different colors." } { "Tag": [ "inequalities" ], "Problem": "Let a,b,c are real numbers such ... $ c^2\\plus{}2(a\\plus{}b)(a\\plus{}c)\\plus{}1 \\le2(a\\plus{}b\\plus{}c)$\r\nProve that ___ $ |b| \\ge |a|$.", "Solution_1": "Solution from school . :D \r\n\r\n\u0397 \u03b4\u03bf\u03c3\u03bc\u03ad\u03bd\u03b7 \u03c0\u03b1\u03af\u03c1\u03bd\u03b5\u03b9 \u03c4\u03b7 \u03bc\u03bf\u03c1\u03c6\u03ae $ c^2\\plus{}c(2a\\plus{}2b\\minus{}2)\\plus{}2a^2\\plus{}2ab\\minus{}2a\\minus{}2b\\plus{}1\\leq 0$ \r\n\r\n\u0397 \u03b4\u03b1\u03ba\u03c1\u03af\u03bd\u03bf\u03c5\u03c3\u03b1 \u03b1\u03c5\u03c4\u03ae\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03ae :wink: \u03ba\u03b1\u03b9 $ D\\equal{}b^2\\minus{}a^2$ qed :)", "Solution_2": "\u03a9\u03c1\u03b1\u03af\u03bf\u03c2 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad ... \u03b5\u03b3\u03ce \u03ad\u03c7\u03c9 \u03bc\u03b9\u03b1 \u03ac\u03bb\u03bb\u03b7 \u03bb\u03cd\u03c3\u03b7.\r\n\u0398\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c6\u03ae\u03c3\u03c9 \u03bb\u03af\u03b3\u03b5\u03c2 \u03bc\u03ad\u03c1\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b5\u03b2\u03ac\u03c3\u03c9 \u03bc\u03b5\u03c4\u03ac :)", "Solution_3": "\u0392\u03b1\u03c1\u03b9\u03ad\u03bc\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03bd\u03b5\u03b2\u03ac\u03c3\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03bc\u03ad\u03c1\u03b5\u03c2 ...\r\n\r\n\u0397 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 :\r\n\r\n\u039c\u03b5\u03c4\u03b1 \u03b1\u03c0\u03cc \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03ba\u03b1\u03c4\u03b1\u03bb\u03ae\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7\u03bd \r\n\r\n$ 2a^2\\plus{}c^2\\plus{}1\\plus{}2(ab\\plus{}bc\\plus{}ca)\\minus{}2(a\\plus{}b\\plus{}c) \\le0$\r\n\r\n\u03cc\u03c0\u03bf\u03c5 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03bf \u03ad\u03bd\u03b1 \u03b1^2 \u03ba\u03b1\u03b9 \u03b1\u03bb\u03bb\u03ac\u03b6\u03bf\u03bd\u03c4\u03ac\u03c2 \u03c4\u03bf\u03c5 \u03bc\u03ad\u03bb\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03b8\u03b1\u03c6\u03b1\u03b9\u03c1\u03cc\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b1 \u03b4\u03c5\u03bf \u03bc\u03ad\u03bb\u03b7 \u03c4\u03bf \u03b2^2 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \r\n\r\n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}2ab\\plus{}2bc\\plus{}2ca\\minus{}2a\\minus{}2b\\minus{}2c\\plus{}1 \\le b^2\\minus{}a^2$, \u03ae\r\n\r\n$ (a\\plus{}b\\plus{}c\\minus{}1)^2\\le b^2\\minus{}a^2$ <=> $ b^2\\minus{}a^2 \\ge0$ \u03ba.\u03bb.\u03c0" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "$ \\odot O_2$ inscribes $ \\odot O_1$ at $ P$, $ AB$ is a chord of $ \\odot O_1$ and is the tangent of $ \\odot O_2$, $ C$ is the point of tangency.\r\n\r\n\r\n$ PA$ meets $ \\odot O_2$ at $ E$, $ PB$ meets $ \\odot O_2$ at $ F$. $ PC$ meets $ \\odot O_1$ at $ G$ and $ EF$ meets $ PG$ at $ D$. $ AD$ meets $ \\odot O_1$ at $ H$.\r\n\r\n\r\nProve that: $ H,F,G$ are collinear.", "Solution_1": "Dear Mathlinkers,\r\nlet Tg the tangnet to O1 at G.\r\n1. Tg // AB // EF (by homothetie or Reim's theorem)\r\n2. DF is the Pascal's line of the degenerated hexagone TgPBAHG.\r\n3. Conclusion : H, F and G are collinear.\r\nSincerely\r\nJean-Louis" } { "Tag": [], "Problem": "Find all positive integers $ x$ that leave a remainder of $ 1$ when divided by $ 2, 3$ and $ 5$.", "Solution_1": "Numbers of the form $ x\\equal{}30m\\plus{}1$ where m is an integer satisfy the conditions of the problem i think." } { "Tag": [], "Problem": "What is the value of $ x$ in the equation $ 2x \\plus{} \\frac{1}{2}x \\plus{} 2(1\\plus{}x) \\equal{} 29$?", "Solution_1": "Expanding gives $ 2x\\plus{}\\frac{1}{2}x\\plus{}2x\\plus{}2\\equal{}29$ or $ 4\\frac{1}{2}x\\equal{}27$. Thus $ x\\equal{}\\boxed{6}$." } { "Tag": [ "puzzles" ], "Problem": "A frog fell down a 20 meter well. Each day the frog climbs 3 meters up. At night it slips down 2 meters from sleeping. At this rate, how many days does it take for the frog to climb out of the well?", "Solution_1": "[hide=\"Yup...\"]\n$17$ (full) days to get to $17$ meters, then the afternoon of the 18th day to reach the top.\n\nSo, like $17 \\frac{1}{2}$ days. [/hide]", "Solution_2": "[hide=\"more specifically\"]At the end of the eighteenth day, he reaches the top (and presumably climbs over the edge so that he doesn't slip back down), so $18$ days.[/hide]", "Solution_3": "[quote=\"roadnottaken\"][hide=\"more specifically\"]At the end of the eighteenth day, he reaches the top (and presumably climbs over the edge so that he doesn't slip back down), so $18$ days.[/hide][/quote]\r\n\r\nWell what do you define as the \"end\" of the day? (This is why I put my answer as I did.) :)", "Solution_4": "I think 17.5 is better..." } { "Tag": [], "Problem": "I can buy $ 3$ tapes for the price of 2 CDs. I paid $ \\$28$ for 2 tapes and 1 CD. How many dollars should I pay for 1 CD?", "Solution_1": "Ok,\r\n \r\nLet t=tapes and c=cd\r\n\r\n28=2t+c\r\n3t=2c\r\n56=4t+2c=7t\r\n\r\nTherefore, t=8.\r\n\r\n28=2(8)+c= 16+c. So, c=12 which is the cost of one cd." } { "Tag": [ "function", "limit", "calculus", "derivative", "topology", "inequalities", "real analysis" ], "Problem": "A function $f: \\mathbb{R}\\to \\mathbb{C}$ is called a Schwarz function if $f$ is infinitely many times differentiable and if\r\n\\[\\lim_{|x| \\to \\infty}\\left|x\\right|^{n}\\left|f(x)\\right| = 0 \\]\r\nfor all positive integers $n$. Let $\\mathcal{S}$ be the set of all Schwarz functions. \r\n\r\nShow that the map $\\mathcal{F}$, which maps each function to its Fourier transform, determines a bijection $\\mathcal{S}\\to \\mathcal{S}$.", "Solution_1": "Arne, this is Schwartz's space, not Schwarz. Also, I think you fogot to add some derivatives to these functions.", "Solution_2": "How do you mean, I forgot some derivatives?", "Solution_3": "it's more than a bijection : it is an homeomorphism (with Swartz's topology)", "Solution_4": "[quote=\"harazi\"]Arne, this is Schwartz's space, not Schwarz. Also, I think you fogot to add some derivatives to these functions.[/quote]\n[quote=\"Arne\"]How do you mean, I forgot some derivatives?\n[/quote]\r\n\\[\\lim_{|x|\\to\\infty}|x|^{n}|f^{(k)}(x)|=0 \\]", "Solution_5": "To Arne: the usual definition of Schwartz's space is the set of $C^{\\infty}$ functions such that $|x|^{k}|f^{(n)}(x)|$ tends to $0$ for all $k$ and $n$. But apparently here you work with another space.", "Solution_6": "But is the result I stated (and which I'm trying to prove) correct?", "Solution_7": "First, a note on spelling: this is Laurent Schwartz (20th century, French), who was a different person than the H.A. Schwarz (19th century, German) who was behind the Cauchy-Schwarz inequality and Schwarz's lemma in complex variables.\r\n\r\nThe defnintion has to be as RIP and harazi stated it: every derivative goes to zero at infinity faster than any power of $x.$ The result would not be true without the correct definition of the space.\r\n\r\nThe proof takes advantage of this: \r\n\r\nIf $g(x)=x^{j}\\left(\\frac{d}{dx}\\right)^{j}f(x),$ then $\\widehat{g}(\\xi)=C\\left(\\frac{d}{d\\xi}\\right)^{k}\\left(\\xi^{j}\\widehat{f}(\\xi)\\right)$\r\n\r\n(The constant $C$ is just a collection of powers of $i$ and of $2\\pi.$ The exact details depend on the details of your definition of the Fourier transform.)\r\n\r\nIn order to claim boundedness and continuity of the transform, we need for the function being transformed to be in $L^{1}.$ We can accomplish that in this proof by investing a couple of extra powers into the decay rate at $\\infty.$\r\n\r\nThe essential thought: the more rapidly the original function decays at $\\infty,$ the smoother its transform is. And the smoother a function is, the more rapidly its transform decays at $\\infty.$ But that pairing is why the definition of the Schwartz space must involve both derivatives and rates of decay on an equal footing." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $G$ be a finitely generated group of $n*n$ complex matrices .then $G$ contains normal subgroup of finite index which contains no nontrivial element of finite order.", "Solution_1": "I like this result :) I know a nice proof of this using $p$-adic specialisations :P", "Solution_2": "hi,\r\nI was never aware of the existence of such a lemmma :blush: ,thanks,I saw on net a proof of this lemma which they call elementary(I wonder how they can say this),neways its here :\r\n[url=http://retro.seals.ch/digbib/view;jsessionid=397B3205E597289E3406EB3E5684E91F?did=c1:42731&p=273]retro.seals.ch/digbib/view;jsessionid=397B3205E597289E3406EB3E5684E91F?did=c1:42731&p=273[/url]", "Solution_3": ":) There is a half-page elementary solution by Cassels, assuming a 5-page elementary $p$-adic specialization theorem (again by Cassels), which is:\r\n\r\nLet $K$ be a field finitely generated over $Q$. Let $S \\subset K$ be a fintie set not containing $0$. Then there exist infinitely many primes $p$ (indeed, a set of positive density) for which there is an embedding $\\gamma: K \\to \\mathbb{Q}_p$ such that $|\\gamma(s)|_p = 1$ for all $s \\in S$; that is, an embedding taking all elements of $S$ to $p$-adic units.\r\n\r\nIt is a very useful result -- it's most striking and famous application is to the Skolem-Mahler-Lech theorem.\r\n\r\n--Vesselin", "Solution_4": "Nice :) Can you show (or give some references) how is it used in a proof of Skolem-Mahler-Lech theorem ?\r\n\r\nAnd also I'd like to see solution to original problem using this result :)" } { "Tag": [ "vector", "LaTeX", "trigonometry", "trig identities", "Law of Cosines" ], "Problem": "$O = (0, 0)$\r\n$A = (206, 2007)$\r\n$B = (2006, 207)$\r\nCompute $cos(AOB)$.\r\n\r\n[hide=\"Hint\"]\nDot product\n[/hide]", "Solution_1": "[hide=\"solution\"]$\\begin{eqnarray*}\\parallel \\vv{a}\\parallel \\cdot\\parallel \\vv{b}\\parallel \\cdot\\cos(AOB)&=&\\vv{a}\\cdot\\vv{b}\\\\ \\sqrt{(206^{2}+2007^{2})(2006^{2}+207^{2})}\\cos(AOB)&=&206\\cdot2006+2007\\cdot207\\\\ \\cos(AOB)&=&\\frac{165737}{\\sqrt{814097\\cdot813377}}\\end{eanarray*}$\nMaybe I made a mistake somewhere... or it's just an awfully computationally heavy problem. BTW, how do you do vectors in LaTeX?[/hide]", "Solution_2": "I'm terrible at latex and this stuff, so bare with me... \r\n\r\n[hide] I would just use the law of cosines and distance formula without using dot product of vectors. Let's see...\n\nDistance between point O and point A = sqrt( 206^2 + 2007^2) \n\nDistance between point O and point B = sqrt (2006^2 + 207^2) \n\nDistance between point A and B = sqrt ( 1800^2 + 1800^2)\n\nThen use the law of cosines to find the cosine of angle AOB\n\nSo... 6480000 = (AO)^2 + (BO)^2 - 2(AO)(BO)cos(AOB)\n\nThen solve...\n[/hide]", "Solution_3": "Couldnt you also find it like this\r\n\r\n[hide]\n$\\cos(AOB) = cos(90-\\arctan(\\frac{206}{2007})-\\arctan(\\frac{207}{2006})) = 0.20367.....$\n\nbut we need completely exact values right, so cant go about it this way?\n[/hide]", "Solution_4": "It is a [b]very[/b] computationally-heavy problem. The first solution was correct." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Let ABCD be a convex quadrilateral and all sides of it are constant. Find a formula for maximum and minumum value of the sum of AC+BD (AC and BD are diagonals of ABCD).", "Solution_1": "I think I found the minimum value of the sum AC +BD . \r\n $ S = AC.BD.sin(AC,BD) \\leq \\frac{(AC+BD)^2}{4} $\r\n $==> AC+BD \\geq 2\\sqrt{S} $ \r\n \"=\" holds AC=BD and AC is perpendicular with BD .", "Solution_2": "In fact your \"=\" can be unattainable. Right?\r\nMoreover, $S$ is not fixed.", "Solution_3": "[quote] Let ABCD be a convex quadrilateral and all sides of it are constant [/quote] \r\n Maybe I don't understand this sentence !!!???", "Solution_4": "I think you should exactly estimate $AC+BD$ in terms of $AB,BC,CD,DA$.", "Solution_5": "[quote][quote]Let $ABCD$ be a convex quadrilateral and all sides of it are constant\n[/quote]\nMaybe I don't understand this sentence !!!???[/quote]\r\nI think that he means that the values of the lenghts of those sides are fixed,no more.", "Solution_6": "[quote=\"Armo\"][quote][quote]Let $ABCD$ be a convex quadrilateral and all sides of it are constant\n[/quote]\nMaybe I don't understand this sentence !!!???[/quote]\nI think that he means that the values of the lenghts of those sides are fixed,no more.[/quote]\r\nHow do you think what did I write above?", "Solution_7": "I just wrote your answer in more details [b]Myth[/b], and note that I answered exactly to the question of [b]nttu[/b]. :)", "Solution_8": ":) :) :) \r\nDon't mind it please!" } { "Tag": [ "ARML", "AMC", "USA(J)MO", "USAMO", "geometry", "Support" ], "Problem": "Not that I have anything to say, but it seems that Las Vegas should have its own topic here (rather than the year-old topic that t0rajir0u dragged up).\r\n\r\nSouthern California will bring four teams this year.", "Solution_1": "NorCal's got Evan O'Dorney. Prepare to be eviscerated. (all a friendly jest :P )", "Solution_2": "Is Evan O'Dorney coming? Is he actually on the ARML team? (And if he is on an ARML team, wouldn't it be SFBA, not NorCal?) What about the other USAMO winners/HM? Is Paul Christiano on the SFBA ARML team? Is there an Oregon team and is Eric Larson on it? Is Sam Elder on the Colorado ARML team?\r\n\r\nJeff Manning is definitely on the Southern California team.", "Solution_3": "From what I heard this weekend at the WA practice, there may be an Oregon team. There's a match between WA and OR next weekend as part of the organizing process.\r\nI also heard that Sam Elder isn't coming this year.", "Solution_4": "Yes, I can confirm that there will be a (somewhat organized) Oregon team this year (since I'm part of it).", "Solution_5": "[quote=\"Kent Merryfield\"]Is Evan O'Dorney coming? Is he actually on the ARML team?[/quote]\r\nHe came to practice and was put on the A team, so I'm assuming he is.", "Solution_6": "Washington and Oregon are having a practice competition next week. Darryl Wu is probably going to be on the A team, by the way :D", "Solution_7": "[quote]Washington and Oregon are having a practice competition next week.[/quote]\r\nIt's a pretty long drive from Eugene or Corvallis to Seattle, or vice versa. Where are you meeting?", "Solution_8": "Portland (approximately halfway). Should be a good warm-up. Most of us are driving down the day before and some of us may join the Oregon team's practice that day.", "Solution_9": "I'm trying to tag along with the SFBA team as a coach, so maybe I'll return to ARML once again :).", "Solution_10": "[quote=\"paladin8\"]I'm trying to tag along with the SFBA team as a coach, so maybe I'll return to ARML once again :).[/quote]\r\n\r\nOh, but you should return to WA team, especially that you missed the opportunity to contribute in senior year 2007 :)", "Solution_11": "[quote=\"t0rajir0u\"]Portland (approximately halfway). Should be a good warm-up. Most of us are driving down the day before and some of us may join the Oregon team's practice that day.[/quote]\r\n\r\nI am just about to send out an update to team. After talking to Oregon coach and Paul, it is confirmed that Saturday's lecture / practice will be canceled due to various conflict (most WA team members can't make to 12noon anyway, I thought you guys who head there on Saturday are just determined to make contributions to US economy). There will be an orientation session instead 8:30 - 9:15am (not much a lecture) by Paul. For team members heading to Portland on Saturday: if this is the first ARML and you have not been to any of the practices before, you may want to be there 8:30am.", "Solution_12": "[quote=\"minghu\"][quote=\"paladin8\"]I'm trying to tag along with the SFBA team as a coach, so maybe I'll return to ARML once again :).[/quote]\n\nOh, but you should return to WA team, especially that you missed the opportunity to contribute in senior year 2007 :)[/quote]\r\n\r\nTrue. I'm not sure they'll let me do that though :P. In any case, it'll be fun to meet up with the WA team again if I end up going. Are you coaching it this year?", "Solution_13": "[quote=\"t0rajir0u\"]Washington and Oregon are having a practice competition next week. Darryl Wu is probably going to be on the A team, by the way :D[/quote]\r\n\r\nCompared to Southern CA and SFBA, Washington has a relatively smaller list of USAMO qualifiers. Therefore it is logical to put all USAMO qualifiers on A team. Don't you envy SFBA's deep pool of talents? (over 60 USAMO qualifiers in the area!)", "Solution_14": "Following up on what minghu said about USAMO qualifiers - here are the number of USAMO qualifiers by ARML region:\r\n\r\nWA: 13\r\nOR: 6, including one HM (Eric Larson)\r\nNorCal: 1\r\nSFBA: 63, including two winners (Evan O'Dorney and Paul Christiano)\r\nSoCal: 19, including one HM (Jeff Manning)\r\nSan Diego: 8\r\nAZ/NV/UT: 5\r\nCO: 6, including one winner (Sam Elder)\r\nNM: 4\r\n\r\nI don't know how many of these people will actually come to ARML. I also don't know how many different teams AZ/NV/UT is. In the case of Southern California, 17 of the 19 will be on the ARML team (11 of them on the A team.)", "Solution_15": "Word has it Evan has joined SFBA. I'm not sure about Paul; he went two years ago, but not last year, and I haven't heard if he's going this year.\r\n\r\nWe have four teams this year, by the way. The last two (now named S and F) will each be shorthanded by three unless transportation works out in such a way that six more people can get to Vegas.\r\n\r\nIn response to the question of AZ/NV/UT, I have no idea if there's a Utah team this year. Tim's created a \"Desert Southwest\" region composed of Arizona, New Mexico, and Nevada. It'll send two teams: One from Arizona, and another with students from both New Mexico and Nevada. (The region was created to allow those states to send as many full teams as possible.)", "Solution_16": "[quote=\"Sly Si\"]We have four teams this year, by the way. The last two (now named S and F) will each be shorthanded by three unless transportation works out in such a way that six more people can get to Vegas.[/quote]\r\nWe have four teams, named SoCal A, B, C, and Gamma. (Or to do that right, $ \\Gamma.$) Unless someone cancels that I don't know about yet, we'll actually have two extra people, 62 students total. We're covering the transportation shortfall with a few extra cars.", "Solution_17": "I'm on the Oregon team, and unfortunately Eric Larson's not doing it. Being the pessimist/realist that I am, I personally think that Washington is probably going to kill us... :|", "Solution_18": "Coach (Manny) said Eric is on ARML roster. Do you mean he won't be there this coming Sunday (May 18) for the intramural?", "Solution_19": "No, Eric decided not to be on the ARML team. He was taken off the roster a few weeks ago.", "Solution_20": "SD is fielding two teams, including its first ever in division A. We will be arriving in the early afternoon, although our plan is to go out for a [i]very[/i] early dinner before the competition begins. (We could call it a late lunch, but most of the kids will have been awake for ten hours by that point.)\r\n\r\nDoes Silas require Sheepsheading? :)", "Solution_21": "Your schedule is similar to ours. You might try looking us up to see who wants to go to dinner together (our guys seem to be partial to the In-N-Out on the edge of campus.) \r\n\r\nAs for Silas, I'm pretty sure he's said elsewhere that he's not coming. We are bringing a couple of recent ARML veterans from other regions: Yasha from North Carolina and Tim from Wisconsin (both now Caltech students).", "Solution_22": "Gah, I would strongly support Sheepsheading, but I'm spending the quarter at Oxford and won't be back until two weeks after ARML. Next year, though, I'm going, even if I have to schedule all my final exams around it.\r\n\r\nBy the way, last I heard, SFBA's bringing ARML vets Nathan (from Washington) and Seva (Alabama), in addition to possibly paladin8.", "Solution_23": "Well, WA did indeed kill us [Team Oregon] yesterday at the Intramural competition. If only Larson was on our team... that'd be soooo sweet.\r\nAs of right now, we have 4 USAMO qualifiers on the team, but none made it to MOP, including Brynmor Chapman. Oh well.", "Solution_24": "Us Nevadans are combining with the New Mexicans for a team. We're probably not going to do very well, but who knows? 5 of us coming from a school for PG kids, two of whom are taking college level math courses.", "Solution_25": "I don't know many of the names on the MOP list ... does anybody know how many of them are from the SFBA team, SoCal, ... etc.?\r\n\r\nI know Washington has 2 people going to MOP (me and Joy, a freshmen).", "Solution_26": "[quote=\"samath\"]I don't know many of the names on the MOP list ... does anybody know how many of them are from the SFBA team, SoCal, ... etc.? [/quote]\r\nFor the California portion of that list, two names (Johnny Li and Zhifan \"Ivan\" Zhang) will be on the Southern California team. The other 8 names belong to the SFBA region; I don't know how many of those 8 are coming to ARML.", "Solution_27": "From my going to practices, I recognize the following names:\r\n\r\nVishal Arul\r\nJohn Boyle\r\nTaylor Han\r\nEvan O'Dorney\r\n\r\nThere may be more, but those people have been at the practices for sure.", "Solution_28": "Paladin8, are you one of the coaches?\r\nI can probably come meet up with you on sunday if you are, and give you an incomplete list of all of the AoPSers going to ARML.", "Solution_29": "Well, here's my version of an incomplete list of AoPSers:\r\n\r\nJAM\r\nlitoxe27182818\r\nIvan Zhang\r\njli\r\nbluejay\r\nZellex\r\nKimby_102\r\nnaitixuy\r\nt\\w/o\r\nBrut3Forc3\r\nruneknightx\r\npythag011\r\nzapi2007\r\nDiffer\r\nmws21\r\nTecnogram888\r\nxsk13\r\n\r\n[i]And,[/i]\r\n\r\nKent Merryfield\r\njmerry\r\nDr. Kevin\r\nzmli\r\nYasha", "Solution_30": "[quote=\"paladin8\"]From my going to practices, I recognize the following names:\n\nVishal Arul\nJohn Boyle\nTaylor Han\nEvan O'Dorney\n\nThere may be more, but those people have been at the practices for sure.[/quote]\r\n\r\nAdd David Zeng and Lynnelle Ye, and your list's complete. As a member of the SFBA team who epic-ly failed USAMO, I know names :P Out of those 6 SFBAers, 4 are AoPSers, as well...\r\n\r\nEDIT: I'm not a coach or anything, but SFBA is sending 4 teams, teams S, F, B, and A, in reverse order of skillz.", "Solution_31": "[quote=\"xscapezaer\"][quote=\"paladin8\"]From my going to practices, I recognize the following names:\n\nVishal Arul\nJohn Boyle\nTaylor Han\nEvan O'Dorney\n\nThere may be more, but those people have been at the practices for sure.[/quote]\n\nAdd David Zeng and Lynnelle Ye, and your list's complete. As a member of the SFBA team who epic-ly failed USAMO, I know names :P Out of those 6 SFBAers, 4 are AoPSers, as well...\n\nEDIT: I'm not a coach or anything, but SFBA is sending 4 teams, teams S, F, B, and A, in reverse order of skillz.[/quote]\r\nnope david and lynnelle arent coming\r\nyeah those 4 are our moppers", "Solution_32": "[quote=\"DiscreetFourierTransform\"]Paladin8, are you one of the coaches?\nI can probably come meet up with you on sunday if you are, and give you an incomplete list of all of the AoPSers going to ARML.[/quote]\r\n\r\nSort of. I've only shown up to the last two practices since I didn't know about it before, but I hope to be going to Vegas with you guys. And yeah I'll probably be there tomorrow (Sunday).", "Solution_33": "[quote=\"Kent Merryfield\"]...the In-N-Out on the edge of campus.[/quote]\r\nYou know, I think I ate there in 2000- the Palos Verdes team I was on knew about it.", "Solution_34": "As for who is going to be there: I count 23 teams from 12 organizations. Here's the list:\r\n\r\nColorado: 2 teams\r\nDesert Southwest: 2 teams (this is NM, AZ, and NV)\r\nNorthern California: 1 team (this is Davis/Sacramento)\r\nOregon: 1 team\r\nSouthern California: 4 teams\r\nSan Diego: 2 teams\r\nSan Francisco Bay Area: 4 teams\r\nUtah: 1 team\r\nWashington: 2 teams\r\nHong Kong: 1 team\r\nPhilippines: 1 team\r\nTaiwan: 2 teams\r\n\r\nOf these 23 teams, four are international \"guest division\" teams, six are Division A teams, and the remaining 13 are Division B teams. The six Division A teams are:\r\n\r\nColorado Granite\r\nSouthern California A\r\nSouthern California B\r\nSan Diego A\r\nSan Francisco Bay Area A\r\nWashington Gold" } { "Tag": [ "geometry", "trapezoid" ], "Problem": "A trapezoid with bases of 6 feet and 8 feet and a height of 4 feet is cut in half. How many square feet are in the area of each half?", "Solution_1": "Area of a trapezoid with bases of $ 6$ feet and $ 8$ feet and a height of $ 4$ feet : $ (6 \\plus{} 8) \\times 4 \\times \\frac {1}{2} \\equal{} 28$ square feet.\r\nThus, $ \\frac {28}{2} \\equal{} \\boxed{14}$ square feet." } { "Tag": [ "calculus", "integration", "function", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "Find the continuous function which satisfies for all real numbers $x$ as follows.\r\n\r\n\\[\\int^x_0 tf(x-t)dt=\\int^x_0 f(t)dt+\\sin x+\\cos x-x-1\\]", "Solution_1": "Linear convolution equation on $[0,\\infty)$: it's time to use Laplace transforms. If $F(s)$ is the Laplace transform of $f(x)$ then:\r\n\r\n$\\frac1{s^2}F(s)=\\frac1sF(s)+\\frac1{s^2+1}+\\frac{s}{s^2+1}-\\frac1{s^2} -\\frac1s.$\r\n\r\nSolving for $F(s)$,\r\n\r\n$F(s)=\\frac{s+1}{(s-1)(s^2+1)}=\\frac1{s-1}-\\frac{s}{s^2+1}.$\r\n\r\nSo $f(x)=e^x-\\cos x.$\r\n\r\nAt least that's the answer for $x\\ge0.$ I haven't considered $x<0$ yet." } { "Tag": [ "probability", "expected value", "combinatorics unsolved", "combinatorics" ], "Problem": "$n, k$ are positive integers. $A_0$ is the set $\\{1, 2, ... , n\\}$. $A_i$ is a randomly chosen subset of $A_{i-1}$ (with each subset having equal probability). Show that the expected number of elements of $A_k$ is $\\dfrac{n}{2^k}$", "Solution_1": "Should $\\dfrac{n}{2k}$ be $\\dfrac{n}{2^k}$?", "Solution_2": "My mistake, I've corrected it, thanks Ravi.", "Solution_3": "I don't really understand... what if $\\frac{n}{2^k}$ is not an integer?", "Solution_4": "It doesn't matter. The average (expected value) doesn't have to be an integer, even if the number of elements is always an integer.", "Solution_5": "this is rather trivial. by linearity of expectation it suffices to show that $P(1\\in A_k)=\\frac{1}{2^k}$. This is seen easily inductively, as $P(1\\in A_k)=P(1\\in A_{k}|1\\in A_{k-1})\\cdot P(1\\in A_{k-1})$ and also noting that $P(1\\in A_{k}|1\\in A_{k-1})=\\frac{1}{2}$, since for any set $X$ containing $1$ the number of subsets of $X$ containing or not containing $1$ is the same.", "Solution_6": "We k-times make a subset of the set A, in each step each number has probability $\\frac 1 2$ to be chosen. So after k steps, it's $\\frac 1 {2^k}$. Because of that, the number of elemtnts has binomial distribution and the expected value is\r\n$E=\\sum_{p=0}^n p{n \\choose p} \\left( \\frac 1 {2^k}\\right)^p\\left( 1-\\frac 1 {2^k}\\right)^{n-p}$, the first element of sum is 0, so \r\n$E=\\sum_{p=1}^n n{n-1 \\choose p-1} \\left( \\frac 1 {2^k}\\right)^p\\left( 1-\\frac 1 {2^k}\\right)^{n-p}$\r\n$E=\\frac n {2^k}\\sum_{p=0}^{n-1}{n-1 \\choose p} \\left( \\frac 1 {2^k}\\right)^p\\left( 1-\\frac 1 {2^k}\\right)^{n-1-p}=\\frac n {2^k} \\left( \\frac 1 {2^k}+\\left(1- \\frac 1 {2^k}\\right)\\right)^{n-1}$, QED", "Solution_7": "Induct on $k$. $k=0$ given. Assume that $E[|A_k|]=\\frac{n}{2^k}$. If $|A_k|=m$ then by linearity of expectation \\[E[|A_{k+1}|]=\\sum_{i=0}^{m}i\\frac{\\dbinom{m}{i}}{2^m}=\\frac{1}{2^m}\\sum_{i=0}^{m}i\\dbinom{m}{i}=\\frac{1}{2^m}(m2^{m-1})=\\frac{m}{2}\\].\n\nfrom the well-known identity $\\sum_{i=0}^{m}i\\dbinom{m}{i}=m2^{m-1}$. Let $p_k(m)$ be the probability that $|A_k|=m$. Then \n\n\\[E[|A_{k+1}|]=\\sum_{m=0}^{n}p_k(m)\\Bigg(\\frac{m}{2}\\Bigg)=\\frac{1}{2}\\sum_{m=0}^{n}p_k(m)m=\\frac{1}{2}E[|A_k|]=\\frac{1}{2}\\Bigg(\\frac{n}{2^k}\\Bigg)\\]\n\nwhich is $\\frac{n}{2^{k+1}}$ so done." } { "Tag": [ "function", "algebra", "domain", "complex numbers" ], "Problem": "If $f(x)=x^2$ and $g(x)=\\sqrt{x}$. Then $f(g(x))=x$, but does the domain restirction of $g(x)$ carry over into $f(g(x))$? I don't think it should. Here's an example, if $x=-1$, then $g(x)=i$ and $f(i)=-1$, thus $f(g(x))=x$ holds true.", "Solution_1": "even though -1 works, the domain restriction still carries over.\r\n\r\nplug in $(\\sqrt{x})^2$ into your calcuator..., there won't be any negative x values..\r\n\r\nits kid of like sayng $\\frac{(x-1)^2}{x-1}$ you can simplify it into x-1 but it still has the domain of x can't be 1", "Solution_2": "Actually the main reason I asked was because my calculator (TI-89) [b]did[/b] give a value.\r\nIn the y= editor I typed $y_1=x^2$, $y_2=\\sqrt{x}$, and $y_3=y_1(y_2(x))$. For $x=-1$, $y_3=-1$.\r\nSince the TI-89 is the only calculator I've tested this on that deals with complex numbers, I assumed the reason other calculators gave an error was just because they couldn't deal with $i$.\r\n\r\nBut thanks for the answer, and I agree completely with your example $\\frac{(x-1)^2}{x-1}$, but I still think $(\\sqrt{x})^2$ is different because instead of involving division by zero, it involves $i$.\r\n\r\nCan anyone show a proof of why this is, or is it just a definition?", "Solution_3": "Does the answer change if I said the domain is $C$ instead of $R$?", "Solution_4": "[quote=\"Jimmy\"]If $f(x)=x^2$ and $g(x)=\\sqrt{x}$. Then $f(g(x))=x$, but does the domain restirction of $g(x)$ carry over into $f(g(x))$? I don't think it should. Here's an example, if $x=-1$, then $g(x)=i$ and $f(i)=-1$, thus $f(g(x))=x$ holds true.[/quote]The domain restriction does carry over, unless $f$ and $g$ are complex-valued functions (Is there a term for this?), as you have treated them, in which case you would normally write $f(z)=z^2$ and $g(z)=\\sqrt{z}$. (Someone correct me if I'm wrong; I don't really know much about functions of a complex variable.)\n\n[quote=\"Jimmy\"]Does the answer change if I said the domain is $Z$ instead of $R$? ( I can't figure out how to type a cursive R for real numbers and the Z that represents complex numers.)[/quote]Actually, $\\mathbf{C}$ is the set of all complex numbers. $\\mathbf{Z}$ is the set of all integers.", "Solution_5": "You can use \\mathbb{} to get that special font (blackboard bold):\r\n\r\n$\\mathbb{C, Z, R, N}$", "Solution_6": "[quote=\"AntonioMainenti\"]Actually, $\\mathbf{C}$ is the set of all complex numbers. $\\mathbf{Z}$ is the set of all integers.[/quote]\r\nThanks I fixed this.", "Solution_7": "[quote=\"Jimmy\"]If $f(x)=x^2$ and $g(x)=\\sqrt{x}$. Then $f(g(x))=x$, but does the domain restirction of $g(x)$ carry over into $f(g(x))$? I don't think it should. Here's an example, if $x=-1$, then $g(x)=i$ and $f(i)=-1$, thus $f(g(x))=x$ holds true.[/quote]\r\nDefining the domains and codomains of functions or mappings clearly distinguishes whether there is a restriction or not. It is useful to write something like $f:\\mathbb{X}\\rightarrow\\mathbb{Y}$ or $f:\\mathsf{F}^n\\rightarrow\\mathsf{F}^m$, where you specify domains, they can tell you more information, rather than guessing what is restricted or what is extended." } { "Tag": [ "inequalities", "function", "limit", "algebra unsolved", "algebra" ], "Problem": "hello!\r\nfind all the functions suchs that :\r\ni)$ \\forall (x,y) \\in \\mathbb{R} : f(x\\plus{}y)\\leq f(x)\\plus{}f(y)$\r\nii)$ \\lim_{x\\rightarrow 0}\\frac{f(x)}{x}\\equal{}1$", "Solution_1": "Then $ f(x)=x$.\r\n[hide=\"Sketch of proof\"]Resolve condition (ii) in terms of the $ \\varepsilon$-$ \\delta$-definition of a limit to see that $ \\lim_{x\\to 0}f(x)=0$. (i) easily gives $ f(0)\\ge 0$. The inequality $ f(0)\\le f(h)+f(-h)$ yields for $ h\\to 0$ the reverse inequality $ f(0)\\le 0$, hence $ f(0)=0$. Then (ii) means that $ f$ is differentiable at $ x=0$ with $ f'(0)=1$.\n\n$ f$ is everywhere differentiable with $ f'\\equiv 1$, this implies $ f(x)=x$. To see this fix $ x_0\\in \\mathbb R$ and use \\[ f(x_0)\\le f(x_0\\pm h)+f(\\mp h)\\le f(x_0)+f(\\pm h)+f(\\mp h)\\] with $ h>0$ to build up the inequalities of the difference quotient \\[ \\frac{f(-h)}{-h}\\le\\frac{f(x_0+h)-f(x_0)}{h}\\le\\frac{f(h)}{h}\\] for $ h>0$. The case $ h<0$ works in exactly the same way. Altogether this implies that $ f$ is differentiable at $ x_0$ with $ f'(x_0)=1$.[/hide]" } { "Tag": [ "algorithm" ], "Problem": "Situation is as follows: we have three towers and $2\n\n(Gauss-Bonnet)Let $\\Sigma$ be a smooth closed oriented surface in $R^3$, then \n\n$$\\int_{\\Sigma}k\\mbox{d}s = 2\\pi\\chi (\\Sigma).$$ \n\n\\item<2-> \n\nHopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has \n\n$$\\int_{M}k\\mbox{d}v =\\frac{1}{2}\\mbox{vol}(S^n)\\chi (M),$$ \n\nwhere $k$ is the Gauss-Kronecker curvature. \n\n\\end{itemize}\n\n\\end{frame}\n\n\n\n\\subsection{AAA1}\n\n\\begin{frame}\n\n\\frametitle{Gauss-Bonnet in 20th Century}\n\n\\begin{itemize} \n\n\\item<1-> \n\n(Gauss-Bonnet)Let $\\Sigma$ be a smooth closed oriented surface in $R^3$, then \n\n$$\\int_{\\Sigma}k\\mbox{d}s = 2\\pi\\chi (\\Sigma).$$ \n\n\\item<2-> \n\nHopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has \n\n\\end{itemize}\n\n\\end{frame}\n\n\n\n\\section{AAAA}\n\n\\subsection{AAA1}\n\n\\begin{frame}\n\n\\frametitle{Gauss-Bonnet in 20th Century}\n\n\\begin{itemize} \n\n\\item \n\n(Gauss-Bonnet)Let $\\Sigma$ be a smooth closed oriented surface in $R^3$, then \n\n$$\\int_{\\Sigma}k\\mbox{d}s = 2\\pi\\chi (\\Sigma).$$ \n\n\\item \n\nHopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has \n\n$$\\int_{M}k\\mbox{d}v =\\frac{1}{2}\\mbox{vol}(S^n)\\chi (M),$$ \n\nwhere $k$ is the Gauss-Kronecker curvature. \n\n\\end{itemize}\n\n\\end{frame}\n\n\\end{document}\n\n[/code]", "Solution_1": "You really ought to post the warnings as we can't guess what they are.\r\nSolution is found [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/a734f9c5b5016fc9/6a0ef006e290f29b?hl=en&lnk=gst&q=beamer+font+warning#6a0ef006e290f29b]here[/url] using a different font or, to use non-standard sizes in the current font [code]\\usepackage[T1]{fontenc} \n\\usepackage{fix-cm}[/code]\r\nYou should not be using \\$\\$...\\$\\$ as that is obsolete. See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=dolldoll]Why use \\ [ ...\\ ] in place of \\$\\$...\\$\\$?[/url]", "Solution_2": "Many thanks!\r\n\r\nFor the beamer, I download 3 packages beamer , pgf and xcolor.\r\n\r\nthe 4 warnings are:\r\n\r\n1. package pgf warning: this package is obsolete and no longer needed on input line 13;\r\n\r\n2. package hyperref warning: option 'pdfpagelabels' is turned off \r\n because \\thepage is undefined\r\n\r\n3. Latex Font warning: Font shape 'OT1/cmss/m/n' in size <4> not available\r\n\r\n4 Latex Font warning: Size substitutions with differences\r\n(Font) up to 1.0pt have occurred.\r\n\r\n\r\n[color=red]\nIf I write the code below:[/color]\r\n[code]\\documentclass{beamer}\n\\usepackage{times }\n\\usefonttheme{professionalfonts}\n\\title{Example Presentation Created with the Beamer Package}\n............\n[/code]\r\n\r\n[color=red]Then the first two warnings still exist[/color]", "Solution_3": "If I write the code below: \r\n[code]\\documentclass{beamer} \n\\usepackage[T1]{fontenc} \n\\usepackage{fix-cm} \n\\title{Example Presentation Created with the Beamer Package} \n............ \n[/code]\r\n\r\nand remove all \\$, the first two warnings still exist", "Solution_4": "1. See [url=http://www.nabble.com/This-package-is-obsolete...-td17642354.html]here[/url]\r\n2. \\documentclass[hyperref={pdfpagelabels=false}]{beamer}" } { "Tag": [ "inequalities", "logarithms" ], "Problem": "$a,$ $b$ and $c$ are non-negative numbers such that $ab+ac+bc=3.$ Prove that:\r\n$(a^{2}+1)(b^{2}+1)(c^{2}+1)\\geq8.$", "Solution_1": "Nice :) \r\n\r\nExpand to get the equivalent \\[a^{2}b^{2}c^{2}+\\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\\right)+\\left(a^{2}+b^{2}+c^{2}\\right)+1 \\geq 8\\]Now let $a+b+c = m$ and $abc = n$. Clearly, $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}= (ab+bc+ca)^{2}-2abc(a+b+c) = 9-2mn$ and $a^{2}+b^{2}+c^{2}= (a+b+c)^{2}-2(ab+bc+ca) = m^{2}-6$. Therefore, we have to prove that $n^{2}+9-2mn+m^{2}-6+1 \\geq 8$, which is equivalent to $|m-n| \\geq 2$. \r\nNow remark that $m^{2}= (a+b+c)^{2}\\geq 3(ab+bc+ca) = 9$, so $m \\geq 3$, and $3 = (ab+bc+ca) \\geq 3\\sqrt[3]{a^{2}b^{2}c^{2}}$, so $1 \\geq n$. Therefore $m\\geq 3 > 1 \\geq n$, so we only need to prove that $m-n \\geq 2$. And that's obvious now.", "Solution_2": "This looks so much like Cauchy...we must be able to use it somehow. \r\n\r\nHm, this looks fairly promising:\r\n\r\n$(a^{2}+1)(b^{2}+1) \\geq (ab+1)^{2}\\\\ (b^{2}+1)(c^{2}+1) \\geq (bc+1)^{2}\\\\ (c^{2}+1)(a^{2}+1) \\geq (ca+1)^{2}$\r\n\r\nThus\r\n\\[(a^{2}+1)(b^{2}+1)(c^{2}+1) \\geq (ab+1)(bc+1)(ac+1) \\]", "Solution_3": "OK, Kurt G\u00f6del.\r\nMy proof:\r\nby Holder:\r\n $(1+1+1+1)(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)\\geq$\r\n$\\geq(ab+ac+bc+1)^{4}=256.$ :)", "Solution_4": "Could you post the statement of Holder's inequality and its proof :?:", "Solution_5": "[quote=\"M4RI0\"]Could you post the statement of Holder's inequality and its proof :?:[/quote]\r\nM4RI0, for you! :lol: \r\nStep1.\r\nLet's $a_{1},$ $a_{2},...,$ $a_{n},$ $\\alpha_{1},$ $\\alpha_{2},...$ and $\\alpha_{n}$ are positive numbers such that \r\n$\\frac{1}{\\alpha_{1}}+\\frac{1}{\\alpha_{2}}+...+\\frac{1}{\\alpha_{n}}=1.$ Then the following inequality ( Young ) holds:\r\n$\\frac{a_{1}^{\\alpha_{1}}}{\\alpha_{1}}+\\frac{a_{2}^{\\alpha_{2}}}{\\alpha_{2}}+...+\\frac{a_{n}^{\\alpha_{n}}}{\\alpha_{n}}\\geq a_{1}a_{2}...a_{n}.$\r\nProof:\r\n$f(x)=\\ln x$ is concave function.\r\nHence, $\\ln\\left(\\frac{a_{1}^{\\alpha_{1}}}{\\alpha_{1}}+\\frac{a_{2}^{\\alpha_{2}}}{\\alpha_{2}}+...+\\frac{a_{n}^{\\alpha_{n}}}{\\alpha_{n}}\\right)\\geq\\frac{1}{\\alpha_{1}}\\cdot\\ln a^{\\alpha_{1}}+\\frac{1}{\\alpha_{2}}\\cdot\\ln a^{\\alpha_{2}}+...+\\frac{1}{\\alpha_{n}}\\cdot\\ln a^{\\alpha_{n}}=$\r\n$=\\ln{a_{1}a_{2}...a_{n}}.$\r\n\r\nStep2.\r\nLet's $a_{ij},$ where $i\\in\\{1,2,...,n\\},$ $j\\in\\{1,2,...,m\\},$ $\\alpha_{1},$ $\\alpha_{2},...$ and $\\alpha_{n}$ \r\nare positive numbers such that $\\frac{1}{\\alpha_{1}}+\\frac{1}{\\alpha_{2}}+...+\\frac{1}{\\alpha_{n}}=1.$\r\n Then the following inequality ( Holder ) holds:\r\n$\\left(a_{11}^{\\alpha_{1}}+a_{12}^{\\alpha_{1}}+...+a_{1m}^{\\alpha_{1}}\\right)^{\\frac{1}{\\alpha_{1}}}\\left(a_{21}^{\\alpha_{2}}+a_{22}^{\\alpha_{2}}+...+a_{2m}^{\\alpha_{2}}\\right)^{\\frac{1}{\\alpha_{2}}}...\\left(a_{n1}^{\\alpha_{n}}+a_{n2}^{\\alpha_{n}}+...+a_{nm}^{\\alpha_{m}}\\right)^{\\frac{1}{\\alpha_{n}}}\\geq$\r\n$\\geq a_{11}a_{21}...a_{n1}+a_{12}a_{22}...a_{n2}+...+a_{1m}a_{2m}...a_{nm}.$\r\nProof:\r\n$\\frac{a_{11}a_{21}...a_{n1}+a_{12}a_{22}...a_{n2}+...+a_{1m}a_{2m}...a_{nm}}{\\left(a_{11}^{\\alpha_{1}}+a_{12}^{\\alpha_{1}}+...+a_{1m}^{\\alpha_{1}}\\right)^{\\frac{1}{\\alpha_{1}}}\\left(a_{21}^{\\alpha_{2}}+a_{22}^{\\alpha_{2}}+...+a_{2m}^{\\alpha_{2}}\\right)^{\\frac{1}{\\alpha_{2}}}...\\left(a_{n1}^{\\alpha_{n}}+a_{n2}^{\\alpha_{n}}+...+a_{nm}^{\\alpha_{m}}\\right)^{\\frac{1}{\\alpha_{n}}}}=$\r\n$=\\sum_{j=1}^{m}\\frac{a_{1j}a_{2j}...a_{nj}}{\\left(a_{11}^{\\alpha_{1}}+a_{12}^{\\alpha_{1}}+...+a_{1m}^{\\alpha_{1}}\\right)^{\\frac{1}{\\alpha_{1}}}\\left(a_{21}^{\\alpha_{2}}+a_{22}^{\\alpha_{2}}+...+a_{2m}^{\\alpha_{2}}\\right)^{\\frac{1}{\\alpha_{2}}}...\\left(a_{n1}^{\\alpha_{n}}+a_{n2}^{\\alpha_{n}}+...+a_{nm}^{\\alpha_{m}}\\right)^{\\frac{1}{\\alpha_{n}}}}=$\r\n$=\\sum_{j=1}^{m}\\left(\\frac{a_{1j}}{\\left(a_{11}^{\\alpha_{1}}+a_{12}^{\\alpha_{1}}+...+a_{1m}^{\\alpha_{1}}\\right)^{\\frac{1}{\\alpha_{1}}}}\\cdot\\frac{a_{2j}}{\\left(a_{21}^{\\alpha_{2}}+a_{22}^{\\alpha_{2}}+...+a_{2m}^{\\alpha_{2}}\\right)^{\\frac{1}{\\alpha_{2}}}}\\cdot...\\cdot\\frac{a_{nj}}{\\left(a_{n1}^{\\alpha_{n}}+a_{n2}^{\\alpha_{n}}+...+a_{nm}^{\\alpha_{n}}\\right)^{\\frac{1}{\\alpha_{n}}}}\\right)\\leq$\r\n$\\leq\\sum_{j=1}^{m}\\left(\\frac{a_{1j}^{\\alpha_{1}}}{\\alpha_{1}\\left(a_{11}^{\\alpha_{1}}+a_{12}^{\\alpha_{1}}+...+a_{1m}^{\\alpha_{1}}\\right)}+\\frac{a_{2j}^{\\alpha_{2}}}{\\alpha_{2}\\left(a_{21}^{\\alpha_{2}}+a_{22}^{\\alpha_{2}}+...+a_{2m}^{\\alpha_{2}}\\right)}+...+\\frac{a_{nj}^{\\alpha_{n}}}{\\alpha_{n}\\left(a_{n1}^{\\alpha_{n}}+a_{n2}^{\\alpha_{n}}+...+a_{nm}^{\\alpha_{n}}\\right)}\\right)=$\r\n$=\\frac{1}{\\alpha_{1}}+\\frac{1}{\\alpha_{2}}+...+\\frac{1}{\\alpha_{n}}=1.$ \r\n\r\n$\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}+\\frac{1}{4}=1.$ \r\nThus, $\\sqrt[4]{4(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)}=$\r\n$=\\left(1^{4}+1^{4}+1^{4}+1^{4}\\right)^{\\frac{1}{4}}\\left(\\left(\\sqrt{ab}\\right)^{4}+\\left(\\sqrt a\\right)^{4}+\\left(\\sqrt b\\right)^{4}+1^{4}\\right)^{\\frac{1}{4}}\\cdot$\r\n$\\cdot\\left(\\left(\\sqrt{a}\\right)^{4}+\\left(\\sqrt{ ac}\\right)^{4}+\\left(\\sqrt c\\right)^{4}+1^{4}\\right)^{\\frac{1}{4}}\\left(\\left(\\sqrt{b}\\right)^{4}+\\left(\\sqrt c\\right)^{4}+\\left(\\sqrt{bc}\\right)^{4}+1^{4}\\right)^{\\frac{1}{4}}\\geq$\r\n$\\geq1\\cdot\\sqrt{ab}\\cdot\\sqrt{a}\\cdot\\sqrt{b}+1\\cdot\\sqrt{a}\\cdot\\sqrt{ac}\\cdot\\sqrt{c}+1\\cdot\\sqrt{b}\\cdot\\sqrt{c}\\cdot\\sqrt{bc}+1\\cdot1\\cdot1\\cdot1=$\r\n$=ab+ac+bc+1=4.$ :)", "Solution_6": "[quote=\"arqady\"]OK, Kurt G\u00f6del.\nMy proof:\nby Holder:\n $(1+1+1+1)(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)\\geq$\n$\\geq(ab+ac+bc+1)^{4}=256.$ :)[/quote]\r\n\r\nOh! this solution is very simple and easy. You are so great!", "Solution_7": "So did you make up the problem with the solution in mind, or the other way around?", "Solution_8": "My solution:\r\n$(a^{2}+1)(b^{2}+1)(c^{2}+1) \\geq (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ \r\n$\\geq (ab+bc+ca)(a+b+c)-(\\frac{ab+bc+ca}{3})^{\\frac{3}{2}}$ \r\n$\\geq (ab+bc+ca)\\sqrt{3(ab+bc+ca)}-(\\frac{ab+bc+ca}{3})^{\\frac{3}{2}}$ $= 8$", "Solution_9": "[quote=\"Ng\u00f4ng\"]My solution:\n$(a^{2}+1)(b^{2}+1)(c^{2}+1) \\geq (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ \n$\\geq (ab+bc+ca)(a+b+c)-(\\frac{ab+bc+ca}{3})^{\\frac{3}{2}}$ \n$\\geq (ab+bc+ca)\\sqrt{3(ab+bc+ca)}-(\\frac{ab+bc+ca}{3})^{\\frac{3}{2}}$ $= 8$[/quote]\r\nNg\u00f4ng, very nice proof! :lol:", "Solution_10": "[quote=arqady]OK, Kurt G\u00f6del.\nMy proof:\nby Holder:\n $(1+1+1+1)(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)\\geq$\n$\\geq(ab+ac+bc+1)^{4}=256.$ :)[/quote]\n\nbut we have to prove that $(a^{2}+1)(b^{2}+1)(c^{2}+1)\\geq8.$....where we proved that in your proof ???\nsorry if this is a stupid question but i am not getting..i think i am missing something simple,can you pls give more details....\n", "Solution_11": "[quote=Understandingmathematics][quote=arqady]OK, Kurt G\u00f6del.\nMy proof:\nby Holder:\n $(1+1+1+1)(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)\\geq$\n$\\geq(ab+ac+bc+1)^{4}=256.$ :)[/quote]\n\nbut we have to prove that $(a^{2}+1)(b^{2}+1)(c^{2}+1)\\geq8.$....where we proved that in your proof ???\nsorry if this is a stupid question but i am not getting..i think i am missing something simple,can you pls give more details....[/quote]\n\n$(a^{2}+1)(b^{2}+1)=\n(a^{2}b^{2}+a^{2}+b^{2}+1)$", "Solution_12": "[quote=shalomrav][quote=Understandingmathematics][quote=arqady]OK, Kurt G\u00f6del.\nMy proof:\nby Holder:\n $(1+1+1+1)(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)\\geq$\n$\\geq(ab+ac+bc+1)^{4}=256.$ :)[/quote]\n\nbut we have to prove that $(a^{2}+1)(b^{2}+1)(c^{2}+1)\\geq8.$....where we proved that in your proof ???\nsorry if this is a stupid question but i am not getting..i think i am missing something simple,can you pls give more details....[/quote]\n\n$(a^{2}+1)(b^{2}+1)=\n(a^{2}b^{2}+a^{2}+b^{2}+1)$[/quote]\n\n\nthanks ...\n" } { "Tag": [], "Problem": "The normal boiling point of methanol(CH3OH) is 64.7 degree celsius, and its molar enthalpy of vaporization is 71.8kJ/mole. Calculate the change in enthalpy when 33.5 grams of liquid methanol is vaporized into a gas at its normal boiling point.", "Solution_1": "The molar mass of methanol is 32.042 g/mol, and so $ n_{methanol} \\equal{} \\frac {33.5\\,g}{32.042\\,g/mol} \\equal{} 1.04\\,mol$. Therefore, \r\n\r\n$ \\Delta H \\equal{} 1.04\\,mol \\cdot 71.8\\,kJ/mol \\equal{} 74.7\\,kJ$." } { "Tag": [ "geometry", "LaTeX", "area of a triangle", "geometry proposed" ], "Problem": "This problem was posted by Levi. Actually, the post contained flawed LaTeX code which forced the compiler into an endless loop, so I had to delete it. Here is the actual problem:\r\n\r\n[color=blue]Consider a line $AB$ and three collinear (in this order) points $X,Y,Z\\not\\in AB$. Then, prove that:\n\n- If the line AB does not intersect the segment XZ, then $XZ\\cdot (AYB)=XY\\cdot (ABZ)+YZ\\cdot (XAB)$.\n- If the line AB intersects the segment XY, then $XZ\\cdot (AYB)=XY\\cdot (ABZ)-YZ\\cdot (XAB)$.\n- If the line AB intersects the segment YZ, then $XZ\\cdot (AYB)=-XY\\cdot (ABZ)+YZ\\cdot (XAB)$.\n\n[b]Remark.[/b] Hereby, the notation (UVW) denotes the area of a triangle UVW.[/color]\r\n\r\n Darij", "Solution_1": "Maybe it is better to say:\r\n\r\n$\\triangle{AYB}=\\lambda \\triangle{AZB}+(1-\\lambda)\\triangle{AXB}$ where $\\lambda = \\frac{XY}{XZ}$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Is there exists a partition of $ Z_\\plus{}$ into three non-empty subsets $ A_1,A_2,A_3$ such that\r\nfor all $ x\\in A_i$ ,$ y\\in A_j$ then $ xy \\plus{} x\\plus{} y \\in A_k$ ( $ i,j,k$ are three pairwise numbers in the set {1,2,3} )", "Solution_1": "Suppose such a partition exists, and add $ 1$ to each of its elements. It is then a partition of $ \\mathbb{Z}_{>1}$ such that for all $ x \\in A_i, y \\in A_j$ we have $ xy \\in A_k$. Say that $ 2 \\in A_1$ without loss of generality. \r\n\r\nSuppose further that some other positive integer $ a \\in A_2$. Then $ 2a \\in A_3, 2a^2 \\in A_1, 4a \\in A_2$ and $ 8a, 2a^3, 4a^2 \\in A_3$. But then $ 4$ is not in $ A_1$ (or else $ 8a \\in A_2$, contradiction), $ 4$ is not in $ A_2$ (or else $ 8a \\in A_1$, contradiction), and $ 4$ is not in $ A_3$ (or else $ 4a \\in A_1$, contradiction). \r\n\r\nWe conclude that $ A_2$ is empty, also a contradiction. Hence no such partition exists." } { "Tag": [ "function", "geometry", "real analysis", "real analysis theorems" ], "Problem": "Prove that:\r\n\r\nA real-valued function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous almost everywhere.", "Solution_1": "continuous almost everywhere means what? continuous on a dense subset?", "Solution_2": "[quote=\"pleurestique\"]continuous almost everywhere means what? continuous on a dense subset?[/quote]\r\n\r\nobviously! jesus christ!", "Solution_3": "Almost everywhere means that the stated property does not hold for at most a set of measure zero. I'm not entirely sure if this is equivalent to what you said. (I'm a beginner in this area) But looking strictly at the definitions I would say no. Sorry if I said a dumb thing.", "Solution_4": "No, they are not equivalent; density is a topological notion of largeness whereas almost everywhere is a measure theoretic notion. For example, for every $t\\in I=[0,1]$, there exists a dense subset of $I$ having (Lebesgue) measure $t$. We can even find uncountable subsets of $[0,1]$ with Lebesgue measure zero.\r\n\r\nAs for the proof of the stated theorem, it can be found in almost any real analysis textbook. Probably on the internet, as well.", "Solution_5": "i've also heard people (professional mathematicians included) use \"almost everywhere\" to mean \"at all but a finite number of places\"... i didn't know the phrase actually had an official definition, heh... or does it?", "Solution_6": "It does have the official definition 'except on a set of measure zero'. The definition which only includes finitely many points (I admit, I've never heard this usage) would thus be subsumed in the measure-theoretic definition, since any set consisting of finitely many points has Lebesgue measure zero." } { "Tag": [ "function", "inequalities", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Here's (part of) a cool problem from Stein's book:\r\n\r\nA sequence $\\{f_{k}\\}$ of measurable functions on $\\mathbb{R}^{d}$ is Cauchy in measure if for every $\\epsilon > 0$,\r\n$m(\\{x : |f_{k}(x)-f_{l}(x)| > \\epsilon \\}) \\to 0$ as $k, l \\to \\infty$.\r\n\r\nWe say that $\\{f_{k}\\}$ converges in measure to a (measurable) function $f$ if for every $\\epsilon > 0$ \r\n$m(\\{x : |f_{k}(x)-f(x)| > \\epsilon \\}) \\to 0$ as $k \\to \\infty$. \r\n\r\nIf $\\{f_{k}\\}$ converges to $f$ in measure, does the sequence also converge to $f$ in $L^{1}$? \r\n\r\nMy additional question (maybe stupid):\r\n- What about $L^{2}$? Do either directions of \r\n\"$f_{k}\\to f$ in $L^{2}$ $\\Leftrightarrow$ $f_{k}\\to f$ in measure\"\r\nhold?", "Solution_1": "Convergence in measure is very weak; it does not even imply that the functions are integrable, let alone close in $L^{p}$ norm. For example, $f_{n}(x)=\\frac{1}{nx}$ converges to $0$ in measure on the interval $[0,1]$.\r\n\r\nConvergence in $L^{p}$ for any $p$ implies convergence in measure by something called Chebyshev's inequality, like \\[m(\\{x : |f_{k}(x)-f(x)| > \\epsilon \\}) \\le \\epsilon^{-1}\\int |f_{k}(x)-f(x)|\\,dm(x)\\]" } { "Tag": [ "vector", "analytic geometry", "algebra", "polynomial", "conics", "Galois Theory", "geometry solved" ], "Problem": "Let $\\triangle ABC$ find $M$ in plane which satisfy equation:\r\n$MA\\vec{MA}+MB\\vec{MB}+MC\\vec{MC}=\\vec{0}$\r\n\r\nIt is really hard with me!!!", "Solution_1": "So you are asking for the points M in the plane of triangle ABC such that the barycentric coordinates of M with respect to triangle ABC are ( MA : MB : MC ).\r\n\r\nI have finally done the computation. For a triangle ABC with a = 2, b = 2, c = 3, the required point M belongs to the families of points with barycentric coordinates\r\n\r\n$\\left(2: 2: 3u\\right)$, where u is a root of the polynomial $162x^4+243x^3+162x^2-56$;\r\n$\\left(325314v^3-83052v^2-143592v-15376: 2: 17880v\\right)$, where v is a root of the polynomial $1749x^4-108x^3-784x^2-168x-16$.\r\n\r\nBoth of the polynomials are irreducible and have the Galois group $S_4$, so their roots cannot be constructed with compass and ruler. Thus, the points M cannot be constructed with compass and ruler either.\r\n\r\nIt gets even worse if triangle ABC is non-isosceles; for a triangle ABC with a = 3, b = 4, c = 5, I obtain a polynomial of degree 10. Its Galois group is $S_{10}$, if I haven't done some stupid mistake, so the point shouldn't have a construction using conics either.\r\n\r\nBy the way, the Galois groups were computed using [url=http://www.math.tu-berlin.de/~kant/kash.html]KASH[/url], a nice program where you can just type\r\n[code]Galois (1337*x^3-23*x+1);[/code]\r\nand you get the Galois group of $1337x^3-23x+1$.\r\n\r\n Darij", "Solution_2": "Hi Darij, I understand your trying to solve it !! Through your word I can see that this problem is really hard. I think the contruction of M by ruler and compass is less important than show the barycentric of M, In Catersian coordinates, this work is equivalent with solving the system equation as you solved.\r\nWe knew the famous problem is Fermat problem, Fermat problem is similar this problem\r\nFirst we can see that $F$ is point that satifies equation \r\n$\\frac{\\vec{FA}}{FA}+\\frac{\\vec{FB}}{FB}+\\frac{\\vec{FC}}{FC}=\\vec{0} (*)$\r\nand then ETC we can see that F has barycentric:\r\n$(a\\sec(A-\\frac{\\pi}{6}),b\\sec(B-\\frac{\\pi}{6}),c\\sec(C-\\frac{\\pi}{6}))$\r\nthis much we can say that we can solved (*) and the root is this barycentric. I think that !!!" } { "Tag": [], "Problem": "In a game of tic-tac-toe, where X and O take turns, or moves, marking spaces in a 3x3 grid. The player who succeeds in placing three respective horizontal, vertical, or diagonal row wins the game. If X goes first, how many sequences of mobes are there that end with X winning in 5 total moves of both players?", "Solution_1": "There are eight total ways to place the X's. Each of these have 6 ways to order the X moves. And for each of these, there are 6 x 5 = 30 ways to move with O's. So there are 8 x 6 x 30 = 1440 total sequences of moves." } { "Tag": [], "Problem": "Use energy conservation arguments to verify that a cylindrical sound (where the wave fronts are cylindrical shells originating from a line source) wave must be written on the form.\r\n\r\n$ \\delta P(r,t)\\equal{}\\frac{\\delta P_0}{\\sqrt{r}} cos(kr\\minus{}wt)$", "Solution_1": "Solution:\r\n\r\n$ P\\equal{}I*2\\pi rl$\r\n\r\n$ I\\equal{}K*A^2$\r\n\r\n$ A\\equal{}\\sqrt{\\frac{I}{K}}\\equal{}\\frac{1}{\\sqrt{r}} \\sqrt{\\frac{P}{2\\pi lK}}\\equal{}\\frac{C}{\\sqrt{r}}$\r\n\r\n$ \\psi\\equal{}Acos(kr\\minus{}wt)\\equal{}\\frac{C}{\\sqrt{r}} cos(kr\\minus{}wt)$" } { "Tag": [ "puzzles" ], "Problem": "A surveyor goes to a house where a woman open the door (S: Surveyor, W: Woman)\r\n[list]S: How many children do you have?\nW: Three\nS: Age of each?\nW: The product is 36 and the sum is equal to the number of my neighbor's house[/list]\nThe surveyor goes to the check the number of the neighbor's house, and then comes back to the woman and says\n[list]S: There is no enough data.\nW: Ohh! Sorry, you are right, sorry. The older plays the piano.[/list]\r\nNow the surveyor has enough data to findout the age of the children. What is the age of each child?", "Solution_1": "Ah, yes. I have seen this problem in the [i]Art and Craft of Problem Solving[/i] and also [i]In Code[/i]. It's a fun problem...I wont give a solution so others who haven't already seen it can try it.", "Solution_2": "[hide]We know from the way the woman phrases her last comment that there is only one oldest child. So we're looking for three numbers which multiply to be 36, as well as sum to a sum that is shared by another triple which also multiplies to 36 but has more than one largest number. 9, 2, 2, satisfies this through 6, 6, 1 and since the question implies that the solution is unique, I won't bother making sure there's no other possibility. The children are 9 years, 2 years, and 2 years old.[/hide]", "Solution_3": "[quote=\"joml88\"]Ah, yes. I have seen this problem in the [i]Art and Craft of Problem Solving[/i] and also [i]In Code[/i]. It's a fun problem...I wont give a solution so others who haven't already seen it can try it.[/quote]\r\n\r\nHeh, I too have seen this problem in the [i]Art and Craft of Problem Solving![/i]", "Solution_4": "[quote=\"joml88\"]Ah, yes. I have seen this problem in the [i]Art and Craft of Problem Solving[/i] and also [i]In Code[/i]. It's a fun problem...I wont give a solution so others who haven't already seen it can try it.[/quote]joml, would you recomend [i]In Code[/i], I mean, I have it here at home, but I have been to lazy to start reading it, and now I have to give it back to the library, then, would you recomend me to check it out again?[quote=\"nr1337\"]Heh, I too have seen this problem in the [i]Art and Craft of Problem Solving![/i][/quote]OK, we all agree it is in the book [i]Art and Craft of Problem Solving[/i]!\r\n\r\nBy the way LynnelleYe, good solution ;)", "Solution_5": "I really enjoyed reading [i]In Code[/i] a lot. A lot of the book would probably be repetitive for you, however. She discusses puzzle, like the ones in this thread, that motivated her early in her life. You probably have seen them all before and could skip that part. She also explains a lot of the basic math and basic cryptology. When I read the book a lot of the math was new to me but if you've seen most of it before (as you probably have) then you can just skim through those parts. The rest of the book is very enjoyable also. I would recommend it as a read. I think that a lot of my joy from reading the book was the puzzles and basic math I hadn't seen before. So this book would probably interest the beginner rather than people with more experience though I still recommend it as a read to anyone.", "Solution_6": "can also be 6, 3, 2 u have 2 know what his neighbors number is", "Solution_7": "[quote=\"Ihatepie\"]can also be 6, 3, 2 u have 2 know what his neighbors number is[/quote]\r\n\r\nIf it's 6,3,2 the neighbor's number is 11 and 6,3,2 is the only solution if that's the case so the first part is enough information and it doesn't fit the puzzle.", "Solution_8": "As delivered here, \r\n\r\nthis is a flawed problem with the line\r\n\r\n\"The older plays the piano.\" It is awkward.\r\n\r\nThe word \"older\" necessarily means there is a choice between\r\n\r\ntwo quantities. Instead, the use of the word \"oldest\" would \r\n\r\nhave meant the choice to be among (potentially) three or \r\n\r\nmore. Even with twins who arrived later on in the family, the \r\n\r\noldest would occupy #1 spot as oldest, and the younger twins \r\n\r\nwould tie at the #3 spot in the age ranking.\r\n\r\nA parent with younger twins and a third child older \r\n\r\nthan the twins would instead say, \"My oldest child plays the \r\n\r\npiano.\"\r\n\r\nAnd if there's not enough logical information to go on from \r\n\r\nthere for this puzzle, then one can redo the puzzle's set-up.", "Solution_9": "box of rocks,\r\n\r\nnot to be rude or anything, but shouldn't it be eldest, as opposed to oldest?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "find all positive integer $n$ satisfies $n$ divide $2^n+1$", "Solution_1": "do you know whether this problem has a solution or not? i mean, there are \"too many\" integers $n$ satisfying your request.\r\nif you solved IMO 2000/5 then you will know that there are infinitely many integers $n$.. for example take $n=3^m$ and then you have $3^{m+1}| 2^{3^m}+1$...\r\n or you can take with that $m$ an integer $t$ that is a product of primes that divide $2^{3^m}+1$, $t$ not divisible by 3, and you will have that $nt| 2^n+1$ and therefor $nt|2^{nt}+1$..", "Solution_2": "I remember this problem appeared on forum before.\r\nOne more example: take an arbitrary $x_0$ s.t. $x_0\\mid 2^{x_0}+1$ and define $x_{n+1}=2^{x_n}+1$." } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Find the condition of $ x$ that $ x^{x^{x^{x^{.^{.^{.}}}}}}$\r\na)Has no limit.\r\nb)Has only 1 limit.\r\nc)Has 2 limits.", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=77942\r\n\r\nPretty informative topic, and the answer to your question is given.\r\n\r\na)$ x > e^{\\frac {1}{e}}$\r\nb)$ x\\in\\left[e^{\\frac {1}{e}},\\frac {1}{e^e}\\right]$\r\nc)$ x < \\frac {1}{e^e}$", "Solution_2": "thanks a lot" } { "Tag": [], "Problem": "In the accompanying numeration system, B * C = B. What is the value of (B * B) * (C * C)?", "Solution_1": "Basically, you take the first term, which is B, and use that as your answer.\r\n\r\nOr:\r\n\r\n(B*B)*(C*C)=B*C=B", "Solution_2": "Hm?\r\nHow about inputting some numbers for B and C?\r\nB=2, and C=1. The expression gives 4 not 2. \r\n\r\n[hide]\n\\[ (B\\cdot B)(C\\cdot C)\\equal{}B^2\\cdot C^2\\equal{}(B\\cdot C)(B\\cdot C)\\equal{}B^2\n\\]\n[/hide]", "Solution_3": "the table was this:\r\n\r\n\r\n * A B C\r\n A B A C\r\n B A C B\r\n C B C A\r\n\r\nso it should have been B*B=C and C*C=A so C*A should have been C..... but the answer says B.", "Solution_4": "[quote=\"GameBot\"]In the accompanying numeration system, B * C = B. What is the value of (B * B) * (C * C)?[/quote]\r\nC*A is B, according to your chart" } { "Tag": [ "AMC 12 A", "AMC" ], "Problem": "Hello:\r\n\r\nI have been using the excellent contest resource that AOPS hosts. ( http://www.artofproblemsolving.com/Forum/resources.php ) The variety of problems present in there is truly amazing.\r\n\r\nAs I was going through it, however, I noticed that some of the problems were missing. For example, none of problems from the 2002 AMC 12 A and B are present. I have found them online, and I was wondering if there was anyway I could upload them onto the resource. I have some free time to do all the uploading, I think :) \r\n\r\nAlso, when I browse through the problems, I have difficulty viewing images. For example, #s 12 and 13 on AMC 12 2006 A have images along with them, but I simply see a \"External_Image\". Is that only a problem with me or are other users also affected by it?\r\n\r\nThanks,\r\n\r\nGautam", "Solution_1": "whoops. I browsed around and I did find something that answered my first question on the AOPS wiki.\r\n\r\nbut what about that second question", "Solution_2": "No, the images don't work for anyone because they were uploaded when AoPS was using a different server. When they switched to a new server the images weren't transferred over. It seems that at some point someone's going to have to upload new images into the galleries. Either that or get the images from the old server back working. However that would be a job for the admins, they're busy, and it would take at least few hours/days to do that." } { "Tag": [ "linear algebra", "matrix", "search", "vector", "algebra", "polynomial", "induction" ], "Problem": "What conditions are necessary or sufficient for a real(or any field that is not closed) matrix to be triangularizable? \r\nAnyone knows or has any problems that have a solution involving triangularizable matrices?\r\nI found these here(loup blanc's posts), but only vaguely understand them:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=144008\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=633174368&t=184775\r\nI would also kindly ask not to use any concepts more advanced than fields (vector spaces and God knows what).\r\nThanks.", "Solution_1": "triangularizable := similar to a triangular matrix = the characteristic polynomials splits completely = the minimal polynomial splits completely", "Solution_2": "The necessary and sufficient condition for a matrix to be triangularizable is simply that all of its eigenvalues belong the field in question. We can also write that as the fact that the characteristic polynomial can be completely factored into linear factors over the field in question. (The characteristic polynomial \"splits\".)\r\n\r\nThe proof isn't that hard. If the matrix is triagularizable, we can read the eigenvalues off of the main diagonal of the triangular form. For the converse, the proof proceeds by induction on the size of the matrix. Try it yourself. Suppose you have one eigenvector, $ Av\\equal{}\\lambda v.$ Extend $ v$ to a basis, which is to say, create an invertible matrix $ P$ with $ v$ as its first column. Now compute $ P^{\\minus{}1}AP.$ What does it look like?", "Solution_3": "Okay.\r\nI have a question:\r\n1.loup blanc wrote in his solution from my first link:\r\n[quote]Proof: $ rank(AB \\minus{} BA)\\leq1$ then A and B are simultaneously triangularizable over $ \\mathbb{C}$.[/quote] I do not understand why this condition is necessary. Is there a difference between \"A is triangularizable\" and \"A and B are simultaneously triangularizable\"?\r\n \r\n\r\nAnd some conclusions. Please correct me if I am wrong.\r\n1.So basically any matrix whose elements belong to a closed field (like $ \\mathbb {C}$, for example) is triangularizable .\r\n2.If its eigenvalues belong to the field it's triangularizable;furthermore if the eigenvalues are distinct then it's diagonalizable.", "Solution_4": "The term \"simultaneously triangularizable\" means that there is some $ M$ such that $ M^{\\minus{}1}AM$ and $ M^{\\minus{}1}BM$ are [b]both[/b] upper triangular. It's a much stronger condition.\r\n\r\nYour conclusions are correct. For the second, I like this statement: A matrix is diagonalizable if and only if its minimal polynomial splits as a product of distinct linear factors. Since the minimal polynomial is a factor of the characteristic polynomial, this happens whenever the characteristic polynomial splits this way." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Find the value of $a$ for which $\\int_{0}^{\\frac{\\pi}{2}}|a\\sin x-\\cos x|\\ dx\\ (a>0)$ is minimized.", "Solution_1": "When $a\\sin(x)-\\cos(x) < 0$, the sign should be reversed. So, letting $c=\\tan^{-1}(\\frac{1}{a})$, (which is the point of sign change of integrand)\r\n\r\n$F(a) = \\int_{0}^{\\pi/2}|a\\sin(x)-\\cos(x)|dx$ \r\n$= \\int_{0}^{c}(a\\sin(x)-\\cos(x))dx+\\int_{c}^{\\pi/2}-(a\\sin(x)-\\cos(x))dx$\r\nCalculating the integral, we find\r\n$F(a) =-2(a\\sin(c)-\\cos(c))+a+1$\r\n\r\nTo find the lowest rung, we solve $\\frac{dF}{da}= 0$ for $a$. Calculating the derivate, we find\r\n$\\frac{1}{2}+\\cos(c) = \\frac{a\\sin(c)-\\cos(c)}{1+a^{2}}$\r\nPlugging $a = \\frac{1}{tan(c)}$, we get\r\n$\\frac{1}{2}+\\cos(c) = 0$\r\nSolving this for $a$ (which should be positive), minima occurs at $\\frac{1}{\\sqrt 3}$.", "Solution_2": "[color=darkblue]Define $x\\ .s.s.\\ y\\Longleftrightarrow xy>0\\ \\vee \\ x=y=0\\ .$([u]same sign[/u]). If $a\\le 0$, then $F(a)=\\int_{0}^{\\pi}{2}(\\cos x-a\\sin x)\\ dx=1-a$ and $\\boxed{\\ \\min_{a\\le 0}F(a)=F(0)=1\\ }\\ .$\nIf $a>0$, then $a\\sin x-\\cos x\\ .s.s.\\ \\tan x-\\frac{1}{a}=\\tan x-\\tan\\left(\\arctan\\frac{1}{a}\\right)\\ .s.s.\\ x-\\arctan\\frac{1}{a}$ and in this case \n$F(a)=\\int_{0}^\\frac{\\pi}{2}\\left|a\\sin x-\\cos x\\right|\\ dx=$ $\\int_{0}^{\\arctan\\frac{1}{a}}(\\cos x-a\\sin x)\\ dx+\\int_{\\arctan\\frac{1}{a}}^\\frac{\\pi}{2}(a\\sin x-\\cos x)\\ dx=$ $-1-a+2\\sqrt{a^{2}+1}\\ .$\nBut $F'(a)=-1+\\frac{2a}{\\sqrt{a^{2}+1}}\\ .s.s.\\ 2a-\\sqrt{a^{2}+1}\\ .s.s.\\ 4a^{2}-(a^{2}+1)=$ $3a^{2}-1\\ .s.s.\\ a-\\frac{\\sqrt 3}{3}$, i.e. $a_{\\mathrm{min}}=\\frac{\\sqrt3}{3}\\ .$\nTherefore, $\\boxed{\\ \\min_{a>0}F(a)=F\\left(\\frac{\\sqrt 3}{3}\\right)=\\sqrt 3-1\\ }\\ .$ In conclusion, $\\min_{a\\in R}F(a)=F\\left(\\frac{\\sqrt 3}{3}\\right)=\\sqrt 3-1\\ .$[/color]", "Solution_3": "Your answers are correct. :) \r\nHere is my solutuon.\r\n\r\nFor $t$ such that $a\\sin t-\\cos t=0\\ \\left(01 $ a polynomial $ P(x) $ commutes with a polynomial\r\n$ Q(x) = x^{n} $, then it is of a form $ P(x) = x^{m} $ for some natural $ m $.", "Solution_1": "I hope with commute you mean $P(Q(x))=Q(P(x))$, because otherwise my solution is wrong...\r\n\r\nProof:\r\nLet be $P(x)=x^m+a_{m-1} x^{m-1} + ... + a_1 x + a_0$\r\nNow assume that there is an $a_i \\neq 0, i deg(P) $, then we will have that $ P(x) $ has at least $ N $ different\r\nroots, so $ P(x) \\equiv 0 $.\r\nTherefore if $ P(x) $ is nonzero polynomial then it has the only root zero,\r\nand because it is monic, then $ P(x) = x^{m} $ for some $ m $.", "Solution_4": "[quote=\"Leva1980\"]We consider only monic polynomials.\n\n Prove that if for $ n>1 $ a polynomial $ P(x) $ commutes with a polynomial\n$ Q(x) = x^{n} $, then it is of a form $ P(x) = x^{m} $ for some natural $ m $.[/quote]\r\nI think you forgot the trivial solution P(x)=0 ;)", "Solution_5": "But I consider only monic polynomials :D !" } { "Tag": [ "factorial", "calculus", "integration" ], "Problem": "Sorry for the lameness, but can anyone show me why 0! = 1?? My teacher refused to explain she said just take it as a given.", "Solution_1": "Here is one way of looking at it.\r\n\r\nn!=n(n-1)(n-2)...(3)(2)(1)\r\nn!/n=(n-1)!\r\n3!/3=2!; 3*2/3=2\r\n2!/2=1!; 2/2=1\r\n1!=1\r\n1!/1=1=0!", "Solution_2": "Given:$1!=1$\r\n$1\\cdot0!=1$, because $n!=n\\cdot(n-1)!$. On top of that, that equation is the identity property of multiplication, which means that $0!=1$ as well.", "Solution_3": "hmm...curse your awesome clarity. :ninja:", "Solution_4": "How many ways can you choose zero things? Exactly one way, right? Just \r\ntake none of them - anything else you do is wrong. :)", "Solution_5": "I understand n! = n (n-1)!. \r\nbut in that case,\r\n0! = 0*(0-1)! and i thought you couldnt do negative factorials. \r\n[quote]On top of that, that equation is the identity property of multiplication, which means that 0!=1 as well.[/quote]\r\nis it just a given?", "Solution_6": "0!=1\r\nIt's just a definition, no proof involved. It's defined that way, so that the combinations and permutations work out and fit all the rules, like nCr=nC(n-r). If 0! were defined otherwise, then we would always have to state something like \"excluding the 0 case\", and we would have to have separate definitions for the 0 case.", "Solution_7": "It simplifies a lot of formulae. For example, we need $0!=1$ to say that $\\binom{n}{n}=\\frac {n!}{n!\\cdot 0!}=1$", "Solution_8": "The simplest explinations are:\r\n\r\n1. 1! = 1*0! = 0! = 1.\r\n2. There is only 1 way to choose 0 objects out of n: do nothing!\r\n\r\nThere's also the calculus way... :D\r\n\r\n\\begin{eqnarray*}\r\n0! &=& \\Gamma (1) \\\\\r\n&=& \\int_0^\\infty t^{1-1} e^{-t}\\, dt\\\\\r\n&=& \\int_0^\\infty e^{-t} \\, dt\\\\\r\n&=& 1\r\n\\end{eqnarray*}" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "let $ f_n$ be the n-th Fibonacci number.\r\n\r\nfor $ q > 1$ let $ L_q(z)$ be the Lambert q-series:\r\n\\[ L_q(z) \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {z^n}{q^n \\minus{} 1} \\equal{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {z}{q^n \\minus{} z}\r\n\\]\r\nin the disk $ |z| < q$.\r\n\r\nprove that there exists 3 numbers $ A,B,q$ of the form $ q_1 \\plus{} q_2\\sqrt {5}$, $ q_1,q_2\\in Q$, such that\r\n\\[ \\sum\\frac {1}{f_n} \\equal{} AL_q(B)\r\n\\]\r\n(probably $ A \\equal{} \\minus{} \\frac {5 \\plus{} \\sqrt {5}}{2}$ , $ B \\equal{} \\frac {1 \\plus{} \\sqrt {5}}{2}$ , $ q \\equal{} \\minus{} \\frac {3 \\plus{} \\sqrt {5}}{2}$ would work fine)", "Solution_1": "no idea? :blush:" } { "Tag": [], "Problem": "does anyone have shortlist 2007 ?please post iff available :D \r\n :lol:", "Solution_1": "[quote=\"gauravpatil\"]does anyone have shortlist 2007 ?please post iff available :D \n :lol:[/quote]\r\nwill not be [b]officially[/b] available till July 2008 :P" } { "Tag": [ "function", "LaTeX", "algebra proposed", "algebra" ], "Problem": "Let $ f$: $ \\mathbb{N^*} \\to \\mathbb{N^*}$ such that:\r\n$ \\forall n \\in \\mathbb{N^*}$: $ f(n\\plus{}f(n))\\equal{}f(n)$\r\n\r\n1)- Prove that if the set $ f(\\mathbb{N^*})$ is finite then f is periodic\r\n2)- Give an example of a such function but not periodic\r\n\r\n\r\nSome one knows where was it proposed?", "Solution_1": "$ 1)$\r\nif we take $ (a_i)$ such that $ a_{i \\plus{} 1} \\equal{} f(a_i) \\plus{} a_i$ and $ a_0 \\equal{} n$\r\nthen $ f(a_{i \\plus{} 1}) \\equal{} f(a_i)$ then $ \\forall i\\in\\mathbb{N}: \\ f(n) \\equal{} f(a_i)$\r\nand $ a_{i \\plus{} 1} \\equal{} f(a_i) \\plus{} a_i \\equal{} f(n) \\plus{} a_i$ gives $ a_i \\equal{} if(n) \\plus{} n$\r\nthen $ \\forall i\\in\\mathbb{N}: \\ f(n) \\equal{} f(n \\plus{} if(n))$\r\nwe suppose that $ f(\\mathbb{N}^*) \\equal{} \\{b_1,b_2,...,b_h\\}$ finite and we take $ T \\equal{} ppcm(b_1,b_2,...,b_h)$\r\nso $ f(n \\plus{} T) \\equal{} f(n)$ because $ \\exists m:$ such that $ T \\equal{} mf(a)$\r\n\r\n$ 2)$\r\nexample:\r\nwe take $ f$ such that $ f(n) \\equal{} 2^{a \\plus{} 1}\\iff 2^a|n\\ and\\ 2^{a \\plus{} 1}\\not|n$\r\nor $ \\forall n,k\\in\\mathbb{N}: \\ f(2^m(2k \\plus{} 1)) \\equal{} 2^{m \\plus{} 1}$\r\n\r\nso if $ n \\equal{} 2^m(2k \\plus{} 1)$\r\n$ f(n \\plus{} f(n)) \\equal{} f(2^m(2k \\plus{} 1) \\plus{} 2^{m \\plus{} 1}) \\equal{} f(2^{m}(2k \\plus{} 3)) \\equal{} 2^{m\\plus{}1} \\equal{} f(n)$", "Solution_2": "Sorry for not typing LATEX, as i'm a new member and i don't have a math career.\r\n\r\nThe first part is quite easy, for the second part, i got the same sample as aviateurpilot did. I just show here the road to this sample:\r\n\r\nIn fact, what we need is an unbounded sequence a1 $ \\frac {AC}{CE'} \\equal{} \\frac {BC}{BE'}$ (3). From (1),(2) and (3) => BF=CF.[/hide]", "Solution_2": "let AB intersect CD in J \r\nAD intersect BC in I \r\nDE intersect BC in N \r\nEC intersect AI in F \r\n\r\nby theorem of Menelai for triangle BCJ and points N, D, E \r\n\r\nBN/NC.CD/DJ.JE/EB=1, but BN=NC, then JD/DC=JE/EB; \r\n\r\nby theorem of Menelai for triangle IDC and points A, J, B \r\n\r\nIA/AD.DJ/JC.CB/BI=1, but IA=IB, then JD/DC=JC/CB; \r\n\r\nJC/CB=JD/DC=JE/EB, then JC/CB=JE/EB then CE is the bisector of angle DCI \r\nLet angle DCI=2a; angle AIB=2b, then angle ABI=90-b, then angle EBC=90+b, angle BCE=angle FCI=a, then angle BEC=90-a-b \r\nangle CAD= angle DCA=angle CDI/2=(180-2b-2a)/2=90-a-b=angle AEC, then \r\n\r\nangle CAD= angle DCA=angle AEC, then (AEC) touthes AI and DC at points A and C.", "Solution_3": "[b]To limes123 [/b](to shorten little bit the proof):\r\n\r\nBy the isosceles triangles ABG and ACD, from triangle ACG we easily get Number of Distinct Prime factors of n.\r\nbig-omega(n) --> Number of Prime Factors (Counting Multiplicity.)\r\n\r\nSo for example ,\r\n\r\n omega(18) = 2.\r\nbig-omega(18) = 3.\r\n\r\nClearly, omega(n) <= big-omega(n).\r\n\r\nNow we have been given the two series that is we have been given omega(n) and also big-omega(n), and we need to calculate\r\nthe series given here -- http://www.research.att.com/~njas/sequences/A143519. \r\n\r\nI am listing out some terms here --\r\n\r\n n omega big-omega Required term.\r\n \r\n 2 1 1 1\r\n 3 1 1 1 \r\n 4 1 2 -1\r\n 5 1 1 1\r\n 6 2 2 -2\r\n 8 1 3 0\r\n 10 2 2 -2 \r\n\r\n\r\nClearly there exists a relation between the required term and the two series omega and bigomega but I am not able to figure that out.\r\n\r\nPlease suggest what the relation can be ?\r\n\r\n\r\nThanks.", "Solution_1": "What do you mean by a \"relation\"? Neither value determines the other for a particular choice of $ n$. Do you mean an asymptotic relation?" } { "Tag": [], "Problem": "should a chromate ion be written as $ \\ce{CrO4^{2\\minus{}}}$ or $ \\ce{CrO4^{\\minus{}2}}$", "Solution_1": "Doesn't really matter. Not sure what the official IUPAC rules are on this, but either one should be accepted. I usually write \\ce{CrO4^{2-}} because the numbers line up nicely.", "Solution_2": "actually what i meant was the former case normally means the Oxidation number of the central atom which here isn't obviuosly $ 2\\minus{}$ .\r\nthat is $ 2\\minus{}$ represents oxidation state and $ \\minus{}2$ net charge", "Solution_3": "That number represents the charge of the ion and not the oxidation state of the metal. To give information about the oxidation state of a (transition) metal we use the Shock number: the oxidation state of the metal written in Roman and between brackets (). Also, it is common practice to write charges as \"2-\" and oxidation numbers as \"-2\".", "Solution_4": "actually we were taught what i wrote" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Can someone help me explain why the only simple, cyclic groups are those of prime order?", "Solution_1": "Cyclic groups are abelian, so a cyclic group of order $ pq$ has normal subgroups of orders $ p, q$.", "Solution_2": "subgroups of a cyclic group of order n correspond to the divisors of n." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Is $ \\pi$ transcendental over $ \\bar{\\mathbb{Q}}$?", "Solution_1": "Yes. If it is then it is algebraic over $ \\mathbb Q$.", "Solution_2": "N.T.TUAN means the right thing but his second sentence can be misunderstood.\r\nAlgebraicness is a tranistive property, thus if $ C|B|A$ are field extensions, then $ C|A$ is algebraic if and only if $ C|B$ and $ B|A$ are algebraic. Especially, being algebraic over $ \\bar{\\mathbb Q}$ is the same as being algebraic over $ \\mathbb Q$ (so iff it is already in $ \\bar{\\mathbb Q}$)." } { "Tag": [ "AMC" ], "Problem": "Call a positive real number [i]special[/i] if it has a decimal representation that consists entirely of digits 0 and 7. For example, $ \\frac{700}{99}\\equal{}7.0707070707...$ and 77.007 are special numbers. What is the smallest [i]n[/i] such that 1 can be written as a sum of [i]n[/i] special numbers?\r\n\r\n(A)7 (B)8 (C)9 (D)10 (E)1 cannot be represented as a sum of finitely many special numbers", "Solution_1": "[hide=\"Hint\"] Phrase it this way: write $ \\frac{1}{7}$ as a sum of numbers that consist only of the digits $ 0$ and $ 1$. What is the decimal expansion of $ \\frac{1}{7}$? [/hide]" } { "Tag": [ "probability", "revivedA9YearOldThread" ], "Problem": "In a class of 28 students, the teacher selects four people at random to participate in a geography contest. What is the probability that this group of four students includes at least two of the top three geography students in the class? Express your answer as a common fraction.", "Solution_1": "[hide=\"Solution\"] $\\frac{_{25}C_{1}\\times_{3}C_{3}+_{25}C_{2}\\times_{3}C_{2}}{_{28}C_{4}}=\\frac{37}{819}$. :? I remember this question. I don't know if this is right, though. [/hide]", "Solution_2": "it's wrong", "Solution_3": "It's right, my answer key of truth says so.\r\n\r\nI missed the problem because I made the mistake of going\r\n\r\n3(ways to select at least 2)*26*25/2=975\r\n\r\nbut that double counts 50 of them as for example selecting 1,2 and then 3 is the same as selecting 2,3 and then 1", "Solution_4": "[quote=\"pgpatel\"]it's wrong[/quote]\r\nI looked, and answer key says it's right.", "Solution_5": "Never mind :D. I for some reason recall the answer being distinctly higher.", "Solution_6": "I think it should be :\r\n\r\n$1 - \\frac{3*C_{4,26}}{C_{4,28}}$", "Solution_7": "what exactly was used to solve the problem (the formula above)?", "Solution_8": "[hide]\nUmm...\nI think that the easiest way to do it is \n$1-\\frac{3\\cdot C_{26,4}}{C_{28.4}}$. I don't know how to calculate that, though.\nThe fraction is the probability that there won't be two top students.[/hide]", "Solution_9": "[quote=\"ajai\"]what exactly was used to solve the problem (the formula above)?[/quote]\r\n\r\nThe reasoning is as follows:\r\n\r\nThere are 3C2 ways to have exactly 2 of the top 3. Then we have to choose another 2 from the 25 people that are not in the top 3, so 25C2 * 3C2. \r\n\r\nLikewise, there are 3C3 ways to have all 3 of the top 3, leaving room for us to choose 1 from the other 25, so 25 C 1. \r\n\r\nThere are 28C4 ways to pick 4 students, so that is the denominator.", "Solution_10": "We count the number of ways to choose a group of four students including at least two of the top three geography students. This is just $\\binom{3}{2}\\cdot \\binom{25}{2} + \\binom{3}{3}\\cdot\\binom{25}{1} = 925$, since we can choose either 2 or 3 of the top students to be in this group. In all, there are $\\binom{28}{4} = 20475$ groups of four students. Thus our desired probability is $\\frac{925}{20475} = \\boxed{\\frac{37}{819}}$.", "Solution_11": "AOPS Wrote this, BTW. I take no credit.", "Solution_12": "We count the number of ways to choose a group of four students including at least two of the top three geography students. This is just $\\binom{3}{2}\\cdot \\binom{25}{2} + \\binom{3}{3}\\cdot\\binom{25}{1} = 925$, since we can choose either 2 or 3 of the top students to be in this group. In all, there are $\\binom{28}{4} = 20475$ groups of four students. Thus our desired probability is $\\frac{925}{20475} = \\boxed{\\frac{37}{819}}$." } { "Tag": [ "trigonometry", "geometry" ], "Problem": "Find the angles of a triangle with vertices at (0,0), (17,6), and (-12,3)\r\n\r\n\r\nthanks,\r\npierre", "Solution_1": "Using the distance formula, we get:\r\n$\\cot{\\frac{\\sqrt{325}}{\\sqrt{850}}}$\r\n$\\cot{\\frac{\\sqrt{850}}{\\sqrt{153}}}$\r\n$\\cot{\\frac{\\sqrt{153}}{\\sqrt{325}}}$", "Solution_2": "[quote=\"croatian_dan\"]Find the angles of a triangle with vertices at (0,0), (17,6), and (-12,3)\n\n\nthanks,\npierre[/quote]\r\nThis should go in HSb not classroom math.", "Solution_3": "Whoops, I'm wrong. It's not a right triangle, never mind. :blush:", "Solution_4": "Try the shoelace formula...", "Solution_5": "[quote=\"philB\"]Try the shoelace formula...[/quote]\r\n\r\nShoelace is for side measurements...not angles...but u can use the Law of Sines/Cosines?", "Solution_6": "The Law of Sines should work...", "Solution_7": "[quote=\"xpmath\"]The Law of Sines should work...[/quote]\r\n\r\n :D I keep forgetting which ones the laws work for :)", "Solution_8": "Rule of Cosines would work too, but I'm too lazy to do it.", "Solution_9": "or one could just do the following, which should go in high school basics.\r\n\r\n[hide]using determinants, one finds that the area is 123/4. but we have that $[ABC] = \\frac{ab\\sin{\\angle C}}{2}$, hence $\\sin{\\angle C}=$ whatever.i'm lazy. you do the calculations.[/hide]", "Solution_10": "[quote=\"Teki-Teki\"]...whatever.i'm lazy. you do the calculations.[/hide][/quote]\r\nExactly. :lol:" } { "Tag": [ "Asymptote", "\\/closed" ], "Problem": "Is there a way to get some of the features (like centroid, orthocenter...ect) that is available on Geogebra on AoPS to your personal Geogebra...", "Solution_1": "I'm not a user of GeoGebra but that of Asymptote. But since they're similar, try looking for packages?", "Solution_2": "GeoGebra is not similar to Asymptote. GeoGebra has a more user-friendly interface, meaning you can just plot points, with your mouse, it's much more visual.\r\n\r\nWith Asymptote, you type in code.\r\n\r\nTo Nandubetax: Can't you just construct them?" } { "Tag": [ "counting", "distinguishability", "combinatorics unsolved", "combinatorics" ], "Problem": "1) How many ways can I make a collection of 20 bottles from 8 different types?\r\n\r\n2) Repeat the same problem with the condition: there must be between 4 and 7 bottles of EACH type in the collection.", "Solution_1": "1) the answer is the number of solutions in non-negative integers to $ x_1\\plus{}x_2\\plus{}\\cdots\\plus{}x_8\\equal{}20$, which is $ \\binom{20\\plus{}8\\minus{}1}{8\\minus{}1}\\equal{}\\binom{27}{7}$ (the number of ways to arrange 20 indistinguishable $ 1$'s and 8-1 indistinguishable bars in a row, the bars being there to divide the 1's into 8 parts).\r\n\r\n2) this is impossible, as there would have to be at least $ 8\\cdot 4 \\equal{} 32$ bottles. maybe the question was something else...", "Solution_2": "Thank you. I should have typed 40 bottles instead. I have reposted the problem." } { "Tag": [], "Problem": "Find the number of $ (a,b,c) \\in N$ such that\r\n$ abc|(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)$", "Solution_1": "(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1\r\nFrom the equation above, we deduce:\r\nab+bc+ca+a+b+c+1=kabc with k is a natural number\r\n=> 1/a+1/b+1/c+1/ab+1/bc+1/ca +1/abc=k\r\nBy assumming that a>=b>=c, we have k is less than or equal to 7/c\r\n :blush: I can't think further at the moment, anybody helps ? :blush:", "Solution_2": "[hide]\n$ k \\le 7/c \\Rightarrow k \\le 7$...i think you could just bash from there\n[/hide]", "Solution_3": "Do not double post:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250991[/url]" } { "Tag": [ "algebra open", "algebra" ], "Problem": "$ \\sum_{r\\equal{}0}^{q}{(^qC_r)(^NC_{n\\minus{}r})(a^{N\\minus{}n\\plus{}r})}$\r\nguys please tell me how to split the T_r to get a telescoping sum \r\nother methods are also invited \r\n :)", "Solution_1": "replies are invited", "Solution_2": "a prompt reply will be appreciated" } { "Tag": [ "vector", "geometry", "3D geometry", "sphere", "geometry unsolved" ], "Problem": "If $P_{i}, (i=1,...,n)$ are points on a unit sphere, then $\\sum_{i\\leq j}|P_{i}P_{j}|^{2}\\leq n^{2}.$", "Solution_1": "anyone? :)", "Solution_2": "If $O$ is center of that sphere, then your ineq is $(OP_{1}+OP_{2}+...+OP_{n})^{2}\\geq 0$. Done! :D", "Solution_3": "Thanks,N.T.TUAN" } { "Tag": [ "MATHCOUNTS", "geometry", "3D geometry", "algebra", "polynomial" ], "Problem": "Well, I wasn't quite sure if this problem was MATHCOUNTS level, but it's from a \"sprint round\" from a local competition. So, it should be. I doubt it's too easy--and I also doubt it's too hard. Here goes:\r\n\r\nGiven these two equations:\r\n$ a \\plus{} b \\equal{} 7$\r\n$ a^3 \\plus{} b^3 \\equal{} 42$\r\n\r\nSolve for \r\n$ \\frac{1}{a} \\plus{} \\frac {1}{b}$\r\n\r\n[hide=\"Answer\"]$ \\frac{21}{43}$[/hide]\r\n\r\nWe tried substituting in $ 7 \\minus{} b$ in for $ a$, but that didn't work. \r\n...I think you factor this somehow.\r\nOur coach says that \"Since it's a sprint round, it wouldn't be this messy. There must be some sort of trick or something.\" \r\n>.< Any ideas?", "Solution_1": "$ \\frac{1}{a}\\plus{}\\frac{1}{b}\\equal{}\\frac{a\\plus{}b}{ab}$\r\nUse that. It's helpful.", "Solution_2": "cube the first equation $ a^3\\plus{}3a^2b\\plus{}3ab^2\\plus{}b^3\\equal{}343$ factor the middle terms and substitute the second in. $ 42\\plus{}3ab(a\\plus{}b)\\equal{}343$ solve for ab. $ ab\\equal{}\\frac{43} {3}$. divide the first equation by ab to get $ \\frac{1} {a} \\plus{} \\frac{1} {b}\\equal{}\\frac{7} {ab}\\equal{}\\frac{7} {\\frac{43} {3}}\\equal{}\\frac{21} {43}$", "Solution_3": "Hmm...cool. :D Thanks.\r\n\r\nIs there a certain way to know when to factor or cube, or do you just mess around with the problem, or do you just know by experience? Just curious. :D", "Solution_4": "This is a type of problem you learn to recognize by experience. Here's the key: [b]you don't have to find $ a$ and $ b$ to find $ \\frac {1}{a} \\plus{} \\frac {1}{b}$.[/b]\r\n\r\nA specific piece of technique: any symmetric polynomial $ f(a, b)$ in two variables (a polynomial such that $ f(a, b) \\equal{} f(b, a)$) can be written as a polynomial in the variables $ u \\equal{} a \\plus{} b, v \\equal{} ab$. These are known as the [i]elementary symmetric polynomials in two variables[/i] (and the generalization to three variables is given by $ a \\plus{} b \\plus{} c, ab \\plus{} bc \\plus{} ca, abc$). Here we have\r\n\r\n$ u \\equal{} 7$\r\n$ u(u^2 \\minus{} 3v) \\equal{} 42$\r\n\r\nand $ \\frac {1}{a} \\plus{} \\frac {1}{b} \\equal{} \\frac {u}{v}$.\r\n\r\n[b]Exercise:[/b] Given that\r\n\r\n$ x \\plus{} y \\equal{} 5$\r\n$ x^2 \\plus{} y^2 \\equal{} 13$\r\n\r\nfind $ x^3 \\plus{} y^3$.\r\n\r\n[hide=\"Hint\"] $ x^2 \\plus{} y^2 \\equal{} u^2 \\minus{} 2v$. [/hide]", "Solution_5": "[hide]\n\\begin{align*} (x + y)^2 & = 25 \\\\\nx^2 + 2xy + y^2 & = 25 \\\\\n2xy & = 12 \\\\\nxy & = 6 \\\\\n(x + y)(x^2 + y^2) & = x^3 + y^3 + xy(x + y) \\\\\nx^3 + y^3 & = (5)(13) - 6(5) = \\boxed{35} \\end{align*}\n\nYay!\n\nOh, and by the way, $ x = 2\\ \\text{and}\\ y = 3$ or $ x = 3\\ \\text{and}\\ y = 2$[/hide]" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "let f an automorphism of C from C \r\n\r\nDo we have for any z in C , $f(e^{z})=e^{f(z)}$ ?", "Solution_1": "If $f$ is continuous, we do; the only continuous automorphisms are the identity and conjugation.\r\n\r\nIf $f$ isn't continuous, it doesn't work. We don't have to send $e$ to a real number, after all.", "Solution_2": "What you mean by automorphism?\r\nDo you mean a comform automorphism, is it $\\frac{\\partial f}{\\partial \\overline{z}}=0$ and bijective?" } { "Tag": [ "function", "quadratics", "analytic geometry" ], "Problem": "I'm confused about Hamiltonians. What is an example of a system where the Hamiltonian is NOT equal to the energy of the system?", "Solution_1": "Possibly I'm misunderstanding your question, but the Hamiltonian is always an energy function for a system: $ H\\equal{}T\\plus{}U$, where $ T$ is kinetic energy and $ U$ the potential energy. \r\n\r\nI think you really want to know is when the Hamiltonian equates total energy $ E$ of the system. A necessary condition for that is that the equations of motion must be independent of the time (not in a strictly sense) but time must not explicity appear in $ H$, i.e.: $ \\frac{\\partial H}{\\partial t}\\equal{}0$.\r\nThis is equivalent to the kinetic energy is an homogeneous quadratic function of the generalized coordinates $ \\dot{q}_{j}$. In this case $ T\\equal{}\\sum_{j,k}a_{jk}\\dot{q}_{j}\\dot{q}_{k}$\r\n\r\nFor an example when $ H$ is not equal to $ E$, consider ths system generated by a pendulum of lenght $ l$ and mass $ m$, where the point supporting the pendulum is moves on a circle of radius $ a$ with constant angular velocity $ \\omega$: $ x\\equal{}acos(\\omega t)\\plus{}bsin(\\phi)$ and $ y\\equal{}asin(\\omega t)\\minus{}bcos(\\phi)$." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a, b, c$ three non-negative real numbers. Prove that the following inequality holds:\r\n\r\n \\[ \\frac{a^{2}+2bc}{b+c}+\\frac{b^{2}+2ca}{c+a}+\\frac{c^{2}+2ab}{a+b}\\geq\\frac{3}{2}(a+b+c). \\]", "Solution_1": "[quote=\"Cezar Lupu\"]Let $a, b, c$ three non-negative real numbers. Prove that the following inequality holds:\n\n \\[ \\frac{a^{2}+2bc}{b+c}+\\frac{b^{2}+2ca}{c+a}+\\frac{c^{2}+2ab}{a+b}\\geq\\frac{3}{2}(a+b+c). \\][/quote]\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=b%2Bc&t=85938\r\n :)", "Solution_2": "[quote=\"Cezar Lupu\"]Let $a, b, c$ three non-negative real numbers. Prove that the following inequality holds:\n\n \\[ \\frac{a^{2}+2bc}{b+c}+\\frac{b^{2}+2ca}{c+a}+\\frac{c^{2}+2ab}{a+b}\\geq\\frac{3}{2}(a+b+c). \\][/quote]\r\n\r\n My proof is here. I don't know if it is as good as SOS of Nameless or not. But anyway, I found it:\r\nAfter expanding, it is equivalent to:\r\n$2\\sum a^2(a-b)(a-c)+ \\sum ac (a-c)^2 \\geq 0$\r\n :D" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $(a_{n})_{n\\geq 1}$ be reals sequence such that $e^{a_{n}}+na_{n}=2(\\forall n\\geq 1)$. Evaluate $\\lim_{n\\to\\infty}n(1-na_{n})$.", "Solution_1": "for $n$ let us consider $f(x)=e^{x}+nx-2$ then from the fact that $(1+1/n)^{n}2>f(\\frac{n-1}{n^{2}})$. So $\\frac{1}{n+1})>a_{n}>\\frac{n-1}{n^{2}}$. That's $1>n(1-na_{n})>\\frac{n}{n+1}$. This show $n(1-na_{n})\\to1$ when $n\\to\\infty$", "Solution_2": "For any $n$ , denote $f_{n}(x)=e^{x}+nx$. We have $f'_{n}(x)=e^{x}+n>0\\forall x\\in\\mathbb{R}$ and $f_{n}(0)=1<2,f_{n}(\\frac{2}{n})=e^{\\frac{2}{n}}+2>2$ therefore exists $a_{n}$ and \r\n$0b^2\\plus{}c^2$.", "Solution_2": "Please, can present your proof of [b]only if part[/b] ? Thanks.", "Solution_3": "squaring 2 sides of the equality we get\r\n$ (a^2\\minus{}b^2\\minus{}c^2)(1\\plus{}\\frac{2a}{\\sqrt{a^2\\minus{}c^2}\\plus{}b}\\plus{}\\frac{2a}{\\sqrt{a^2\\minus{}b^2}\\plus{}c}\\plus{}\\frac{2a^2}{\\sqrt{a^4\\minus{}a^2c^2\\minus{}a^2b^2\\plus{}b^2c^2}\\plus{}bc})\\equal{}0$ so we complete the solution", "Solution_4": "[quote=\"Virgil Nicula\"]Let $ ABC$ be a triangle for which $ a\\ge b$ and $ a\\ge c$ . Prove that exists the equivalence \n\n$ \\left(\\sqrt {a \\plus{} b} \\plus{} \\sqrt {a \\minus{} b}\\right)\\cdot \\left(\\sqrt {a \\plus{} c} \\plus{} \\sqrt {a \\minus{} c}\\right) \\equal{} (a \\plus{} b \\plus{} c)\\sqrt 2\\ \\Longleftrightarrow\\ A \\equal{} 90^{\\circ}$ .[/quote]\r\n$ A \\equal{} 90^\\circ \\Rightarrow a^2\\equal{}b^2\\plus{}c^2$. After squaring we get $ 2(a\\plus{}\\sqrt{a^2\\minus{}b^2})(a\\plus{}\\sqrt{a^2\\minus{}c^2}) \\equal{}$ $ (a\\plus{}b\\plus{}c)^2 \\iff 2(a\\plus{}c)(a\\plus{}b) \\equal{} (a\\plus{}b\\plus{}c)^2 \\iff a^2\\equal{}b^2\\plus{}c^2$. So $ A \\equal{} 90^{\\circ} \\Longrightarrow \\left(\\sqrt{a\\plus{}b}\\plus{}\\sqrt{a\\minus{}b}\\right)\\cdot\\left(\\sqrt{a\\plus{}c}\\plus{}\\sqrt{a\\minus{}c}\\right) \\equal{} (a\\plus{}b\\plus{}c)\\sqrt 2$\r\nThe other direction:\r\nLet $ f(x) \\equal{} \\sqrt{a\\plus{}x}\\plus{}\\sqrt{a\\minus{}x}$. Then $ f'(x) \\equal{} \\frac{1}{2\\sqrt{a\\plus{}x}} \\minus{} \\frac{1}{2\\sqrt{a\\minus{}x}} < 0$.\r\nFix $ a,b$. Then $ LHS$ is decreasing as function of $ c$, since $ LHS \\equal{} f(b)f(c), f'(x)<0$, and $ RHS$ is increasing as function of $ c$. Hence there is at most one solution. $ a^2\\equal{}b^2\\plus{}c^2$ is a solution by the argument given above. And we are done.", "Solution_5": "[quote=\"Virgil Nicula\"][color=darkred]Let $ ABC$ be a triangle for which $ a\\ge b$ si $ a\\ge c$ . Prove that \n\n$ \\left(\\sqrt {a \\plus{} b} \\plus{} \\sqrt {a \\minus{} b}\\right)\\cdot \\left(\\sqrt {a \\plus{} c} \\plus{} \\sqrt {a \\minus{} c}\\right) \\equal{} (a \\plus{} b \\plus{} c)\\sqrt 2\\ \\Longleftrightarrow\\ A \\equal{} 90^{\\circ}$ .[/color] [/quote]\n\n[color=darkblue][b][u]Proof[/u].[/b] Denote $ X \\equal{} (a \\plus{} b)(a \\plus{} c)$ , $ Y \\equal{} (a \\minus{} b)(a \\minus{} c)$ , $ Z \\equal{} (a \\minus{} b)(a \\plus{} c)$ , $ T \\equal{} (a \\plus{} b)(a \\minus{} c)$ . \n\nProve easily that $ 2p^2 \\minus{} X \\equal{} 2(p \\minus{} a)^2 \\minus{} Y \\equal{} 2(p \\minus{} b)^2 \\minus{} Z \\equal{} 2(p \\minus{} c)^2 \\minus{} T \\equal{} \\frac {b^2 \\plus{} c^2 \\minus{} a^2}{2} .$ Thus \n\n$ \\left(p\\sqrt 2 \\minus{} \\sqrt X\\right)\\ .s.s.\\ \\left[(p \\minus{} a)\\sqrt 2 \\minus{} \\sqrt Y\\right]\\ .s.s.\\ \\left[(p \\minus{} b)\\sqrt 2 \\minus{} \\sqrt Z\\right]\\ .s.s.\\ \\left[(p \\minus{} c)\\sqrt 2 \\minus{} \\sqrt T\\right]\\ .s.s.\\ \\frac {b^2 \\plus{} c^2 \\minus{} a^2}{2}$ .\n\nTherefore, $ \\left(p\\sqrt 2 \\minus{} \\sqrt X\\right) \\plus{} \\left[(p \\minus{} a)\\sqrt 2 \\minus{} \\sqrt Y\\right] \\plus{}$ $ \\left[(p \\minus{} b)\\sqrt 2 \\minus{} \\sqrt Z\\right] \\plus{} \\left[(p \\minus{} c)\\sqrt 2 \\minus{} \\sqrt T\\right]\\ .s.s.\\ \\frac {b^2 \\plus{} c^2 \\minus{} a^2}{2}$ , \n\ni.e. $ (a \\plus{} b \\plus{} c)\\sqrt 2 \\minus{} \\left(\\sqrt {a \\plus{} b} \\plus{} \\sqrt {a \\minus{} b}\\right)\\cdot \\left(\\sqrt {a \\plus{} c} \\plus{} \\sqrt {a \\minus{} c}\\right)\\ .s.s.\\ \\frac {b^2 \\plus{} c^2 \\minus{} a^2}{2}$ . In conclusion, \n\n$ A\\ \\begin{array}{c} < \\\\\n\\ \\equal{} \\\\\n\\ > \\end{array}\\ 90^{\\circ}\\ \\Longleftrightarrow\\ b^2 \\plus{} c^2\\ \\begin{array}{c} > \\\\\n\\ \\equal{} \\\\\n\\ < \\end{array}\\ a^2\\ \\Longleftrightarrow$ $ \\underline{\\overline{\\left\\|\\ \\left(\\sqrt {a \\plus{} b} \\plus{} \\sqrt {a \\minus{} b}\\right)\\cdot \\left(\\sqrt {a \\plus{} c} \\plus{} \\sqrt {a \\minus{} c}\\right)\\ \\begin{array}{c} < \\\\\n\\ \\equal{} \\\\\n\\ > \\end{array}\\ (a \\plus{} b \\plus{} c)\\sqrt 2\\ \\right\\|}}$ .\n\n[b]Remark.[/b] I used the notation $ X\\ \\mathrm{.s.s.}\\ Y\\ \\Longleftrightarrow\\ XY > 0$ or $ X \\equal{} Y \\equal{} 0$ ([b]s[/b]ame [b]s[/b]ign).[/color]", "Solution_6": "\\[ \\left(\\sqrt{a\\plus{}b}\\plus{}\\sqrt{a\\minus{}b}\\right)\\cdot\\left(\\sqrt{a\\plus{}c}\\plus{}\\sqrt{a\\minus{}c}\\right)\\equal{}(a\\plus{}b\\plus{}c)\\sqrt 2\\Leftrightarrow2(a\\plus{}\\sqrt{a^2\\minus{}b^2})(a\\plus{}\\sqrt{a^2\\minus{}c^2})\\equal{}(a\\plus{}b\\plus{}c)^2\\]\r\n\r\nIf $ A\\equal{}90^{\\circ}$, then \\[ 2(a\\plus{}\\sqrt{a^2\\minus{}b^2})(a\\plus{}\\sqrt{a^2\\minus{}c^2})\\equal{}2(a\\plus{}c)(a\\plus{}b)\\equal{}2a^2\\plus{}2ab\\plus{}2bc\\plus{}2ca\\equal{}a^2\\plus{}b^2\\plus{}c^2\\plus{}2ab\\plus{}2bc\\plus{}2ca\\equal{}(a\\plus{}b\\plus{}c)^2\\]\r\n\r\nIf $ A\\ne90^{\\circ}$, consider two cases. (i) $ a^2>b^2\\plus{}c^2$, then \\[ 2(a\\plus{}\\sqrt{a^2\\minus{}b^2})(a\\plus{}\\sqrt{a^2\\minus{}c^2})>2(a\\plus{}c)(a\\plus{}b)\\equal{}2a^2\\plus{}2ab\\plus{}2bc\\plus{}2ca>a^2\\plus{}b^2\\plus{}c^2\\plus{}2ab\\plus{}2bc\\plus{}2ca\\equal{}(a\\plus{}b\\plus{}c)^2\\] (ii) $ a^2 m \\minus{} 1$ there are few connected graphs, since any connected graph contains at least $ m \\geq n \\minus{} 1$ edges (when a tree). \r\nAlso, the fact that the $ d_i$'s represent the degree structure of a graph adds restrictions to just having $ \\sum_{i \\equal{} 1}^n d_i \\equal{} 2m$; later on, in your double sum, should we assume the $ d_i$'s are degrees of some connected graph? If not, why even speak about graphs?", "Solution_4": "yes i meant $ m\\ge n \\minus{} 1$ sorry. it's corrected already\r\n\r\n [quote=\"mavropnevma\"]\nAlso, the fact that the $ d_i$'s represent the degree structure of a graph adds restrictions to just having $ \\sum_{i \\equal{} 1}^n d_i \\equal{} 2m$; later on, in your double sum, should we assume the $ d_i$'s are degrees of some connected graph? [/quote]\n\nthe sum of the degrees of a grph without loops is twice the edge number. Since the graphs are connected, there are no vertices of zero degree\n\n [quote=\"mavropnevma\"]\nIf not, why even speak about graphs?[/quote]\r\nof course, one can formulate the problem without speaking about graphs" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be the positive real numbers.Prove that:\r\n$ \\frac{ab^3}{c}\\plus{}\\frac{bc^3}{a}\\plus{}\\frac{ca^3}{b}\\plus{}3abc$\u2265$ 2(a^2b\\plus{}b^2c\\plus{}c^2a)$", "Solution_1": "If you divide both sides by $ abc$, and after substituting $ x \\equal{} \\dfrac{a}{b}, y \\equal{} \\dfrac{b}{c}$ and $ z \\equal{} \\dfrac{c}{a}$ the inequality becomes equivalent to:\r\n\r\n$ x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 3\\ge 2(xy \\plus{} yz \\plus{} zx)$, for $ xyz \\equal{} 1$.\r\n\r\nBut this is an application of the following one:\r\n\r\n[quote]Let $ a,b,c\\ge 0$, then\n\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} 2abc \\plus{} 1\\ge 2(ab \\plus{} bc \\plus{} ca)$[/quote]\r\n\r\nTry to prove it. It's easy enough.", "Solution_2": "[quote=\"SUPERMAN2\"]Let $ a,b,c$ be the positive real numbers.Prove that:\n$ \\frac {ab^3}{c} \\plus{} \\frac {bc^3}{a} \\plus{} \\frac {ca^3}{b} \\plus{} 3abc$\u2265$ 2(a^2b \\plus{} b^2c \\plus{} c^2a)$[/quote]\r\n\r\nIt's equivolent to:\r\n$ a^2b^4\\plus{}b^2c^4\\plus{}a^4c^2 \\plus{}3a^2b^2c^2 \\ge 2a^3b^2c\\plus{}2ab^3c^2\\plus{}2a^2bc^3$\r\nLet's get $ x\\equal{}(a^2b^4)^{\\frac{1}{3}}$, $ y\\equal{}(b^2c^4)^{\\frac{1}{3}}$, $ z\\equal{}(a^4c^2)^{\\frac{1}{3}}$\r\nThen inequality equiovolent to:\r\n$ x^3\\plus{}y^3\\plus{}z^3\\plus{}3xyz \\ge 2(xy)^{\\frac{3}{2}}\\plus{}2(yz)^{\\frac{3}{2}}\\plus{}2(xz)^{\\frac{3}{2}}$\r\nwich is right becouse by Schur ineqality we have:\r\n$ x^3\\plus{}y^3\\plus{}z^3\\plus{}3xyz \\ge \\sum_{sym}{x^2y} \\ge 2(xy)^{\\frac{3}{2}}\\plus{}2(yz)^{\\frac{3}{2}}\\plus{}2(xz)^{\\frac{3}{2}}$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "a,b,c > 0\r\n\r\nprove that :\r\n\r\nhttp://www.monsterup.com/image.php?url=upload/1238886362528.gif", "Solution_1": "[quote]$ \\sqrt {\\frac {{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\frac {{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\frac {{2c}}{{c \\plus{} a}}} \\le 3$[/quote]\r\n\r\n$ S^2 \\equal{} ( {\\sqrt {\\frac {{2a}}{{a \\plus{} b}}} \\plus{} \\sqrt {\\frac {{2b}}{{b \\plus{} c}}} \\plus{} \\sqrt {\\frac {{2c}}{{c \\plus{} a}}} })^2 \\le (2 \\plus{} 2 \\plus{} 2)( {\\frac {a}{{a \\plus{} b}} \\plus{} \\frac {b}{{b \\plus{} c}} \\plus{} \\frac {c}{{c \\plus{} a}}}) \\le 9$\r\n\r\n$ \\Leftrightarrow \\frac {a}{{a \\plus{} b}} \\plus{} \\frac {b}{{b \\plus{} c}} \\plus{} \\frac {c}{{c \\plus{} a}} \\le \\frac {3}{2}$", "Solution_2": "[quote=\"L_Euler\"]\n$ \\Leftrightarrow \\frac {a}{{a \\plus{} b}} \\plus{} \\frac {b}{{b \\plus{} c}} \\plus{} \\frac {c}{{c \\plus{} a}} \\le \\frac {3}{2}$[/quote]\r\nwhich is wrong. Try $ a \\equal{} b \\equal{} 1$ and $ c\\rightarrow0^ \\plus{} .$ :wink:\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c \\in [1,2]$. Prove or disprove that\r\n\\[ \\sqrt{\\frac{(2a\\minus{}b)(2a\\minus{}c)}{b^2\\plus{}c^2}} \\plus{}\\sqrt{\\frac{(2b\\minus{}c)(2b\\minus{}a)}{c^2\\plus{}a^2}} \\plus{}\\sqrt{\\frac{(2c\\minus{}a)(2c\\minus{}b)}{a^2\\plus{}b^2}} \\ge \\frac{3\\sqrt{2}}{2}\\]\r\nEquality holds if and only if $ a\\equal{}b\\equal{}c$ or $ (a,b,c)\\equal{}(2,1,1)$.\r\n\r\nI think this is a very hard inequality (at least with me :(). I have tried to prove it many times but I failed. :(", "Solution_1": "Maybe VasC can help me?", "Solution_2": "I do not dare to try it even.....maybe arqady or vasc can help us..... :roll:" } { "Tag": [ "linear algebra", "matrix", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "Solve sytem equation with x,y,z>0: $ \\left\\{\\begin{matrix}\r\nz^2\\plus{}2xyz\\equal{}1 & \\\\ \r\n3x^2y^2\\plus{}3y^2x\\equal{}1\\plus{}x^3y^4 & \\\\ \r\nz\\plus{}zy^4\\plus{}4y^3\\equal{}4y\\plus{}6y^2z & \r\n\\end{matrix}\\right.$", "Solution_1": "i have a solution for your problem\r\nsetting $ y\\equal{}tana (0 < a <\\frac{\\pi}{2})$\r\nfrom the third equation we get $ z\\equal{}tan4a$, from the first equation we get $ xy\\equal{}\\frac{1}{tan8a}$, substituting to the second we get $ tana\\equal{}tan24a$ and we can easily solve this equation", "Solution_2": "Why that typical substitution?", "Solution_3": "The last equation is the easiest one to solve for one variable (it is linear in $ z$), so\r\n\\[ z(1 \\plus{} y^4 \\minus{} 6y^2) \\equal{} 4y \\minus{} 4y^3\\implies z \\equal{} \\frac {4y \\minus{} 4y^3}{1 \\minus{} 6y^2 \\plus{} y^4} \\equal{} \\frac {\\binom41y \\minus{} \\binom43y^3}{\\binom40 \\minus{} \\binom42y^2 \\plus{} \\binom44y^4}.\\]\r\nThis should remind you of [url=http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Tangents]this tangent sum identity[/url].\r\n\r\nIt might also remind you of $ \\cos{4x} \\plus{} i\\sin{4x} \\equal{} (\\cos{x} \\plus{} i\\sin{x})^4$." } { "Tag": [], "Problem": "In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:\r\n\\[ \\text{(A)}\\ 5 \\qquad \\text{(B)}\\ 7 \\qquad \\text{(C)}\\ 10 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 14 \\]", "Solution_1": "[quote=\"MCrawford\"]In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:\n\\[ \\text{(A)}\\ 5 \\qquad \\text{(B)}\\ 7 \\qquad \\text{(C)}\\ 10 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 14 \\][/quote]\r\n[hide]\n$ x $ is the number of cows, $ y $ is the number of chickens.\n\n$ 4x+2y=14+2(x+y) $\n\n$ 4x+2y=14+2x+2y $\n\n$ 2x=14 $\n\n$ x=7 $ cows. \n\nSo the answer is $ (B) $[/hide]", "Solution_2": "[hide=\"Sol\"]\n$ 4a+2b=14+2a+2b $\n$ 2a=14 $\n$ a=7 $\nNo. of cows=$ 7 $\n$ \\boxed{\\text{(B)}} $\n[/hide]", "Solution_3": "[hide]Let the number of cows be $ x $ and the number of chickens be $ y $.\n\n$ (4x+2y)-14=2(x+y) $\n\n$ 4x+2y-14-2x-2y=0 $\n\n$ 2x=14 $\n\n$ x=7 $\n\nAnswer: $ \\boxed{\\text{(C)}} $[/hide]", "Solution_4": "[quote=\"YooSam\"][hide]Let the number of cows be $x$ and the number of chickens be $y$.\n\n$(4x+2y)-14=2(x+y)$\n\n$4x+2y-14-2x-2y=0$\n\n$2x=14$\n\n$x=7$\n\nAnswer: $\\boxed{\\text{(C)}}$[/hide][/quote]\r\nI hope you mean $\\text{(B)}$", "Solution_5": "[quote=\"nat mc\"][quote=\"YooSam\"][hide]Let the number of cows be $x$ and the number of chickens be $y$.\n\n$(4x+2y)-14=2(x+y)$\n\n$4x+2y-14-2x-2y=0$\n\n$2x=14$\n\n$x=7$\n\nAnswer: $\\boxed{\\text{(C)}}$[/hide][/quote]\nI hope you mean $\\text{(B)}$[/quote]\r\n\r\n :blush: Oops, I was going to typed B since I got the right answer, but don't know why I typed C :blush: ." } { "Tag": [ "integration", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "By considering $\\int\\frac{x^{n}}{1-x^{8}}dx$ or otherwise prove that \r\n\r\n \\[ \\pi=\\sum_{i=0}^{\\infty}\\frac{1}{16^{i}}\\left(\\frac{4}{8i+1}-\\frac{2}{8i+4}-\\frac{1}{8i+5}-\\frac{1}{8i+6}\\right). \\]", "Solution_1": "IOW, you want the proof of the BBP formula.\r\n\r\nWell, here it goes :\r\nFor any integer $k$, we have :\r\n$\\sum_{i=0}^{\\infty} \\frac 1{16^i(8i+k)} = \\sqrt{2}^k \\sum_{i=0}^{\\infty} \\left[\\frac{x^{k+8i}}{8i+k} \\right]_0^{1/\\sqrt{2}}$\r\n$= \\sqrt{2}^k \\sum_{i=0}^{\\infty} \\int_0^{1/\\sqrt{2}} x^{k-1+8i}dx$\r\n$= \\sqrt{2}^k \\int_0^{1/\\sqrt{2}} \\frac{x^{k-1}}{1-x^8}dx$.\r\nSo :\r\n$\\sum_{i=0}^{\\infty}\\frac 1{16^i}\\left(\\frac 4{8i+1}-\\frac 2{8i+4}-\\frac 1{8i+5}-\\frac 1{8i+6}\\right)$\r\n$= \\int_0^{1/\\sqrt{2}} \\frac{4\\sqrt{2} - 8x^3 - 4\\sqrt{2}x^4 - 8x^5}{1-x^8}dx$,\r\nhence, if we let $y = \\sqrt{2}x$, and simplifying :\r\n$= \\int_0^1 \\frac{16(y-1)}{y^4-2y^3+4y-4}dy$\r\n$= \\int_0^1 4 \\frac{2-y}{y^2 - 2y + 2}dy + \\int_0^1 4 \\frac y{y^2-2}dy$\r\n$= \\int_0^1 \\frac{4 - 4y}{y^2-2y+2}dy + \\int_0^1 \\frac 4{1 + (y-1)^2}dy + \\int_0^1 4 \\frac y{y^2-2}dy$\r\n$= [-2 \\ln{(y^2-2y+2)} + 4 \\arctan{(y-1)} + 2 \\ln{(2-y^2)}]_0^1$\r\n$= \\pi$.\r\n\r\n:)" } { "Tag": [ "group theory", "abstract algebra", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "Let G be a group and g,h in G. Define the commutator of g and h as [g,h]= gh(hg)^-1. Then define the commutator subgroup, denoted [G,G], of G as the subgroup generated by all the commutators of elements of G, i.e. [G,G]=<{[g,h]: g,h in G}>.\r\n\r\n(a) What does an element of [G,G] look like?\r\nHint: it is not enough to only consider elements of the form [g,h].\r\n *DONE*\r\n\r\n(b) Prove that [G,G] is a normal subgroup of G.\r\n*DONE*\r\n\r\n(c) Prove that G/[G,G] is an Abelian group. \r\n\r\nI can't get part (c)....is it asking to show that ab=ba for some a,b in [G,G] where a and b are commutators?", "Solution_1": "What you want to prove is that $ gh[G,G] \\equal{} hg[G,G]$ equivalent to $ hgh^{\\minus{}1}g^{\\minus{}1}[G,G] \\equal{} [G,G]$ equivalent to $ hgh^{\\minus{}1}g^{\\minus{}1} \\in [G,G]$", "Solution_2": "For your first expression, are g and h just elements of G?", "Solution_3": "Yes.", "Solution_4": "Is the product: $ g[G,G]h[G,G]g^{\\minus{}1}[G,G]h^{\\minus{}1}[G,G]$ not a commutator?\r\nAnd if it is then call it $ [u,v]$ with $ u$ and $ v$ in G so that we have \r\n$ gh[G,G] \\equal{} [u,v](hg[G,G]) \\equal{} ([u,v][G,G])hg \\equal{} [G,G]hg \\equal{} hg[G,G]$.", "Solution_5": "Where is your first equality from? (it implies gh[G,G]=hg[G,G])", "Solution_6": "Prove that the union of intervals [1,n] from n=1 to n=infinity is all of N.\r\n\r\n\r\nDo I use induction on this? Archimedes? (This question is before the section of Archimedes though). I need help on how to start it!", "Solution_7": "Start a new thread!", "Solution_8": "Woops! Sorry...new to the site and didn't realize I did that..thanks!" } { "Tag": [ "geometry", "congruent triangles" ], "Problem": "Start with four congruent triangles. In one step, you may take any triangle and cut it in two with the altitude from the right angle. Prove that you can never get rid of congruent triangles", "Solution_1": "I don't know what you mean by 'get rid of.' Obviously if you cut one, then they are different. Perhaps the point of the problem is that the cut makes two triangles similar to the first...", "Solution_2": "it basically means that you will always have congruent triangles. Think invariance.", "Solution_3": "[quote=\"mdk\"]you may take any triangle and cut it in two with the altitude from the right angle. [/quote]\r\n\r\nSo are all four triangles right?", "Solution_4": "yes, yes they are.", "Solution_5": "[hide=\"hint\"]\nRemember the problem with the amoeba which starts at (0, 0), and splits to create one amoeba at (1, 0) and one at (0, 1)? And then every amoeba thereafter can split to put one amoeba above it and one to the right? Okay...now how does this apply to our problem?[/hide]\n\n[hide=\"bigger hint\"]If you start with four amoebas in the bottom left corner, show that you'll always have some two amoebas on the same lattice point.[/hide]" } { "Tag": [ "function" ], "Problem": "Propanol(C3H7OH) melts at -126.5 degree Celsius and boils at 97.4 degree Celsius. Draw a qualitative sketch of how the entropy changes as propanol at 150.0 degree Celsius is cooled to a solid at -150.0 degree Celsius. The entire process occurs at a constant pressure of 1 atm", "Solution_1": "Quantitatively, since the pressure is constant and considering a reversible process between phase transitions, \r\n\r\n$ \\Delta S \\equal{} \\frac{\\Delta H}{T}$.\r\n\r\nQuantitatively, I think it's obvious that $ \\Delta S$ must decrease.", "Solution_2": "[quote=\"Carcul\"]Quantitatively, since the pressure is constant and considering a reversible process between phase transitions, \n\n$ \\Delta S \\equal{} \\frac {\\Delta H}{T}$.\n\nQuantitatively, I think it's obvious that $ \\Delta S$ must decrease.[/quote]\r\nCan you explain a little bit more detailed in the final conclusion?I think delta H is a constant and T increases, so delta S decreases? By the way, the problem asks for the sketch, so does that mean we have a graph of the function shown above?", "Solution_3": "$ \\Delta H$ can only be considered constant at phase transitions. The liquid state is more ordered than the gaseous state, and the solid state is more ordered than the liquid state, so, given the interpretation of Entropy of a system as its degree of order, on this process $ S_{syst}$ must decrease." } { "Tag": [ "LaTeX", "function", "integration", "calculus", "inequalities", "derivative", "real analysis" ], "Problem": "[url]http://www.mojoimage.com/free-image-hosting-view-00.php?id=11132.GIF[/url]", "Solution_1": "why no one answer the question? give some idea!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_2": "A significant fraction of the posters here - myself included - will refuse to click through to any problem posted as an external link. So I don't know what your problem is or whether or not it is of any interest.\r\n\r\nIf you want to speak to the people here, type your problem out in this forum. Learning a little $ \\text{\\LaTeX}$ would help.", "Solution_3": "suppose $ f(t)$ is a nonnegative continuous concave function on[a,b] not identically zero.\r\n\r\ndefine $ F \\equal{} \\frac {2}{b \\minus{} a}\\int_a^b f(x)dx$. and $ \\phi (x)$ is a convex function on [0,F]\r\n\r\nprove $ \\frac {1}{F}\\int_0^F\\phi(x)dx \\ge \\frac {1}{b \\minus{} a}\\int_a^b \\phi(f(t)) dt$\r\n\r\n\r\nso i suppose the first thing to notice is that F is twice the average of a convex function over the interval,\r\nbut is it possible that a point exists such that f(x)>F in [a,b]? if so.. isnt the integral not really defined?", "Solution_4": "it is impossible the case happen", "Solution_5": "no one can solve the question!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_6": "Well, Kent told you why, didn't he? ;)\r\n\r\nThe problem is not difficult at all (it is next to impossible to ask a difficult question about an arbitrary convex function of one variable :)). Without loss of generality, $ [a,b] \\equal{} [0,1]$. Note that adding a linear function to $ \\psi$ doesn't change anything. Now, every convex function is a linear function plus a weighted average of $ (x \\minus{} t)_ \\plus{}$, so it will suffice to prove the result for $ \\psi(x) \\equal{} (x \\minus{} t)_ \\plus{}$ with $ t\\in[0,2\\bar f]$. The inequality in question then reduces to $ \\int_0^1 (f \\minus{} t)_ \\plus{} \\le \\frac {(2\\bar f \\minus{} t)^2}{4\\bar f}$. Now, if you look at this picture,\r\n[asy]size(200,200,IgnoreAspect);\nD((-0.3,0)--(1.3,0),black);\nD((0,0)--(0,2),black);\nreal a=0.7;\npair[]Q=IPs(D(MP(\"t\",(0,a),W)--(1,a)),D((0,0)--(0.3,2)--(1,0),red));\nD((0,0.5)..Q[0]..(0.5,0.95)..Q[1]..(1,0.2),blue+linewidth(1));\nD((1,0)--(1,0.2),black);\nD(Q[0]--Q[1],orange+linewidth(1));\nMP(\"s\",(Q[0]+Q[1])/2);\nMP(\"0\",(0,0)); MP(\"1\",(1,0));[/asy]\r\nyou'll realize after a minite of thinking that $ \\int_0^1(f \\minus{} t)_ \\plus{}$ can be estimated by both $ s^2\\bar f$ and $ \\bar f \\minus{} \\frac {s \\plus{} 1}2 t$. The first estimate increases with $ s$ and the second decreases with $ s$. These estimates are equal for $ s \\equal{} s_0 \\equal{} 1 \\minus{} \\frac t{2\\bar f}$, which gives exactly the value we want. That's it (I leave the details to you to figure out).", "Solution_7": "thank you Kent Merryfield and fedja, sorry, I am learning mathmaitcal analysis by myself and I am not familiar with latex and i will follow your advice to learn latex.\r\n\r\ni still have a question: every convex function is a linear function plus a weighted average of ??? did you mean it can be approximated?", "Solution_8": "Either approximated, or represented as $ \\psi(x) \\equal{} ax \\plus{} b \\plus{} \\int_I (x \\minus{} t)_ \\plus{} \\,d\\mu(t)$ with respect to some non-negative measure $ \\mu$ on $ I$: either interpretation is fine (at least, inside the interval $ I$: you may have some stupid problems with the endpoints if $ \\psi$ is discontinuous or has infinite derivatives there, but they do not really matter here)", "Solution_9": "$ \\int _{0} ^{ \\infty } f(x)dx$\r\n\r\nI am leaning write fomular with latex.", "Solution_10": "if is discontinuous or has infinite derivatives there, but they do not really matter here ????????????????????\r\n\r\n\r\n\r\nI still have trouble with this point?????????????", "Solution_11": "dear fedja, would you please give an explanation about the case when if the function is discontinuous or has infinite derivatives", "Solution_12": "Sorry for the delay. Note that $ \\psi$ can be represented as a limit of an [i]increasing[/i] sequence of good (say, piecewise linear) convex functions on $ (0,2\\bar f)$ and that, since $ f$ is not identically $ 0$, the sets $ \\{f\\equal{}0\\}$ and $ \\{f\\equal{}2\\bar f\\}$ have measure $ 0$. Thus, the Monotone Convergence Theorem does the trick.", "Solution_13": "how can we construct such and [b]increasing[/b] sequencesse. why the second set has measure 0 ???" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Let {$ a_n$} be a strictly increasing sequence of positive integers. Can the series\r\n\r\n$ (1 \\minus{} a_1/a_2) \\plus{} (1 \\minus{} a_2/a_3) \\plus{} (1 \\minus{} a_3/a_4) \\plus{} ...$\r\n\r\never converge?\r\n\r\nPlease provide a proof.", "Solution_1": "yes. see for example\r\n$ a_1\\equal{}3$ and $ a_{i\\plus{}1}\\equal{}\\frac{i^2}{i^2\\minus{}1}a_i$", "Solution_2": "Well that doesn't really work. Say you wanted a_2, the next term. You would need to set i = 1. But... that would be 1/0 * 3... bad :P", "Solution_3": "Yes there is! Take any function $ f$ for which we have:\r\n\r\n$ \\sum^\\infty_{n = 1} f(n) < \\infty$\r\n\r\nand\r\n\r\n$ f(n) > 0$\r\n\r\nNow, we can set:\r\n\r\n$ 1 - \\frac {a_{2n - 1}}{a_{2n}} = f(n) \\, \\Rightarrow \\, a_n = a_0 \\prod ^n _{i = 1}\\frac {1}{1 - f\\left(\\frac {n}{2}\\right)}$\r\n\r\nand there you go. You can try a simple example:\r\n\r\n$ f\\left(\\frac {n}{2}\\right) = \\frac {1}{(n + 1)^2}$\r\n\r\nand then we get\r\n\r\n$ a_n = a_0 \\prod ^n _{i = 1}\\frac {1}{1 - \\frac {1}{(i + 1)^}} = a_0 \\prod ^n _{i = 1}\\frac {(i + 1)^2}{(i + 1)^2 - 1} = 2 a_0 \\frac {(n + 1)!^2}{n!(n + 2)!} = 2 a_0 \\frac {n + 1}{n + 2}$\r\n\r\nFrom this we have\r\n\r\n$ 1 - \\frac {a_{2n - 1}}{a_{2n}} = \\frac {1}{(2n + 1)^2}$\r\n\r\nand I suppose no further explanation is needed! ;)", "Solution_4": "I know it seems I'm playing advocatus diaboli here, but one simple question: how can you guarantee the \"integerness\" of these sequences, milin? As the problem states, the sequence is made of positive integers. For instance, in your example $ a_0$ has to be divisible with every prime form $ \\mathbb{N}$... Maybe I'm overseeing something?", "Solution_5": "It's obvious that a bounded sequence will never work, so the examples posted thus far are ridiculous.\r\n\r\nOn the other hand, it's enough to find any sequence of positive real numbers that tends to infinity and satisfies the given condition. This can certainly be arranged.\r\n\r\nedit: Ohps, I forgot what the condition was and thought of a different one.", "Solution_6": "Theorem: A series $ \\sum x_n$ with no term equal to $ \\pm 1$ converges absolutely if and only if $ \\prod (1\\plus{}x_n)$ (or $ \\prod (1\\minus{}x_n)$ converges to a nonzero value.\r\n\r\nIn our case, we use the latter form: $ \\frac{a_1}{a_2}\\cdot\\frac{a_2}{a_3}\\cdot\\frac{a_3}{a_4}\\cdots$.\r\n\r\nThe product telescopes. If $ a_n\\to\\infty$, it diverges to zero and the series diverges to $ \\infty$. If $ a_n$ stays bounded and converges to some limit $ L$, the product converges to $ \\frac{a_1}{L}$ and the series converges.\r\nThe second case is impossible with a strictly increasing sequence of positive integers, so the series diverges.", "Solution_7": "would it be possible for you to give a proof of that theorem or provide a link to where i can find one. thanks.", "Solution_8": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=204755]here[/url]." } { "Tag": [ "geometry", "3D geometry", "sphere", "tetrahedron", "pyramid", "Pythagorean Theorem" ], "Problem": "Three mutually tangent spheres of radius $ 1$ rest on a horizontal plane. A sphere of radius $ 2$ rests on them. What is the distance from the plane to the top of the larger sphere?\r\n\r\n$ \\textbf{(A)}\\ 3 \\plus{} \\frac {\\sqrt {30}}{2} \\qquad \\textbf{(B)}\\ 3 \\plus{} \\frac {\\sqrt {69}}{3} \\qquad \\textbf{(C)}\\ 3 \\plus{} \\frac {\\sqrt {123}}{4}\\qquad \\textbf{(D)}\\ \\frac {52}{9}\\qquad \\textbf{(E)}\\ 3 \\plus{} 2\\sqrt2$", "Solution_1": "Wow. Interesting problem, but very subtle solution.\r\n[hide]\nLet the centers of the smaller spheres be $A$, $B$ and $C$. Let the center of the bigger sphere be $D$. $s=AD=BD=CD=2+1=3$. $AB=BC=AC=1+1=2$. This forms tetrahedron $ABCD$. We need to find the height $h$ of tetrahedron $ABCD$. Using $s$ as hypotenuse of a right triangle, and the distance $d$ from a vertex of the base $ABC$ to the center of $ABC$, we have $d=2\\sqrt{3}/3$. Using pythagorean theorem we find that $h=\\frac{\\sqrt{69}}{3}$. Therefore we conclude that the distance from the bottom to top is $h+1+2=\\boxed{\\frac{\\sqrt{69}}{3}+3}$, which is $\\boxed{B}$.[/hide]", "Solution_2": "Interesting, but I did that a different way, using orthogonal views\r\n\r\n[hide]\nIf you were to draw the \"top view\" of the configuation [looking from the top straight down to the bottom] you would draw the 3 radius 1 circles with their centers making an equalateral triangle and a big circle with radius 2 concentric with the \"triangle.\"\n\nThe \"front view\" of the configuration [looking from the front through the shape], you would draw 2 circles of radius 1 and a circle radius 2 on top [becuase they are mutually tangent]\n\nLooking at are top view, we can calculate the distance between the center of the radius-1 circles to the center of the radius-2 circle, or the centroid of the formed triangle. Becuase the triangle is formed by connecting the circles of radius 1, it is equalateral with side length 2. The distance to the centroid would thus be $\\frac{2\\sqrt3}{3}$.\n\nNow we use the front view. Lets find the distance between the center of the radius-1 circle and the center of the radius-2 circle. Since they are tangent to each other, the distance is 3.\n\nBecuase the views are orthogonal, we notice that the center of the radius-2 circle is directly above the centroid of the equalateral triangle. Since they form a right triangle [in 3-d space], we can find the length from the centroid of the triangle to the center of the radius-2 circle. Through phythagorean theorem, we have the the required distance to be $=\\sqrt{3^2-(\\frac{2\\sqrt3}{3})^2}=\\frac{\\sqrt{69}}{3}$\n\nBy knowing that distance, to make it fit the requried distance, we add 1 for the radius 1 circle, and add 2 for the radius 2 circle to find the distance from the top of the configuration to the absolute bottom of the confuguration. Therefore the answer is $\\frac{\\sqrt{69}}{3} +3$[/hide]\r\n\r\nAlthough my method is rather complicated, I think it can be extended to solve similar problems, say a radius-2 sphere resting on 4 tangent radius-1 spheres.", "Solution_3": "[hide=\"hint:\"]connect the centers of the spheres to create a pyramid that you can find the height of[/hide]", "Solution_4": "[hide]Connecting the centers of the spheres, we get a pyramid with an equilateral base of side $2$ and the other $3$ edges of side $3.$ Finding the height of this is trivial... it yields $\\frac{\\sqrt{69}}{3}$. Add this to the radius of one of the lower spheres and the radius of the upper sphere to get $3+\\frac{\\sqrt{69}}{3}.$[/hide]", "Solution_5": "[b]Solution:[/b] Ugh. These problems. 3D Geo = Take cross sections. Take the cross section joining the center of the big sphere and two of the smaller spheres. Let those centers be $X, Y, Z$ respectively. Let the foot of the altitude from $X$ be $M$. Notice that $XY = 2 + 1 = 3$ and that $YM = 1$. So we have that $XM= \\sqrt{8}$. Now, let the bottom of the large sphere be called $O$. Call the distance from $O$ to the plane $h$. Notice that we have $(1+h)^2 + (\\frac{\\sqrt{3}}{3})^2 = {\\sqrt{8}}^2$ (we get the $\\frac{\\sqrt{3}}{3}$ value as that is the distance from $O$ to $M$ and we can exploit the equilateral triangle formed from the centers of all three congruent smaller spheres). Solving, we have that $1+h = \\sqrt{\\frac{23}{3}}$. However, notice that the question is asking for the distance from the plane to the top of the sphere, so we want to find $4+h$. Hence our answer is $\\boxed{B}$." } { "Tag": [ "function", "factorial", "complex numbers" ], "Problem": "How can I solve the following problem:\r\n\r\n$ \\sum_{n \\equal{} 0}^\\infty \\sum_{k \\equal{} 0}^{n} [{2n\\choose2k} {(2k\\plus{}1)}^{\\minus{}x}] / 2^{n\\plus{}1}$\r\n\r\n\r\nNote: The denominator, 2^(n+1) is inside of the infinite series and outside of the inner series, though come to think of it, I don't think it matters.\r\n\r\nNote: x is a constant.", "Solution_1": "I don't know what to do for $ x \\neq 0$...\r\n\r\nBut for $ x\\equal{}0$:\r\n\r\n$ (1\\minus{}1)^{2n} \\equal{} \\binom{2n}{0} \\minus{}\\binom{2n}{1} \\plus{} \\binom{2n}{2} \\minus{} \\cdots \\plus{} \\binom{2n}{2n}$ so $ \\binom{2n}{0} \\plus{} \\binom{2n}{2} \\plus{} \\cdots \\plus{} \\binom{2n}{2n} \\equal{} \\binom{2n}{1} \\plus{} \\binom{2n}{3} \\plus{} \\cdots \\plus{} \\binom{2n}{2n\\minus{}1}$. Since $ (1\\plus{}1)^{2n} \\equal{} 2^{2n} \\equal{} \\binom{2n}{0} \\plus{} \\binom{2n}{1} \\plus{}\\binom{2n}{2} \\plus{} \\cdots \\plus{} \\binom{2n}{2n}$, we have $ 2^{2n} \\equal{} 2(\\binom{2n}{0} \\plus{} \\binom{2n}{2} \\plus{} \\cdots \\plus{} \\binom{2n}{2n})$ or $ \\binom{2n}{0} \\plus{} \\binom{2n}{2} \\plus{} \\cdots \\plus{} \\binom{2n}{2n} \\equal{} 2^{2n\\minus{}1}$.\r\n\r\nSo for $ x\\equal{}0$ your summation becomes $ \\sum_{n\\equal{}0}^\\infty \\frac{2^{2n\\minus{}1}}{2^{n\\plus{}1}} \\equal{} \\sum_{n\\equal{}0}^\\infty 2^{n\\minus{}2}$ which clearly diverges.", "Solution_2": "x is a complex number with a real part greater than zero.\r\n\r\nIs there a way to simplify the sums to an easy function of x.\r\n\r\nIs there a way to simplify the sums at all?", "Solution_3": "This looks suspiciously similar to the right-hand side of the identity\r\n\r\n $ [\\sum_{n \\equal{} 0}^\\infty f(n)x^n]*[\\sum_{n \\equal{} 0}^\\infty g(n)x^n] \\equal{} \\sum_{n \\equal{} 0}^\\infty [\\sum_{k \\equal{} 0}^n f(k)g(n \\minus{} k)]x^n$", "Solution_4": "[quote=\"epkid08\"]x is a complex number with a real part greater than zero.\n\nIs there a way to simplify the sums to an easy function of x.\n\nIs there a way to simplify the sums at all?[/quote]\r\n\r\nOh I think I got it!\r\n\r\nWhat if you split up the binomial coefficient, and factor the (2n)! out of the inner series (can you even do this?), so the inner series is left with:\r\n\r\n$ \\frac {1} {k!(n \\minus{} k)!(2k \\plus{} 1)^x}$\r\n\r\nSo...\r\n\r\n$ g(a) \\equal{} \\frac {1}{a!}$\r\n$ f(a) \\equal{} \\frac {1}{a!(2a \\plus{} 1)^x}$\r\n\r\nbut the problem is the function outside of the inner sum would not be x^n, it would be $ (2n!)(2^{ \\minus{} (2n \\plus{} 1)})$. I'm sure it's fine that it's -(2n+1) and not n, but what about (2n!)?\r\n\r\nAny ideas?", "Solution_5": "I don't think there's any chance this is going to be nice. $ (2k \\plus{} 1)^{\\minus{}x}$ is just not a nicely-behaved expression.", "Solution_6": "[quote=\"JBL\"]I don't think there's any chance this is going to be nice. $ (2k \\plus{} 1)^{ \\minus{} x}$ is just not a nicely-behaved expression.[/quote]\r\n\r\nProbably not later on, but I want to split it up so I can simplify the other part further, so I need to find out how to simplify $ \\frac{(2n)!}{2^{n\\plus{}1}}$ into something like $ x^n$.", "Solution_7": "Well I found this handy identity:\r\n\r\n$ cosh(z) \\equal{} \\sum_{n\\equal{}0}^{\\infty} \\frac{z^{2n}}{(2n)!}$\r\n\r\nAs close as I am, I'm off by one number.\r\n\r\nDoes anyone have ideas on how to find...\r\n\r\n$ \\sum_{n\\equal{}0}^{\\infty} \\frac{z^{2n\\plus{}1}}{(2n)!}$\r\n\r\n...in terms of...\r\n\r\n$ cosh(z)$\r\n\r\n?", "Solution_8": "Try $ z \\text{ cosh } z$. But that doesn't help with the summation you gave at all.", "Solution_9": "[quote=\"t0rajir0u\"]Try $ z \\text{ cosh } z$. But that doesn't help with the summation you gave at all.[/quote]\r\n\r\nLet's say \r\n\r\n$ \\sum_{n \\equal{} 0}^{\\infty} \\frac {z^{2n \\plus{} 1}}{(2n)!} \\equal{} zcosh(z)$\r\n\r\nCouldn't we then say\r\n\r\n$ \\sum^{\\infty}_{n \\equal{} 0} \\frac {(2n)!}{2^{2n \\plus{} 1}} \\sum^n_{k \\equal{} 0} \\frac {1} {k!(n \\minus{} k)!(2k \\plus{} 1)^x} \\equal{} \\sum^{\\infty}_{n \\equal{} 0} \\frac {1}{2cosh(2)} \\sum^n_{k \\equal{} 0} \\frac {1} {k!(n \\minus{} k)!(2k \\plus{} 1)^x}$\r\n\r\nAnd further more\r\n\r\n$ \\sum^{\\infty}_{n \\equal{} 0} \\frac {(2n)!}{2^{2n \\plus{} 1}} \\sum^n_{k \\equal{} 0} \\frac {1} {k!(n \\minus{} k)!(2k \\plus{} 1)^x} \\equal{} \\frac {\\sum^{\\infty}_{n \\equal{} 0} \\sum^n_{k \\equal{} 0} \\frac {1} {k!(n \\minus{} k)!(2k \\plus{} 1)^x}}{2cosh(2)}$\r\n\r\n?\r\n\r\nEdit: On second thought, I don't think it's \"legal\" to do that. Or is it okay since it's an infinite sum?", "Solution_10": "- First of all, information about a sum $ \\sum a_n$ doesn't tell you anything about $ \\sum \\frac{1}{a_n}$. You don't know anything about the form of $ \\sum \\frac{(2n)!}{2^{2n\\plus{}1}}$, and for good reason - this sum diverges horribly and is totally unrelated to $ \\text{cosh } 2$.\r\n\r\n- The original factorization is invalid. In your product you'll have terms where the $ (2n)!$ doesn't match up with the factorials on the bottom (which aren't correct in the factorization either).", "Solution_11": "[quote=\"t0rajir0u\"]- First of all, information about a sum $ \\sum a_n$ doesn't tell you anything about $ \\sum \\frac {1}{a_n}$. You don't know anything about the form of $ \\sum \\frac {(2n)!}{2^{2n \\plus{} 1}}$, and for good reason - this sum diverges horribly and is totally unrelated to $ \\text{cosh } 2$.\n\n- The original factorization is invalid. In your product you'll have terms where the $ (2n)!$ doesn't match up with the factorials on the bottom (which aren't correct in the factorization either).[/quote]\r\n\r\nDo you think summation by parts will get me anywhere?", "Solution_12": "The standard trick for manipulating double summations is to change the order of summation, which is essentially a generalization of summation by parts. What is the source of this problem? Knowing whether a solution exists is crucial to finding it (if it exists).", "Solution_13": "[quote=\"t0rajir0u\"]The standard trick for manipulating double summations is to change the order of summation, which is essentially a generalization of summation by parts. What is the source of this problem? Knowing whether a solution exists is crucial to finding it (if it exists).[/quote]\r\n\r\nWell actually I don't know if a solution exists. I do know that it diverges, but I wanted to simplify it to find the rate at which it diverges. I also know that the finite differences between terms n+1 and n, approach zero as n approaches infinity.", "Solution_14": "Finding asymptotics and bounds is a very different (and much simpler) question than computing a closed form. You should probably have asked that question to start with." } { "Tag": [], "Problem": "If n is a prime number, what is the smallest composite number produced by $n^{2}-n-1$?", "Solution_1": "[quote=\"Phoenix257\"]If n is a prime number, what is the smallest composite number produced by $n^{2}-n-1$?[/quote]\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=prime&t=134338", "Solution_2": "[hide]n=13, 169-13-1=155=composite\n\n[/hide]-jorian" } { "Tag": [ "AMC", "AIME", "geometry", "USA(J)MO", "USAMO", "geometric transformation", "reflection" ], "Problem": "Is it just me or AIME II was harder AIME I?", "Solution_1": "i was careless on AIME 1 but i thought both were close. They were both alot hareder than previous years though.", "Solution_2": "I would argue that AIME II is not necessarily HARDER mathematically. However, the way it is so cleverly designed gives loads and loads of room for counting errors. (What a bummer :-( )\r\n\r\nIn any case, I opened up the thing and it was like...\"When did I sign up for the verbal SAT?\"", "Solution_3": "it was a little wordy and i wouldve liked more pictures and geometry. if i remember correctly i dont think the first one had too many geometry problems either. :(", "Solution_4": "[quote=\"SirErnest\"]it was a little wordy and i wouldve liked more pictures and geometry. [/quote]\n\nA \"little\" wordy huh? #5 and #6 were each about the size of a paragraph in a history term paper.\n\n[quote] if i remember correctly i dont think the first one had too many geometry problems either. :([/quote]\r\n\r\nYou remember incorrectly--it was practically all geometry. ;)", "Solution_5": "Since when is more geometry better? I for one am unable to think in terms of similar triangles. Had to solve the paper folding problem with a couple of pythagoras equations. They should have more of those questions everybody hates, like #15 on the AIME1 and AIME2. Those are easier =P.", "Solution_6": "I thought AIME II was harder, but the two tests were roughly comparable in difficulty. However, I only took the II, so that opinion probably has a heavy tilt to it.", "Solution_7": "AIME 1 was easier, in my opinion...but i dunno, not by much, both were not that hard mathematically, the stupid errors definitely took their toll\r\n\r\nAIME1 had 8 geometry questions i believe...AIME2 only had like 3...", "Solution_8": "[quote=\"**********\"]\nAIME1 had 8 geometry questions i believe...AIME2 only had like 3...[/quote]\r\n\r\nlet's recount shall we...\r\n\r\n1, 3, 7, 11, 12, 13\r\n\r\n6 geometry problems... :-)", "Solution_9": "[quote=\"MithsApprentice\"][quote=\"**********\"]\nAIME1 had 8 geometry questions i believe...AIME2 only had like 3...[/quote]\n\nlet's recount shall we...\n\n1, 3, 7, 11, 12, 13\n\n6 geometry problems... :-)[/quote]\r\n\r\ni never claimed to be good at counting, hehe...3 isnt geometry...so ur down to 5...regardless, its still less than the aime1", "Solution_10": "[quote=\"**********\"][quote=\"MithsApprentice\"][quote=\"**********\"]\nAIME1 had 8 geometry questions i believe...AIME2 only had like 3...[/quote]\n\nlet's recount shall we...\n\n1, 3, 7, 11, 12, 13\n\n6 geometry problems... :-)[/quote]\n\ni never claimed to be good at counting, hehe...3 isnt geometry...so ur down to 5...regardless, its still less than the aime1[/quote]\r\n\r\n3 is too geometry!...it's 3-D geometry...but still geometry...", "Solution_11": "I thought the II was a lot harder than the I.", "Solution_12": "A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?", "Solution_13": "[quote=\"tokenadult\"]A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?[/quote]\r\n\r\nWhat do you mean? The problems for both tests are all available in the two appropriate math jam transcripts.", "Solution_14": "[quote=\"tokenadult\"]A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?[/quote]\r\n\r\nWe took one officially and the other unofficially for practice. :-)\r\n\r\nTest questions are released 24 hours after the test, meaning that you can post and discuss them and stuff.", "Solution_15": "[quote=\"Fierytycoon\"]What do you mean? The problems for both tests are all available in the two appropriate math jam transcripts.[/quote]\r\n\r\nWhat I mean is that the experience of taking the test \"for keeps\" and discussing the other test in a math jam is different. Taking the real test that counts on the real day is always a different experience from taking a practice test. I noticed when people were discussing the AMC 10/12 A and B tests the same phenomenon seemed to occur: most people thought the test that they took was the harder test, if they only took one. In that instance, people could take both the A and B versions of the same test in the same year, as a few AoPSers reported that they did, and I think the consensus is that the overall statistical result matches people's perceptions that the A tests were harder (especially for the 10), but no one is allowed to take both AIMEs for real in one year, so how could anyone match results between the two tests?", "Solution_16": "What I did was take the AIME A test just like I would be taking the actual thing. (My official score is the B score) Comparing that score with my actual score, I feel that the B is more difficult. I realize it is probably different taking it as a mock test (probably doing less checking and recalculations) than the actual one. However, I still believe that an almost-fair comparison can be taken that way.", "Solution_17": "[quote=\"MithsApprentice\"]What I did was take the AIME A test just like I would be taking the actual thing. (My official score is the B score) Comparing that score with my actual score, I feel that the B is more difficult. I realize it is probably different taking it as a mock test (probably doing less checking and recalculations) than the actual one. However, I still believe that an almost-fair comparison can be taken that way.[/quote]\r\n\r\nI got a dismal 3 on the AIME B (several stupid mistakes, most notably on 14), and I believe that I would have gotten around 8 on the AIME A. However, I'm Canadian, so this matters little, as I qualified for (and wrote) the CMO earlier this year. For that matter, I royally messed up the CMO, too. ::shrugs::", "Solution_18": "I took the AIME1 and skimmed the AIME2, and it definately looked harder. And I don't think I'm unique in that belief, so if the people who took the 1 and the people who took the 2 thought it was harder, it likely was. Granted, I'm lazy and hate reading long math problems. I put off question 5 of the AIME1 (or whatever that one was about the 2 day math competition) until there were <20 minutes at the very end because I didn't want to read the whole thing.", "Solution_19": "Well, I didn't take either officially, but I looked at most of the problems on each test, and I thought that the second AIME was significantly easier. There weren't too many questions on the second AIME that were just messy computations (more or less) the way there were on the first one. At least, I tend to think of the geometry problems around #10 on the first AIME as being messy computations that I mess up far too often. On the second AIME, only #12 was a messy computation. At least, of the ones I have seen so far, only #12 fits that description.", "Solution_20": "[quote=\"tokenadult\"]A given test taker can only take one AIME per year, so where is everyone getting the impression that one test is harder than the other this year? How would anyone know?[/quote]\r\n\r\nA lot of the people who officially took the II practiced with the I and have some idea of what their score would have been. Also, kids who have been doing this forever have established something of a pecking order which tends to be consistant. If kids A, B, and C consistantly do better than kids X, Y, Z, but on this particular occasion X,Y,Z take the first test and do better than A,B,C taking the second test, it suggests a difference in the tests.", "Solution_21": "I think describing which AIME was \"harder\" is a bit misleading. If you like number theory, the second AIME was easier.\r\n\r\nI personally felt like the \"middle\" problems were a bit harder on the second AIME while the \"hard\" problems were easier on the second AIME (mostly because I thought problem 15 from the first AIME was enormously time consuming). It wouldn't surprise me if students who are the kind who attempt 7 or 8 problems thought the second was harder while those who are the kind who attempt 10-15 thought the second was easier.\r\n\r\nI personally made several little errors as I worked through the first 2/3 of the second AIME. I misread at least two problems and made a couple of small algebra mistakes all of which I had to go back and fix. (This is a lesson to organize your work. I was not being very organized at all. From what I can see, organization makes even more of a difference now than back when I was taking the AIME.)\r\n\r\nOverall, I think the committee did a great job of balancing the difficulty for the kinds of students who would be fighting for 7 or 8 correct (where the AIME floor is likely to be most years).", "Solution_22": "I only took the AIME2 as a real test, but...\r\n\r\nI think it was easier to mess up on it.\r\nI solved 14 problems but made 4 stupid mistakes - most notably on #5, where i made the same stupid mistake a lot of people made...:(\r\nAnd #14, where i fluctuated between 100 and 109 only to realize 2 minutes after test was over that it was 108... However, #15 was easier than AIME 1.\r\n\r\nOn AIME1, problems 1-14 were all about the same difficulty, while 15 was a lot harder.", "Solution_23": "Ok, well that's not true. There was a big difference in difficulty among the first 14 problems, though compared to the 15, I suppose they seem relatively equal.", "Solution_24": "Well, compared to the 2003 test, for example, I think there is no doubt that #1 on 2004 was much harder and #14 was much easier, so the difference between 1 and 14 was significantly less than it has been in previous years.", "Solution_25": "[quote=\"MithsApprentice\"]What I did was take the AIME A test just like I would be taking the actual thing. (My official score is the B score) [/quote]\r\n\r\nI'm at least as confused as tokenadult about this. I thought you were *required* to take the AIME-I, unless there was an unavoidable conflict, such as your school being closed. So how could anybody take the AIME-I under real test conditions, but then take the AIME-II for their official score?", "Solution_26": "I assume by \"real test conditions\" he means that he put himself in a room for three hours with a pencil, paper, and the test, and acted as if he was taking the exam officially.", "Solution_27": "[quote=\"JBL\"]I assume by \"real test conditions\" he means that he put himself in a room for three hours with a pencil, paper, and the test, and acted as if he was taking the exam officially.[/quote]\r\n\r\ncorrect..everything except the actual answer sheet...\r\n\r\nIt was spring break week...", "Solution_28": "The AIME in general is getting easier compare to 90s. It is harder than those in 80s. However, there happened to be fewer pictures and more words. I misread #6 on AIME I and did not understand #14 at all. \r\n\r\nAlso, the easy problems are getting harder, the hard problems are getting easier.", "Solution_29": "[quote=\"MithsApprentice\"][quote=\"JBL\"]I assume by \"real test conditions\" he means that he put himself in a room for three hours with a pencil, paper, and the test, and acted as if he was taking the exam officially.[/quote]\n\ncorrect..everything except the actual answer sheet...\n\nIt was spring break week...[/quote]\r\n\r\nBut you're homeschooled, so you can take Spring Break whenever you want. You took Spring Break during AIME I week on purpose, so that you'd have more time to prepare for AIME II?", "Solution_30": "[quote=\"rcv\"] I'm at least as confused as tokenadult about this. I thought you were *required* to take the AIME-I, unless there was an unavoidable conflict, such as your school being closed. [/quote]\r\n\r\nWell, you didn't have to provide \"proof\" that the school was on spring break. The difference in price was an incentive to take the 1st date if you could. Also, planning to do the 1st date gives you a fall-back if someone gets sick or has an emergency.\r\n\r\nI had all 5 of my group's qualifiers take the 2nd date because just one of them had an AMC-sanctioned conflict and I didn't want to deal with exam arrangements twice. I feel a little bad about that now, because I think the II date was harder. I had 2 kids who had qualified for the USAMO last year who just missed this year after taking the II. One of them had qualified for each of the past 3 years. I can't help but think they would be in if I had had everyone take the I. Overall, I think it's better to do the first date if you can. Any benefit from extra studying time is trivial.", "Solution_31": "In last two years the AIME II is also harder than AIME I. I think they made the AIME II harder than AIME I. So taking AIME I is always a good choice.", "Solution_32": "I thought last year's AIME II was easier than the I. I havent taken a look at this year's II, but I don't think that its ever just \"easier\" or \"harder.\" Depends on what kind of problems you are better at.", "Solution_33": "[quote=\"JP\"]The AIME in general is getting easier compare to 90s. It is harder than those in 80s. However, there happened to be fewer pictures and more words. I misread #6 on AIME I and did not understand #14 at all. \n\nAlso, the easy problems are getting harder, the hard problems are getting easier.[/quote]\r\n\r\nI partially agree with this. The easier questions are slightly harder now and most of the problems past number 10 are a shade easier. The geometry seems easier to me, but that might be because I wasn't big on geometry and have learned more since.\r\n\r\nThe tests in the 90s varied a lot. If you look at 1994 or 1995 you might think they were hard. If you look at some of the others, they might look a little easier than of late.\r\n\r\nI think that there are now more long word problems that are easier to stumble over and more comp sci algorithmic problems. I think this reflects the times and makes the AIME a little different, though perhaps not particularly easier or harder.", "Solution_34": "By looking at the [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimestats.html]score distribution[/url], I would conclude that the AIME II was definitely harder. I only took the AIME I.", "Solution_35": "My theory about that was that since most people will take the AIME1, you'd probably have most of the super-geniuses take the first one and you might have a higher average test-taking ability on the first one, so that could skew the average score somewhat, but looking at the average AMC scores of the people taking AIME (which was [i]higher[/i] on the AIME2, with a significantly lower average AIME score), it does look like the second one was harder. Of course, even with statistics, this is still subjective, because people are better in different specializations. I got a 9 taking the AIME1 in practice (with a 12-13 after adding in questions I solved minus a stupid mistake) and got a 10 on the AIME2, with a 14 after disregarding stupid mistakes.", "Solution_36": "That's very interesting. I had just about the same thing. I didn't make any mistakes on the AIME I test which I took informally, though, and I would've then gotten a 13 if it were official :-(. But I definitely can't say I'm not satisfied with 11." } { "Tag": [ "calculus", "derivative" ], "Problem": "Hey, i have a simple calc question for an implicit differentiation problem. \r\n\r\nFind the second derivative of y^2=x^3. \r\n\r\nThe book says the answer is 3x/4y, but I keep getting a more complicated result. I thought this problem would be just another easy one, i guess i must just be making an algebra error of something.\r\n\r\nCan someone do the problem showing steps, thanks.", "Solution_1": "What answer did you get? What was your method?", "Solution_2": "i simply differentiated both sides with respect to x to solve for the first derivative and for this i got dy/dx=3x^2/2y.\r\n\r\nthen i tried to do the derivative of this with the quotient rule, but i got a long answer:\r\n\r\nd^2y/dx^2=((6x)(2y)-(3x^2)(2(dy/dx)))(2y)^2\r\n\r\nAfter simplifying this it didnt go down to 3x/4y. I am baffled bc/ implicit differentiation isn't supposed to be hard.\r\n\r\nplease help", "Solution_3": "Think about what the second derivative is: the derivative of the first derivative.\r\n\r\n[tex]y^2=x^3[/tex]\r\n[tex]\\frac{d}{dx}(y^2)=\\frac{d}{dx}(x^3)[/tex]\r\n[tex]\\frac{d}{dx}(y^2)=?[/tex]\r\nlet [tex]g(x)=y[/tex] and [tex]f(x)=x^2[/tex]. So, [tex]\\frac{d}{dx}f(g(x))=f'(g(x)) \\cdot g'(x)=2y \\cdot \\frac{dy}{dx}[/tex]\r\n\r\nSo now you have: [tex]2y \\cdot \\frac{dy}{dx} = 3x^2[/tex]\r\n\r\nSolve for [tex]\\frac{dy}{dx}[/tex] to get: [tex]\\frac{dy}{dx}= \\frac{3x^2}{2y}[/tex]\r\n\r\nNow take the derivative of both sides, taking the derivative of the first derivative which is dy/dx:\r\n\r\n[tex]\\frac{d}{dx}(\\frac{dy}{dx})=\\frac{d^2y}{dx^2}=\\frac{d}{dx}(\\frac{3x^2}{2y})[/tex]\r\n\r\nWhat is [tex]\\frac{d}{dx}(\\frac{3x^2}{2y})[/tex]?\r\n\r\nApply the quotient rule, let [tex]f(x)=3x^2[/tex] and [tex]g(x)=2y[/tex], since:\r\n\r\n[tex]\\frac{d}{dx}(\\frac{f(x)}{g(x)})=\\frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}[/tex]\r\n\r\nYou get:\r\n\r\n[tex]f'(x)=6x[/tex] and [tex]g'(x)=2 \\cdot \\frac{dy}{dx}= 2 \\cdot (\\frac{3x^2}{2y})[/tex].\r\n\r\nThus:\r\n[tex]\\displaystyle\\frac{d^2y}{dx^2}=\\frac{(6x)(2y)-(3x^2)(\\frac{3x^2}{y})}{4y^2}[/tex]\r\n\r\nSimplify the RHS to:\r\n\r\n[tex]\\displaystyle\\frac{d^2y}{dx^2}=\\frac{12xy^2-9x^4}{4y^3} [/tex]\r\n\r\nHmm, so either I am wrong or your textbook is, and I think we both know where the error most likely lies =).\r\n\r\nPS It's summer, why are you learning calculus? I ask b/c I am doing the same thing.", "Solution_4": "Bunch of errors, editing it.", "Solution_5": "ya i noticed quite a few, i dont wanna say anything until i give u a chance to fix it tho", "Solution_6": "i got ((12xy^2)-(9x^4))/4y^3", "Solution_7": "So basically all we differ on is the exponent of y on the bottom?", "Solution_8": "Okay, I realized I added wrong. I agree w/what you said.", "Solution_9": "ya, i got cubed because u multiply by y/y to get rid of the fraction in the numerator, which should make 4y^3, not 4y^5 in the denominator.\r\n\r\nthanks for the confirmation of my answer tho, im takin calc in a crappy community college over the summer, and i have im leavin before the class ends so it might as well be an independent study, and i guess in just not so confident of my calc skills yet. thanks", "Solution_10": "Hm, let me give it a try. Keep in mind that I hate calculus. ;)\r\n\r\n[tex]\\displaystyle 2y{{dy}\\over{dx}}=3x^2[/tex], so\r\n[tex]\\displaystyle{{dy}\\over{dx}}={{3x^2}\\over{2y}}={{3}\\over{2}}x^2y^{-1}[/tex]\r\n[tex]\\displaystyle{{d^2y}\\over{dx^2}}={{3}\\over{2}}\\left(2xy^{-1}+x^2(-1)y^{-2}{{dy}\\over{dx}}\\right)[/tex]; substitute in dy/dx to get\r\n[tex]\\displaystyle ={{3}\\over{2}}\\left({{2x}\\over{y}}-{{x^2}\\over{y^2}}\\left({{3x^2}\\over{2y}}\\right)}\\right)[/tex]; substitute in y2 to get\r\n[tex]\\displaystyle ={{3}\\over{2}}\\left({{2x}\\over{y}}-{{x^2}\\over{x^3}}\\left({{3x^2}\\over{2y}}\\right)}\\right)[/tex]\r\n[tex]\\displaystyle ={{3}\\over{2}}\\left({{2x}\\over{y}}-{{3x}\\over{2y}}\\right)}\\right)[/tex]\r\n[tex]\\displaystyle ={{3x}\\over{4y}}~.[/tex]", "Solution_11": "[quote=\"brian216\"]i got ((12xy^2)-(9x^4))/4y^3[/quote]\r\n\r\nWhy don't you sub in y2 = x3? That changes the expression into 3x/4y as required...", "Solution_12": "[tex]\\frac{d^2y}{dx^2}=\\frac{12xy^2-9x^4}{4y^3}[/tex]\r\n[tex]\\frac{d^2y}{dx^2}=\\frac{12x(y^2)-9x^4}{4y(y^2)}=\\frac{12x^4-9x^4}{4x^3y}=\\frac{3x}{4y}[/tex]\r\n\r\nWhy do you get rid of some y's but not others? Both answers are right...\r\n\r\nBrian I am taking calculus over the summer too, at University of Rhode Island... You class might not be that great since we did implicit differentiation on second day. Either that or my class is just super fast. Anyway, good luck! If you are trying to skip out of HS calc I would study a little on your own.", "Solution_13": "Well, I can change the original equation into y = x3/2, but then it wouldn't be implicit differentiation any more. So I just change the even powers of y in hope that it yields an expression that looks nice. Comparatively, 3x/4y is obviously simpler than (12xy2 - 9x4) / 4y3.\r\n\r\nOf course I can change the remaining y too, but then I get a fraction in my exponent.", "Solution_14": "ya, i knew i was just being stuid algebra-wise. i forgot about substituting from the original expression.\r\n\r\nthanks guys", "Solution_15": "You can do this without the quotient rule, as well:\r\n\r\n[tex]y^2 = x^3[/tex]\r\n[tex]2y\\frac{dy}{dx}=3x^2[/tex]\r\nThen we can differentiate that again, to get\r\n[tex]2\\left(\\frac{dy}{dx}\\right)^2 + 2y\\frac{d^2y}{dx^2}=6x[/tex]\r\n\r\nThen we can substitute in for [tex]\\frac{dy}{dx}[/tex] and solve that. It makes for the limiting of fractions, which usually makes life easier." } { "Tag": [ "LaTeX" ], "Problem": "What's the best way to incorporate Geometer Sketchpad images into latex files? I can't import .gs pictures directly, so is there some gs-->.png converter?", "Solution_1": "See Valentin's answer at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=339812#339812" } { "Tag": [ "linear algebra", "matrix", "function", "algebra", "rational function" ], "Problem": "use cramers rule to solve the following problem\r\n\r\n2x-y=4\r\n3x+y-z=10\r\ny+z=3", "Solution_1": "\\[ \\\\2x-y+0z=4\\\\3x+y-z=10\\\\0x+y+z=3 \\]\r\n$x=\\frac{\\left|\\begin{matrix}4&-1&0\\\\10&1&-1\\\\3&1&1\\end{matrix}\\right|}{\\Delta}$ $y=\\frac{\\left|\\begin{matrix}2&4&0\\\\3&10&-1\\\\0&3&1\\end{matrix}\\right|}{\\Delta}$ $z=\\frac{\\left|\\begin{matrix}2&-1&4\\\\3&1&10\\\\0&1&3\\end{matrix}\\right|}{\\Delta}$ $\\Delta=\\left|\\begin{matrix}2&-1&0\\\\3&1&-1\\\\0&1&1\\end{matrix}\\right|$\r\n\r\nAnd the result follows.\r\n\r\nMasoud Zargar", "Solution_2": "In practice, it is essentially always faster to use Gaussian elimination. What is Cramer's rule actually good for? Theoretical applications: the solution is a rational function of the various coefficients, which we could write down if we needed to.", "Solution_3": "Cramer's rule is handy to have when the size of the matrix is very small but the coefficients are not simple - I'm thinking in particular of cases in which the coefficients are algebraic expressions rather than numbers. But I think my definition of \"very small\" stops at $3\\times 3$ unless the matrix is quite sparse." } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "A nice one:\r\nFp={M s.t. M^p=In} where M are n*n real matrices.\r\nFind the accumulation points of Fp.", "Solution_1": "In $M_n(C)$ think matrices with eigenvalues of modulus 1", "Solution_2": "That's pretty clear, but we have to determine which ones. I'm not really sure (this is sort of new to me), but I think we can even say that the eigenvalues of the matrices in the adherence are roots of $x^p-1$. Is this true?", "Solution_3": "He did mention they were \"real matrices\", so it's $M_n(\\mathbb R)$." } { "Tag": [ "geometry" ], "Problem": "Four rectangular paper strips of length $10$ and width $1$ are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?\n[asy]\nsize(120);\ndraw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1));\ndraw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)--(4,6)--(4,5)--(5,5)--(5,4)--(4,4));[/asy]\n$ \\textbf{(A)}\\ 36 \\qquad \\textbf{(B)}\\ 40 \\qquad \\textbf{(C)}\\ 44 \\qquad \\textbf{(D)}\\ 98 \\qquad \\textbf{(E)}\\ 100 $", "Solution_1": "first of all, there are four parts that are overlapped by two strips. Each of them are 1 by 1 squares. so, we have some area, which is going to be $1+1+1+1 = 4$. \r\n\r\nNow, since we have taken out the overlapped areas, each of the strips are going to decrease its length by $2$, since $1$ has been taken out from each ends. therefore, the final area is $4((10-2)(1)) + 4 = 32 + 4 = 36$.\r\n\r\nthus, the answer is $A$.", "Solution_2": "[hide]\n$A$, since they overlap perpendicularly, the area lost is $1+1+1+1=4$\nSince each of the strips area is 10, $4(10)-4=40-4=36$.[/hide]" } { "Tag": [ "number theory", "greatest common divisor", "number theory proposed" ], "Problem": "aind all triple $(a,b,c)$ such that:\r\n$a^2b,b^2c,c^2a|a^3+b^3+c^3$", "Solution_1": "I'm wondering if $a,b,c$ are positive integers or not ?", "Solution_2": "yes $a,b,c$ are positive integers", "Solution_3": "Note that $(a, b, c)$ works iff $(ka, kb, kc)$ works, so consider only triples $(a, b, c)$ with their $GCD$ equal to $1$. We'll show that any such triple not only has $(a, b, c)=1$ but also $(a, b)=(b, c)=(a, c)=1$. Indeed, suppose two of them are divisible by some prime $p$, say $a, b$. Then $p \\mid a^2b \\mid a^3+b^3+c^3$, and since $p \\mid a, b$ we have $p \\mid c^3 \\Rightarrow p \\mid c$ contradicting $(a, b, c)=1$.\r\n\r\nThen $a, b, c$ are pairwise coprime. The conditions given imply $a^2, b^2, c^2 \\mid a^3+b^3+c^3$ which in turn implies $(abc)^2 \\mid a^3+b^3+c^3$ (since $a, b, c$ are pairwise coprime), and if this is satisfied then the original condition is too, so the problem becomes symmetric in $a, b, c$. Then WLOG assume $a \\leq b \\leq c$. Then $a^2b^2c^2 \\leq a^3+b^3+c^3 \\leq ac^2+bc^2+c^3$, so $a^2b^2 \\leq a+b+c$. Thus $c \\geq a^2b^2-a-b$.\r\n\r\nNow, remember that $c^2 \\mid a^3+b^3$.\r\n\r\nSuppose $a \\leq 2$. Then $c \\geq 4b^2-2b \\geq 2b^2$, so a number that is at least $4b^4$ divides $a^3+b^3$ which is at most $2b^3$, absurd. Then $a=1$. We now have $c \\geq b^2-b-1, c^2 \\mid b^3+1$. However, for $b \\geq 4$, it's easy to see that $c^2>b^3+1$. Then $b \\leq 3$.\r\n\r\nIf $b=1$ we have $c^2 \\mid c^3+2$, so $c^2 \\mid 2, c=1$, and $(1, 1, 1)$ works. If $b=2$ we have $4c^2 \\mid c^3+9$. Then $c^2 \\mid c^3+9$, so $c^2 \\mid 9$. Since $a \\leq b \\leq c$ we have $c \\leq 2$ so $c=3$, and $(1, 2, 3)$ works. If $b=3$ we have $9c^2 \\mid c^3+28$, so $c^2 \\mid c^3+28, c^2 \\mid 28$. The largest square dividing $28$ is $4$ so $c \\leq 20 [/b] ;\r\n[b]Prove that : [/b]\r\n $ \\int_{0}^{\\infty} {(A+xI)^{n}.e^{-x}.dx} = n! (I+\\sum_{k=1}^{n} \\frac {A^{k}}{k!})$\r\n[b]I is identity nxn matrix !!![/b]\r\n[b]\"..Do math , love math to love native place , love country..\"[/b]", "Solution_1": "$ \\int_{0}^{\\plus{}\\infty} {x^{n}e^{\\minus{}x}}\\equal{}n!$\r\n[b]I am sorry because I do'nt write gamma function[/b]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a, b, c$ positive reals such that $ 0 \\leq a, b, c \\leq 1$. Prove that\r\n\r\n$ (a^2b \\plus{} b^2c \\plus{} c^2a)(ab^2 \\plus{} bc^2 \\plus{} ca^2)\\geq (a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} 1)^2$", "Solution_1": "[quote=\"Ligouras\"]Let $ a, b, c$ positive reals such that $ 0 \\leq a, b, c \\leq 1$. Prove that\n\n$ (a^2b \\plus{} b^2c \\plus{} c^2a)(ab^2 \\plus{} bc^2 \\plus{} ca^2)\\geq (a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} 1)^2$[/quote]\r\nI am sorry, but your inequality is not true for $ a\\equal{}b\\equal{}c \\rightarrow 0.$\r\n\r\nBy the way, HAPPY NEW YEAR!" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "$AH$ is a height of $\\triangle ABC$ and $AD$ is a bisector of $\\angle BAC$. Let $E$ be a projection of $D$ onto $AC$. $AB$ and $EH$ intersect at point $N$. Let $M$ be the midpoint of $BC$. $MN$ is perpendicular to $AD$. Given $\\angle BAC=\\alpha$ find $\\angle ABC$.", "Solution_1": "The answer is (without a proof):\r\n[hide]\n$\\angle ABC = 90^{\\circ} - \\frac{\\alpha} {4}$\nor\n$\\angle ABC = \\angle BAC = 45^{\\circ}$\n[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose G is a group of order $ 5203$ which equals $ 11^2*43$.\r\n\r\nNeed to find values of $ n_{11}$ and $ n_{43}$ and find out if G is simple. \r\n\r\nThe factors of $ 5203$ are: $ 1, 11, 43, 121$ and $ 5203$.\r\n\r\nBy Sylow's theorems: $ n_p$ divides $ 5203$ and $ n_p \\equal{} 1$ (mod p), i.e. $ n_{11} \\equal{} 1$ (mod 11) and $ n_{43}\\equal{}1$ (mod 43)\r\n\r\nThen $ n_{11}$ and $ n_{43}$ both equal $ 1$. \r\n\r\nIs this correct or have I missed anything?\r\n\r\nThank you for your help!", "Solution_1": "All good. All groups of this order are abelian.\r\n\r\nSteve", "Solution_2": "[quote=\"SD5225272\"]All good. All groups of this order are abelian.\n\nSteve[/quote]\r\n\r\nThank you!" } { "Tag": [ "function", "geometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Find all positive integers $n$ such that $PA^{n}+PB^{n}+PC^{n}$ does not depend upon $P$.", "Solution_1": "When $P = A, B, C$ the function $f(n) = PA^{n}+PB^{n}+PC^{n}$ takes on the value $2s^{n}$ where $s$ is the side length of the triangle.\r\n\r\nOn the other hand, when $P$ is midway between any two of $A, B, C$ then $f(n)$ takes the value $(2+2^{n}) r^{n}$ where $r$ is the circumradius.\r\n\r\nClearly if we are to have $2s^{n}= (2+2^{n})r^{n}$ then $\\frac{s}{r}= \\sqrt{3}= \\sqrt[n]{1+2^{n-1}}$ which holds for $n = 2, 4$ and nothing else (I think).", "Solution_2": "t0rajir0u,\r\n\r\nI know $n=4$ also holds, value of $PA^{4}+PB^{4}+PC^{4}$ is $18R^{4}$. I will check your solution later :)", "Solution_3": "Surely Math Relec =Mathematical reflection? you should write clearly your source! :D This is a problem in Mathematical reflection", "Solution_4": "This problem is proposed by Oleg Mushkarov.\r\n\r\nI have known the reverse of this problem from Japanese university entrance exam as follows, so I have studied the very Our problem before.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=754146#754146]Here is a Link.[/url]", "Solution_5": "[quote=\"N.T.TUAN\"]t0rajir0u,\n\nI know $n=4$ also holds[/quote]\r\n\r\nSorry, I made an error. It's been corrected. :oops:", "Solution_6": "[quote=\"gemath\"]This is a problem in Mathematical reflection[/quote]\r\nYes, my good friend :D", "Solution_7": "Hint\r\nIf $f(P)=PA^{n}+PB^{n}+PC^{n}$ does not depend $P$ forall $P\\in (O)$ then $f(A)=f(M)\\Rightarrow n=2$ or $n=4$ with $M$ is midpoint of small arc $BC$\r\nIf $n=2 ,n=4$ then really $f(P)=const$ :)" } { "Tag": [ "factorial", "search", "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "2^n = a ! + b ! + c !", "Solution_1": "It has been posted recently. Search for it :wink:", "Solution_2": "[quote=\"sax\"]2^n = a ! + b ! + c ![/quote]\r\n\r\nWlog say $ a\\geq b\\geq c$ Then $ c! |2^n$ and so either $ c \\equal{} 1$, either $ c \\equal{} 2$\r\n\r\n1) $ c \\equal{} 1$\r\n$ n > 0$ and so $ a! \\plus{} b!$ must be odd and so $ b \\equal{} 1$ and we want $ 2^n \\equal{} a! \\plus{} 2$\r\nIf $ a\\geq 4$, RHS$ \\equal{} 2\\pmod 4$ and so $ n \\equal{} 1$ which is impossible.\r\nSo $ a < 4$ and the only solutions $ 8 \\equal{} 3! \\plus{} 1! \\plus{} 1!$ and $ 4 \\equal{} 2! \\plus{} 1! \\plus{} 1!$\r\n\r\n2) $ c \\equal{} 2$\r\nSo $ 2^{n \\minus{} 1} \\equal{} \\frac {a!}{2} \\plus{} \\frac {b!}{2} \\plus{} 1$\r\nIf $ b\\geq 4$, RHS$ \\equal{} 2\\pmod 4$ and so $ n\\leq 2$ which is impossible. So either $ b \\equal{} 2$, either $ b \\equal{} 3$\r\n\r\n2.1) $ b \\equal{} 2$\r\n$ 2^{n \\minus{} 1} \\equal{} \\frac {a!}{2} \\plus{} 2$\r\nIf $ a\\geq 6$, RHS$ \\equal{} 2\\pmod 4$ and so $ n\\leq 2$ which is impossible. So $ a\\in\\{2,3,4,5\\}$ and no solution\r\n\r\n2.2) $ b \\equal{} 3$\r\n$ 2^{n \\minus{} 1} \\equal{} \\frac {a!}{2} \\plus{} 4$\r\nIf $ a\\geq 6$, RHS$ \\equal{} 4\\pmod 8$ and so $ n\\leq 3$ which is impossible. So $ a\\in\\{3,4,5\\}$ and two solutions : $ a \\equal{} 4$ and $ a \\equal{} 5$\r\n\r\n3) Synthesis of solutions :\r\n$ 4 \\equal{} 2! \\plus{} 1! \\plus{} 1!$\r\n$ 8 \\equal{} 3! \\plus{} 1! \\plus{} 1!$\r\n$ 32 \\equal{} 4! \\plus{} 3! \\plus{} 2!$\r\n$ 128 \\equal{} 5! \\plus{} 3! \\plus{} 2!$", "Solution_3": "thanks, but why c! divides 2^n?", "Solution_4": "[quote=\"sax\"]thanks, but why c! divides 2^n?[/quote]\r\n\r\nuhhh, this is the first line of my demo. I hope you'll not encounter problems at each line :wink: \r\n\r\nSince $ a\\geq b\\geq c$, we know that $ c! | a!$ and $ c! | b!$ and so $ c! | a!\\plus{}b!\\plus{}c!$ and so $ c!|2^n$", "Solution_5": "ok, thanks :o" } { "Tag": [ "analytic geometry", "graphing lines", "slope", "trigonometry" ], "Problem": "Two slopes with the Name of $M$ and $m$ (With masses $M$ ane $m$)are on each other . There is No friction at all , Anywhere. ( Look at the Picture)\r\n\r\nWhat is the accerelation of $M$ at the time zero ?\r\n\r\n\r\n\r\n\r\nPS. Please note that the usual approach of decomposing the gravity force doesnt work.", "Solution_1": "[quote=\"lomos_lupin\"]Two slopes with the Name of $M$ and $m$ (With masses $M$ ane $m$)are on each other . There is No friction at all , Anywhere. ( Look at the Picture)\n\nWhat is the accerelation of $M$ at the time zero ?\n\n\n\n\nPS. Please note that the usual approach of decomposing the gravity force doesnt work.[/quote]\r\n\r\nHmm.. why not? Isn't the force just $mg \\cos t \\sin t$ where $t$ is the angle of the slope M, which means the acceleration is $\\frac{m}{M}\\sin 2t$?", "Solution_2": "Not exactly, object $m$ is accerelating itselt, the force it exerts on object $M$ doesnt decompose like that.\r\n\r\nThats would have worked if we had friction.", "Solution_3": "this problem is unsolvable because the center of mass of m needs to be known because it certain positions torque would be caused but at others it wouldnt. If there is no torque, the force on $m$ is $mg\\sin(\\theta)$ and the force on $M$ is $\\frac{mgsin(2\\theta)}{2}$ This can only be done if one knows the angle of the incline of $M$", "Solution_4": "the answer\r\n$a_{M}=\\frac{mgcosa sina}{M(1-\\frac{m}{M}sin^{2}a})$\r\nno tongue. :huh:" } { "Tag": [ "limit", "logarithms", "real analysis", "real analysis unsolved" ], "Problem": "Here is a limit, for those interested in challenges! :) \r\n$\\lim_{n\\to\\infty}\\sqrt[n^{2}]{\\binom{n+1}{1}\\binom{n+2}{2}\\binom{n+3}{3}... \\binom{2n}{n}}$.", "Solution_1": "$L=\\lim_{n\\to \\infty}\\frac{n(\\ln (n+1)-\\ln 1)+(n-1)(\\ln (n+2)-\\ln 2)+\\cdots+(\\ln (2n)-\\ln n)}{n^{2}}$\r\nBy $Stolz's$Theorem:\r\nWe have:\r\n$L=\\lim_{n\\to \\infty}\\frac{\\ln (2n+2)+2\\ln (2n+1)+\\cdots+2\\ln (n+2)-(n+1)\\ln(n+1)-\\ln (n)-\\cdots-\\ln 1}{2n+1}$\r\nThen by it again:\r\n$L=\\lim_{n\\to \\infty}\\frac{\\ln (2n+4)+2\\ln (2n+3)-(n+4)\\ln(n+2)+n\\ln(n+1)}{2}$\r\n$L=\\lim_{n\\to \\infty}\\frac{\\ln (2n+4)+2\\ln (2n+3)+\\ln(2n+2)-4\\ln(n+2)+\\ln (\\frac{n+1}{n+2})^{n}}{2}$\r\n$L=2\\ln 2+\\frac{1}{2e}$\r\nSo the answer is $e^{\\ln 2+\\frac{1}{2e}}=2e^{\\frac{1}{2e}}$\r\nHope I am right." } { "Tag": [ "LaTeX", "inequalities", "inequalities unsolved" ], "Problem": "prove that \r\n\r\n$(1-\\frac{1}{n^2})^n (1+\\frac{1}{n}) <1$\r\n\r\n\r\nfor $n$ integer( n#0)", "Solution_1": "if any one can rewrite this exercice with LaTEX\r\nTHNKS\r\n\r\n[color=red]Note by Megus: I rewrited it, as you asked but next time you won't be so lucky -try to learn LaTeX - it's surely not hard[/color]", "Solution_2": "It is just AM-GM: above inequality is equivalent to following:\r\n\r\n$\\sqrt[2n+1]{(n+1)^{n+1}(n-1)^n} < \\frac {n(2n+1) +1}{2n+1} < n$", "Solution_3": "As $n$ is integer, we have : $1-\\frac{1}{n^2}>=0$ and $(1+\\frac{1}{n})>=0$.\r\n\r\nWe can thus use the AM-GM :\r\n\r\n$\\sqrt [n+1] {(1-\\frac{1}{n^2})^n (1+\\frac{1}{n})}=< \\frac {n-n*\\frac{1}{n^2}+1+\\frac{1}{n}}{n+1}=1$\r\n\r\nThen $(1-\\frac{1}{n^2})^n (1+\\frac{1}{n})=< 1$\r\n\r\nBut we have the equality only if $1-\\frac{1}{n^2}=1+\\frac{1}{n}$ only if $n=-1$ and we check that this solution is false. Thus :\r\n\r\n$(1-\\frac{1}{n^2})^n (1+\\frac{1}{n})< 1$" } { "Tag": [ "pigeonhole principle" ], "Problem": "Consider $ 9$ positive integer numbers which are divisors of $ 30^{2009}$ .\r\nProve that one can find two of them such that their product a square .", "Solution_1": "$ 30^{2009}\\equal{}2^{2009}3^{2009}5^{2009}$. Pick a set of 9 divisors of this number. They will each be of the form $ 2^a3^b5^c$. By the Pigeonhole Principle, there will be a subset of 5 numbers such that, for every element, a has the same parity. Any two of these numbers multiplied will be of the form $ 2^{2x}3^y5^z$. Similarly, within this set, there will be a subset of 3 numbers such that, for every element, b has the same parity. Any two of these numbers multiplied will be of the form $ 2^{2x}3^{2y}5^z$. Finally, within this set, there must be a pair of numbers such that, for both numbers, c has the same parity. These two numbers multiplied will be of the form $ 2^{2x}3^{2y}5^{2z}$, which is a perfect square, as desired.", "Solution_2": "[hide=\"Slightly different but same concept\"]\nThere are 2 cases for the parity of a, for the parity of b, and for c, for a total of $ 2^3\\equal{}8$ so out of 9, two must be the same by the Pigeonhole Principle[/hide]" } { "Tag": [], "Problem": "in this game there are 3 rows one with 1 stick, one with 3, and the other with 5.\r\n[hide=\"for you visual people it looks like this\"]\n\n|\n|||\n|||||\n[/hide]\nyou can remove as many sticks from one row as you would like during your turn. the point is to avoid removing the last stick.\ndescribe a winning stategy if you go first.\n\n[hide=\"here is what i have\"]\nif you remove three stick from the row with 5 then it is always possible to win. I dont know how to prove this with math or anything i just figured it out with trial and error[/hide]", "Solution_1": "how many peoples play per game?", "Solution_2": "Two players. And this game is more commonly called Nim.", "Solution_3": "Yes I have heard of NIM too.\r\nI think it is also possible to win removing one stick from row 2. Again I did not do mathematical calculations, but checked the actual possibilities.\r\n\r\nIf he picks up the other two, you pickup the entire row 3 and win.\r\nIf he picks up one of them, you pick up four from row3 and win again.\r\nIf he picks up three from row3, you pickup row1 and win.\r\nIf he picks up four from row3, you pick up one from row2 and win.\r\nIf he picks up the entire row3, you pick up the rest of row2 and win.\r\nIf he picks up row1, you pick up three from row3 and win.\r\n\r\nand so on.", "Solution_4": "No no this is not done. It was wrong. Sorry! :blush:", "Solution_5": "we could play this game on aim or sometin\r\n\r\nill teach it to mi friends\r\n\r\nit sounds fun", "Solution_6": "I know a game like this. You count up to 100, whoever gets to 100 first wins. But, you can only count 4 numbers at a time (e.g. 1, 2, 3, 4).", "Solution_7": "[quote=\"pascal_1623\"]I know a game like this. You count up to 100, whoever gets to 100 first wins. But, you can only count 4 numbers at a time (e.g. 1, 2, 3, 4).[/quote]\r\n\r\n[hide=\"And the strategy...\"]\nIf the other person says x, you say 5-x. Since 100 is divisible by 5, the second player always wins.\n[/hide]", "Solution_8": "first player wins if you are allowed to count max n at a time where n + 1 is not divisible by 100.", "Solution_9": "[url]http://www.transience.com.au/pearl3.html[/url]\r\n\r\nTry this game, it's similar.", "Solution_10": "Ouch, is there the limit of the levels?, I reach level 11 (100%) and it really hurts to make a lot of LP (Losing position) (Four piles, I wonder if the highest level consist of 7 piles :maybe: )", "Solution_11": "Probably not, computers can generate stuff like that.\r\n\r\nHowever, I'm pretty sure there is some modeleing thing you can do.\r\n\r\nIf you have two piles left, you just have to make them equal, and you win." } { "Tag": [ "function", "limit", "real analysis", "real analysis solved" ], "Problem": "Let $f$ be a continious function from $\\mathbb{R}\\to\\mathbb{R}$ such that $|f(x)-f(y)|\\geq|x-y|$ for all $x,y$. Prove that the range of $f$ is all of $\\mathbb{R}$", "Solution_1": "$f$ is injective, so assume WLOG it's strictly increasing. if it has a real limit $\\ell$ when $x$ goes to $\\infty$, then for some $\\varepsilon>0,c>0$ we have $|f(x)-f(y)|<\\varepsilon$ whenever $x,y>c$, which is a contradiction. This means that $\\lim_{x\\to\\infty}f(x)=\\infty$. In the same way we show that $\\lim_{x\\to-\\infty}f(x)=-\\infty$.", "Solution_2": "what about the following pb:\r\nlet $K$ be a metric compact set.\r\nLet $f$ be a continuous function from $K$ to $K$ such that $d(f(x);f(y)) \\geq d(x,y)$ for all $x,y$ in $K$. Is $f$ a bijection ?", "Solution_3": "[quote=\"alekk\"]what about the following pb:\nlet $K$ be a metric compact set.\nLet $f$ be a continuous function from $K$ to $K$ such that $d(f(x);f(y)) \\geq d(x,y)$ for all $x,y$ in $K$. Is $f$ a bijection ?[/quote]\r\n\r\nYes, I think it does follow that $f$ is surjective (and, of course, it's also injective).\r\n\r\nLet $K_n=f^{(n)}(K)$. $(K_n)_n$ is a descending (i.e. $K_{n+1}\\subseteq K_n$) sequence of compact sets, so $\\cap_{n\\ge 0}K_n$ is non-void. Call that intersection $F$. $F$ is a compact subset of $K$ satisfying $f(F)=F$. Assume now that $F\\ne K$. There is a point $x\\in K\\setminus F$ s.t. $d(x,F)=t>0$. We have $d(f^{(n)}(x),f^{(n)}(F))=d(f^{(n)}(x),F)\\ge t,\\ \\forall n\\ge 0$. However, $f^{(n)}(x)\\in K_n,\\ \\forall n\\ge 0$. Let $F'$ be the set of points which have distance $\\ge t$ from $F$. $F'$ is a compact subset of $K$. From what we have above we see that $K_n\\cap F'\\ne\\emptyset,\\ \\forall n\\ge 0$. This means that $F\\cap F'\\ne\\emptyset\\ (*)$, which is absurd ($(*)$ follows from $F\\cap F'=\\bigcap_{n\\ge 0}(K_n\\cap F')$).\r\nIs it Ok? :?", "Solution_4": "Now, to finish this topic, prove that $f$ is actually an isometry, i.e., $d(f(x),f(y))=d(x,y)$ for all $x,y\\in K$. :)", "Solution_5": "There is a proof of this in post number $12$, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=isometry&t=18269]here[/url]. Is your solution different, fedja?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be positive real numbers *edited :wink: *.\r\n\r\nProve that\r\n\r\n$ \\sqrt {\\frac {\\left(\\sum \\frac 1 a \\right)\\cdot \\left(\\sum a\\right)}{2}}\\geq \\sum \\sqrt [4]{\\frac {a^2}{(a \\plus{} b)(a \\plus{} c)}}$", "Solution_1": "[quote=\"Georg-A.\"]Let $ a,b,c$ be positive real numbers such that $ abc \\equal{} 1$.\n\nProve that\n\n$ \\sqrt {\\frac {\\left(\\sum \\frac 1 a \\right)\\cdot \\left(\\sum a\\right)}{2}}\\geq \\sum \\sqrt [4]{\\frac {a^2}{(a \\plus{} b)(a \\plus{} c)}}$[/quote]\r\nThe condition $ abc\\equal{}1$ is irrelevant..\r\n$ LHS \\ge \\sqrt{\\frac{9}{2}}$ by cauchy-schwarz.\r\nAnd using $ \\sqrt{x}\\plus{}\\sqrt{y}\\plus{}\\sqrt{z} \\ge \\sqrt{3}\\sqrt{x\\plus{}y\\plus{}z}$ on the $ RHS$ we only have to prove:\r\n$ \\sqrt{\\frac{9}{2}} \\ge \\sqrt { 3\\sum_{cyc} \\sqrt{\\frac{a^2}{(a\\plus{}b)(a\\plus{}c)}}}$\r\nBut $ \\sum_{cyc} \\sqrt{\\frac{a^2}{(a\\plus{}b)(a\\plus{}c)}} \\equal{} \\sum_{cyc} \\sqrt{\\frac{a}{a\\plus{}b} \\cdot \\frac{a}{a\\plus{}c}} \\le \\sum_{cyc} \\frac{\\frac{a}{a\\plus{}b}\\plus{}\\frac{a}{a\\plus{}c}}{2} \\equal{} \\frac{3}{2}$, and we are done :)", "Solution_2": "[quote=\"Mathias_DK\"]\nThe condition $ abc \\equal{} 1$ is irrelevant..\n[/quote]\r\n\r\nOf course you`re right - i had changed the inequality not noticing that the condition becomes unnecessary :blush: \r\n\r\nAnyway, thanks for your solution :)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "there are N points in the plane such that the total number of pairwise distances of these N points is at most n. Prove that $ N <\\equal{} (n \\plus{} 1)^2$.", "Solution_1": "We shall have maximum number of pairwise distances if all the points on the plane are distinct. Hence we shall have $ N\\choose2$ distances. By the condition given, $ N\\choose 2$=n\r\nSo we have to prove $ N<\\equal{}$ $ N\\choose 2$ $ N\\choose 2$ + 1 + 2$ N\\choose 2$ and this I feel is trivial.", "Solution_2": "I don't come here too often, so I haven't been able to keep an eye. I hope you didn't post this before the deadline, mnrc_2.\r\n\r\nAssume $ N > (n\\plus{}1)^2$. Suppose $ N \\equal{} (n\\plus{}1)^2 \\plus{} 1$.\r\n\r\nConsider a point P out of those given, which is a \"boundary\" point, i.e. a line through P has all the other $ (n\\plus{}1)^2$ points on one side of it (some may be on the line itself).\r\n\r\nApply the pigeon hole principle to prove that at least n + 3 of these remaining points must lie at the same distance from P. Obviously, from the given and the assumed conditions, these points all lie on a semicircle with center P.\r\n\r\nIt is now easy to show that these last n + 3 points define at least n + 2 distinct distances among them, which is a contradiction.", "Solution_3": "The same reasoning works for $ n^2\\plus{}n$ instead of $ (n\\plus{}1)^2$.\r\n\r\nNote that a boundary point is just a vertex of the convex hull.\r\n\r\nPierre.", "Solution_4": "Yup, it is. Yes, and I forgot to mention the rather obvious fact that whenever you prove it for $ N \\equal{} (n\\plus{}1)^2 \\plus{}1$, you can obviously prove it for any N greater than that. That's VERY trivial.", "Solution_5": "But, as far as I know, $ n^2\\plus{}n$ is not greater than $ (n\\plus{}1)^2$ :wink: \r\n\r\nPierre.", "Solution_6": "Well, lol, I was referring to MY earlier solution, not yours :wink: \r\n\r\nBecause I had forgotten to extend my proof to cover all possible N's given in the problem, so ...\r\n\r\nYes, of course the reasoning holds true for any N greater than $ n^2 \\plus{} n$.\r\n\r\nRik", "Solution_7": "i don't understand any of the 2 proofs. please elaborate a bit further?", "Solution_8": "Ah, that sweet Erdos distance problem again! :). Alex Iosevich put a nice expository paper about it online: [url]http://www.math.missouri.edu/%7Eiosevich/erdoshs.pdf[/url]" } { "Tag": [ "geometry" ], "Problem": "A circle is inscribed in a triangle. If the area of the triangle is 15 units squared and the area of the circle is 25pi, what is the area of the part of the triangle that is not blocked by the circle? \r\n(I wish I had a diagram. :noo:)\r\n\r\nnote: there is a few typos,", "Solution_1": "How can you inscribe a circle with an area of $ 25\\pi$ into a triangle with an area of $ 15$? :spam: <-Exactly.", "Solution_2": "[quote=\"007math\"]How can you inscribe a circle with an area of $ 25\\pi$ into a triangle with an area of $ 15$? :spam: <-Exactly.[/quote]\r\n\r\nOops, typo. It's supposed to be triangle inscribed in circle!\r\nAnd what you're caculating is the part of the circle that is showing from behind the triangle. :noo: I hate typos", "Solution_3": "In that case, wouldn't it just be $ 25 \\pi \\minus{}15$?" } { "Tag": [ "geometry", "rectangle" ], "Problem": "The equiangular convex hexagon $ ABCDEF$ has $ AB \\equal{} 1$, $ BC \\equal{} 4$, $ CD \\equal{} 2$, and $ DE \\equal{} 4$. The area of the hexagon is\r\n\r\n$ \\textbf{(A)}\\ \\frac{15}{2}\\sqrt{3}\\qquad \r\n\\textbf{(B)}\\ 9\\sqrt{3}\\qquad \r\n\\textbf{(C)}\\ 16\\qquad \r\n\\textbf{(D)}\\ \\frac{39}{4}\\sqrt{3}\\qquad \r\n\\textbf{(E)}\\ \\frac{43}{4}\\sqrt{3}$", "Solution_1": "[hide=\"Click for solution\"]\nWe extend $ \\overline{FA}$ and $ \\overline{CB}$ to meet at $ X$, $ \\overline{BC}$ and $ \\overline{ED}$ to meet at $ Y$, and $ \\overline{DE}$ and $ \\overline{AF}$ to meet at $ Z$. The interior angles of the hexagon are $ 120^\\circ$, so the triangles $ \\triangle XYZ$, $ \\triangle ABX$, $ \\triangle CDY$, and $ \\triangle EFZ$ are equilateral. Since $ AB=1$, $ BX=1$. Since $ CD=2$, $ CY=2$. Thus, $ XY=7$ and $ YZ=7$. Since $ YD=2$ and $ DE=4$, $ EZ=1$. The area of the hexagon can thus be found by subtracting the areas of the three small triangles from the area of the large triangle:\n\n$ 7^2 \\left (\\frac{\\sqrt{3}}{4} \\right ) -1^2 \\left (\\frac{\\sqrt{3}}{4} \\right )-2^2 \\left ( \\frac{\\sqrt{3}}{4} \\right )-1^2 \\left (\\frac{\\sqrt{3}}{4} \\right )=\\frac{43\\sqrt{3}}{4}$, or answer choice $ \\boxed{\\textbf{(E)}}$.\n[/hide]", "Solution_2": "Or, you can inscribe the hexagon in a rectangle whose sides contain AB and DE, and the other two sides contain C and F. Let the corners of the rectangle be G,H,J, and K, where GEF,DHC,CJB, and AKF are 30-60-90 right triangles. Let GE=x, AK=y. Since GH=JK, we have\r\n$ x \\plus{} 2 \\equal{} y$\r\nAlso, $ GF \\equal{} \\sqrt {3} \\cdot x$ and $ FK \\equal{} \\sqrt {3} \\cdot y$, so since GK=HJ, we have\r\n$ x \\plus{} y \\equal{} 3$\r\nSolving, x=$ \\frac{1}{2}$ and AK=$ \\frac {5}{2}$. The area of the hexagon is the area of the rectangle minus the area of the four right triangles, which is $ \\frac {43\\sqrt {3}}{4}$" } { "Tag": [], "Problem": "I just looked at this forum, and I know I posted the 20,000th user topic, where in the world is it?????????? Why did 10,000th user replace it?", "Solution_1": "I haven't seen it but maybe it's because 10000th User's was posted a while ago? Does it even matter? Also, is it really necessary to make a topic over something stupid like this?", "Solution_2": "There is already a thread discussing the 20,000th user. hello, please address future questions like this to me via PM." } { "Tag": [ "inequalities", "logarithms", "inequalities solved" ], "Problem": "$x_1, x_2 \\cdots x_n \\in R^+$. Prove:\r\n\r\n\\[x_1^{x_1} \\cdot x_2^{x_2} \\cdots x_n^{x_n} \\geq (x_1 \\cdot x_2 \\cdots x_n) ^ {\\frac{x_1+x_2+\\cdots+x_n}{n}}\\]", "Solution_1": "I'm not sure,Maybe I can use Muirhead.", "Solution_2": "take log and chebyshev.", "Solution_3": "Oops. :blush: I can't use Muirhead now.", "Solution_4": "Well, I think we can use the most basic method... \r\nAssume $x_1 \\leq x_2 \\leq \\cdots \\leq x_n$. Then tranform the inequality into: $ (\\frac{x_i}{x_{i+1}})^{(x_i-x_{i+1})}$ or sth like that.... \r\nCan we prove it like that?", "Solution_5": "I am pretty sure we can, yes.\r\n\r\nAnyway, the Chebyshef is also very nice... suppose WLOG that the given numbers are in decreasing order, then by Chebyshef, since\r\n\r\n\\[\\ln{x_1} \\geq \\ln{x_2} \\geq ... \\geq \\ln{x_n}\\] we have\r\n\r\n\\[x_1 \\ln{x_1} + x_2 \\ln{x_2} + ... + x_n \\ln{x_n} \\geq \\frac{1}{n}\\left(x_1 + x_2 + ... + x_n\\right)\\left(\\ln{x_1} + \\ln{x_2} + ... + \\ln{x_n}\\right)\\]\r\n\r\nand hence \\[e^{\\textrm{RHS}} \\leq e^{\\textrm{LHS}}\\] from which the result follows.", "Solution_6": "Or Jensen. By the way this is a well-known inequality for sure. ;)" } { "Tag": [ "MATHCOUNTS", "percent", "number theory", "prime numbers" ], "Problem": "What percent of two-digit positive integer is prime?\r\n\r\nThis is just to get you thinking about memorizing all the prime numbers.", "Solution_1": "[hide]I think 23 1/3 percent)[/hide]", "Solution_2": "[hide]11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97. 99-10=89+1=90 21/90=7/30. 7/30*100=23 1/3. \n\n\n\nFootnote. It's really handy to know all the prime numbers under 200.[/hide]", "Solution_3": "It's even handier to know which numbers ``look prime'' but aren't: the two most important of those are probably 91 and 301.", "Solution_4": "Yup yup, correct, just to get you thinking...just watch out for those divisible by 3 or 7 and you should be fine for all prime under 100. 3, 7, 11 and 13 if under 200 :)" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "how many $\\alpha \\in [0,\\frac{\\pi}{2}]$ satisfied the equation $\\cos \\alpha \\cos 2\\alpha \\cos 4\\alpha=\\frac18$?", "Solution_1": "Take the graph .I think there are three roots of this equation.", "Solution_2": "The double-angle identity $\\sin 2x = 2 \\sin x \\cos x$ can help. Multiplying both sides of the original equation by $8 \\sin \\alpha$ gives \\[\\sin \\alpha (2 \\cos \\alpha) (2 \\cos 2\\alpha) (2 \\cos 4 \\alpha) = \\sin \\alpha \\, .\\] By applying the double-angle identity three times, the left-hand side is $\\sin 8 \\alpha$. So we get the equation \\[\\sin 8 \\alpha = \\sin \\alpha \\, .\\] From here, it should be smooth sailing. We have to remember that $\\alpha = 0$ is an extraneous solution.", "Solution_3": "[quote=\"Ravi B\"] \\[\\sin 8 \\alpha = \\sin \\alpha \\, .\\] [/quote]\r\nGive $7\\alpha =2\\pi k, k\\not=0$ or $9\\alpha =\\pi (2k+1)$. Therefore 5 solutions.", "Solution_4": "Basically, after you get sin(8a) = sin(a), you need to find the values in the interval such that 8a = a + 2n(pi)", "Solution_5": "Don't forget about the supplementary case $8 \\alpha = \\pi-\\alpha+2n\\pi$. That's why Rust wrote two equations.", "Solution_6": "[quote=\"Ravi B\"]Don't forget about the supplementary case $8 \\alpha = \\pi-\\alpha+2n\\pi$. That's why Rust wrote two equations.[/quote]\r\nThank you all for the nice proof. :)" } { "Tag": [ "integration", "real analysis", "real analysis unsolved" ], "Problem": "Consider:\r\n\r\n3(3 + x^2)(dx/dt) = 2((1 + x^2)^2)e^(-(t^2)).\r\n\r\nIf x(0) <= 1 show that there is a M > 0 st |x(t)| < M for t >=0.", "Solution_1": "[quote=\"joao1\"]Consider:\n\n3(3 + x^2)(dx/dt) = 2((1 + x^2)^2)e^(-(t^2)).\n\nIf x(0) <= 1 show that there is a M > 0 st |x(t)| < M for t >=0.[/quote]\r\n\r\nSolved the differential equation we got:\r\n$3arctgx+\\frac{3x}{2(1+x^{2})}=\\int_{0}^{t}{e^{-y^{2}}}dy+K$ so \r\ncouse $x(0)\\le 1$ we have $3arctgx+\\frac{3x}{2(1+x^{2})}\\le \\int_{0}^{t}{e^{-y^{2}}}dy+\\frac{3(\\pi+1)}{4}\\le \\frac{3\\pi+2\\sqrt{\\pi}+3}{4}$ where from we got that $|x(t)|1$.\r\nYour answer looks like the triangle should be right, so $ 1\\plus{}x^2\\equal{}x^4\\implies (x^2)^2\\minus{}x^2\\minus{}1\\equal{}0\\implies x^2\\equal{}\\frac{1\\plus{}\\sqrt5}2$, but the square of your answer is $ \\frac{3\\minus{}\\sqrt5}2$, which isn't exactly the same...", "Solution_2": "From the conditions of the problem, the 2 equal sides are smallest in one, and biggest in other triangle, call them a and b, and call the other sides c and d.\r\n\r\nFirst: c,a,b\r\nSecond: a,b,d sorted by length\r\n\r\n$ \\frac bd\\equal{}\\frac ab\\equal{}\\frac ca$\r\n\r\n$ d\\equal{}\\frac{b^2}{a},c\\equal{}\\frac{a^2}{b}$\r\n\r\nThe ratio is $ \\frac ba$, and it needn't to be your fraction.", "Solution_3": "I know I've seen this problem somewhere, but I can't seem to remember where...\r\n\r\nThe problem should be 'Prove that the ratio of the sides that correspond lies in between $ \\frac {\\sqrt {5} \\minus{} 1}{2}$' and $ \\frac {\\sqrt {5} \\plus{} 1}{2}$.", "Solution_4": "Well, let's try it now. If a=b, it's trivial. \r\nc+a>b\r\n$ \\frac{a^2}{b}\\plus{}a>b$\r\n$ a^2\\plus{}ab>b^2$\r\n\r\nSimilarly $ a\\plus{}b>\\frac{b^2}{a}$\r\n\r\n$ a^2\\plus{}ab>b^2$, same as above\r\n\r\nNow solve the quadratics for equality case, and it immediately follows that interval of the solutions is between the solutions of the quadratics.", "Solution_5": "It's from Canada, 2007:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229080[/url]", "Solution_6": "Sorry, I forgot to copy the second part.", "Solution_7": "I don't think one single inequality will suffice. It can be shown through case work that after normalizing the smaller triangle to side length of 1 for the smallest side, the triangles must be $ 1,x,x^2$ and $ x,x^2,x^3$ for $ x > 1$.\r\nThen through the inequalities $ 1 \\plus{} x > x^2,x^2 \\plus{} x > 1,x^2 \\plus{} 1 > x$ we have\r\n$ x^2 \\minus{} x \\minus{} 1 < 0\\implies x < \\frac {1 \\plus{} \\sqrt5}2$\r\n$ x^2 \\plus{} x \\plus{} 1 > 0\\implies x > \\frac { \\minus{} 1 \\plus{} \\sqrt5}2$\r\n$ x^2 \\minus{} x \\plus{} 1 > x^2 \\minus{} 2x \\plus{} 1 \\equal{} (x \\minus{} 1)^2 > 0$ which is always true.\r\n\r\nI posted the following in another thread (it's a trig \"marathon\") but no one responded\r\n\r\nHere's a quick trig identity I made up to verify\r\n\r\n$ (\\sin x \\plus{} \\sin y)(\\sin x \\minus{} \\sin y) \\equal{} \\sin(x \\plus{} y)\\sin(x \\minus{} y)$\r\n\r\nBy the way this problem was already solved, but someone might feel free to find a synthetic geometric proof for it:\r\n\r\nSquares $ ABMN$ and $ BCPQ$ are constructed externally on the sides $ AB$ and $ BC$ of the triangle $ ABC$. If $ BD$ is a median in $ \\triangle ABC$, prove that its extension is a height in $ \\triangle BMQ$", "Solution_8": "The sine problem:\r\n\r\nLet $ \\sin x\\equal{}a,\\cos x\\equal{}b,\\sin y\\equal{}c,\\cos y\\equal{}d$\r\n\r\n$ \\text{LHS}\\equal{}a^2\\minus{}c^2$\r\n\r\n$ \\text{RHS}\\equal{}(ad\\plus{}bc)(ad\\minus{}bc)\\equal{}a^2d^2\\minus{}b^2c^2\\equal{}a^2(1\\minus{}c^2)\\minus{}(1\\minus{}a^2)c^2\\equal{}a^2\\minus{}c^2$\r\n\r\nNow the 2 sides are equal. \r\n\r\nThe second is well known, search for it.", "Solution_9": "Hehe I made up that trig identity just to trick the newbies into thinking $ \\sin(x\\plus{}y)\\equal{}\\sin x\\plus{}\\sin y$ :P \r\nBy the way I tried to search using the keywords \"square side median construct externally triangle\" but the search didn't return anything... so Bugi could you just post your solution (using synthetic geometry)?\r\nAlso feel free to post the next problem :!:", "Solution_10": "Nothing?\r\nhttp://www.mathlinks.ro/viewtopic.php?search_id=1921350132&t=116424" } { "Tag": [ "inequalities", "geometry", "3D geometry", "calculus", "derivative", "symmetry", "algebra" ], "Problem": "What about this pb?\r\na,b,c positive reals less than 1.\r\nprove: a2+b2+c2 <= a2b+b2c+c2a+1\r\ni have a nice sol (::)", "Solution_1": "is the inequality like so:\r\na^2+b^2+c^2<=a^2b+b^2c+c^2a+1?", "Solution_2": "yes", "Solution_3": "a^2 (1-b)+ b^2(1-c) +c^2(1-a) <=a(1-b) +b(1-c) +c(1-a) = a+b+c -ab-bc-ac (1)\r\nBut (1-a)(1-b)(1-c)>=0, then we get 1>=abc+a+b+c-ab-bc-ca>= a+b+c -ab-bc-ac (2)\r\nCombining (1) and (2),we have: a^2+b^2+c^2<=a^2b+b^2c+c^2a+1\r\n :D \r\nEasy!!!", "Solution_4": "or you can notice that\r\nf(a,b,c)=a^2 (1-b)+ b^2(1-c) +c^2(1-a)\r\nis convex in each variable and we are done", "Solution_5": "I found a nice solution by considering a, b, c segments on the sides of a unit cube, so, for example, a^2(1-b) is a volume, but it would be really hard for me to explain it, because I have no pictures.", "Solution_6": "Define f(a,b,c) = a^2(1-b) + b^2(1-c) + c^2(1-a)\r\nThe problem is equivalent to asking us to prove that f(a,b,c) <= 1\r\nfor positive reals all less than 1. I'll prove the slightly \r\nstronger case where a,b,c <= 1.\r\nTaking partial derivatives:\r\ndf/da = 2a(1-b) - c2\r\nd2f/da2 = 2(1-b) > 0 (except when b = 1, when it equals 0)\r\nBy cyclic symmetry, we see that d2f/db2 > 0 and d2f/dc2 >0 all \r\nexcept possibly at edge points in the domain\r\nLet a1,b1,c1, be points where f achieves a max.\r\nSuppose one of these three points occurs at a value other than 0 or 1.\r\nSuppose it's a1\r\nThen d2f/da2(a1,b1,c1) > 0\r\nThat means that either (of both) of f(0,b1,c1) and f(1,b1,c1) is \r\ngreater than f(a1,b1,c1) Contradiction. Analogous contradctions \r\nwill be had if the point not = 0 or 1 is b1 or c1.\r\nTherefore, a1 = 0 or 1, b1= 0 or 1, and c1 = 0 or 1\r\nSo there are eight possibilities for (a1,b1,c1):\r\n(0,0,0)\r\n(0,0,1)\r\n(0,1,0)\r\n(0,1,1)\r\n(1,0,0)\r\n(1,0,1)\r\n(1,1,0)\r\n(1,1,1)\r\nPlug each of them into the formula for f, and you'll see that the \r\nmax of f is indeed <= 1.\r\n\r\n\r\ncheers!! :D" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that : If $ x,y,z,t \\geq 0$ then :\r\n1) $ 3(x^2\\plus{}y^2\\plus{}z^2\\plus{}t^2)\\plus{}4\\sqrt{xyzt} \\geq (x\\plus{}y\\plus{}z\\plus{}t)^2$ \r\n2) $ \\frac{x^3}{(x\\plus{}y)^3}\\plus{}\\frac{y^3}{(y\\plus{}z)^3}\\plus{}\\frac{z^3}{(z\\plus{}x)^3} \\geq \\frac{3}{8}$", "Solution_1": "[quote=\"QuyBac\"]Prove that : If $ x,y,z,t \\geq 0$ then :\n1) $ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \n2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\geq \\frac {3}{8}$[/quote]\r\n\r\n$ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \r\n\r\nsubstitution $ x\\equal{}x^2,y\\equal{}y^2,z\\equal{}z^2,t\\equal{}t^2$\r\n\r\nwe only prove\r\n\r\n$ 3(x^4 \\plus{} y^4 \\plus{} z^4 \\plus{} t^4) \\plus{} 4{xyzt} \\minus{}(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2)^2\\geq 0....(**)$ \r\n\r\nassume $ x\\geq y \\geq z \\geq t$\r\n\r\nsubstitution $ x\\equal{}a\\plus{}b\\plus{}c\\plus{}t,y\\equal{}b\\plus{}c\\plus{}t,z\\equal{}c\\plus{}t$ $ (a,b,c\\geq 0)$\r\n\r\n$ (**) \\Longleftrightarrow (8b^2\\plus{}6c^2\\plus{}8ab\\plus{}4ac\\plus{}8bc\\plus{}6a^2)t^2\\plus{}(4c^3\\plus{}16a^2c$ $ \\plus{}8b^3\\plus{}4ac^2\\plus{}8a^3\\plus{}20abc\\plus{}12ab^2\\plus{}20b^2c\\plus{}20a^2b\\plus{}8bc^2)t$\r\n\r\n$ \\plus{}8a^2c^2\\plus{}8a^3b\\plus{}2b^4\\plus{}20a^2bc\\plus{}8b^3c\\plus{}4ab^3\\plus{}8b^2c^2\\plus{}2a^4$ $ \\plus{}12ab^2c\\plus{}8abc^2\\plus{}8a^3c\\plus{}10a^2b^2\\geq 0$\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\equal{} \\sum \\frac {x^4}{x(x \\plus{} y)^3}$ $ \\geq \\frac {(x^2 \\plus{} y^2 \\plus{} z^2)^2}{x(x \\plus{} y)^3 \\plus{} y(y \\plus{} z)^3 \\plus{} z(z \\plus{} x)^3}$\r\n\r\nwe only prove\r\n\r\n$ \\frac {(x^2 \\plus{} y^2 \\plus{} z^2)^2}{x(x \\plus{} y)^3 \\plus{} y(y \\plus{} z)^3 \\plus{} z(z \\plus{} x)^3}\\geq \\frac {3}{8}......(*)$\r\n\r\n\r\nassume $ x \\equal{} max{(x,y,z)}$\r\n\r\n$ (*)\\Longleftrightarrow (5x^2 \\plus{} ( \\minus{} 4y \\plus{} 2z)x \\plus{} 3y^2 \\plus{} 9z^2 \\minus{} 7yz)(x \\minus{} y)(x \\minus{} z) \\plus{}$ $ (5y^2 \\minus{} 2yz \\plus{} 5z^2)(y \\minus{} z)^2\\geq 0$\r\n\r\neasy prove\r\n\r\n$ 5x^2 \\plus{} ( \\minus{} 4y \\plus{} 2z)x \\plus{} 3y^2 \\plus{} 9z^2 \\minus{} 7yz\\geq 0$\r\n\r\n$ 5y^2 \\minus{} 2yz \\plus{} 5z^2\\geq 0$", "Solution_2": "[quote=\"QuyBac\"]Prove that : If $ x,y,z,t \\geq 0$ then :\n1) $ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \n[/quote]\n\nWe assume that $ x \\ge y \\ge z \\ge t$\n$ f(x;y;z;t): \\equal{} 3({x^2} \\plus{} {y^2} \\plus{} {z^2} \\plus{} {t^2}) \\plus{} 4\\sqrt {xyzt} \\minus{} {(x \\plus{} y \\plus{} z \\plus{} t)^2}$\nSo\n$ f(x;y;z;t) \\minus{} f\\left( {\\sqrt {xy} ;\\sqrt {xy} ;z;t} \\right) \\equal{} 2{\\left( {\\sqrt x \\minus{} \\sqrt y } \\right)^2}\\left( {x \\plus{} y \\minus{} z \\minus{} t \\plus{} 2\\sqrt {xy} } \\right) \\ge 0$\n$ \\Rightarrow f(x;y;z;t) \\ge f\\left( {\\sqrt {xy} ;\\sqrt {xy} ;z;t} \\right)$\nBy S.M.V theorem, we must to prove that \n$ f(a;a;a;t) \\ge 0$\n$ \\Leftrightarrow 3\\left( {9{a^2} \\plus{} {t^2}} \\right) \\plus{} 4\\sqrt {{a^3}t} \\ge {\\left( {3a \\plus{} t} \\right)^2}$\nEasy to prove\n\n[quote]2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\geq \\frac {3}{8}$[/quote]\r\nThis ineq can be proved by Cauchy-Schwarz and Schur :lol: :lol:", "Solution_3": "[quote=\"QuyBac\"]Prove that : If $ x,y,z,t \\geq 0$ then :\n1) $ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \n2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\geq \\frac {3}{8}$[/quote]\r\n\r\n$ \\sum \\frac {a^3}{(c \\plus{} b)^3} \\ge \\frac {(\\sum \\frac {a}{b \\plus{} c})^3}{9} \\ge^{Nessbit}\\frac {(\\frac {3}{2})^3}{9} \\equal{} \\frac {3}{8}$", "Solution_4": "[quote=\"bigbang195\"][quote=\"QuyBac\"]Prove that : If $ x,y,z,t \\geq 0$ then :\n1) $ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \n2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\geq \\frac {3}{8}$[/quote]\n\n$ \\sum \\frac {a^3}{(c \\plus{} b)^3} \\ge \\frac {(\\sum \\frac {a}{b \\plus{} c})^3}{9} \\ge^{Nessbit}\\frac {(\\frac {3}{2})^3}{9} \\equal{} \\frac {3}{8}$[/quote]\r\nSorry but your solution is wrong, since you proved $ \\sum_{cyc}\\frac{a^3}{(b\\plus{}c)^3}\\geq \\frac 38$ which is weaker than the original inequality.\r\n\r\nHowever, I think that the second can also be proved using Mixing Variables method, but I don't have a proof yet.", "Solution_5": "[quote=\"Potla\"][quote=\"bigbang195\"][quote=\"QuyBac\"]Prove that : If $ x,y,z,t \\geq 0$ then :\n1) $ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \n2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\geq \\frac {3}{8}$[/quote]\n\n$ \\sum \\frac {a^3}{(c \\plus{} b)^3} \\ge \\frac {(\\sum \\frac {a}{b \\plus{} c})^3}{9} \\ge^{Nessbit}\\frac {(\\frac {3}{2})^3}{9} \\equal{} \\frac {3}{8}$[/quote]\nSorry but your solution is wrong, since you proved $ \\sum_{cyc}\\frac {a^3}{(b \\plus{} c)^3}\\geq \\frac 38$ which is weaker than the original inequality.\n\nHowever, I think that the second can also be proved using Mixing Variables method, but I don't have a proof yet.[/quote]\r\n\r\n$ \\sum \\frac {a}{b \\plus{} c} \\ge \\frac{3}{2}$ this inequality is right !", "Solution_6": "[quote=\"QuyBac\"]Prove that : If $ x,y,z,t \\geq 0$ then :\n1) $ 3(x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} t^2) \\plus{} 4\\sqrt {xyzt} \\geq (x \\plus{} y \\plus{} z \\plus{} t)^2$ \n2) $ \\frac {x^3}{(x \\plus{} y)^3} \\plus{} \\frac {y^3}{(y \\plus{} z)^3} \\plus{} \\frac {z^3}{(z \\plus{} x)^3} \\geq \\frac {3}{8}$[/quote]\r\n\r\nFor (1), it is Turkevici's Inequality. And it is [b]Problem 6.34 - page 301[/b] in new book (where it is presented with three nice and simple proofs):\r\n\r\nVasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009." } { "Tag": [ "ARML", "induction", "function" ], "Problem": "You want to measure every possible weight from 1 to 100 with 5 different weights using a scale. If you can place weights on either side, what weights should you choose?", "Solution_1": "[hide=\"Solution\"] Balanced ternary notation dictates the choices $1, 3, 9, 27, 81$. For more information:\n\nhttp://en.wikipedia.org/wiki/Balanced_ternary [/hide]", "Solution_2": "[hide]1, 3, 9, 27, 81\nReason: Any number from 1 to 100 can be expressed as a five-digit base-3 number with each digit as a -1, 0, or 1.[/hide]", "Solution_3": "This reminds me of the ARML power round...", "Solution_4": "That's exactly what I was thinking - ARML power", "Solution_5": "This is such a well-known problem that I'm surprised it was even on the ARML this year. However, you should still prove that powers of 3 work, and nothing else works (is balanced terniary a citable fact?).", "Solution_6": "How do you prove all numbers can be expressed in balanced ternary? How many balanced-trits does it take to represent a number n?", "Solution_7": "The question is equivalent to the following: How would you convert an unbalanced ternary number into a balanced one? This is quite easy to answer.\r\n\r\nMasoud Zargar", "Solution_8": "[quote=\"JavaMan\"]How do you prove all numbers can be expressed in balanced ternary?[/quote]\r\n\r\nInduction is a good way to go.", "Solution_9": "[quote=\"boxedexe\"]The question is equivalent to the following: How would you convert an unbalanced ternary number into a balanced one? This is quite easy to answer.\n\nMasoud Zargar[/quote]\r\n\r\nWell this is easy. you can find unique representation if the division theorem holds (so you just repeatedly apply it and you get something unique, and with some induction on the way).\r\n\r\nSo everything boils down to showing that division theorem holds for balance ternary. But note that the only difference between balance ternary from the regular one is that instead of writing:\r\n\r\n$a = 3b+2$\r\n\r\nyou write:\r\n\r\n$a = 3(b+1)-1$\r\n\r\nSo there is a bijection for cases of division theorem between regular and balanced ternary. So since we have unique $a$ and $b$ in the regular version, we must also have unique $a$ and $b$ in the balanced version. Therefore every integer is representable and unique in the balanced ternary system.", "Solution_10": "[quote=\"boxedexe\"]The question is equivalent to the following: How would you convert an unbalanced ternary number into a balanced one? This is quite easy to answer.[/quote]\r\n\r\nAs I wrote in the Power Round, this has a nice interpretation in terms of generating functions. The generating function for the positive integers in base three is\r\n\r\n$(1+x+x^{2})(1+x^{3}+x^{6})(1+x^{9}+x^{18})... = 1+x+x^{2}+x^{3}+...$\r\n\r\nWhich illustrates that all naturals have a unique ternary representation (this expansion is easily proven by multiplying by $1-x$). Dividing each factor by its middle term,\r\n\r\n$(x^{-1}+1+x^{1})(x^{-3}+1+x^{3})(x^{-9}+1+x^{9})... = ...+x^{-2}+x^{-1}+1+x+x^{2}+x^{3}+...$\r\n\r\nWhich illustrates that all integers have a unique balanced ternary representation. ;)" } { "Tag": [ "inequalities", "integration", "inequalities unsolved" ], "Problem": "Let $ a,b,c>0$ and $ \\sum \\frac{1}{a}\\equal{}1$. Prove that\r\n\\[ \\sum \\frac{a}{b(b\\plus{}1)^2}\\geq \\frac{243}{(a\\plus{}b\\plus{}c)(a\\plus{}b\\plus{}c\\plus{}3)^2}.\\]", "Solution_1": "by Minkowski's ineq we get:\r\n$ (a \\plus{} b \\plus{} c)(a \\plus{} b \\plus{} c \\plus{} 3)^2 \\ge (\\sum_{}^{} \\sqrt [3]{a(a \\plus{} 1)^2})^3$\r\nso it's enough to prove:\r\n$ (\\sum_{}^{} \\frac {a}{b(b \\plus{} 1)^2}) (\\sum_{}^{} \\sqrt [3]{b(b \\plus{} 1)^2})^3 \\ge 3^5$\r\nor \r\n$ (\\sum_{}^{} \\frac {a}{b(b \\plus{} 1)^2})^ \\frac {1}{4} (\\sum_{}^{} \\sqrt [3]{b(b \\plus{} 1)^2})^\\frac {3}{4} \\ge 3^\\frac {5}{4}$\r\nusing Holder and inequality between power means we have:\r\n$ L \\ge \\sum_{}^{} a^ \\frac {1}{4} \\ge 3 ( \\frac {3}{\\sum_{}^{} \\frac {1}{a} })^ \\frac {1}{4} \\equal{} 3^ \\frac {5}{4}$\r\ndone :lol:", "Solution_2": "Another way is to notice that LHS is\r\n\\[ 2\\int_{0}^1 (\\sum \\frac {ax}{(x \\plus{} b)^3})dx.\\]" } { "Tag": [], "Problem": "There's a cool forum up that's solely dedicated to college and graduate school. You can ask your questions about the SAT, ACT, AP tests, GRE, etc. If you have any questions about this stuff, then you should join today!\r\n\r\n[url]http://s215234424.onlinehome.us/index.php[/url]", "Solution_1": "There's also CollegeConfidential, though that place is unbelievably neurotic, and I tend to avoid it at all costs.", "Solution_2": "Well, this forum is relatively new (I think) and there's not very many people. And the administrator is very friendly and will help you and answer any of your questions." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "This isn't really a problem from a book or anything, just something that I can't figure out. Suppose we have two disjoint compact sets D and K in R^n. How can I show that the distance between D and K, that is the infimum of $|x-y|$ taken over all x in D and all y in K, is strictly larger than zero?", "Solution_1": "$d: K \\to \\mathbb{R}^+$, the distance from a point $x \\in K$ to $D$, is a continuous function on the compact $K$. So it reaches its minimum to a point $x_0 \\in K$, and of course $dist(x_0,D)>0$ because $D$ is closed and $x_0$ does not belong to $D$.", "Solution_2": "[quote=\"Kalle\"]This isn't really a problem from a book or anything, just something that I can't figure out. Suppose we have two disjoint compact sets D and K in R^n. How can I show that the distance between D and K, that is the infimum of $|x-y|$ taken over all x in D and all y in K, is strictly larger than zero?[/quote]One simple way could be like this. Form the product $K \\times D \\subseteq \\mathbb{R}^n \\times \\mathbb{R}^n$. The product space is compact, and the map $K \\times D \\ni (x,y) \\mapsto |x-y| \\in \\mathbb{R^+}$ is continuous. Hence the image of that map is compact and in particular closed. So if the infimum of $|x-y|$ were zero, there'd exists $x' \\in K$, $y' \\in D$ w. $|x' - y'| = 0$, contradicting $K \\cap D = \\emptyset$.", "Solution_3": "Thanks! I get it now :)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ x_1,x_2,...,x_n$ such that\r\n$ 0 < x_1\\leq x_2\\leq x_3\\leq...\\leq x_n$\r\n and $ P \\equal{} x_2^k(x_1 \\minus{} x_3) \\plus{} x_3^k(x_2 \\minus{} x_4) \\plus{} ..... \\plus{} x_1^k(x_n \\minus{} x_2)$\r\nProve that\r\n$ P\\geq 0$ if $ k > 1$\r\n$ P\\leq 0$ if $ 0 < k < 1$", "Solution_1": "Rewrite the expression as:\r\n\\[ (x_{n\\minus{}1}\\minus{}x_1)x_{n\\minus{}1}^k \\plus{} (x_n\\minus{}x_2)x_n^k \\minus{} \\sum_{i\\equal{}1}^{n\\minus{}2} {(x_{i\\plus{}2}\\minus{}x_i)x_{i\\plus{}1}^k}\\]\r\n\\[ \\equal{} (\\sum_{i\\equal{}1}^{n\\minus{}2} (x_{i\\plus{}1}\\minus{}x_i))x_{n\\minus{}1}^k \\plus{} (\\sum_{i\\equal{}2}^{n\\minus{}1} (x_{i\\plus{}1}\\minus{}x_i))x_n^k \\minus{} \\sum_{i\\equal{}1}^{n\\minus{}2} {(x_{i\\plus{}2}\\minus{}x_{i\\plus{}1}\\plus{}x_{i\\plus{}1}\\minus{}x_i)x_{i\\plus{}1}^k}\\]\r\n\\[ \\equal{}(\\sum_{i\\equal{}1}^{n\\minus{}2} (x_{i\\plus{}1}\\minus{}x_i))(x_{n\\minus{}1}^k \\minus{} x_{i\\plus{}1}^k) \\plus{} (\\sum_{i\\equal{}2}^{n\\minus{}1} (x_{i\\plus{}1}\\minus{}x_i))(x_{n}^k \\minus{} x_i^k)\\]" } { "Tag": [ "abstract algebra", "group theory", "superior algebra", "superior algebra open" ], "Problem": "Find a group of finite presentation that has $(\\mathbb{Q},+)$ as a subgroup. \r\nSuch a group exists if you were doubting it.", "Solution_1": "Do you mean \"finitely generated\"?", "Solution_2": "A presentation for a group si $G=$ where $S$ is a set (the generators of $G$) and $R$ is a set of finite length words (they are relations between the generators) in the elements (letters) of $S$.\r\n\r\nA finite presentation means $S$ and $R$ are finite. \r\nFinitely generated means just that $S$ is finite.\r\n\r\nFor example:\r\n$\\mathbb{F}_2=$ is the free group on two generators with no realtions.\r\n$\\mathbb{Z}^n=$ is the free abelian group on $n$ generators.\r\n\r\nA presentation for $(\\mathbb{Q},+)$ is $$. The correspondence is $\\frac 1{n!}=a_n$.\r\nNotice that for this presentation $S$ and $R$ are both infinite.\r\nThere is no finite presentation for $(\\mathbb{Q},+)$ because it is not finitely generated. But this is not what I'm asking for, is it?", "Solution_3": "If you say that such a group exists, then why are you putting it in the [i]Open Questions[/i] subforum?", "Solution_4": "[quote=\"perfect_radio\"]If you say that such a group exists, then why are you putting it in the [i]Open Questions[/i] subforum?[/quote]\r\nThe proof of the existence is not constructive. It is an open problem to give an example, but it is known that it must exist." } { "Tag": [ "inequalities", "linear algebra", "matrix", "induction", "algebra", "polynomial", "complex numbers" ], "Problem": "Show that for any square matrix A exist an unitary matrix C such that $CA{C^{-1}}$ is upper triangular", "Solution_1": "use induction\r\npick a random eigenvalue of the matrix\r\nnow take an eigenvector of it, and extend it to a basis\r\nthe first column will be completely zero in your new matrix, except for the very first element, which will be the eigenvalue\r\nnow you can use induction", "Solution_2": "or use Gram-Schmidt process ..", "Solution_3": "[quote=\"alekk\"]or use Gram-Schmidt process ..[/quote]\r\n\r\n1. How to pronounce Schmidt? I totally don't know how to pronounce a word with the string Schm\r\n\r\n2. Do we need the property that the characteristic polynomial of A splits?", "Solution_4": "Definitely , because two unitary equivalent (or simply equivalent for that matter) have the same characteristic polynomial, and an upper or lower triangle matrix obviously has a splitting char poly because you can just read the eigenvalues of the diagonal\r\n\r\nBut why do you ask? I didn't know unitary matrices was defined in any other field other than the complex numbers? Could you explain?", "Solution_5": "Alekk, can you elaborate? ;)", "Solution_6": "[quote=\"fredbel6\"]Definitely , because two unitary equivalent (or simply equivalent for that matter) have the same characteristic polynomial, and an upper or lower triangle matrix obviously has a splitting char poly because you can just read the eigenvalues of the diagonal\n\nBut why do you ask? I didn't know unitary matrices was defined in any other field other than the complex numbers? Could you explain?[/quote]\r\n\r\noh, sorry I thought in R you can call it unitary/orthogonal.", "Solution_7": "Take an eigenvector $X_1$ coressponding to an eigenvalue of the matrix $A$,called $a_1$.Then construct an orthonormate basis $X_1,...,X_n$(here you use the schmidt diag process).The matrix $A$ written in this base has $a_1$ in the first position of the first column,and 0 elsewhere on that column.It rests to see why the basis transformation matrix it is an unitary matrix(i.e ...$Matrix_BX=$unitary,where $B$ is a certain basis of $C^n$ and $X$ is a orthonormate basis of $C^n$).After that you can continue the process inductively choosing the $(n-1)\\cdot(n-1)$sub-matrix of $A$,and so on...\r\nI hope it's ok.", "Solution_8": "[quote=\"Soarer\"]\noh, sorry I thought in R you can call it unitary/orthogonal.[/quote]\r\n\r\nIn $\\mathbb{R}$ I believe we call it orthogonal, and in $\\mathbb{C}$ we call it unitary.\r\nWell, I understand your question now.\r\n\r\nIn $\\mathbb{C}$ every polynomial splits, but in $\\mathbb{R}$ that is not true, and then Schur's theorem should be corrected : every matrix with a splitting characteristic polynomial over the reals is orthogonally equivalent to an upper diagonal matrix" } { "Tag": [ "algorithm", "combinatorics proposed", "combinatorics" ], "Problem": "Let $n>2$ be an integer. In a country there are $n$ cities and every two of them are connected by a direct road. Each road is assigned an integer from the set $\\{1, 2,\\ldots ,m\\}$ (different roads may be assigned the same number). The [i]priority[/i] of a city is the sum of the numbers assigned to roads which lead to it. Find the smallest $m$ for which it is possible that all cities have a different priority.", "Solution_1": "We claim that the answer is 3. If we reduce each of the numbers assigned on the roads by 1, then the priority of each city is reduced by $ n\\minus{}1$, so all city still have distinct priorities. Now, each road is assigned an integer from $ \\{0,1,2,\\ldots,m\\minus{}1\\}$. If $ m\\minus{}1\\equal{}0$, all roads have the same priority. If $ m\\minus{}1\\equal{}1$, each road is assigned either 0 or 1. The greatest possible priority of a city is $ n\\minus{}1$ and the smallest possible priority is $ 0$. So the priorities must be $ 0,1,2,\\ldots,n\\minus{}1$. All roads from the city of priority $ 0$ must be assigned 0, while all roads from the city of priority $ n\\minus{}1$ must be assigned $ 1$. Considering the road connecting these two cities, we get a contradiction. Thus $ m\\minus{}1\\ge2$. We prove that $ m\\minus{}1\\equal{}2$ is possible.\r\n\r\nWe will find an algorithm to assign each road an integer from $ \\{1,2,3\\}$. We divide into two cases:\r\n\r\n(i) $ n$ is even ($ n\\equal{}2k$)\r\nLet the cities be $ C_1,C_2,\\ldots,C_n$ such that the priority of $ C_i$ is less than that of $ C_j$ for all $ i=3 becouse if m<=2 thus there are at least tow road that has a same number and ...\nnow we prove by apriori that if m=3 its possible to put number on roads suppose that for n=k it is possible now consider another country with n roads put the smallest priority of n city in priority of new city and suppose that its k(k>n). now k=x1+x2+...+xn(x i=1,2,3)and the biggest number of road connect to city which has a biggest priority and ....if k=n-1 so we cant write k=x1+...+xn .now consider k=3n.it will be possible.if its not tru help me ", "Solution_3": "Obviously for $m=2$ our priority can have just $n-1$ different value we have $n$ vertices => contradiction \n\nNow we prove $m=3$ is our suitable solution. for this we should just build this graph for any $n$. \n\n[b]obvious claim [/b]: about each graph at least one of the states is true: \n[u]state 1[/u]: the minimum of priorities isn't equal to $n-1$\n[u]state 2[/u]: the maximum of priorities isn't equal to $3(n-1)$\n\n[b]proof[/b]: if about one graph none of the states is true so the priority of one of the vertex for example $A$ is $n-1$ and the priority of $B$ is $3(n-1)$ so all of the contact edges of $A$ are $1$ and all of the edges of $B$ are $3$. Consider the edge between $A$ and $B$=>contradiction \n\nNow just we should use induction. Our conditions are true for $n=3,4,...,k-1$ We want to prove for $n=k$\nso cosider the graph when $n=k-1$ .\n if the [u]state 1[/u] is true for this graph we just add one vertex in this graph and put the number of the edges between this new vertex and the others $1$ \nif the [u]state 2[/u] is true for this graph add one vertex and put new edges $n-1$\nSo we are done :agent: " } { "Tag": [ "HMMT" ], "Problem": "Team Penguin can only do it at 6:30 PM EST Thursday. Is that time O.K. for SagginWolley?", "Solution_1": "NO, ZENITH XAN GETS HOME AT 22:30PM EST ON THURSDAYS.\r\n\r\nWEDNESDAY 17:30-18:30 EST OR SUNDAY FROM 12:00 TO 20:00 EST OR 21:00-23:59 EST\r\n\r\nTUESDAY AND MONDAY TIMES MIGHT BE AVAILABLE BUT WILL BE HIGHLY LIMITED.", "Solution_2": "wait, we can do the tournament on sunday?\r\n\r\nIf that's the case, I will be home from HMMT and we are going to be home all day...\r\n\r\n\r\nAny time on Sunday should be good, let me check with abacadaea and zolojetto...", "Solution_3": "we will try every hour i guess because my parents are always planning something. 12:00 is good. 13:00-15:00 we are usually doing things. erm, post which hours between 12:00 and 20:00 on sunday are good for you. i will probably be able to work around most things.", "Solution_4": "[quote=\"abacadaea\"]we will try every hour i guess because my parents are always planning something. 12:00 is good. 13:00-15:00 we are usually doing things. erm, post which hours between 12:00 and 20:00 on sunday are good for you. i will probably be able to work around most things.[/quote]\r\nzolojetto has told me that any time on Sunday will be good for him. Any time on Sunday is good for me too.", "Solution_5": "OK. @SagginWolley: How is Sunday 4 O'clock? That is the best time for Penguin." } { "Tag": [ "geometry", "trigonometry", "MATHCOUNTS", "AMC", "AIME", "ratio" ], "Problem": "I know everyone is involved into chapter problems, but here's a problem from the handbook of 2003-2004 (the only problem I haven't solved!):\r\n\r\nThere are two non congruent triangles. Each have one side with 10 cm, another side 12cm, and both have the angle facing the 10cm side measuring 45 degrees. In square centimeters, what is the sum of the areas of the two triangles?", "Solution_1": "[hide]\nUse Law of Sines.\n\n58.45+10.47=68.92[/hide]", "Solution_2": "dasherm: I believe your answer is incorrect.\r\n\r\nEverybody: This is MATHCOUNTS. Please solve this without using trigonometry.\r\n\r\nThere is a wonderful solution. Anybody who finds it on their own will feel a great satisfaction. Even though this is a request for help, please hide your solutions.", "Solution_3": "I drew a picture-\r\n\r\n[hide]I got 72 as the answer. First I realised that though the two triangles are not congruent, they still have the same area. Next what i did was draw a perpendicular line and that made a right icoceles triangle. Then I did 2x 2 =144. So x= :sqrt: 72. The area of the triangle then is 36. The area of both triangles is 72.[/hide]", "Solution_4": "[quote=\"GoBraves\"][hide]... the two triangles ... have the same area. ...[/hide][/quote]\r\nI am intrigued by your solution, even though I don't think the assertion quoted above is correct", "Solution_5": "Here's what I figure. Since the figures are noncongruent, there are only two possible ways that you can come up with the figure, with the 12 being the longer side (creating an obtuse triangle I think), and the twelve being the shorter side (like in gobraves' drawing... I'm still working on a solution tho...", "Solution_6": "hockey's on the right track, after i did it, I found the problem from that handbook(on my hard drive)\r\n\r\nNice picture Braves, but I think that's only one of the triangles", "Solution_7": "[quote=\"rcv\"]dasherm: I believe your answer is incorrect.\n\nEverybody: This is MATHCOUNTS. Please solve this without using trigonometry.\n\nThere is a wonderful solution. Anybody who finds it on their own will feel a great satisfaction. Even though this is a request for help, please hide your solutions.[/quote]\r\n\r\nSo trig is not permited in mathcounts?", "Solution_8": "[quote=\"kidwithshirt\"][quote=\"rcv\"]dasherm: I believe your answer is incorrect.\n\nEverybody: This is MATHCOUNTS. Please solve this without using trigonometry.\n\nThere is a wonderful solution. Anybody who finds it on their own will feel a great satisfaction. Even though this is a request for help, please hide your solutions.[/quote]\n\nSo trig is not permited in mathcounts?[/quote]\r\n\r\nIts permitted, any method of solving is permitted (except cheating), but all of the problems can be solved without trig.", "Solution_9": "30(sqrt 3) + 50. I'm guessing by making two triangles that are not congruent. Please correct me if I'm wrong. Is anybody even listening?", "Solution_10": "jason, I'm listening. That's not the right answer. The answer can be found with using so little math it is quite elegant!! The only math I used was probably that of an average 5th or 6th grade class. It's seeing how to use it that makes it special :)", "Solution_11": "[quote=\"frost13\"]The answer can be found with using so little math it is quite elegant!![/quote]\r\nYes! I am sure frost13 and I are thinking of the same solution.\r\n\r\nThis was a request for help. If somebody doesn't solve this in the next day or two, frost13 or I will have to post our elegant solutions, thus depriving you guys of finding it for yourselves.", "Solution_12": "Hopefully rcv will mean the \"next day\" part, not the \"or two\" part", "Solution_13": "[quote=\"GoBraves\"][quote=\"kidwithshirt\"][quote=\"rcv\"]dasherm: I believe your answer is incorrect.\n\nEverybody: This is MATHCOUNTS. Please solve this without using trigonometry.\n\nThere is a wonderful solution. Anybody who finds it on their own will feel a great satisfaction. Even though this is a request for help, please hide your solutions.[/quote]\n\nSo trig is not permited in mathcounts?[/quote]\n\nIts permitted, any method of solving is permitted (except cheating), but all of the problems can be solved without trig.[/quote]\r\n\r\n...and no one wants to do something like sine of 42 in the middle of the sprint round... Just like the AIMEs are designed for if you have Pre-Calc knowledge, the Mathcounts problems are designed for anyone who has taken Algebra 1 and has a bit more knowledge than that..", "Solution_14": "Also, using trigonometry ratios when solving a problem relating to 30-60-90 or 45-45-90 triangles is much harder than just memorizing the special triangles relationships.", "Solution_15": "Yup... My dad tried a mathcounts problem with a 30:60:90 problem and went into all this trig... and it was so much more extensive than knowing the ratios and avoiding trig.", "Solution_16": "Too bad they can't have a trigonometry table at the competitions for those who don't know their special triangles :lol: ...", "Solution_17": "[hide=\"Here is my answer.\"]\nI drew out both triangles, ontop of each other. Then, I moved the obtuse triangle to the other side and flipped it. Since we knew that the angle opposite the side length 10, we have a 45-45-90 right triangle, with base and height 12. THerefore, the area is:\n$\\frac{12\\cdot 12}{2}=72$\nMy answer is 72.\n[/hide]", "Solution_18": "[quote=\"smiley\"][hide=\"Here is my answer.\"]\nI drew out both triangles, ontop of each other. Then, I moved the obtuse triangle to the other side and flipped it. Since we knew that the angle opposite the side length 10, we have a 45-45-90 right triangle, with base and height 12. THerefore, the area is:\n$\\frac{12\\cdot 12}{2}=72$\nMy answer is 72.\n[/hide][/quote]\r\n\r\nI don't exactly get your solution... where did you put in the ten? Can you label out your pic?", "Solution_19": "[quote=\"hockeyadi23\"][quote=\"smiley\"][hide=\"Here is my answer.\"]\nI drew out both triangles, ontop of each other. Then, I moved the obtuse triangle to the other side and flipped it. Since we knew that the angle opposite the side length 10, we have a 45-45-90 right triangle, with base and height 12. THerefore, the area is:\n$\\frac{12\\cdot 12}{2}=72$\nMy answer is 72.\n[/hide][/quote]\n\nI don't exactly get your solution... where did you put in the ten? Can you label out your pic?[/quote]\r\n\r\nSmiley's got it, here is a similar picture with one less line and with lables. The two original triangles have been combined on their common edge of 10 to form one isoceles triangle with sides 12 and two 45 degree angles making in a right isoceles.", "Solution_20": "Thx.. I see it now..", "Solution_21": "that's a really nice solution... I would've never thought of that." } { "Tag": [ "probability", "percent", "LaTeX" ], "Problem": "Mary and Jerry have three kids: Harry, Larry, and Steve. If three family members are picked at random, what is the probability that they are all male? Express your answer as a percentage.", "Solution_1": "[quote=\"Treething\"]Mary and Jerry have three kids: Harry, Larry, and Steve. If three family members are picked at random, what is the probability that they are all male? Express your answer as a percentage.[/quote]\r\n\r\n[hide]$\\frac{\\binom{4}{3}}{\\binom{5}{3}}=\\frac{4}{10}=\\boxed{40}$percent[/hide]", "Solution_2": "SplashD, your answer is \"[Unparseable or potentially dangerous latex formula. Error 6]\" ;)\r\n\r\n[hide]\nThe answer is $\\frac{{4 \\choose 3}}{{5 \\choose 3}}=\\frac{2}{5}=40 \\%$\n[/hide]\r\n\r\nOh, and why can't I quote SplashD?\r\n\r\n[size=75][color=darkred]Fixed your LaTeX\n-nebula42[/color][/size]", "Solution_3": "The quote option is hidden under one of the hide tags in his signature.\r\n\r\n[hide=\"answer\"]There are $\\binom43$ ways to choose 3 boys and $\\binom53$ total ways to choose them, making the answer $\\frac{\\binom43}{\\binom53}=\\frac{4}{10}=\\frac25=40\\%$[/hide]\r\n\r\nmathnerd, you need to have \\frac{{4\\choose3}{{5\\choose3}} or what I have :)", "Solution_4": "[hide=\"Assuming mary is female and jerry, harry larry, and steve are male\"]$\\frac{\\binom43}{\\binom53}=\\frac25=40\\%$[/hide]Treething said [b]%[/b]", "Solution_5": "[hide]The answer is $\\frac{\\binom{4}{3}}{\\binom{5}{3}}=\\frac{2}{5}=40\\%$.\n\n$40\\%$[/hide]", "Solution_6": "[quote=\"mathnerd314\"]SplashD, your answer is \"[Unparseable or potentially dangerous latex formula. Error 6]\" ;)\n[/quote]\r\n\r\nYeah, I put % in $LaTeX$ instead of \\%.", "Solution_7": "[quote=\"mathnerd314\"]SplashD, your answer is \"[Unparseable or potentially dangerous latex formula. Error 6]\" ;)\n\n[hide]\nThe answer is $\\frac{4 \\choose 3}{5 \\choose 3}=\\frac{2}{5}$\n[/hide]\n\nOh, and why can't I quote SplashD?[/quote]\r\n\r\nAnd your answer has an error.\r\n\r\nYou can quote SplashD, just click in his signature, \"Click here to reveal hidden content\", or something like that, and then the quote will come." } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Consider an entire function $ F$ that is real valued on the segment $ [0,1]$. Prove that it is real-valued on the whole real axis.", "Solution_1": "Its Taylor expansion at $ z\\equal{}1/2$ has real coefficients." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "we have a equation:$x^8+9y^8=10z^4$.How many solutions does it have satisfise:\r\n$x,y,z\\in Z^+$ and $gcd(x,y,z)=1$?\r\n :( $\\leftarrow$ I unsolve it ,help me.", "Solution_1": "It is equavalent $x^8-y^8=10(z^4-y^8)$. It had onlu solutution x=y=z=1.", "Solution_2": "[quote=\"Rust\"]It is equavalent $x^8-y^8=10(z^4-y^8)$. It had onlu solutution x=y=z=1.[/quote]\r\nI don't understand your solution. :(", "Solution_3": "Can you post your solution?" } { "Tag": [], "Problem": "For how many real values of x is \r\nsquare root (120 \u2212 square root (x) an integer?\r\n\r\nA. 3 B. 6 C. 9 D. 10 E. 11", "Solution_1": "hello, is your term $ \\sqrt{120\\minus{}\\sqrt{x}}$?\r\nSonnhard.", "Solution_2": "Post Deleted", "Solution_3": "Assuming he meant $ \\sqrt{120\\minus{}\\sqrt{x}}$:\r\n\r\nWe pretty much want $ 120 \\minus{} \\sqrt{x}$ to equal a square, so when square rooted we would achieve an integer. The squares of $ 0\\minus{}10$ are below $ 120$, so therefore we have $ \\boxed{11}$ possible values of $ x$ where $ x$ is a real and $ \\sqrt{120\\minus{}\\sqrt{x}}$ is an integer." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Maximize the following expressions\r\n\r\n$a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd \\qquad a,b,c,d\\in [0,1]$\r\n\r\nand\r\n\r\n$\\sum_{sym}a^{2}b \\qquad a,b,c >0, a+b+c=1$", "Solution_1": "[quote=\"venatrix\"]Maximize the following expressions\n\n$a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd \\qquad a,b,c,d\\in [0,1]$\n[/quote]\nLet $A(a,b,c,d)=a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd.$\nSince, $\\frac{\\partial^{2}A}{\\partial a^{2}}=\\frac{\\partial^{2}A}{\\partial b^{2}}=\\frac{\\partial^{2}A}{\\partial c^{2}}=\\frac{\\partial^{2}A}{\\partial d^{2}}=2>0$ then\n$\\max_{\\{a,b,c,d\\}\\subset[0,1]}A(a,b,c,d)=\\max\\{A(1,1,1,1),A(1,1,1,0),A(1,1,0,0),A(1,0,0,0),A(0,0,0,0)\\}=$\n$=\\max\\{-2,0,1,1,0\\}=1.$\n[quote=\"venatrix\"]\n$\\sum_{sym}a^{2}b \\qquad a,b,c >0, a+b+c=1$[/quote]\r\nSee the Sung-yoon Kim's post for $\\sum_{cyc}a^{2}b$ here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960\r\nFor $\\sum_{sym}a^{2}b$ it's a trivial question: $\\sum_{sym}a^{2}b\\leq\\frac{1}{4}.$ :wink:" } { "Tag": [ "IMO", "Brocard", "angles", "geometry", "Triangle", "geometric inequality", "imo 1991" ], "Problem": "Let $ \\,ABC\\,$ be a triangle and $ \\,P\\,$ an interior point of $ \\,ABC\\,$. Show that at least one of the angles $ \\,\\angle PAB,\\;\\angle PBC,\\;\\angle PCA\\,$ is less than or equal to $ 30^{\\circ }$.", "Solution_1": "A solution was given in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=16171#p16171]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273 post #6[/url] by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=694]Johann Peter Dirichlet[/url]:\r\n\r\n\"Let ABC be a triangle and X an interior point of ABC. Show that at least one of the angles XAB, XBC, XCA is less than or equal to 30 degrees.\"\r\n\r\nThe solution is obvious with the Brocard point! The anterior messages contains all necessary theorems. P is the Brocard poin of the triangle ABC. If X=P then aren't nothing to prove. If X<>P then X is in the interior of one of the triangles PAB,PBC,PCA (wlog PAB).Then the problem is obvious (connect the point X to point A and verify the angle).", "Solution_2": "Here's a solution that I found right now. It is very late at my place, so I will skip the obvious parts.\r\n\r\n[i]Solution.[/i] Assume to the contrary that all the angles are greater than $ 30^{\\circ}$. Let $ PA\\equal{}x,PB\\equal{}y,PC\\equal{}z$. Then it follows that $ x^2\\equal{}z^2\\plus{}b^2\\minus{}2bz\\cos\\angle PCA>z^2\\plus{}b^2\\minus{}bz\\sqrt{3}$. Similarly, we find anologous inequalities. Adding them, we obtain $ a^2\\plus{}b^2\\plus{}c^2<\\sqrt{3}(bz\\plus{}ay\\plus{}cx)$. But $ [ABC]\\equal{}\\frac{1}{2}\\sum cx\\sin\\angle PAB>\\frac{ay\\plus{}bz\\plus{}cx}{4}$. Thus, $ a^2\\plus{}b^2\\plus{}c^2<4\\sqrt{3}[ABC]$. This is a contradiction to a well-known inequality (an old IMO problem that can be easily proven using Heron's formula along with AM-GM). $ \\Box$", "Solution_3": "Let $ d(P,AB)\\equal{}x$ , $ d(P,BC)\\equal{}y$ , $ d(P,CA)\\equal{}z$\r\nlemma. $ PA \\plus{} PB \\plus{} PC \\geq 2(x \\plus{} y \\plus{} z)$\r\n\r\nAt least one of $ PA\\geq 2x$ , $ PB\\geq 2y$ , $ PC\\geq 2z$ holds by lemma.\r\nif we assume $ PA\\geq 2x$ , $ \\frac {x}{PA}\\leq\\frac {1}{2}$ and $ \\angle{PAB}\\leq 30^{\\circ}$", "Solution_4": "[hide=\"Solution\"]Let $ \\angle PAC \\equal{} \\gamma$, $ \\angle PCB \\equal{} \\angle \\alpha$, $ \\angle PBA \\equal{} \\angle \\beta$, $ \\angle PAB \\equal{} \\alpha'$, $ \\angle PBC \\equal{} \\beta'$, and $ \\angle PCA \\equal{} \\gamma'$. \n\nWe will use contradiction. Suppose that $ \\alpha, \\beta, \\gamma > 30^{\\circ}$. Now, observe that $ \\sin \\alpha, \\sin \\beta, \\sin \\beta > 1/2$, since $ 30^{\\circ} < \\alpha, \\beta, \\gamma < 150^{\\circ}$. Now, observe that $ 1/8 < \\sin \\alpha \\sin \\beta \\sin \\gamma \\equal{} \\sin \\alpha' \\sin \\beta' \\sin \\gamma'$ by Ceva's. Squaring both sides, $ 1/64 < (\\sin \\alpha' \\sin \\beta' \\sin \\gamma')^2 \\equal{} \\sin \\alpha \\sin \\alpha' \\sin \\beta \\sin \\beta' \\sin \\gamma \\sin \\gamma'$. Now, observe that $ \\minus{}\\ln \\sin x$ is convex, so by Jensen's, \n\n$ \\minus{}(\\ln(\\sin \\alpha \\sin \\alpha' \\sin \\beta \\sin \\beta' \\sin \\gamma \\sin \\gamma'))/6 \\geq \\minus{}\\ln(\\sin(\\frac{\\alpha \\plus{} \\alpha' \\plus{} \\beta \\plus{} \\beta' \\plus{} \\gamma \\plus{} \\gamma'}{6}))$\n\nwhich is equal to $ \\minus{}\\ln(1/2)$. Rearranging, we get that $ \\sin \\alpha \\sin \\alpha' \\sin \\beta \\sin \\beta' \\sin \\gamma \\sin \\gamma' \\leq 1/64$, which is a contradiction. [/hide]", "Solution_5": "erdos-mordell inequality (when converted to trig form0 gives an immediate contradiction when we assume that all marked angles are at least 30.", "Solution_6": "Here's a very simple solution that I found for this problem:\n\nIf we draw a Triangle and interior point P, we label the vertices A, B, C anyway we want WLOG. Since the labeling of the vertices of the triangle can be interchanged, it suffices to show that at least one of the 6 angles $ \\,\\angle PAB,\\;\\angle PBC,\\;\\angle PCA,\\;\\angle PBA ,\\;\\angle PCB ,\\;\\angle PAC\\,$ is less than or equal to 30. We prove this statement by contradiction. First, we assume that all six of these angles are greater than 30. Then, the sum of these angles must be greater than 180, which is a contradiction since the sum of these six angles must be 180. Thus, at least one of these 6 angles must be less than or equal to 30. We can label the vertices A, B, and C so that either $ \\,\\angle PAB,\\;\\angle PBC,\\;\\angle PCA\\,$ is one of these six angles.\n\nCan someone look through and see if I made any mistakes in this proof?", "Solution_7": "Assume that $\\,\\angle MAB, \\,\\angle MBC, \\,\\angle MCA$ are all greater than $30^{\\circ}$.\n\nBy $\\textbf{sine Ceva},$\n\n$\\sin(MAC)\\sin(MBA)\\sin(MCB)$ \n= $\\sin(MAB)\\sin(MBC)\\sin(MCA) > sin^3(30^{\\circ})=\\frac{1}{8}$.\n\nAlso, from other side, $\\,\\angle MAC + \\,\\angle MBA + \\,\\angle MCB < 180^{\\circ} -3*30^{\\circ} = 90^{\\circ}$.\n\nNote that the function $f(x)=\\log(\\sin(x))$ is concave on $[0,\\pi]$. Applying Jensen, we get a contradiction here. ", "Solution_8": "[i]Trivialized by Jensen's[/i]\n\nBut more trivialized by Erdos-Mordell inequality:\n\nLet the projections of $M$ onto sides $BC,AC,$ and $AB$ be $X,Y,$ and $S$, respectively. By Erdos-Mordell, $$MA+MB+MC \\geq 2(MX+MY+MS).$$ This implies that at least one of $MA \\geq 2MS$, $MB \\geq 2MX$, or $MC \\geq 2MY$ must be true. WLOG, let $MA \\geq 2MS$. Since in right $\\triangle SAM$ we have $\\sin SAM \\leq \\frac{MS}{MA}=\\frac{1}{2}$, which implies $\\angle SAM \\leq 30^\\circ$ and we are done. $\\mathbb{S.A.M.!}$", "Solution_9": "Same as everyone else.\n\n[hide=Solution]Denote the feet of the perpendiculars from $P$ to $BC, AC, AB$ by $X,Y,Z$ respectively. Then Erdos-Mordell gives $PA+PB+PC \\geq 2(PX+PY+PZ)$, so at least one of the inequalities $PA \\geq 2MZ, PB \\geq 2PX, PC \\geq 2PY$ holds true. This means at least one of the inequalities $\\sin \\angle PAB \\leq \\frac{1}{2}, \\sin \\angle PBC \\leq \\frac{1}{2}, \\sin \\angle PCA \\leq \\frac{1}{2}$ holds true, so at least one of the angles is less than or equal to $30$ as desired.[/hide]" } { "Tag": [ "trigonometry" ], "Problem": "Given triangle ABC. Prove that $ 8\\cos A\\cos B\\cos C\\le\\cos (A\\minus{}B)\\cos (B\\minus{}C)\\cos (C\\minus{}A)$", "Solution_1": "This was posted not long ago\r\nIsn't It from an on-going competition!!", "Solution_2": "T/6 379\r\n\r\nN.N.Trung, are you Vietnamese? If so, you should have common sense. \r\n\r\nShould be locked!\r\n\r\nJapanese Communities Moderator\r\n\r\nkunny", "Solution_3": "I think too, it should be locked." } { "Tag": [ "group theory", "abstract algebra", "Support", "superior algebra", "superior algebra unsolved" ], "Problem": "Suppose we have a $ k$-cycle in the group $ S_{n}$ where $ k\\leqslant n$. Is there any reasonable way to calculate the number of distinct subgroups of $ S_{n}$ generated by all $ k$-cycles?\r\n\r\nFor example, if we take all $ \\left(abcde\\right)\\in S_{12}$, how many of the subgroups $ \\left<\\left(abcde\\right)\\right>$ are distinct? I know that there are $ \\binom{12}{4}$ such $ 5$-cycles, but they don't all generate distinct subgroups.", "Solution_1": "Let's say that the support of a cycle is the set of its non-fixed elements. Any two $ k$ cycles with distinct supports generate distinct subgroups.\r\n\r\nTwo $ k$ cycles with the same support generate the same subgroup iff one of them is the $ p$th power of the other, where $ p\\in \\mathbb Z_k^*$. There are $ \\varphi(k)$ such values $ p$. Thus we get $ \\frac{(k\\minus{}1)!}{\\varphi(k)}$ distinct subgroups from $ k$-cycles with fixed support. \r\n\r\nSince there are $ \\binom{n}{k}$ possible supports, there are $ \\frac{(k\\minus{}1)!}{\\varphi(k)}\\binom{n}{k}$ distinct subgroups generated by $ k$-cycles. Right? :maybe:" } { "Tag": [], "Problem": "Anyone play Halo 2 on Xbox Live in here? It would be pretty cool to play a game with some AoPSers. If anyone wants to add me to their friend list my Gamertag is cyclotomic.", "Solution_1": "Man, I wish I had an XBOX. Do you have HALO 1 for the PC?", "Solution_2": "Nope, sorry.", "Solution_3": "Ah Halo 2 is so different...grrr", "Solution_4": "Halo 2 + Xbox Live = Awesome\r\n\r\nMy handle is Pitcany.", "Solution_5": "Want it, except I'd feel guilty about having it and wasting the money because I hardly ever get to play Halo2 as it is with school and everything. \r\n\r\nOh well, at least there's still LAN parties.", "Solution_6": "i have halo 2, but not with x-box live. u haf 2 pay extra for that cause its connected to the world wide web. but i never play it anyways....my brother does. :lol:", "Solution_7": "halo 2 is awesome but i can't get xbox live cuz i have stupid dial up and my xbox is not compatible anymore as i had some modifications done to it. o well. i'll just go to a friends house to play online.", "Solution_8": "[b][i][u]i have halo 2 but i dont have live.. i wish i did..[/u][/i][/b]", "Solution_9": "Yo EFUZZY.\r\nI got Halo 1, and i'm willing to play. Add Divine eggnog to your aim list, and im me when im online. My username is [hide=\"click text\"]sh!tbr!ck[/hide]. I stopped cussing a long time ago, but that is my best file[/hide]", "Solution_10": "Has anyone found the scarab gun?", "Solution_11": "how do you get Halo 1 demo online version for PC ?\r\n\r\nI really want to try it...........\r\n\r\nnever played halo before except one time when I went over my friend's house....\r\n\r\nwell, it was halo 1 though.....", "Solution_12": "try bittorrent?", "Solution_13": "sheepwarrior , you sound like my chinese friend Raymond...\r\n\r\nwell, I'll try...it never worked forme though...", "Solution_14": "could you give me the site address?bittorrent..." } { "Tag": [ "Putnam", "function", "number theory", "relatively prime", "college contests" ], "Problem": "As for the solution to this problem, it used height functions, and it says that for every next composed function, the height of the solutions are at least greater than 2 plus the height of the previous composed function. But how does that mean that the intersection is empty? I just don't see it. And what is a height function? Please help me. Please? :blush:", "Solution_1": "Hey,\r\n[quote=\"sexynsmartjenny\"]\nAs for the solution to this problem, it used height functions, and it says that for every next composed function, the height of the solutions are at least greater than 2 plus the height of the previous composed function. But how does that mean that the intersection is empty? I just don't see it. And what is a height function? Please help me. Please? :blush:\n[/quote]\r\nIt would be helpful to know what the original problem was, \r\nPutnam 2001 B-4\r\nLet $ {S}$ denote the set of rational numbers different from $ {{\\{}{{ - 1, 0, 1}}{\\}}}$. Define $ {{f: S}{\\to}{S}}$ by $ {{f(x)} = x-\\frac1x}$. Prove or disprove that\r\n$ {\\bigcap_{n = 1}^{\\infty}}{{f}^{(n)}}{(S)} = {\\emptyset}$\r\nwhere $ {{f}^{(n)}}$ denotes $ {f}$ composed with itself $ {n}$ times.\r\n\r\nAs for what a height function is, I'm not all that sure myself. In the [i]Art and Craft of Problem Solving[/i], 2nd Edition, Paul Zeitz touches on the concept but doesn't elaborate. In anycase, I found these websites pretty useful for some background information,\r\n\r\nhttp://www.math.brown.edu/~jhs/HtSurveyMSRIJan06.pdf\r\nhttp://planetmath.org/encyclopedia/HeightFunction.html\r\n\r\nThanks, \r\n\r\n-APSStudent", "Solution_2": "I doubt that any of those links contain information relevant to this problem. As for the solution: since $ x\\minus{}1$ and $ x$ are relatively prime, a rational number in $ S$ must have an infinite number of prime factors, which is absurd. (Maybe this is where the \"height\" comes in: the number of prime factors?)", "Solution_3": "Keyword missing: the set of numbers [i]different[\\i] from $ \\{ \\minus{} 1,0,1\\}$.\r\n\r\nThe idea: start with all rational numbers except those three- $ 1\\mapsto 0\\mapsto \\infty$ and $ \\minus{} 1\\mapsto 0\\mapsto\\infty$. Apply the function to get $ f(S)$, which is contained in $ S$. Apply it again, getting $ f^2(S)\\subset f(S)$. Repeat. No $ f^k(S)$ is empty, but each $ q$ appears only in a finite subset of them.\r\n\r\nThe height idea in the proof: show that $ f(r)$ is always more complicated than $ r$ in some sense for rational $ r$ other than the initial excluded list.", "Solution_4": "Is the function well-defined?\r\n\r\nIsn't $ f(1/2) \\equal{} \\frac {1/2 \\minus{} 1}{1/2} \\equal{} \\frac { \\minus{} 1/2}{1/2} \\equal{} \\minus{} 1$?\r\n\r\nEdit: Internet tells me that the correct $ f$ is $ f(x) \\equal{} x \\minus{} 1/x$.\r\n\r\nEdit 2: Really neat idea, jmerry.\r\n\r\nEdit 3: Jenny, the \"height function\" here must be the following $ H$ or something similar: If $ a/b$ is fully reduced (i.e. $ \\gcd(a,b)\\equal{}1$), then $ H(a,b)\\equal{}\\max(a,b)$.", "Solution_5": "I also recall seeing the height function $ H(a, b) \\equal{} a \\plus{} b$ in Tate-Silverman (unless I'm remembering incorrectly)." } { "Tag": [], "Problem": "Farz konid yek donbale az takjomleiha mesle $\\{x^{\\zeta_i}\\}$(Har yek az $\\zeta_i$ ha yek bordar dar $(\\mathbb N\\cup\\{0\\})^n$ hastand va $x^{(a_1,\\dots,a_n)}=(x_1^{a_1},\\dots,x_n^{a_n}$) Sabet konid ke yek $N$ vojood darad ke har yek az aazaye donbaleh bar yeki az $n$ jomleye avval bakhshpazir bashad.\r\n\r\nMasalan agar $n=1$ masaleh hamoon asle khoshtartibi ast.", "Solution_1": "bebakhshid vali in joor lem ha dar che ketabi hast va dar che jahayi mishe masale haye marboote ro peyda kard?", "Solution_2": "aghaye hatami in mas'ale ba nazariye ramzi hal nemishe? :)", "Solution_3": "Na rabti nadare rahat mitoonid esteghra bezanid", "Solution_4": "[quote=\"mehdi\"]bebakhshid vali in joor lem ha dar che ketabi hast va dar che jahayi mishe masale haye marboote ro peyda kard?[/quote]\r\ninjoor lem ha mokhtasse afrade khafani chon ostad hatamie. :D", "Solution_5": "pas hatman bayad esmesho gozasht \"Hatami's lemma\" chon male ishoone(jeddi migam)", "Solution_6": "vali aghaye hatami man ye esbat ba nazariye ramzi barash didam. :)", "Solution_7": "Khoshal misham age esbate to ra bebinam oon ra befrest. :D\r\n\r\nVali baz ham pishnahad mikonam be esteghra rooye $n$ fekr kon", "Solution_8": "Baad az 5 mah hich kas natooneste (ya nakhaste) rahe halli baraye in masaleh befreste. :!: :!:", "Solution_9": "mitavan farz kard ozve tekrari vojood nadarad.\r\nmasale ra ba esteghra rooye $n$ sabet mikonim.\r\nbaraye $n=1$ ke vazeh ast.(asle khosh tartibi)\r\nfarz konim baraye $n-1$ hokm dorost bashad hokm ra be ezaye $n$ natije migirim.\r\nhameye azaye donbale ra bedoone moallefeye avval dar nazar migirim.\r\ntebghe farze esteghra $N_1$ vojood darad ke moallefeye dovvom ta $n$ome har yek az aaza bar yeki az $N_1$ jomleye avval bakhshpazir bashad.be hamin tartib ba hazfe moallefehaye dovom ta $n$om aadade $N_2,\\dots,N_n$\r\nmoshabeh vojood darand. $N$ ra bozorgtarin dar beine anha migirim.\r\nfarz konim $a_i$ bozorgtarin tavane moallefeye $i$ome $N$ ozve avvale donbale bashad. tamame azaye donbale be shekle $(x_1^{b_1},\\dots,x_n^{b_n})$ ke $b_i \\le a_i$ be ezaye har $i$ ra dar nazar migirim.(tebghe farz az har kodam hadde aksar yeki vojood darad)\r\nhal $M$ ra dar nazar migirim ke az $N$ bozorgtar bashad va hameye in aadad ra dar bar girad.(chon teedade in no aadad motanahist pas $M$ vojood darad)\r\nin $M$ sharayete masale ra erza mikonad. pas hokm sabet shod. :) :lol:" } { "Tag": [], "Problem": "What is the arithmetic mean of the first million odd numbers?", "Solution_1": "We have the first and last number are $ 1$ and $ 1,000,000(2)\\minus{}1$. Thus our average is $ \\boxed{1000000}$.", "Solution_2": "In more detail...\n\nWe know that the sum of odd integers are $ N^2$ where N is the number of odd integers. We can then say:\n\n$ 1,000,000^2\\linebreak \\linebreak \\frac{1,000,000^2}{1,000,000}$\n\n$ \\text{Thus, our answer is} \\boxed{1,000,000}$" } { "Tag": [], "Problem": "If $ 4^x \\minus{} 4^{x \\minus{} 1} \\equal{} 24$, then $ (2x)^x$ equals:\r\n\r\n$ \\textbf{(A)}\\ 5\\sqrt{5}\\qquad \r\n\\textbf{(B)}\\ \\sqrt{5}\\qquad \r\n\\textbf{(C)}\\ 25\\sqrt{5}\\qquad \r\n\\textbf{(D)}\\ 125\\qquad \r\n\\textbf{(E)}\\ 25$", "Solution_1": "[hide=\"Solution\"]$ \\frac {3\\cdot 4^x}{4} \\equal{} 24 \\Rightarrow 4^x \\equal{} 2^{2x} \\equal{} 32 \\Rightarrow x \\equal{} \\frac {5}{2}$\n\n$ (2x)^x \\equal{} (5)^{5/2} \\equal{} 25\\sqrt {5}, \\boxed{\\text{C}}$[/hide]", "Solution_2": "hello, simplifying your equation we get $ 4^x(1\\minus{}\\frac{1}{4})\\equal{}24$, this is equivalent with $ 4^x\\equal{}32$, this is equivalent with $ 2^{2x}\\equal{}2^5$ or $ x\\equal{}\\frac{5}{2}$, from here we get $ (2\\cdot \\frac{5}{2})^{\\frac{5}{2}}\\equal{}5^{\\frac{5}{2}}\\equal{}25\\sqrt{5}$\r\nSonnhard." } { "Tag": [ "geometry", "incenter", "circumcircle", "projective geometry", "geometry unsolved" ], "Problem": "In a triangle $ABC$ with $\\angle B < 45^{\\cdot}$, $D$ is a point on $BC$ such that the incenter of $\\Delta ABD$ coincides with the circumcenter $O$ of $\\Delta ABC$. Let $P$ be the intersection point of the tangent lines to the circumcircle $O'$ of $\\Delta AOC$ at points $A$ and $C$. The lines $AD$ and $CO$ meet at $Q$. The tangent to $O'$ at $O$ meets $PQ$ at $X$. Prove that $XO = XD$.", "Solution_1": "It's a nice problem :) Of course if incenter of $ABD$ and circumcenter of $ABC$ coincide then $ABC$ is isosceles with $BA=BC$ We have also that $ACO$ is isosceles with $OA=OC$. This means that tangent to $C'$ in $O$ ( it will be denoted by $l$) is parallel to $AC$ and also that tangents to $C'$ in $A$ and $C$ intersects on line $BO$. As $\\angle B <45$ , this intersection point lies on the other side of $AC$ ( with respect to $B$) As there is only one point $R$ on line $l$ such that $RD=RO$ We may formulate our statement as follows: We take point $R$ on $l$ such that $RD=RO$. We have to prove that $P,Q,R$ are collinear. \r\n\r\nWe will prove now that points $A,C,D,O$ are concyclic. Indeed, from symetry we obtain \\[ \\angle DCO= \\angle OAB \\] also as $AO$ is bisector of $\\angle DAB$ we obtain also \\[ \\angle OAB= \\angle DAO \\] so finally \\[ \\angle DCO=\\angle DAO \\].\r\nAnd points $A,C,D,O$ are concyclic. Let $C\\\"$ be the common circle passing through these points. As $RD=RO$ and $RO$ is tangent to $C\\\"$, also $RD$ is tangent to $C\\\"$. So we may say that $R$ is an intersection point of tangents to $C\\\"$ in points $D$ and $O$. Let lines $AC$ and $DO$ intersects in $W$. We will prove that both $R$ and $P$ lies on the polar line of $C\\\"$ ( with respect to $W$). Then our statement would be obvious, cos' $Q$ lies on this polar line. Let $WW_1$ and $WW_2$ be lines tangent to $C\\\"$ in $W_1$ and $W_2$ respectively. Consider quadrangle $AW_1CW_2$. It's diagonal $AC$ and tangents to $C\\\"$ in $W_1$ and $W_2$ intersects in one point. (now we use \"magic\" theorem.) This means that also diagonal $W_1W_2$ and tangents in $A$ and $C$ intersects in one point, so points $W_1W_2P$ are colinear , and point $P$ lies on the polar line $W_1W_2$. In analogical way, by considering quadrangle $W_1OW_2D$ we may obtain that $R$ lies on the polar line $W_1W_2$ which means that points $R,Q,P$ are colinear, which ends the proof of our statement. \r\n\r\nIn this proof, I used the following theorem : If $XYZT$ is cyclic and diagonal $XZ$ , tangent in $Y$ and tangent in $T$, concurrent, then also concurrent diagonal $YT$, tangent in $X$ and tangent in $Z$. \r\n\r\nNice isn't it ;)" } { "Tag": [ "Ring Theory", "number theory", "relatively prime", "greatest common divisor", "least common multiple", "number theory unsolved" ], "Problem": "Let $ T$ be a finite set of positive integers, satisfying the following conditions:\n1. For any two elements of $ T$, their greatest common divisor and their least common multiple are also elements of $ T$.\n2. For any element $ x$ of $ T$, there exists an element $ x'$ of $ T$ such that $ x$ and $ x'$ are relatively prime, and their least common multiple is the largest number in $ T$. \n\nFor each such set $ T$, denote by $ s(T)$ its number of elements. It is known that $ s(T) < 1990$; find the largest value $ s(T)$ may take.", "Solution_1": "Let $ N$ denote the largest element of $ T$. By condition 2, $ 1 \\in T$ and every element of $ T$ is a divisor $ d$ of $ N$ such that $ \\frac {N}{d}$ is relatively prime to $ d$. This requires that if $ p | d$ then $ p^k \\parallel{} d$ where $ p^k \\parallel{} N$. Thus we can suppose WLOG that $ N$ is squarefree (reducing the powers of each prime to $ 1$ does not change the structure of the set). \n\nWe can now give $ T$ the structure of a subalgebra of a [url=http://en.wikipedia.org/wiki/Boolean_algebra_%28structure%29]Boolean algebra[/url], from which it will follow that $ s(T)$ is a power of $ 2$. In particular, let $ N$ be a product of $ n$ primes $ p_1 p_2 ... p_n$ and associate with each $ d \\in T$ the $ n$-tuplet $ (d_1, d_2, ... d_n), d_i \\in \\mathbb{Z}_2$ such that $ d \\equal{} p_1^{d_1} ... p_n^{d_n}$. The first statement says that $ T$ is closed under intersection and union and the second that $ T$ is closed under inverse. $ T$ is therefore the subalgebra of a finite Boolean algebra (which is the Boolean algebra of all divisors of $ N$), and it must therefore have order dividing $ 2^n$. (Equivalently, $ T$ can be given the structure of a subring of a Boolean ring.)\n\nHence $ \\text{max}(s(T)) \\equal{} 1024$.", "Solution_2": "Who can help me?Thanks", "Solution_3": "[quote=\"hungkg\"]Who can help me?Thanks[/quote]\nHelp what ? t0rajir0u has been post the solution." } { "Tag": [ "trigonometry", "Pythagorean Theorem", "geometry", "geometry proposed" ], "Problem": "There are two triangles: $\\triangle{ABC}$ and $\\triangle{XYZ}$. If $\\angle{B}=\\angle{Y}$, prove:\r\n\\[\\frac{\\triangle{ABC}}{\\triangle{XYZ}}=\\frac{a^2+c^2-b^2}{x^2+z^2-y^2}\\]\r\n\r\nAlso if $\\angle{B}+\\angle{Y}= 180^{o}$, then prove:\r\n\\[\\frac{\\triangle{ABC}}{\\triangle{XYZ}}=-\\frac{a^2+c^2-b^2}{x^2+z^2-y^2}\\]\r\n\r\n[b][size=150]NOTE: Try to prove it without trig and Pythagorean theorem or any other stuff related to these.[/size][/b]", "Solution_1": "anyone?", "Solution_2": "Those Can be solved wiht cosinus law" } { "Tag": [ "absolute value" ], "Problem": "Given the equation: $ (x^2 \\plus{} 1)(y^2 \\plus{} 2)(z^2 \\plus{} 3)(t^2 \\plus{} 4)(n^2 \\plus{} 5) \\equal{}$ $ 32.xyztn\\sqrt{120}$(x,y,z,t,n are real numbers)\r\nFind the number of solutions of this equations.", "Solution_1": "AM-GM tears down the problem.\r\n\r\nSo $ \\boxed{1}$ solution exists (which is, obviously, $ x\\equal{}1,y\\equal{}\\sqrt{2},z\\equal{}\\sqrt{3},t\\equal{}2,n\\equal{}\\sqrt{5}$).", "Solution_2": "[quote=\"MathAndKnowledge\"]AM-GM tears down the problem.\n\nSo $ \\boxed{1}$ solution exists (which is, obviously, $ x \\equal{} 1,y \\equal{} \\sqrt {2},z \\equal{} \\sqrt {3},t \\equal{} 2,n \\equal{} \\sqrt {5}$).[/quote]\r\nBe careful! x,y,z,t,n are real numbers :wink:", "Solution_3": "I know I'm asking a stupid question, but how do you apply AM-GM???", "Solution_4": "[quote=\"1=2\"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]\r\nI guess the reason you are afraid of not using AM-GM is just because x,y,z,t,n are real numbers. So, I will help you out with just this \"cheap\" hint: Why don't you use the absolute value first, multiply them, and then compare with the RHS? :wink: . The whole point of this problem is mainly lying on the combinatorics part though. I think we have a general formula for the number of solutions of the n-variable equation as well. :)", "Solution_5": "[quote=\"1=2\"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]\r\n\r\nIt's AM-GM on every term, but since they are real numbers it is easier to use SOS\r\n\r\n$ (x\\minus{}1)^2 \\ge 0 \\Rightarrow x^2 \\plus{} 1 \\ge 2x$\r\n$ (y\\minus{}\\sqrt{2})^2 \\ge 0 \\Rightarrow y^2 \\plus{} 2 \\ge 2\\sqrt{2}y$...\r\n\r\nNow multiplying all of these solutions we get\r\n\r\n$ (x^2\\minus{}1)(y^2\\minus{}2)(z^3\\minus{}3)(t^4\\minus{}4)(4^n\\minus{}n) \\ge 32xyztn\\sqrt{120}$", "Solution_6": "[quote=\"ocha\"][quote=\"1=2\"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]\n\nIt's AM-GM on every term, but since they are real numbers it is easier to use SOS\n\n$ (x \\minus{} 1)^2 \\ge 0 \\Rightarrow x^2 \\plus{} 1 \\ge 2x$\n$ (y \\minus{} \\sqrt {2})^2 \\ge 0 \\Rightarrow y^2 \\plus{} 2 \\ge 2\\sqrt {2}y$...\n\nNow multiplying all of these solutions we get\n\n$ (x^2 \\minus{} 1)(y^2 \\minus{} 2)(z^3 \\minus{} 3)(t^4 \\minus{} 4)(4^n \\minus{} n) \\ge 32xyztn\\sqrt {120}$[/quote]\r\nI think you mean $ (x^2 \\plus{} 1)(y^2 \\plus{} 2)(z^2 \\plus{} 3)(t^2 \\plus{} 4)(n^2 \\plus{} 5) \\ge 32xyztn\\sqrt {120}$? That is a strange typo. :huh:\r\n\r\nAnd, since no one has actually said it yet, we have $ x\\equal{}\\pm1, y\\equal{}\\pm \\sqrt{2}, z\\equal{}\\pm \\sqrt{3}, t\\equal{}\\pm2, n\\equal{}\\pm\\sqrt{5}$, so we have $ 2^5\\equal{}32$ solutions total.", "Solution_7": "[quote=\"Brut3Forc3\"][quote=\"ocha\"][quote=\"1=2\"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]\n\nIt's AM-GM on every term, but since they are real numbers it is easier to use SOS\n\n$ (x \\minus{} 1)^2 \\ge 0 \\Rightarrow x^2 \\plus{} 1 \\ge 2x$\n$ (y \\minus{} \\sqrt {2})^2 \\ge 0 \\Rightarrow y^2 \\plus{} 2 \\ge 2\\sqrt {2}y$...\n\nNow multiplying all of these solutions we get\n\n$ (x^2 \\minus{} 1)(y^2 \\minus{} 2)(z^3 \\minus{} 3)(t^4 \\minus{} 4)(4^n \\minus{} n) \\ge 32xyztn\\sqrt {120}$[/quote]\nI think you mean $ (x^2 \\plus{} 1)(y^2 \\plus{} 2)(z^2 \\plus{} 3)(t^2 \\plus{} 4)(n^2 \\plus{} 5) \\ge 32xyztn\\sqrt {120}$? That is a strange typo. :huh:\n\nAnd, since no one has actually said it yet, we have $ x \\equal{} \\pm1, y \\equal{} \\pm \\sqrt {2}, z \\equal{} \\pm \\sqrt {3}, t \\equal{} \\pm2, n \\equal{} \\pm\\sqrt {5}$, so we have $ 2^5 \\equal{} 32$ solutions total.[/quote]\r\n@ocha: You can't multiply both sides without assuring that both sides are positive. Your solution has that \"flaw\".\r\n@Brut3Forc3: The whole point here is not that simple! Did you notice something strange about the number of variables? If you didn't let me tell you this: IT'S ODD!. :wink: . Your answer is not correct!", "Solution_8": "Noooooo!\r\nOkay, pick signs for $ x,y,z,t$, then $ n$'s sign is fixed, so $ 16$.", "Solution_9": "[quote=\"Brut3Forc3\"]Noooooo!\nOkay, pick signs for $ x,y,z,t$, then $ n$'s sign is fixed, so $ 16$.[/quote]\r\nThat's correct. :) \r\nThe formula for n-variables is:$ 2^{(n \\minus{} 1)}$." } { "Tag": [ "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ \\mu \\equal{} \\Sigma_{i \\equal{} 1}^n x_i dx_i$\r\nShow that there is an n-1 form $ \\omega$ defined on $ R^n \\minus{} {0}$ with the property that $ \\mu \\wedge \\omega \\equal{} dx_1 \\wedge ... \\wedge dx_n$", "Solution_1": "Let $ \\omega_i\\equal{}dx_1\\wedge\\dots\\wedge dx_{i\\minus{}1}\\wedge dx_{i\\plus{}1}\\wedge\\dots \\wedge dx_n$. Let $ F(x)\\equal{}\\sum_j x_j^2$. Now just put $ \\omega\\equal{}\\sum_i (\\minus{}1)^{i\\minus{}1}\\frac {x_i}{F(x)}\\omega_i$. This construction is by no means unique, of course." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Dear nguovin, can you please help me to make this inequality stronger? (I think it is a bit strong)....... :blush: \r\nProve That, for positive reals $ a, b, c$ we have \r\n$ \\frac{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}{8abc}\\ge\\frac{\\sqrt{\\prod_{cyc}((a^2\\plus{}b)(a\\plus{}b^2))}}{(a\\plus{}bc)(b\\plus{}ca)(c\\plus{}ab)}$\r\n\r\nHope you will enjoy it and help me strengthen it! :roll: :lol:", "Solution_1": "$ (a\\plus{}bc)(b\\plus{}c)\\equal{}b(a\\plus{}c^2)\\plus{}c(a\\plus{}b^2) \\geq\\ 2 \\sqrt{bc(a\\plus{}c^2)(a\\plus{}b^2)}$\r\nI think you used my old hint :P . \r\nBut thank you very much for your gift, Potla. I like this ineq :)", "Solution_2": "[quote=\"nguoivn\"]$ (a \\plus{} bc)(b \\plus{} c) \\equal{} b(a \\plus{} c^2) \\plus{} c(a \\plus{} b^2) \\geq\\ 2 \\sqrt {bc(a \\plus{} c^2)(a \\plus{} b^2)}$\nI think you used my old hint :P . \nBut thank you very much for your gift, Potla. I like this ineq :)[/quote]\r\nMany many thanks for your interest, and of course I used that hint as I liked it tremendously.....\r\n\r\nHowever, it was too easy to be dedicated to YOU; :rotfl: \r\nBut you did not answer my question \"Can we make it stronger?\"......... :oops:" } { "Tag": [ "algebra", "function", "domain", "vector", "geometry", "3D geometry", "sphere" ], "Problem": "So here are some functions of the following types...\r\n\r\nf: R -> R^2 (curves in the plane)\r\nf: R -> R^3 (curves in space)\r\nf: R^2 -> R (functions f(x,y) of 2 vars)\r\nf: R^3 -> R: (functions f(x,y,z) of 3 vars)\r\nf: R^2 -> R^2 (vector fields v(x,y) in the plane)\r\n\r\nThe question is - why are curves in the plane of the form R -> R^2? My intuition tells me R^2 -> R^2 (since after all, curves in the plane are based on x and y coordinates...). And R^2 is a cartesian product of two sets. For any curve, I'd expect x AND y input values...", "Solution_1": "Think parametric: $ (x(t),y(t))$.\r\n\r\nAlso, this is about domains and codomains, not domains and ranges. Space-filling curves are pathological.", "Solution_2": "Okay yeah.\r\n\r\nSo would f(t) = be a curve of form R -> R^2?", "Solution_3": "Yes, that's a curve. What about it?", "Solution_4": "[quote=\"Simfish\"]So would f(t) = be a curve of form R -> R^2?[/quote]\r\n\r\nThink about functions in terms of input and output. In this case, your input is some real number t. Your output is some set of ordered pairs (t^3 - t,t^2), such as (0,0), (0,1), (6,4), etc.\r\n\r\nI like to think about functions like R->R^2 somewhat like embeddings. The idea is that you imagine you're an ant walking along the curve. It seems to you like the world is just a line. It's a curvy line, but it's still just a line, not like a plane or \"space\". You can only go forwards and backwards (one direction and its opposite). That means that it's \"one-dimension\". It happens to be sitting inside a plane (two dimensions), but the curve itself only has one dimension (forward/backward). That \"one dimension\" is the t value. It goes up or down (forward/backward).\r\n\r\nA function R^2 -> R^3 is like a piece of paper in space. You can bend the paper and wiggle it around, but if you're an ant on the page, you can go in two directions: forward/backward and left/right. That's like a sphere (like Earth), N/S and E/W. You make a function like that by having an INPUT of two variables, say t and s, and an output of three real numbers (an ordered triple), like this: f(t,s) = (t^2, s - 2t, st^3). You can play with t to make your ant move forwards/backwards, and you can play with s to make your ant move left/right." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "An integer n>1 and a prime p are such that n divides p-1 and p devides $ n^{3}-1$.Show that 4p+3 is the square of an integer", "Solution_1": "[quote=\"tuanct\"]An integer n>1 and a prime p are such that n divides p-1 and p devides $ n^{3}-1$.Show that 4p+3 is the square of an integer[/quote]\r\n\r\nThat's wrong.\r\n\r\nWith for instance $ n=2$ and $ p=7$\r\n\r\n$ 2$ divides $ 6$, $ 7$ divides $ 2^{3}-1$ but $ 4*7+3=31$ is not a perfect square.", "Solution_2": "I thank that the correct question is:\"Show that 4p-3 is the square of an integer and not 4p+3\"\r\nthis problem is already posted." } { "Tag": [ "LaTeX", "inequalities proposed", "inequalities" ], "Problem": "prove that a*(n+1)/a*n+b*n +b*(n+1)/b*n+c*n +c*(n+1)/(c*n+a*n) >=a+b+c/2\r\n\r\n\r\nfor positive reel numbers :mad: :D", "Solution_1": "Please use latex and explain your problem clearly.", "Solution_2": "If you mean\r\n$\\sum_{cyc}\\frac{a^{n+1}}{a^n+b^n}\\ge\\frac{a+b+c}2$ I think it is not true for $n>6$, see\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=22937 and \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=382832.\r\n\r\nOn the contrary, for $\\sum_{cyc}\\frac{a^{n+1}}{b^n+c^n}\\ge\\frac{a+b+c}2$,\r\nwe can prove easily \r\n$\\sum_{cyc}\\frac{a^{n+1}}{b^n+c^n }\\ge \\sum_{cyc}\\frac{a^{n}}{ b^{n-1} +c^{n-1}}$\r\nwith the same simple trick (let's call it \"cyclical SOS\") as [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=455429]this one[/url].", "Solution_3": "I think its not true for n>4", "Solution_4": "[quote=\"Albanian Eagle\"]I think its not true for n>4[/quote]\r\nI think you are wrong. See here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113&postorder=asc\r\npost 10. ;)" } { "Tag": [ "vector", "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ p$ be a prime number and $ G$ be a pro-$ p$-group, I want to prove that if $ H^1(G)\\equal{}Hom(G,\\mathbb{Z}/p\\mathbb{Z})\\equal{}0$ then $ G\\equal{}\\{1\\}$. Little hint, please, Thanks in advance for your help.", "Solution_1": "This is just a property of finite p-groups: \r\n\r\nFor a (pro-) p-group G, Hom(G,Z/pZ) = Hom( G/\u03a6(G), Z/pZ ) is the dual vector space to G/\u03a6(G), so has dimension equal to minimum number of generators of G.\r\n\r\nHere \u03a6(G) = G^p[G,G] is the (closure of the) subgroup generated by all g^p and [g,h] for g,h in the (pro-) p-group G." } { "Tag": [], "Problem": "when u cut onions in to small pieces i was told that the cell wall breaks, but what happens when u put the onion juice into a solution with detergent?? isnt that where the cell wall really breaks?? --(to extract DNA)\r\n\r\nif this is confusing and ur willing to help just give a message for further explanaations...\r\n\r\nthank you", "Solution_1": "I thinkyou're getting the cell wall confused with the cell membrane. DNA in plant cells is enclosed not only by a cell wall but also by two separate membranes - a cellular one and a nuclear one. It's been a while since I've done this, so I might be wrong, but I believe the detergent is helpful because it breaks down those membranes allowing more DNA to escape. Cutting the onions beforehand is necessary because it ruptures cell walls and cell membranes and allows more cells to be exposed to the detergent. Cutting the onions with a knife would be unlikely to break very many of the nuclear membranes inside." } { "Tag": [ "algebra", "polynomial", "number theory proposed", "number theory" ], "Problem": "Polynomial division\r\n\r\na)\tFor what values of k is x-2 a factor x^4 \u2013 5x^3 + 3x + k in Q[x]?\r\n\r\nb)\tFor what values of k is x+1 a factor of x^4 + 2x^3 \u2013 3x^2 + kx + 1 in Z5[x]", "Solution_1": "[b]Lemma:[/b] $ x \\minus{} a$ is a factor of a polynomial $ P(x) \\in K[x]$ (where $ K$ is a field) if and only if $ P(a) \\equal{} 0$." } { "Tag": [ "ceiling function", "logarithms", "floor function", "function", "algebra" ], "Problem": "Let $2^{2005}$ have $x$ digits, and $5^{2005}$ have $y$ digits. Find $x+y$", "Solution_1": "2005 log2 +2005 log5=2005\r\nso there're 2002 digits", "Solution_2": "that is close, [hide=\"answer\"]but consider that you are taking $\\lceil 2005\\log{2} \\rceil+\\lceil 2005*\\log{5} \\rceil$, so the actual answer is $2006$[/hide]", "Solution_3": "Okay, now with this do you round up the values to integers, or do you make it the closest integer?\r\nAlso, is there any way to see without a calculator the value of\r\n$2005*\\log(2)$\r\nor \r\n$2005*\\log(5)$\r\nThanks in advance", "Solution_4": "[quote=\"ljh\"]2005 log2 +2005 log5=2005\nso there're[b] 2002 [/b]digits[/quote]\r\nThere may be typo here, $10^{2005}$ does have $2006$ digits ..The answer is in fact $2006$ \r\n\r\nOne way to thnk about this, is that the number of digits in a product (say $m \\cdot n$) is:\r\n either sum of the digits of $m$ and $n$ (if there is a \"carry over\") \r\n Or One less than the sum of digits (If there is no carry over- in the last digit) \r\n\r\nSo sum of digits, is number of digits in the product ($2^{2005} \\cdot 5^{2005}= 10^{2005}$) or $2006$ since there is a carry over.\r\n(The method is essentially same as ljh's) \r\n\r\nAlso one \"easy fact \" to remember is $2^{10} = 1024 = little\\_bit\\_more\\_than \\ 10^3$ so $log 2$ is a little more than $0.3$ (about $0.30103$)..\r\nOnce you know $log 2$ you can estimate $log 4$ (=2*log 2) or $log 5$ (=(1-log 2)) ... etc", "Solution_5": "[hide]$2005\\log2+2005\\log5=\\log(2^{2005}\\cdot{5^{2005}})=\\log(10^{2005})=2005$. Since $10^{2005}$ has a 1 before its 2005 zeros, there are 2006 digits. [/hide]", "Solution_6": "just to be a bit more rigirous\r\n\r\n[hide]from my previous post, we have\n\n$\\lceil 2005\\log{2} \\rceil+\\lceil 2005*\\log{5} \\rceil$\n\nusing $\\lceil x \\rceil = -\\lfloor -x \\rfloor$\n\n$=-\\lfloor -2005\\log{2} \\rfloor-\\lfloor -2005*\\log{5} \\rfloor$\n\nthen using $\\lfloor x \\rfloor = x - \\{x\\}$\n\n$=2005\\log{2}+2005\\log{5}+\\{ -2005\\log{2} \\}+\\{ -2005*\\log{5} \\}$\n$=2005+\\{ -2005\\log{2} \\}+\\{ -2005*\\log{5} \\}$\n\nsince the top statement was an integer, the bottom statement (being equal to the top) must be an integer, also by the definition of the floor function, we know that $0\\le \\lfloor x \\rfloor <1$ so\n\n$0\\le \\lfloor x \\rfloor + \\lfloor y \\rfloor<2$, and we know that they have fractional parts, so the only possiblilty is that part =1, so\n\n$=2005+1=\\boxed{2006}$[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "If $ a,b,c$ be non-negative real numbers and $ abc\\equal{}1$ Prove that \\[ \\sum_{cyc}\\frac{a}{(\\sqrt{ab}\\plus{}\\sqrt{a}\\plus{}\\sqrt{b}\\plus{}1)^2} \\geq \\frac{3}{16}\\]", "Solution_1": "[quote=\"apollo\"]If $ a,b,c$ be non-negative real numbers and $ abc \\equal{} 1$ Prove that\n\\[ \\sum_{cyc}\\frac {a}{(\\sqrt {ab} \\plus{} \\sqrt {a} \\plus{} \\sqrt {b} \\plus{} 1)^2} \\geq \\frac {3}{16}\n\\]\n[/quote]\r\nBy Cauchy-Schwarz inequality, we have:\r\n$ \\sum \\frac{a}{(\\sqrt {ab} \\plus{} \\sqrt {a} \\plus{} \\sqrt {b} \\plus{} 1)^2}\\equal{}\\sum \\frac{a}{(\\sqrt{a}\\plus{}1)^2(\\sqrt{b}\\plus{}1)^2} \\ge \\sum \\frac{a}{4(a\\plus{}1)(b\\plus{}1)}$\r\nSo, we need to prove:\r\n$ \\sum \\frac{a}{(a\\plus{}1)(b\\plus{}1)} \\ge \\frac{3}{4}$\r\n$ \\Leftrightarrow 4(a(c\\plus{}1)\\plus{}b(a\\plus{}1)\\plus{}c(b\\plus{}1)) \\ge 3(a\\plus{}1)(b\\plus{}1)(c\\plus{}1)$\r\n$ \\Leftrightarrow a\\plus{}b\\plus{}c\\plus{}ab\\plus{}bc\\plus{}ca \\ge 6$\r\nWhich is clearly true by AM-GM: $ a\\plus{}b\\plus{}c\\plus{}ab\\plus{}bc\\plus{}ca \\ge 6\\sqrt[6]{(abc)^3}\\equal{}6$\r\nEquality holds if and only if $ a\\equal{}b\\equal{}c\\equal{}1$", "Solution_2": "Putting $ x\\equal{}\\frac{a}{b},y\\equal{}\\frac{b}{c},z\\equal{}\\frac{c}{a}$ the initial inequality is equivalent to\r\n\r\n$ \\sum_{cyc} \\frac{xz}{(\\sqrt x \\plus{} \\sqrt y)^2(\\sqrt x \\plus{} \\sqrt y)^2} \\geq \\frac{3}{16}$\r\n\r\nNow using $ Q\\minus{}A$ for the denominator we get:\r\n\r\n$ \\sum_{cyc} \\frac{xz}{(\\sqrt x \\plus{} \\sqrt y)^2(\\sqrt x \\plus{} \\sqrt y)^2} \\geq \\sum_{cyc} \\frac{xz}{4(x \\plus{} y)(y \\plus{} z)}$\r\n\r\nNow again a substitution $ x\\plus{}y\\equal{}2p, y\\plus{}z\\equal{}2r, z\\plus{}x\\equal{}2s$ gives the equivalent inequality\r\n\r\n$ \\sum_{cyc} \\frac{(p\\plus{}s\\minus{}r)(s\\plus{}r\\minus{}p)}{pr} \\geq 3$, and this is schur.." } { "Tag": [], "Problem": "Seventy-two 50-cent pieces are lined side-by-side on a table. Joe replaces every second 50-cent piece with a quarter and leaves. Jane replaces every third remaining 50-cent piece with a dime and leaves. Jim replaces every fourth remaining 50-cent piece with a nickel and leaves. What is the number of dollars in the amount of money left on the table? Express your answer to the nearest cent.", "Solution_1": "[hide]\nSo you start with 72 half dollars.\n36 are changed into quarters.\n36 left. \n12 are changed into dimes\n24 left\n6 are changed into nickels.\n18 left\n\nSo 18 half dollars, 36 quarters, 12 dimes, 6 nickels\n$\\$9+\\$9+\\$1.20+\\$0.30=\\$19.50$[/hide]" } { "Tag": [ "geometry", "geometric transformation", "reflection", "calculus", "calculus computations" ], "Problem": "When real numbers $ a,\\ b$ vary subject to $ a^2 \\plus{} b^2 \\equal{} 1$, find the region in which the line $ (1 \\plus{} a)x \\plus{} (1 \\minus{} a)y \\equal{} b$ can be passed.", "Solution_1": "kunny,could you provide any hints?\r\nI looked the problem for a while and I can't figure it out :(", "Solution_2": "all such lines have their intercepts of the same sign. moreover, the lines with negative $ b$ are just reflections across the origin of the lines with positive $ b$. so assume that $ b>0$. then the product of the $ x$- and $ y$- intercepts is constant ($ 1$) and the $ x$-intercept is $ \\frac{b}{1\\plus{}a}\\equal{}\\sqrt{\\frac{1\\minus{}a}{1\\plus{}a}}$, which takes on all positive values. from this we gather that the lines sweep out the entire 2nd and 4th quadrants and the region inside the curves in the 1st and 3rd quadrants with the property that all their tangent lines have intercepts of product $ 1$. (by \"inside\" i mean the side with the origin.)\r\n\r\nso what are these curves? it's easy to check that the two branches of $ y\\equal{}\\frac{1}{4x}$ satisfy the above properties, so they must be the curves defining our region.", "Solution_3": "[quote=\"new_member\"]kunny,could you provide any hints?\nI looked the problem for a while and I can't figure it out :([/quote]\r\n\r\nSuppose that the line passe through the point $ (1,\\ 1)$, $ a,\\ b$ would satisfy $ b\\equal{}2$ and $ a$ is arbiterary real numbers, but these don't satisfy the condition $ a^2\\plus{}b^2\\equal{}1$, thus the line will not pass the point $ (1,\\ 1)$.", "Solution_4": "[quote=\"kunny\"][quote=\"new_member\"]kunny,could you provide any hints?\nI looked the problem for a while and I can't figure it out :([/quote]\n\nSuppose that the line passe through the point $ (1,\\ 1)$, $ a,\\ b$ would satisfy $ b \\equal{} 2$ and $ a$ is arbiterary real numbers, but these don't satisfy the condition $ a^2 \\plus{} b^2 \\equal{} 1$, thus the line will not pass the point $ (1,\\ 1)$.[/quote]\r\n\r\ni think he understood the problem statement. he just wanted a hint towards a solution.", "Solution_5": "[quote=\"pluristiq\"][quote=\"kunny\"][quote=\"new_member\"]kunny,could you provide any hints?\nI looked the problem for a while and I can't figure it out :([/quote]\n\nSuppose that the line passe through the point $ (1,\\ 1)$, $ a,\\ b$ would satisfy $ b \\equal{} 2$ and $ a$ is arbiterary real numbers, but these don't satisfy the condition $ a^2 \\plus{} b^2 \\equal{} 1$, thus the line will not pass the point $ (1,\\ 1)$.[/quote]\n\ni think he understood the problem statement. he just wanted a hint towards a solution.[/quote]\r\n\r\nWe have 3 Solution at least. I was at loss whether this problem should be posted in the forum or not.\r\nBecause the problem can be solved in a basic solution. I should have posted the problem in Inetermidate section." } { "Tag": [ "ratio" ], "Problem": "The ratio of the measures of the angles of a triangle is 3:2:1. Given that the shortest side of the triangle is 12 meters long, what is the number of meters in the longest side of the triangle?", "Solution_1": "Someone please post how to do this problem!", "Solution_2": "We have $ 3x \\plus{} 2x \\plus{} x \\equal{} 180\\implies x \\equal{} 30$, which says that our triangle is a $ 30$-$ 60$-$ 90$ right triangle. Therefore, the longest side is the hypotenuse and therefore $ 2(12) \\equal{} \\boxed{24}$." } { "Tag": [ "limit", "logarithms", "algebra unsolved", "algebra" ], "Problem": "Let $x_n$ be root of equation: $x^n=x^2+x+1$.Find $\\displaystyle \\lim_{n \\rightarrow \\infty}n(x_n-1)$", "Solution_1": "Short answer: [hide]The limit is ln(3).[/hide]", "Solution_2": "It is clear $x\\to 1$, so $x^{2}+x+1\\to 3$, therefore it is impossible that $x>1+\\frac{\\ln(3+\\varepsilon)}{n}$ or $x<1+\\frac{\\ln(3-\\varepsilon)}{n}$ for large $n$.", "Solution_3": "If $x_{n}=-1$ for even n?", "Solution_4": "Clearly he considers positive root.", "Solution_5": "We have $x_{n}\\longrightarrow 1$ and $x_{n}^{n}=x_{n}^{2}+x_{n}+1$, hence $x_{n}^{n}\\longrightarrow 3$. Therefore, $n\\ln x_{n}\\longrightarrow \\ln 3$. Since $\\lim_{n\\rightarrow+\\infty}\\frac{\\ln x_{n}}{x_{n}-1}=1$, it follows that $\\lim_{n\\rightarrow+\\infty}n(x_{n}-1)=\\ln 3.$", "Solution_6": "Can be proved, that \\[\\lim_{n\\to \\infty }n^{2}(x_{n}-1-\\frac{\\ln3}{n}) =\\frac{3\\ln 3 }{2}.\\]" } { "Tag": [], "Problem": "A juggler throws a ball straight up into the air with a speed of 10m/s, with what speed would she need to throw a second ball helf a second later, starting from the same position as the first, in order to hit the first ball at the top of its trajectory?", "Solution_1": "$ v_{f}\\equal{}10\\minus{}gt\\Rightarrow t\\equal{}g/10$ seconds for the first ball to reach the top of its path. This happens at a height of $ 0\\equal{}100\\minus{}2gd\\Rightarrow d\\equal{}50/g$ meters. The second ball must reach a height of $ 50/g$ meters in $ (g\\minus{}5)/10$ seconds. So\r\n$ \\frac{50}{g}\\equal{}v_{0}\\frac{g\\minus{}5}{10}\\plus{}g\\frac{g\\minus{}5}{20}$\r\nThen you can solve for $ v_{0}$. I would check my math though, I did this pretty quickly.", "Solution_2": ":P $ 27.5 m/s$", "Solution_3": "Did you go based on my work or independently? If you used my work, check it.", "Solution_4": "I got 12.35 m/s", "Solution_5": "$ h \\equal{} 10t \\minus{} 4.9 t^2$ for the original equation.\r\nSolving for the maximum, it occurs at $ \\frac {5}{4.9}$ and has a height of $ \\frac {25}{4.9}$.\r\n\r\nIn terms of a time of reference, the second ball should hit the other ball $ \\frac {2.55}{4.9}$ seconds after it is thrown.\r\n\r\nSetting another equation in the form of $ h \\equal{} st \\minus{} 4.9 t^2$, we have a new equation (remember we are solving for $ s$:\r\n\r\n[hide=\"go through the algebra...\"]$ \\frac {25}{4.9} \\equal{} s \\cdot \\frac {2.55}{4.9} \\minus{} 4.9 \\cdot \\frac {2.55^2}{4.9^2}$. Multiplying both sides $ 4.9$, we get\n[hide=\"and...\"]$ 25 \\equal{} 2.55s \\minus{} 2.55^2$; solving for $ s$ we get\n$ s \\equal{} \\frac {25 \\plus{} 2.55^2}{2.55}$\n\n$ \\equal{} \\boxed{12 m/s}$ (using 2 significant figures). I got the same answer as kysuke[/hide][/hide]\r\n\r\n\r\nThe key to many problems like this is being able to set the right equation with the right variables. If you understand what's being asked, you should be able to solve it.", "Solution_6": "i m rusted :( \r\ncalculation mistake :maybe:" } { "Tag": [ "real analysis", "real analysis theorems" ], "Problem": "Quick question for a project: has the completeness of the real numbers been proven or is it axiomatic, and if it has been proven where can I find a proof? Thank you.", "Solution_1": "This might be of interest:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17848[/url]", "Solution_2": "There are at least three available constructions of the real numbers. Each has its own strengths and weakness, and all involve some fairly tedious arguments.\r\n\r\n1. Dedekind cuts. A number of analysis textbooks do go through this construction, and it is the oldest. Dedekind made it explicit, but you could argue that it is foreshadowed by the Eudoxus defenition of proportion as written in Euclid's [i]Elements.[/i]\r\n\r\n2. As equivalence classes of Cauchy sequences of rational numbers.\r\n\r\n3. As decimal (or whatever base you prefer) expansions.\r\n\r\nThe tedium arises in each construction because you need to prove that you have constructed a complete ordered field. That's quite a few axioms to verify.\r\n\r\nOne can also prove that given any two complete ordered fields, one can find an order-preserving isomorphism between them. This theorem then proves that the real numbers are essentially unique.\r\n\r\nAll of these constructions are abstract and non-intuitive. I might even go so far as to call them \"artificial.\" Don't read too much into the word \"real\" - this is a product of human abstract thought." } { "Tag": [], "Problem": "Cau cau. Dostaly se mi do rukou jedny od kamarada, ktery byl minuly rok v Jankach. Tak jednu z nich posilam.\r\n\r\nMirek", "Solution_1": "Soust\u0159ed\u011bn\u00ed v J\u00e1nk\u00e1ch je fajn, Doc. \u0160im\u0161a n\u00e1m loni rozdal spoustu dobr\u00fdch p\u0159\u00edklad\u016f z teorie \u010d\u00edsel, ty nejlep\u0161\u00ed uk\u00e1zal na p\u0159edn\u00e1\u0161ce. Doc. \u0160vr\u010dek p\u0159edn\u00e1\u0161el n\u011bkter\u00e9 zaj\u00edmav\u00e9 v\u011bty z planimetrie (Menelaova, Cevova, Van Aubelova), Dr. Leischner na p\u0159edn\u00e1\u0161ce odvodil Ptolemaiovu nerovnost, Bretschneiderovo pravidlo a vzorce pro obsahy \u010dty\u0159\u00faheln\u00edk\u016f (obecn\u00fdch). Dr. Zhouf mluvil o sou\u010dtech aritmetick\u00fdch a jin\u00fdch posloupnost\u00ed. \r\nA p\u0159edn\u00e1\u0161ka doktora Hor\u00e1ka zahrnuj\u00edc\u00ed n\u011bkolik planimetrick\u00fdch p\u0159\u00edklad\u016f ... :)" } { "Tag": [ "Support" ], "Problem": "(1)The people living in a small village are supporters of either teams A, B, C or D. In terms of political parties, they are supporters of either parties E, F or G. Some of these people are liars while the rest are truth tellers. The liars lie at every answer they give, while the truth tellers always tell the truth. \r\nThe questions (first four questions were asked to all of the villagers, next three were asked to all of the male villagers, and the last three were asked to all of the female villagers) and the number of people that answered \"yes\" are as follows:\r\n\r\nDo you support A ? (50 Yes answers) \r\nDo you support B ? (60 Yes answers) \r\nDo you support C ? (70 Yes answers) \r\nDo you support D ? (80 Yes answers) \r\n\r\nDo you support E ? (25 Yes answers)\r\nDo you support F ? (30 Yes answers) \r\nDo you support G ? (35 Yes answers)\r\n\r\nDo you support E ? (30 Yes answers)\r\nDo you support F ? (40 Yes answers) \r\nDo you support G ? (50 Yes answers)\r\n\r\nHow many people are living in this village?", "Solution_1": "[hide=\"answer\"]160[/hide]\n\n[hide=\"solution\"] We do a system of equation.\nNotation is natural:\n$ T_{a}$ - number of truth tellers supporters of A, etc\n$ T_{e}^{m}$ - number of truth tellers male supporters of E, etc\n\n$ T_{a} \\plus{} L_{b} \\plus{} L_{c} \\plus{} L_{d} \\equal{} 50$\n$ L_{a} \\plus{} T_{b} \\plus{} L_{c} \\plus{} L_{d} \\equal{} 60$\n$ L_{a} \\plus{} L_{b} \\plus{} T_{c} \\plus{} L_{d} \\equal{} 70$\n$ L_{a} \\plus{} L_{b} \\plus{} L_{c} \\plus{} T_{d} \\equal{} 80$\nA sum: $ T \\plus{} 3L \\equal{} 260$\n\n$ T_{e}^{m} \\plus{} L_{f}^{m} \\plus{} L_{g}^{m} \\equal{} 25$\n... (i'am lazy)\n$ L_{e}^{f} \\plus{} L_{f}^{f} \\plus{} T_{g}^{f} \\equal{} 50$\nA sum: $ T \\plus{} 2L \\equal{} 210$\n\nThe rest is clear: $ T \\plus{} L \\equal{} 160$\n[/hide]", "Solution_2": "thank you !" } { "Tag": [ "limit", "logarithms" ], "Problem": "Find\r\n$\\lim_{x \\to 1} \\frac{\\frac{xlnx}{x-1}-1}{x-1}$\r\ndon't use hopital", "Solution_1": "we have\r\n\r\n$\\lim_{x \\to +\\infty}\\left(1+\\frac{1}{x}\\right)^x=e$\r\n\r\nfrom which\r\n\r\n$\\lim_{x \\to 0^+}(1+x)^{\\frac{1}{x}}=e$\r\n\r\nfrom which\r\n\r\n$\\lim_{x \\to 0^+}\\frac{\\ln(1+x)}{x}=1$\r\n\r\nand from which\r\n\r\n$\\lim_{x \\to 1}\\frac{\\ln(x)}{x-1}=1$\r\n\r\nby some substitutions.\r\n\r\nso in our limit:\r\n\r\n$\\lim_{x \\to 1}\\frac{x\\frac{\\ln x}{x-1}-1}{x-1}=\\lim_{x \\to 1}\\frac{x\\cdot 1 -1}{x-1}=1$", "Solution_2": "Everything works until your last step. It is unfortunately not valid in general to take a limit of only some of the occurences of $x$ but not others.\r\n\r\nI would personally solve this with a [hide]Taylor series,[/hide] but there is probably another way.", "Solution_3": "[quote=\"mathe\"]Find\n$\\lim_{x \\to 1} \\frac{\\frac{xlnx}{x-1}-1}{x-1}$\ndon't use hopital[/quote]\r\n\r\nhopital??", "Solution_4": "He means [url=http://mathworld.wolfram.com/LHospitalsRule.html]L'Hospitals Rule[/url]. (In fact, the question can be solved reasonably easily with it -- I assume the author is looking for a distinct solution.)", "Solution_5": "[quote=\"JBL\"]He means [url=http://mathworld.wolfram.com/LHospitalsRule.html]L'Hospitals Rule[/url]. (In fact, the question can be solved reasonably easily with it -- I assume the author is looking for a distinct solution.)[/quote]\r\n\r\nOhh I see thanks..." } { "Tag": [ "logarithms", "induction", "algebra proposed", "algebra" ], "Problem": "$ 0[a,b]$ such that $ f(x)f(f(x)) \\equal{} x^2$ for all $ x$ in $ [a,b]$", "Solution_1": "$ g(x): \\equal{}\\ln f(e^x): [\\ln a,\\ln b]\\to [\\ln a,\\ln b]$ with $ g(x)\\plus{}g(g(x))\\equal{}2x$ for all $ x\\in [\\ln a,\\ln b]$.\r\nThen by induction \r\n$ g^n(x): \\equal{}\\underbrace{g\\circ \\cdots\\circ g}_{n\\text{ times}}(x)\\equal{}{2x\\plus{}g(x)\\over 3}\\plus{}(\\minus{}2)^n{x\\minus{}g(x)\\over 3}$\r\nand $ g^n(x)\\in [\\ln a,\\ln b]$ for all $ x\\in [\\ln a,\\ln b]$ and $ n\\in\\mathbb{N}$.\r\nSo $ g(x)\\equal{}x$ for all $ x\\in [\\ln a,\\ln b]$\r\nand $ f(x)\\equal{}x$ for all $ x\\in [ a, b]$.", "Solution_2": "See here : http://www.mathlinks.ro/viewtopic.php?t=193350" } { "Tag": [ "logarithms", "trigonometry" ], "Problem": "1. Which of the following is equivalent to $ \\log{\\left(\\frac {\\sin{(2a)}}{2}\\right)}$.\r\n\r\n$ (1) \\ \\log{(\\sin{a})}$\r\n\r\n$ (2) \\ \\frac {1}{2}\\log{(2\\sin{a})}$\r\n\r\n$ (3) \\ \\log{(\\sin{a})}\\log{(\\cos{a})}$\r\n\r\n$ (4) \\ \\log{(\\sin{a})} + \\log{(\\cos{a})}$\r\n\r\nCould you please show work to both of these problems so that I can learn how to do them?\r\n\r\nThanks.\r\n\r\n[EDIT] Sorry, but I typed the wrong the question.", "Solution_1": "Hmmm...\r\n\r\nThese are interesting questions. Does anyone have a solution?", "Solution_2": "[quote=\"314.math\"]Hmmm...\n\nThese are interesting questions. Does anyone have a solution?[/quote]\r\n\r\ni dunno\r\nthe only time i've seen trig involved with logs is in integrals...\r\nand that path really does not look optimistic", "Solution_3": "well, this may seem like spam. i have no idea how to do this specific problem, but here is a logarithmic and trigonometric problem. \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=409912#409912", "Solution_4": "the log property log_n(x)^y=ylog_n(x) will get you the answer for number 1\r\n\r\nThen tan x=sinx/cosx", "Solution_5": "[hide]$ \\log(tan^2(y))$ can be rewritten as $ \\log(sin^2y) - \\log(cos^2y)$\nThis is the same as: $ 2\\log(siny) - 2\\log(cosy)$\n\n $ 2(\\log(siny) - \\log(cosy))$[/hide]\n\nEdit: The problem that the above solution corresponded to was removed.\n\nSolution to New #1\n\n[hide]$ \\log{\\left(\\frac {\\sin{(2a)}}{2}\\right)}$ can be written as\n\n$ \\log{sin(2a)} - \\log{2}$ using properties of logs. Using trig identity $ sin(2a) = 2sin(a)cos(a)$ we can rewrite as\n\n$ \\log{2sin(a)} - \\log{2} + \\log{cos(a)}$\n\nSimplifying, we get $ \\log{sin(a)} + \\log{cos(a)}$[/hide]", "Solution_6": "[quote=\"cognos599\"]well, this may seem like spam. i have no idea how to do this specific problem, but here is a logarithmic and trigonometric problem. \n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=409912#409912[/quote]\r\n\r\nThat doesn't help at all.\r\n\r\nWhat Illos has is correct.\r\n\r\n[hide=\"Classic Log Properties\"]\n\nJust note that:\n\n$ \\log(ab) \\equal{} \\log a \\plus{} \\log b$\n$ \\log(a/b) \\equal{} \\log a \\minus{} \\log b$\n\nThose two were only things you needed to know for this problem. [/hide]", "Solution_7": "Well in that case then, these problems are extremely simple...finding an equal expression can mean just multiplying it by 2/2 or something :P", "Solution_8": "How did you get from $ \\log{2sin(a)} \\minus{} \\log{2} \\plus{} \\log{cos(a)}$\r\n\r\nto\r\n\r\nSimplifying, we get $ \\log{sin(a)} \\plus{} \\log{cos(a)}$?\r\n\r\nThanks.", "Solution_9": "[hide]\nDefinition\n$ \\sin \\left( 2 \\alpha \\right) \\equal{} 2 \\sin \\alpha \\cos \\alpha$.\n\nThen\n$ \\log \\left( \\frac {\\sin \\left( 2 \\alpha \\right) }{2} \\right) \\equal{} \\log \\left( \\frac {2 \\sin \\alpha \\cos \\alpha}{2} \\right) \\equal{} \\log \\sin \\alpha \\plus{} \\log \\cos \\alpha$.\nWhich is $ (4)$.\n[/hide]", "Solution_10": "[quote=\"elcric\"]How did you get from $ \\log{2sin(a)} - \\log{2} + \\log{cos(a)}$\n\nto\n\nSimplifying, we get $ \\log{sin(a)} + \\log{cos(a)}$?\n\nThanks.[/quote]\r\n\r\n$ \\log{2sin(a)}$ is the same as $ \\log{2} + \\log{sin(a)}$.\r\n\r\nPutting this back into the original, we get $ \\log{2} + \\log{sin(a) - \\log{2} + \\log{cos(a)} = \\log{sin(a) + \\log{cos(a)}}}$" } { "Tag": [ "calculus", "integration", "logarithms", "function", "calculus computations" ], "Problem": "Any help is greatly appreciated. How would you do:\r\n\r\n$\\int{\\frac{x}{x^{4}+2x^{2}+1}}$", "Solution_1": "The denominator is a perfect square. Write it as such, and look for a substitution.", "Solution_2": "Thanks! Further problem I'm having: I cannot figure out the mistake in one of these two methods of integration.\r\n\r\n$\\int{\\frac{1}{2x+4}}=\\frac{1}{2}\\ln{(2x+4)}$\r\n\r\n$\\int{\\frac{1}{2x+4}}=\\frac{1}{2}\\int{\\frac{1}{(x+2)}}=\\frac{1}{2}\\ln{(x+2)}$\r\n\r\nClearly there is a problem, since these give $\\frac{1}{2}\\ln{(2x+4)}=\\frac{1}{2}\\ln{(x+2)}$", "Solution_3": "considering constants of integration, two result is same. ($\\frac{1}{2}\\ln{(2x+4)}-\\frac{1}{2}\\ln{(x+2)}= \\ln\\sqrt{2}$, so they differ by only a constant function as expected)", "Solution_4": "[quote=\"bthings_2000\"]Clearly there is a problem, since these give $\\frac{1}{2}\\ln{(2x+4)}=\\frac{1}{2}\\ln{(x+2)}$[/quote]\r\nI'm repeating what sos440 said, but just for emphasis: that's not what \"these give.\" What you can actually derive from this result is that \r\n\r\n$\\frac{1}{2}\\ln{(2x+4)}=\\frac{1}{2}\\ln{(x+2)}+C,$ which is true. Both computations are correct." } { "Tag": [ "MATHCOUNTS", "geometry", "LaTeX", "FTW", "3D geometry", "octahedron", "email" ], "Problem": "HI EVERYONE. I WAS WONDERING WHETHER ANY OF YOU WOULD BE WILLING TO JOIN MY MOCK MATHCOUNTS TOURNAMENT... I NEED HELPERS CUZ IM NOT REALLY GOOD WITH ATTACHMENTS AND STUFFS AND I NEED MORE PROBLEMS!!! \r\n\r\n[b]I AM GOING TO HOLD THIS IN 2-3 WEEKS WHEN PEOPLE HAVE FINISHED SMASHER\"S AND MAYBACH's[/b]\r\n\r\n[hide=\"Participants\"]Please put your names here and copy and paste if interested. Thanks[/hide]", "Solution_1": "I'll be a helper :)", "Solution_2": "Sure cricket I'll be in your mock mc.", "Solution_3": "K cool. Barbie can you walk me through attaching pdf's and stuff??", "Solution_4": "[hide=\"Helpers(1)\"]\nBarbieRocks\n[/hide]\n[hide=\"Participants(2)\"]\nagentcx\nMaybach \n[/hide]\r\n\r\nI know how to create PDFs using open office. I could show you if you PM me(I still want to be a participant though.)", "Solution_5": "[hide=\"Participants(3)\"]agentcx \nMaybach \nBluecarneal[/hide]\n[hide=\"Helpers(1)\"]\nBarbieRocks [/hide]", "Solution_6": "[hide=\"Participants(3)\"]agentcx \nMaybach \nBluecarneal\nPowerOfPi[/hide]\n[hide=\"Helpers(1)\"]\nBarbieRocks [/hide]", "Solution_7": "Wow...we are really getting too many tourneys around here.\r\nBut since CRICKET229 kindly joined mine, :wink: \r\nIll be a helper.\r\n\r\n[hide=\"Participants(4)\"]agentcx \nMaybach \nBluecarneal \nPowerOfPi[/hide]\n\n[hide=\"Helpers(2)\"]KingSmasher3\nBarbieRocks[/hide]", "Solution_8": "I probably could help but I don't think I should because I suck at computer stuff. \r\n[hide]agentcx\nMaybach\nBluecarneal\nPowerOfPi\nThe Hobbit[/hide]\n\n\n[hide]KingSmasher3\nBarbieRocks[/hide]", "Solution_9": "Okay cool, King and Barbie if you can please pm your suggestions and other stuffs. Also I am kind of new to making pdfs and such (like making a Word Document into a PDF file) so I'll take any suggestions or help in that too.\r\n\r\nMy School Round will most likely be out after Jan 5th.", "Solution_10": "[hide=\"Participants (5)\"]agentcx \nMaybach \nBluecarneal \nPowerOfPi\nbbgun34[/hide]\n\n[hide=\"Helpers (2)\"]KingSmasher3 \nBarbieRocks[/hide]", "Solution_11": "As soon as I have at least 10 participants, we will start so never mind about the January thing. I just did a quick rough draft of sprint round. Looks easy at first and gets pretty hard at the end but dont worry chapter will be easier i hope. King and Barbie I'll send you my sprint and target when I'm done, please critique my stupid questions and add your own.", "Solution_12": "No problem;\r\nAlso, [color=red]please don't mind my due dates because mine is a [b]biannual [/b]competition, a.k.a. its continuous.[/color]", "Solution_13": "If possible, I would like to help. I can easily create PDFs, and type up the problems, except for the geometry ones. If I can't help, I would like to participate. BTW, Cricket, don't make the problems to hard LOL.", "Solution_14": "KK thats fine. I will pm you and king and barbie my rough school sprint", "Solution_15": "Hey, you people need to submit. Cuz if you dont submit this you will not be participating in the State Competition :D", "Solution_16": "Can we not do the team? cause really no one is doing it.", "Solution_17": "you didn;t mark mine...\r\n= =", "Solution_18": "sorry, is it too late for me to sign up?", "Solution_19": "Yup. It seems as if this thread is basically dead.", "Solution_20": "[quote=\"explogabloger\"]sorry, is it too late for me to sign up?[/quote]\r\n\r\nThough, it is too late for you to sign up, you can still take the test, and pm Cricket for the answers so that you can check how you did. Look back in the previous pages (of this thread) to find copies of the test.", "Solution_21": "Just asking here (and sorry for reintroducing topic):\n\nWho wants to take a national sprint and target (and maybe team). Anyone can take it, however it will be due fairly soon so.....heres the participants list. Only up to 16 may join so sign up quickly.\n\n[hide=\"Participants(0)\"]\n\n[/hide]\n[hide=\"Helpers(0)\"]\n\n[/hide]\n\nI need 2 helpers as well. So please sign up if interested. \n\nThanks!\n~CRICKET", "Solution_22": "Sign up! !!! /joins", "Solution_23": "I will help.\n\n[hide=\"Helpers\"]\n$/LaTeX$\n[/hide]\n\n[hide=\"Participants\"]\nksun48\n[/hide]\n\nPlease copy the lists.\n\nCricket, feel free to PM me or email me, you know my email.", "Solution_24": "Not too much excitement so far. \n\nWhere should I advertise?\n\nOh, and Latex I couldnt find the template on agmath so where can I find it?", "Solution_25": "Signature!!! Best place.\nAlso the template is [url]http://www.agmath.com/169763.html[/url]\nUnder the Warm-ups.", "Solution_26": "Oops. This thread didn't have too much excitement so I quit going to this for some time. At least one topic is back. :D \n\nI join.", "Solution_27": "Hmm so not much excitement so I just made a target.....\n\nhere It is:", "Solution_28": "I'm assuming (for the DoodleJump question) you don't get points for altitude?", "Solution_29": "Lol...nope. You would in a regular doodle jump game though." } { "Tag": [ "probability", "calculus", "derivative", "expected value" ], "Problem": "In a game, a player tosses a biased coin repeatedly until two successive tails occur, when the game terminates. For each head which occurs the player wins $ \\$ 1$. If $ E$ is the expected number of tosses of the coin in the course of a game, and $ p$ is the probability of a head, explain why\r\n\\[ E \\equal{} p(1 \\plus{} E) \\plus{} (1 \\minus{} p)p(2 \\plus{} E) \\plus{} 2(1 \\minus{} p)^2 ,\r\n\\]\r\nand hence determine $ E$ in terms of $ p$. Find also, in terms of $ p$, the expected winnings in the course of a game.\r\n\r\nA second game is played, with the same rules, except that the player continues to toss the coin until $ r$ successive tails occur. Show that the expected number of tosses in the course of a game is given by the expression $ \\frac {1 \\minus{} q^r}{pq^r}$, where $ q \\equal{} 1 \\minus{} p$.", "Solution_1": "I don't know how to post a tree-diagram, but we can consider three possibilities:\r\n\r\n$ {H}$: game starts again; probabilty $ \\equal{} p$,; expected number of games $ \\equal{} 1 \\plus{} E$.\r\n$ {TH}$: game starts again; probability $ \\equal{} (1 \\minus{} p)p$; expected number of games $ \\equal{} 2 \\plus{} E$,\r\n$ {TT}$: games ends; probability $ \\equal{} (1 \\minus{} p)^{2}$; expected number of games $ \\equal{} 2$.\r\n\r\nThese three sequences are mutually exclusive and collectively exhaustive, so the expected number of games is:\r\n\r\n$ E \\equal{} p(1 \\plus{} E) \\plus{} (1 \\minus{} p)p(2 \\plus{} E) \\equal{} 2(1 \\minus{} p)^{2}$\r\n\r\n$ \\equal{} p \\plus{} pE \\plus{} 2p \\minus{} 2p^{2} \\plus{} (p \\minus{} p^{2})E \\plus{} 2 \\minus{} 4p \\plus{} 2p^{2}$\r\n\r\nTherefore $ (1 \\minus{} 2p \\plus{} p^{2})E \\equal{} 2 \\minus{} p$\r\n\r\n(My backslash key doesn't work: I'll copy one and get back to you.)\r\n\r\nTherefore $ E \\equal{} \\frac {2 \\minus{} p}{(1 \\minus{} p)^{2}}$\r\n\r\nExpected winnings, $ W \\equal{} p(1 \\plus{} W) \\plus{} (1 \\minus{} p)p(2 \\plus{} W) \\plus{} 0.(1 \\minus{} p)^{2}$\r\n\r\nThis gives $ W \\equal{} \\frac {p(2 \\minus{} p)}{(1 \\minus{} p)^{2}}$\r\n\r\nHave to eat. Will try to get back later. Hope this helps.\r\n\r\nBack again.\r\n\r\nSecond game:\r\n\r\n$ E\\equal{}p(1\\plus{}E) \\plus{} pq(2\\plus{}E) \\plus{} pq^{2}(3\\plus{}E) \\plus{} ... \\plus{} pq^{r\\minus{}1}(r\\plus{}E) \\plus{} rq^{r}$\r\n\r\n$ \\equal{} p(1 \\plus{} 2q \\plus{} 3q^{2} \\plus{} .. \\plus{} rq^{r\\minus{}1}) \\plus{}pE(1\\plus{}q\\plus{}q^{2} \\plus{} ... \\plus{} q^{r\\minus{}1}) \\plus{} rq^{n}$\r\n\r\n$ \\equal{}p\\frac{d\\sum_{n\\equal{}1}^{r}q^{n}}{dq} \\plus{}pE\\frac{(1\\minus{}q^{r})}{1\\minus{}q} \\plus{} rq^{r}$\r\n\r\n$ \\equal{}p\\frac{d}{dq}\\frac{(q(1\\minus{}q^{r})}{(1\\minus{}q)} \\plus{}pE\\frac{(1\\minus{}q^{r})}{1\\minus{}q} \\plus{} rq^{n}$\r\n\r\n$ \\equal{}\\frac{p(1\\minus{}(r\\plus{}1)q^{r} \\plus{} rq^{r\\plus{}1}}{(1\\minus{}q)^{2}} \\plus{} pE\\frac{(1\\minus{}q^{r})}{1\\minus{}q}\\plus{} rq^{n}$\r\n\r\n$ \\equal{}\\frac{(1\\minus{}(r\\plus{}1)q^{r} \\plus{} rq^{r\\plus{}1})}{p} \\plus{} E(1\\minus{}q^{r}) \\plus{} rq^{n}$\r\n\r\n$ \\Rightarrow q^{r}E \\equal{} \\frac{(1\\minus{}(r\\plus{}1)q^{r} \\plus{} rq^{r\\plus{}1})}{p} \\plus{} rq^{r}$\r\n\r\n$ \\Rightarrow pq^{r}E \\equal{} 1 \\minus{} (r\\plus{}1)q^{r} \\plus{} rq^{r\\plus{}1} \\plus{} rpq^{r}$\r\n\r\n$ \\equal{} 1 \\minus{} rq^{r} \\minus{} q^{r} \\plus{}rq^{r\\plus{}1} \\plus{} rpq^{r}$\r\n\r\n$ \\equal{}1\\minus{}rq^{r}(1\\minus{}q\\minus{}p) \\minus{} nq^{r}$\r\n\r\n$ \\equal{}1\\minus{}q^{r}$\r\n\r\n$ \\Rightarrow E \\equal{} \\frac{1\\minus{}q^{r}}{pq^{r}}$\r\n\r\nI hope this helps you Aidan. Having seen my rating drop, I contacted orl who assures me it is often vindictiveness. But I almost left mathlinjks.ro permanently and returned to my first love: languages. Being able (hopefully) to help you has renewed my hope and faith in humanity.", "Solution_2": "Thanks, AndrewTom! Second part helped loads. I forgot to \"mask\" $ 1 \\plus{} 2q \\plus{} 3q^2 \\plus{} \\ldots$ as the derivative of $ q \\plus{} q^2 \\plus{} q^3 \\plus{} \\ldots$ so I couldn't proceed. As for languages.. they're great too :) But, keep up the math. Best language ever!" } { "Tag": [ "linear algebra", "matrix", "algorithm", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\n\r\nI wrote this problem after inspired by a similar problem written by my friend. I don't have an elegant solution though, so please post yours :D\r\n\r\nRing $R = \\{a+bi | a,b \\in \\mathbb{Z}\\}$ has an ideal $M = (3+i)R$.\r\n1. Prove that $M$ is not maximal, and find all the maximal ideals that contain $M$\r\n2. Show that $Z/M$ is isomorphic to $\\mathbb{Z}_{2}\\times \\mathbb{Z}_{5}$. Find /describe the isomorphism explicitly.", "Solution_1": "$\\mathcal{R}$ is an euclidean ring.\r\n1) Take one of the divisors of $3+i$: $1+i,1-i,2-i,1+2i$.\r\n2) $\\mathcal{R}\\sim\\mathbb{Z}^{2}$ and the product by $3+i$ is isomorphic to the product \\[\\begin{pmatrix}a\\\\b\\end{pmatrix}\\rightarrow{M}\\begin{pmatrix}a\\\\b\\end{pmatrix}{=}\\begin{pmatrix}3&-1\\\\1&3\\end{pmatrix}\\begin{pmatrix}a\\\\b\\end{pmatrix}\\]\r\nStandard result: if $M\\in{\\mathcal{M}_{n}(\\mathbb{Z})}$ has a non zero det then there exist $P,Q\\in{GL_{n}(\\mathbb{Z})}$ s. t. $PMQ=diag(d_{1},\\cdots,d_{n})$ where $d_{i}>0,d_{i}|d_{i+1}$ (there exists only one such diag matrix and there exists an algorithm which gives explicitly a solution $P,Q$). Moreover the group $\\mathbb{Z}^{n}/M(\\mathbb{Z}^{n})$ is isomorphic to the product group $\\mathbb{Z}_{d_{1}}\\cdots\\mathbb{Z}_{d_{n}}$.\r\nHere \\[M=\\begin{pmatrix}3&-1\\\\1&0\\end{pmatrix}\\begin{pmatrix}1&0\\\\0&10\\end{pmatrix}\\begin{pmatrix}1&3\\\\0&1\\end{pmatrix}\\] and $\\mathbb{Z}^{2}/M(\\mathbb{Z}^{2})$ is isomorpic to $\\mathbb{Z}_{10}$. \\[M(\\mathbb{Z}^{2})=\\begin{pmatrix}3&-1\\\\1&0\\end{pmatrix}\\begin{pmatrix}1&0\\\\0&10\\end{pmatrix}(\\mathbb{Z}^{2})=\\{\\begin{pmatrix}3x-10y\\\\x\\end{pmatrix}|x,y\\in\\mathbb{Z}\\}\\]. If we choose $x=0$ then $\\mathbb{Z}^{2}/M(\\mathbb{Z}^{2})$ is represented by $\\{0,i,\\cdots,9i\\}$ with the relation $a+ib\\approx{i((b-3a) mod 10)}$. It remains to prove the isomorphism for the ring structure:\r\nfor example: $(2+7i)(9+5i)\\approx{i(-2i)}\\approx{-6i}$ and in the other hand $-17+73i\\approx{4i}$.\r\nCAUTION: the product * over $\\mathbb{Z}_{10}$ isn't the standard product: we have seen that $1*(-2)=-6$ and more generally $a*b=3ab$. Here the unit is $-3$. $gcd(3,10)=1$ thus this ring is isomorphic to the standard ring $\\mathbb{Z}_{10}$ ." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $p$ be an odd prime number and $2^{a}-1=1\\cdot 3\\cdot 5\\cdot ...\\cdot p$.\r\nProve that $a$ is divisible by $p-1$.\r\n\r\n[i]Edited by Myth[/i]", "Solution_1": "who can help me slove this problem?thanks" } { "Tag": [ "function", "induction", "linear algebra" ], "Problem": "Hey Guys...\r\n\r\nI am entering my 2nd year in B.Sc. Math. I have some questions regarding the courses I am taking. I do the routine: take notes.. do all practice questions... \"try\" to start studyin well in advance for midterms and finals but like most of you know.. it\u2019s pretty hard to do with a full course load every term. I would just like to know how to tackle these courses. I know what the courses are about 'cause i've looked at the course outlines and the texts a bit. \r\n\r\n[b]Courses:[/b] Calc 3, Calc 4, Linear Algebra, Mathematical Proof(Sets and functions; induction; cardinality; properties of the real numbers; sequences, series, and limits. Logic, structure, style, and clarity of proofs emphasized throughout... and btw this course has the highest failure rate in the dept for some reason) \r\n\r\nIn addition to these 4 math courses... I am taking chem n some comp sci courses as well (planning to go into math/biochem program). \r\n\r\n\r\nThanks again !!!", "Solution_1": "This forum's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]FAQ about Preparing for College Math Study[/url] can be helpful for your questions. Follow the links, and practice the advice on each link.", "Solution_2": "[quote=\"punjm\"]I am entering my 2nd year in B.Sc. Math. I have some questions regarding the courses I am taking.\n\n...\n\nMathematical Proof ... this course has the highest failure rate in the dept for some reason) [/quote]\r\n\r\nYou didn't actually ask any questions, though.\r\n\r\nMost likely, the Mathematical Proof class has a high failure rate because math as a rigorous, proof-based field seems quite different than math as conventionally taught through calc 2, and lots of people don't know how to make the adjustment." } { "Tag": [ "logarithms", "FTW" ], "Problem": "What is the least integer value of $ n$ for which $ \\left( \\frac{2}{3}\\right)^n < \\frac{1}{5}$?", "Solution_1": "Use trial and error (that's what I did), and you find that $ \\frac{16}{81} < \\frac{1}{5}$, so the answer is $ \\boxed{4}$.", "Solution_2": "hello, taking the logarithm of both sides we have $ n\\cdot \\ln(\\frac {2}{3}) < \\ln(\\frac {1}{5})$. From this we have $ n > \\frac {\\ln(\\frac {1}{5})}{\\ln(\\frac {2}{3})} \\equal{} \\frac { \\minus{} \\ln(5)}{\\ln(2) \\minus{} \\ln(3)}\\approx 3.969362295916118$. So the answer is $ \\boxed{4}$. \r\nSonnhard.", "Solution_3": "i believe you mean $ \\frac{\\minus{}\\ln(5)}{\\ln(2)\\minus{}\\ln(3)}.$", "Solution_4": "Even though using natural logarithms may be the right way, for FTW, I recommend my way.\r\n\r\nAnd don't you need a calc for $ ln x$?\r\n\r\nAt least I do... :oops:", "Solution_5": "yes, your way is more effecient than Dr. Sonnhard Graubner's.\r\n\r\ni was just pointing out Dr's mistake." } { "Tag": [ "inequalities", "combinatorics unsolved", "combinatorics" ], "Problem": "try to solve the problem 3 / 4th sel test 2003, when the alphabet has only 3 letters.", "Solution_1": "What is the \"problem 3 / 4th sel test 2003\" or where is it possible to get it ? Thanks.", "Solution_2": "I think he means the Third problem from the fourth Romanian Selection test 2003.\r\nThe Romanian TST van be found at http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26\r\n\r\nTHe original problem is :\r\nA word consists of n letters from the alphabet {a, b, c, d}. A word is called convoluted if it has two consecutive identical blocks of letters. For example, caab and cababdc are convoluted, but abcab is not. Prove that the number of non-convoluted words with n letters is greater than 2^n.\r\n\r\nSo I think he means we should look for a similar inequality (and prove it of course) when the alphabet has only three letters.", "Solution_3": "As far as I remember this problem for four letters is in at least three threads on forum, but I never see solution.\r\n\r\nAnyone, solve it at last! ;)" } { "Tag": [ "search", "counting", "derangement", "combinatorics unsolved", "combinatorics" ], "Problem": "there are 4 balls of different colours and 4 boxes of colours same as those of the balls.find the number of ways in which the balls one each in a box could be placed such that a ball does not go to a box of its own colour", "Solution_1": "For 4 you might as well list everything by hand if you don't know the theory. For the theory, search \"derangement\" or \"rencontres.\"", "Solution_2": "hello\r\nWe have actually discussed this question in our institute when we were in xi th.\r\n[hide=\"Solution\"]\n$ A_{i}$= no. of ways in which we can place a ball with color $ i$ in the box of that color only.\nNo. of required ways=No. of ways these balls can be placed- $ n(A_{1}$ union $ A_{2}$ union $ A_{3}$ union $ A_{4})$\n=$ 4!\\minus{}(4(3!)\\minus{}6(2!)\\plus{}4\\minus{}1)\\equal{}9$[/hide]\r\nthank u", "Solution_3": "Kabi, yes you are correct but it would have been nice if you had inroduced the genral derangements principle to aadil:\r\nThat is $ D_n \\equal{} n! ( 1\\minus{}\\frac{1}{1!}\\plus{} \\frac{1}{2!} \\minus{} \\frac{1}{3!}\\plus{} \\frac{1}{4!}\\minus{}............\\plus{} (\\minus{}1)^n \\frac{1}{n!})$.", "Solution_4": "total number of ways of placing balls in boxes = $ 4!$.\r\n\r\nnumber of permutations in which a ball goes into a box of same colour = 1.\r\n\r\nso, required number of ways = $ 4!$ - $ 1$\r\n\r\n = $ 23$ :maybe:", "Solution_5": "Hello\r\n\r\n[quote=\"Agr_94_Math\"]Kabi, yes you are correct but it would have been nice if you had inroduced the genral derangements principle to aadil:\nThat is $ D_n \\equal{} n! ( 1 \\minus{} \\frac {1}{1!} \\plus{} \\frac {1}{2!} \\minus{} \\frac {1}{3!} \\plus{} \\frac {1}{4!} \\minus{} ............ \\plus{} ( \\minus{} 1)^n \\frac {1}{n!})$.[/quote]\ni thought he would love a basic method. Anyways thank u. :) \n[quote=\"Attila the Hun\"]required number of ways =$ 4!\\minus{}1$[/quote]\r\nAren't u forgetting other cases? :( Kindly edit ur Answer.\r\n\r\nThank u." } { "Tag": [ "number theory", "relatively prime" ], "Problem": "Prove that if $ p$ is prime $ (a,p) \\equal{} (n,p\\minus{}1)\\equal{}1$, then the congruence $ ax^n \\equal{} b (mod p)$ always has exactly one solution.", "Solution_1": "[hide]Suppose $ y$ is also a number satisfying $ ay^n \\equiv b \\mod p$. Then $ ax^n \\equiv ay^n \\mod p$, and since $ (a,p) \\equal{} 1$, $ x^n \\equiv y^n \\mod p$.\nThe case where $ x \\equiv 0 \\mod p$ is trivial. \nSuppose then that $ x$ is a unit mod $ p$. Let $ u$ be the inverse of $ x$. Then $ y^nu^n\\equiv (yu)^n \\equiv 1 \\mod p$. If $ yu$ is not $ 1\\mod p$ then the order of $ yu$ must not be relatively prime with $ p \\minus{} 1$, which is not the case. So, $ yu \\equiv 1 \\mod p$. That is, $ u$ is the inverse of both $ x$ and $ y$, or $ x\\equiv y$.[/hide]" } { "Tag": [ "calculus", "derivative", "inequalities", "function", "geometry", "inequalities proposed", "133" ], "Problem": "But I failed to find a nice solution. My solution was a very dirty and terrible solution such as $\\frac{\\partial}{\\partial c}(\\frac{\\partial}{\\partial c}(\\frac{\\partial}{\\partial b}(\\frac{\\partial f}{\\partial c}(a,b,c))))$.\r\nTry to find a nice solution!\r\n\r\nProblem : Let $a,b,c>0$.\r\nProve that\r\n$\\frac{a}{\\sqrt{b^{2}+15ca}}+\\frac{b}{\\sqrt{c^{2}+15ab}}+\\frac{c}{\\sqrt{a^{2}+15bc}}\\geq \\frac{3}{4}$.", "Solution_1": "It took you long time :lol: .\r\nI remind this is from thread:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=72297\r\n\r\nI failed to solve it no matter how I tried.Partial derivatives too complicated for me :( .\r\nI'll PM few mathlinkers too see what they think about this ultra-difficult inequality.Can you post your ugly solution anyway?", "Solution_2": "Thanks for the reminder.\r\nMy solution is well outside frames of a high school math.\r\nHowever,I vaguelly recall that I found the way soon after author first posted it on mathlinks.\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=595604#595604]Here[/url] you can see outline of such approach.\r\nI tried to adjust it for mathlinkers with at least basic knowledge in analysis.\r\nI couldn't find an \"elementar\" solution for this problem,if that is what was meant by \"nice\" solution.", "Solution_3": "Is $\\frac{\\partial^{2}F}{\\partial x^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial y^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial z^{2}}(1,1,1)=0$ true?\r\nIf $K=15$, $\\frac{\\partial^{2}F}{\\partial x^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial y^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial z^{2}}(1,1,1)=\\frac{507}{2048}$,\r\nso $\\frac{\\partial^{2}F}{\\partial x^{2}}+\\frac{\\partial^{2}F}{\\partial y^{2}}+\\frac{\\partial^{2}F}{\\partial z^{2}}=0$ is not true.\r\n\r\nMy ugly solution:\r\n\r\nI can't post dirty calculations, sorry! :) \r\n\r\n(1) Let $F(a,b,c)=\\sum_{cyc}\\frac{a}{\\sqrt{b^{2}+15ca}}$.\r\nIf $a\\geq b\\geq c\\geq$, $F(a,b,c)\\geq F(b,a,c)$, (by calculation)\r\nso if we prove $F(a,b,c)\\geq \\frac{3}{4}$ when $a\\leq b\\leq c$, we also prove the original inequality.\r\n\r\n(2) Let $f(a,b,c)=\\frac{a}{\\sqrt{b^{2}+15ca}}(\\frac{b}{\\sqrt{c^{2}+15ab}}+\\frac{c}{\\sqrt{a^{2}+15bc}})^{2}$,\r\nthen $f(a,b,c)\\leq \\frac{4}{27}(F(a,b,c))^{3}$,\r\nso if we prove $f(a,b,c)\\geq \\frac{1}{16}$, $F(a,b,c)\\geq \\frac{3}{4}$ is also proven.\r\n\r\n(3) Let $k=15$, then $\\frac{1}{16}=\\frac{4}{(1+k)^{3/2}}$,\r\nso $f(a,b,c)=(k+1)^{3/2}a(a^{2}b^{2}+kb^{3}c+c^{4}+kabc^{2}+2bc\\sqrt{a^{2}+kbc}\\sqrt{c^{2}+kab})$\r\n$-4\\sqrt{b^{2}+kca}(a^{2}+kbc)(c^{2}+kab)$.\r\n\r\n(4) Let $f_{1}(a,b,c)=(k+1)^{3/2}a(2a^{2}b+3kb^{2}c+(k+2)ac^{2}+4kbc\\sqrt ac)-\\frac{4b}{\\sqrt{b^{2}+kca}(a^{2}+kbc)(c^{2}+kab)}$\r\n$-4\\sqrt{b^{2}+kca}(ka^{3}+kc^{3}+2k^{2}abc)$,\r\n$f_{2}(a,b,c)=(k+1)^{3/2}a(a^{2}+3kbc+2kc^{2})-\\frac{2kca}{(b^{2}+kca)^{3/2}}$,\r\n$f_{3}(a,b,c)= (k+1)^{3/2}a(3b+4c)-\\frac{6ac^{2}(a^{2}+kbc)}{(b^{2}+kca)^{3/2}}$\r\n$-\\frac{12bc^{2}+8kab^{2}+2k^{2}a^{2}c}{\\sqrt{b^{2}+kca}}-4ka\\sqrt{b^{2}+kca}$,\r\n$f_{4}(a,b,c)=(1+k)^{3/2}-\\frac{4kb^{3}+6k^{2}abc+6kc^{3}}{(b^{2}+kca)^{3/2}}$.\r\n\r\n(5) Then $\\frac{\\partial}{\\partial b}f(a,b,c)\\geq f_{1}(a,b,c)$,\r\n$\\frac{\\partial}{\\partial b}f_{1}(a,b,c)\\geq 2f_{2}(a,b,c)$,\r\n$\\frac{\\partial}{\\partial c}f_{2}(a,b,c)\\geq kf_{3}(a,b,c)$,\r\n$\\frac{\\partial}{\\partial b}f_{3}(a,b,c)\\geq akf_{4}(a,b,c)$.\r\n(These are not hard. We can prove them easily)\r\n\r\n(6) Therefore, if we prove $f(a,a,c)\\geq 0, f_{1}(a,a,c)\\geq 0, f_{2}(a,b,0)\\geq 0, f_{3}(a,a,c)\\geq 0, f_{4}(a,b,c)\\geq 0$, the original inequality is proven.\r\nWe can prove $f(a,a,c)\\geq 0, f_{1}(a,a,c)\\geq 0, f_{2}(a,b,0)\\geq 0, f_{3}(a,a,c)\\geq 0, f_{4}(a,b,c)\\geq 0$ by dirty calculations.", "Solution_4": "Thanks for showing me your method.If anybody can,it is iurra who cracks such inequalities in more elementar way. I think I heard he is a mastermind in inequalities (and not only there :) ).\r\nI don't understand this comment of yours:\r\n[quote=\"Parkdoosung\"]Is $\\frac{\\partial^{2}F}{\\partial x^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial y^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial z^{2}}(1,1,1)=0$ true?\nIf $K=15$, $\\frac{\\partial^{2}F}{\\partial x^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial y^{2}}(1,1,1)+\\frac{\\partial^{2}F}{\\partial z^{2}}(1,1,1)=\\frac{507}{2048}$,\nso $\\frac{\\partial^{2}F}{\\partial x^{2}}+\\frac{\\partial^{2}F}{\\partial y^{2}}+\\frac{\\partial^{2}F}{\\partial z^{2}}=0$ is not true.\n\n[/quote]\r\nI don't know if esti is right or wrong,but do you mean here.\r\nIf x=y=z then $F(x,y,z)=\\frac{x}{\\sqrt{x^{2}+Kx^{2}}}+\\frac{x}{\\sqrt{x^{2}+Kx^{2}}}+\\frac{x}{\\sqrt{x^{2}+Kx^{2}}}=\\frac{3}{\\sqrt{1+K}}=constant$\r\nRemember that K is any fixed positive constant and not only that \"$\\triangle F=0$\" but even $\\frac{\\partial F}{\\partial x}+\\frac{\\partial F}{\\partial y}+\\frac{\\partial F}{\\partial z}=0$ is true for T(1,1,1) and arbitrary choice of K. Is that Right?", "Solution_5": "$\\frac{\\partial^{2}F}{\\partial x^{2}}(x,y,z)= \\frac{675xz^{2}}{4(y^{2}+15zx)^{5/2}}$\r\n$-\\frac{15z}{(y^{2}+15zx)^{3/2}}+\\frac{3x^{2}z}{(x^{2}+15yz)^{5/2}}-\\frac{z}{(x^{2}+15yz)^{3/2}}+\\frac{675y^{3}}{4(15xy+z^{2})^{5/2}}$.\r\n$\\frac{\\partial^{2}F}{\\partial x^{2}}(1,1,1)= \\frac{169}{2048}$.", "Solution_6": "Are you sure you did the derivative right?\r\nI can't verify that becouse I'm not much experienced in finding partial derivatives with so complicated terms. \r\n\r\nIf x=y=z than function in 3 variables F(x,y,z) becomes function in one variable F(x).\r\nAnd that is constant.\r\nI know how to do derivative of single function variable.\r\nFor F(x)= constant ,first derivative is 0,second is 0,third is 0 etc. \r\nIf you did all correctly,where I went wrong?I can't find mistake.\r\nWhat is going on? :wallbash:", "Solution_7": "Yes, my derivative is right.\r\nF(x,y,z) does not become function in one variable F(x) although x=y=z.\r\nFirst, calculate $\\frac{\\partial^{2}F}{\\partial x^{2}}F(x,y,z)$ and see y and z is constant.\r\nNext, use $y=z=x$.", "Solution_8": "If x=y=z than F(x,y,z)=constant.\r\nI'm sure about that.\r\nAnd the derivative of constant = [size=150]0[/size].\r\nSorry,I think u made mistake somewhere along the line,but I can't tell where. :(", "Solution_9": "Eros88, $F(x,y,z)$ is a constant when $x=y=z$ does not imply its partial derivative at $x=y=z=t$ is 0!", "Solution_10": "[quote=\"Soarer\"]Eros88, $F(x,y,z)$ is a constant when $x=y=z$ does not imply its partial derivative at $x=y=z=t$ is 0![/quote]\r\nOh,I think I'm begining to understand .But still I have difficulties .\r\nHow much is :$\\frac{\\partial^{2}F}{\\partial x^{2}}+\\frac{\\partial^{2}F}{\\partial y^{2}}+\\frac{\\partial^{2}F}{\\partial z^{2}}$ for F(x,y,z);x,y,z>0?\r\nCan you do that math for me Soarer,please?", "Solution_11": "if you mean the original function, no thanks, I don't like monster computations. :D\r\n\r\nJust want to give a simple example on eros88's problem.\r\nConsider $F(x,y) = x-y$. Its partial derivative of x at (1,1) is 1, although $F(x,x)$ is constant.", "Solution_12": "Thanks Soarer for this clarification.\r\nI understand now.\r\nFunny thing : that also indicates that Parkdoosung could be wrong in his reasoning too (about sum of 3 partial derivatives).You got me thinking.\r\nBloody partial derivatives!I have to learn them good as soon as possible. \r\nThey are very important.", "Solution_13": "[i]Calculus`VectorAnalysis`[/i]\r\n\r\n[b]F[X,Y,Z] :=X/Sqrt[Y^2+kXZ]+Y/Sqrt[Z^2+kXY]+Z/Sqrt[X^2+kYZ];\n\n\nGrad[F[X,Y,Z],Cartesian[X,Y,Z]]\nDiv[%]\nLaplacian[F][/b]\r\n\r\n\r\nOutputs are:\r\n[code]\n{-kXZ/(2(Y^2+kXZ)^3/2 +1/(Y^2+kXZ)^1/2 -XZ/(X^2+kYZ)^3/2 -kY^2/2(kXY+Z^2)^3/2,\n-XY/(Y^2+kXZ)^3/2 -kZ^2/(2(X^2+kYZ)^3/2 -kXY/(2(kXY+Z^2)^3/2 +\n1/(kXY+Z^2)^1/2,\n-kX^2/2(Y^2+kXZ)^3/2 -kYZ/2(X^2+ kYZ)^3/2 + 1/(X^2+kYZ)^1/2 -YZ/(kXY+Z^2)^3/2}\n\n{0}\n\n{0}[/code]\r\n \r\n\r\n\r\n\r\n[quote=\"Soarer\"]if you mean the original function, no thanks, I don't like monster computations. :D\n\nJust want to give a simple example on eros88's problem.\nConsider $F(x,y) = x-y$. Its partial derivative of x at (1,1) is 1, although $F(x,x)$ is constant.[/quote]\r\nI'm also one lazy a## when it comes to monster computations,as can be seen from above :wink: .The results can be checked by hand althought I didn't do that.\r\nFrom the discussion ,seems that Parkdoosung and Eros88 haven't mastered multivariable calculus yet.\r\nNot even to mention VectorFields theory.\r\nThis area of math becomes triviality once you get used to it.You can access the sketch of the proof from my first post in thread.I didn't post all the intersteps,and application of the harmonic function theorems.Who wants to do that and write strict step by step proof, all the necessary and even more can be found in the links I recommended.\r\nI don't wan't to discuss the solution to the problem anymore.\r\nOnly,if somebody finds something new to nail down the inequality in \"elementar\" way , maybe then I'll become interested again.\r\nNothing guarantees us that every math problem in the world can be solved in elementar way.\r\nWhat for mathematicians invented high level math tools ?I think the answers are clear even to a nonexpert.", "Solution_14": "It seems that esti have mastered multivariable calculus, but esti doesn't know very easy calculus and how to use computer programs.\r\nBefore studying harmonic function, you must study one variable calculus.", "Solution_15": "[quote=\"Eros88\"]Thanks Soarer for this clarification.\nI understand now.\nFunny thing : that also indicates that Parkdoosung could be wrong in his reasoning too (about sum of 3 partial derivatives).You got me thinking.\nBloody partial derivatives!I have to learn them good as soon as possible. \nThey are very important.[/quote]\r\n\r\nFrankly speaking, I think there is no reason that the laplacian of the original function is constantly 0, after computing round round F wrt x once by hand.", "Solution_16": "[code]\n{-kXZ/(2(Y^2+kXZ)^3/2) +1/(Y^2+kXZ)^1/2 -XZ/(X^2+kYZ)^3/2 -kY^2/(2(kXY+Z^2)^3/2),\n-XY/(Y^2+kXZ)^3/2 -kZ^2/(2(X^2+kYZ)^3/2) -kXY/(2(kXY+Z^2)^3/2) +\n1/(kXY+Z^2)^1/2,\n-kX^2/(2(Y^2+kXZ)^3/2) -kYZ/(2(X^2+ kYZ)^3/2) + 1/(X^2+kYZ)^1/2 -YZ/(kXY+Z^2)^3/2}\n\n\n[/code]\r\nIf esti's computer output is to be trusted first partial derivatives written clean are:\r\n$\\frac{\\partial F}{\\partial x}=-\\frac{kxz}{2(y^{2}+kxz)^{\\frac{3}{2}}}+\\frac{1}{(y^{2}+kxz)^{\\frac{1}{2}}}-\\frac{xz}{(x^{2}+kyz)^{\\frac{3}{2}}}-\\frac{ky^{2}}{2(kxy+z^{2})^{\\frac{3}{2}}}$\r\n$\\frac{\\partial F}{\\partial y}=-\\frac{xy}{(y^{2}+kxz)^{\\frac{3}{2}}}-\\frac{kz}{2(x^{2}+kyz)^{\\frac{3}{2}}}-\\frac{kxy}{2(kxy+z^{2})^{\\frac{3}{2}}}+\\frac{1}{(kxy+z^{2})^{\\frac{1}{2}}}$\r\n$\\frac{\\partial F}{\\partial z}=-\\frac{kx^{2}}{2(y^{2}+kxz)^{\\frac{3}{2}}}-\\frac{kyz}{2(x^{2}+kyz)^{\\frac{3}{2}}}+\\frac{1}{(x^{2}+kyz)^{\\frac{1}{2}}}-\\frac{yz}{(kxy+z^{2})^{\\frac{3}{2}}}$\r\n \r\nIs that is OK guys?\r\nSoarer,will you do carefully the second partial derivatives in $x$, $y$ and $z$ variable please?It's long trail of calculations I know,but \"to belive\" is one thing and to know is another!I want to know.", "Solution_17": "[quote=\"Parkdoosung\"]But I failed to find a nice solution. My solution was a very dirty and terrible solution such as $\\frac{\\partial}{\\partial c}(\\frac{\\partial}{\\partial c}(\\frac{\\partial}{\\partial b}(\\frac{\\partial f}{\\partial c}(a,b,c))))$.[/quote]\r\nI think, a similar inequality is the following:\r\nLet $a,b,c>0$. Prove that\r\n$\\frac{a}{\\sqrt{b^{2}+3c^{2}}}+\\frac{b}{\\sqrt{c^{2}+3a^{2}}}+\\frac{c}{\\sqrt{a^{2}+3b^{2}}}\\geq \\frac{3}{2}$.", "Solution_18": "[quote=\"Vasc\"][quote=\"Parkdoosung\"]But I failed to find a nice solution. My solution was a very dirty and terrible solution such as $\\frac{\\partial}{\\partial c}(\\frac{\\partial}{\\partial c}(\\frac{\\partial}{\\partial b}(\\frac{\\partial f}{\\partial c}(a,b,c))))$.[/quote]\nI think, a similar inequality is the following:\nLet $a,b,c>0$. Prove that\n$\\frac{a}{\\sqrt{b^{2}+3c^{2}}}+\\frac{b}{\\sqrt{c^{2}+3a^{2}}}+\\frac{c}{\\sqrt{a^{2}+3b^{2}}}\\geq \\frac{3}{2}$.[/quote]\r\nSimilar but not the same.\r\nI think Parkdoosung ineq. is a little harder than this one .\r\nDo you have elementar proofs for these inequalities (I mean without analysis)?\r\nAnd what about my question about sum :$\\frac{\\partial^{2}F}{\\partial x^{2}}+\\frac{\\partial^{2}F}{\\partial y^{2}}+\\frac{\\partial^{2}F}{\\partial z^{2}}$? , (x,y,z or a,b,c is just a matter of notation),for Parkdoosung ineq.,when does the sum equals 0?", "Solution_19": "I have not an elementary solution to these inequalities. :D \r\nBut a nice similar inequality is\r\n$\\frac{a}{\\sqrt{3b^{2}+ca}}+\\frac{b}{\\sqrt{3c^{2}+ab}}+\\frac{c}{\\sqrt{3a^{2}+bc}}\\geq \\frac{3}{2}$.", "Solution_20": "[quote=\"Parkdoosung\"](1) Let $F(a,b,c)=\\sum_{cyc}\\frac{a}{\\sqrt{b^{2}+15ca}}$.\nIf $a\\geq b\\geq c\\geq$, $F(a,b,c)\\geq F(b,a,c)$[/quote]\r\nWhy this is true?", "Solution_21": "[quote=\"Vasc\"]If $x,y,z$ are positive numbers, then $\\frac{x}{\\sqrt{3y^{2}+zx}}+\\frac{y}{\\sqrt{3z^{2}+xy}}+\\frac{z}{\\sqrt{3x^{2}+yz}}\\geq \\frac{3}{2}$.[/quote]\r\nUsing the Holder inequality, \r\n\r\n$\\left[\\sum_{cyc}x(z+x)^{3}(3y^{2}+zx)\\right]\\left(\\sum_{cyc}\\frac{x}{\\sqrt{3y^{2}+zx}}\\right)^{2}\\geq\\left[\\sum_{cyc}{x(z+x)}\\right]^{3}$,\r\n\r\nwe have to prove\r\n\r\n$4\\left[\\sum_{cyc}{x(z+x)}\\right]^{3}\\geq9\\sum_{cyc}x(z+x)^{3}(3y^{2}+zx)$.\r\n\r\n\r\n[hide=\"Proof.\"]Without losing generality, we may assuming $x=\\min\\{x,y,z\\}$.\n\n$4\\left[\\sum_{cyc}{x(z+x)}\\right]^{3}-9\\sum_{cyc}x(z+x)^{3}(3y^{2}+zx)$.\n\n$\\equiv F(x,y,z)=F(x,x+s,x+t)$\n\n$=216(s^{2}-st+t^{2})x^{4}+2(164s^{3}-39s^{2}t-21st^{2}+164t^{3})x^{3}$\n\n$+6(26s^{4}+27s^{3}t-34s^{2}t^{2}+33st^{3}+26t^{4})x^{2}$\n\n$+3(13s^{5}+24s^{4}t-22s^{3}t^{2}+8s^{2}t^{3}+15st^{4}+13t^{5})x$\n\n$+4s^{6}+12s^{5}t-12s^{4}t^{2}+s^{3}t^{3}-3s^{2}t^{4}+3st^{5}+4t^{6}\\geq0$,\n\nwhich is clearly true for \n\n$164s^{3}-39s^{2}t-21st^{2}+164t^{3}=39(s+t)(s-t)^{2}+125s^{3}+18st^{2}+125t^{3}\\geq0$\n\nand\n\n$4s^{6}+12s^{5}t-12s^{4}t^{2}+s^{3}t^{3}-3s^{2}t^{4}+3st^{5}+4t^{6}$\n\n$=t(s-t)^{2}(10s^{3}+8s^{2}t+7st^{2}+2t^{3})+4s^{6}+2s^{5}t+s^{2}t^{4}+2t^{6}\\geq0$.[/hide]", "Solution_22": "[quote=\"Vasc\"]Let $x,y,z>0$. Prove that $\\frac{x}{\\sqrt{y^{2}+3z^{2}}}+\\frac{y}{\\sqrt{z^{2}+3x^{2}}}+\\frac{z}{\\sqrt{x^{2}+3y^{2}}}\\geq \\frac{3}{2}$.[/quote]\r\nUsing the Holder inequality, \r\n\r\n$\\left[\\sum_{cyc}x(2x+y)^{3}(y^{2}+3z^{2})\\right]\\left(\\sum_{cyc}\\frac{x}{\\sqrt{y^{2}+3z^{2}}}\\right)^{2}\\geq\\left[\\sum_{cyc}{x(2x+y)}\\right]^{3}$,\r\n\r\nwe have to prove\r\n\r\n$4\\left[\\sum_{cyc}{x(2x+y)}\\right]^{3}\\geq9\\sum_{cyc}x(2x+y)^{3}(y^{2}+3z^{2})$.\r\n\r\n\r\n[hide=\"Proof.\"]Without losing generality, we may assuming $x=\\min\\{x,y,z\\}$.\n\n$4\\left[\\sum_{cyc}{x(2x+y)}\\right]^{3}-9\\sum_{cyc}x(2x+y)^{3}(y^{2}+3z^{2})$.\n\n$\\equiv F(x,y,z)=F(x,x+s,x+t)$\n\n$=810(s^{2}-st+t^{2})x^{4}+27(53s^{3}+17s^{2}t-56st^{2}+53t^{3})x^{3}$\n\n$+9(101s^{4}+146s^{3}t-111s^{2}t^{2}-73st^{3}+101t^{4})x^{2}$\n\n$+9(31s^{5}+48s^{4}t+33s^{3}t^{2}-88s^{2}t^{3}-st^{4}+31t^{5})x$\n\n$+32s^{6}+48s^{5}t+48s^{4}t^{2}-8s^{3}t^{3}-150s^{2}t^{4}+39st^{5}+32t^{6}\\geq0$,\n\nwhich is clearly true for \n\n$101s^{4}+146s^{3}t-111s^{2}t^{2}-73st^{3}+101t^{4}$\n\n$=t(99s+87t)(s-t)^{2}+101s^{4}+47s^{3}t+2st^{3}+14t^{4}$,\n\n$31s^{5}+48s^{4}t+33s^{3}t^{2}-88s^{2}t^{3}-st^{4}+31t^{5}$\n\n$=\\frac{(3s-2t)^{2}(14s^{3}+56s^{2}t+94st^{2}+32t^{3})+91s^{5}+s^{3}t^{2}+st^{4}+89t^{5}}{7}$,\n\n$32s^{6}+48s^{5}t+48s^{4}t^{2}-8s^{3}t^{3}-150s^{2}t^{4}+39st^{5}+32t^{6}$\n\n$=\\frac{(4s-3t)^{2}(12s^{4}+36s^{3}t+65s^{2}t^{2}+74st^{3}+18t^{4})+t^{2}(4s^{4}+4s^{3}t+3s^{2}t^{2}+30t^{4})}{6}$.[/hide]\r\nIf $x, y, z$ are positive numbers, then the following inequality\r\n\\[\\frac{x}{\\sqrt{y^{2}+kz^{2}}}+\\frac{y}{\\sqrt{z^{2}+kx^{2}}}+\\frac{z}{\\sqrt{x^{2}+ky^{2}}}\\geq \\frac{3}{\\sqrt{1+k}}\\]\r\nholds if and only if $0\\leq k\\leq\\frac{49-9\\sqrt{17}}{32}=0.37162654279503297\\cdots\\vee k\\geq\\frac{49+9\\sqrt{17}}{32}=2.6908734572049670\\cdots$.\r\n\r\nWith equality if $k=\\frac{49-9\\sqrt{17}}{32},x=0,y=1,z=\\frac{\\sqrt{17}-1}{4}=0.78076923076923076\\cdots$\r\n\r\nor $k=\\frac{49+9\\sqrt{17}}{32},x=0,y=1,z=\\frac{\\sqrt{17}+1}{4}=1.2807881773399014\\cdots$.", "Solution_23": "[quote=\"Ji Chen\"]\n\n\nWithout losing generality, we may assuming $x=\\min\\{x,y,z\\}$.\n\n$4\\left[\\sum_{cyc}{x(z+x)}\\right]^{3}-9\\sum_{cyc}x(z+x)^{3}(3y^{2}+zx)$.\n\n$\\equiv F(x,y,z)=F(x,x+s,x+t)$\n\n$=216(s^{2}-st+t^{2})x^{4}+2(164s^{3}-39s^{2}t-21st^{2}+164t^{3})x^{3}$\n\n$+6(26s^{4}+27s^{3}t-34s^{2}t^{2}+33st^{3}+26t^{4})x^{2}$\n\n$+3(13s^{5}+24s^{4}t-22s^{3}t^{2}+8s^{2}t^{3}+15st^{4}+13t^{5})x$\n\n$+4s^{6}+12s^{5}t-12s^{4}t^{2}+s^{3}t^{3}-3s^{2}t^{4}+3st^{5}+4t^{6}\\geq0$,\n\nwhich is clearly true for \n\n$164s^{3}-39s^{2}t-21st^{2}+164t^{3}=39(s+t)(s-t)^{2}+125s^{3}+18st^{2}+125t^{3}\\geq0$\n\nand\n\n$4s^{6}+12s^{5}t-12s^{4}t^{2}+s^{3}t^{3}-3s^{2}t^{4}+3st^{5}+4t^{6}$\n\n$=t(s-t)^{2}(10s^{3}+8s^{2}t+7st^{2}+2t^{3})+4s^{6}+2s^{5}t+s^{2}t^{4}+2t^{6}\\geq0$.[/quote]\r\n\r\n\r\nSo one term of 4-degree polynom is bigger or even than zero.\r\nI can`t see how that makes for example the term\r\n\r\n3(13s^5+24s^4t-22s^3t^2+8s^2t^3+15st^4+13t^5)x\r\n\r\nto be bigger or even than zero? Could somebody lighten me a little bit?", "Solution_24": "To Ji chen : \r\nCan you show your Comment about Value worth of k ? I think if it true , then it isn't actual easy", "Solution_25": "[quote=\"Ji Chen\"][quote=\"Vasc\"]Let $ x,y,z > 0$. Prove that $ \\frac {x}{\\sqrt {y^{2} \\plus{} 3z^{2}}} \\plus{} \\frac {y}{\\sqrt {z^{2} \\plus{} 3x^{2}}} \\plus{} \\frac {z}{\\sqrt {x^{2} \\plus{} 3y^{2}}}\\geq \\frac {3}{2}$.[/quote]\nIf $ x, y, z$ are positive numbers, then the following inequality\n\\[ \\frac {x}{\\sqrt {y^{2} \\plus{} kz^{2}}} \\plus{} \\frac {y}{\\sqrt {z^{2} \\plus{} kx^{2}}} \\plus{} \\frac {z}{\\sqrt {x^{2} \\plus{} ky^{2}}}\\geq \\frac {3}{\\sqrt {1 \\plus{} k}}\n\\]\nholds if and only if \n\n$ 0\\leq k\\leq\\frac {49 \\minus{} 9\\sqrt {17}}{32} \\equal{} 0.37162654279503297\\cdots$\n\n$ \\vee k\\geq\\frac {49 \\plus{} 9\\sqrt {17}}{32} \\equal{} 2.6908734572049670\\cdots$.\n\nWith equality if $ k \\equal{} \\frac {49 \\minus{} 9\\sqrt {17}}{32},x \\equal{} 0,y \\equal{} 1,z \\equal{} \\frac {\\sqrt {17} \\minus{} 1}{4} \\equal{} 0.78076923076923076\\cdots$\n\nor $ k \\equal{} \\frac {49 \\plus{} 9\\sqrt {17}}{32},x \\equal{} 0,y \\equal{} 1,z \\equal{} \\frac {\\sqrt {17} \\plus{} 1}{4} \\equal{} 1.2807881773399014\\cdots$.[/quote]\r\n\r\nIf $ x,y,z > 0,\\frac {49 \\minus{} 9\\sqrt {17}}{32}\\leq k\\leq\\frac {49 \\plus{} 9\\sqrt {17}}{32}$, then the infimum of $ \\frac {x}{\\sqrt {y^{2} \\plus{} kz^{2}}} \\plus{} \\frac {y}{\\sqrt {z^{2} \\plus{} kx^{2}}} \\plus{} \\frac {z}{\\sqrt {x^{2} \\plus{} ky^{2}}}$ is $ \\frac {2}{\\sqrt [4]{ k}}$.", "Solution_26": "[quote=\"Vasc\"]Let $ x,y,z > 0$. Prove that $ \\sqrt {\\frac {x}{y \\plus{} 3z}} \\plus{} \\sqrt {\\frac {y}{z \\plus{} 3x}} \\plus{} \\sqrt {\\frac {z}{x \\plus{} 3y}}\\geq \\frac {3}{2}$.[/quote]\r\nUsing the Holder inequality, \r\n\r\n$ \\sum_{cyc}x^{2}(y \\plus{} 3z)(x \\plus{} 2y \\plus{} z)^{3}\\left(\\sum_{cyc}\\sqrt {\\frac {x}{y \\plus{} 3z}}\\right)^{2}\\geq\\left[\\sum_{cyc}{x(x \\plus{} 2y \\plus{} z)}\\right]^{3}$,\r\n\r\nwe have to prove\r\n\r\n$ 4\\left[\\sum_{cyc}{x(x \\plus{} 2y \\plus{} z)}\\right]^{3}\\geq9\\sum_{cyc}x^{2}(y \\plus{} 3z)(x \\plus{} 2y \\plus{} z)^{3}$.\r\n\r\n\r\n[hide=\"Proof.\"]Without losing generality, we may assuming $ x \\equal{} \\min\\{x,y,z\\}$.\n\n$ 4\\left[\\sum_{cyc}{x(x \\plus{} 2y \\plus{} z)}\\right]^{3} \\minus{} 9\\sum_{cyc}x^{2}(y \\plus{} 3z)(x \\plus{} 2y \\plus{} z)^{3}$\n\n$ \\equiv F(x,y,z) \\equal{} F(x,x \\plus{} s,x \\plus{} t)$\n\n$ \\equal{} 432(s^2 \\minus{} st \\plus{} t^2)x^4 \\plus{} 4(161s^3 \\plus{} 78s^2t \\minus{} 129st^2 \\plus{} 161t^3)x^3$\n\n$ \\plus{} 3(103s^4 \\plus{} 254s^3t \\minus{} 115s^2t^2 \\minus{} 22st^3 \\plus{} 103t^4)x^2$\n\n$ \\plus{} 6(10s^5 \\plus{} 46s^4t \\plus{} 23s^3t^2 \\minus{} 37s^2t^3 \\plus{} 7st^4 \\plus{} 10t^5)x$\n\n$ \\plus{} 4s^6 \\plus{} 27s^5t \\plus{} 39s^4t^2 \\minus{} 9s^3t^3 \\minus{} 33s^2t^4 \\plus{} 9st^5 \\plus{} 4t^6\\geq0$,\n\nwhich is clearly true for \n\n$ 103s^4 \\plus{} 254s^3t \\minus{} 115s^2t^2 \\minus{} 22st^3 \\plus{} 103t^4$\n\n$ \\equal{} t(99s \\plus{} 83t)(s \\minus{} t)^2 \\plus{} 103s^4 \\plus{} 155s^3t \\plus{} 45st^3 \\plus{} 20t^4$,\n\n$ 10s^5 \\plus{} 46s^4t \\plus{} 23s^3t^2 \\minus{} 37s^2t^3 \\plus{} 7st^4 \\plus{} 10t^5$\n\n$ \\equal{} (s \\minus{} t)^2(10s^3 \\plus{} 7s^2t \\plus{} 27st^2 \\plus{} 10t^3) \\plus{} 59s^4t$,\n\n$ 4s^6 \\plus{} 27s^5t \\plus{} 39s^4t^2 \\minus{} 9s^3t^3 \\minus{} 33s^2t^4 \\plus{} 9st^5 \\plus{} 4t^6$\n\n$ \\equal{} \\frac {t(s \\plus{} t)(s \\plus{} 4t)(4s \\plus{} t)(3s \\minus{} 2t)^2 \\plus{} s^3(16s^3 \\plus{} 72s^2t \\plus{} 15st^2 \\plus{} 11t^3)}{4}$.[/hide]", "Solution_27": "[quote=\"Ji Chen\"]If $ x, y, z$ are positive numbers, then the following inequality\n\n$ \\sqrt {\\frac {x}{y + kz}} + \\sqrt {\\frac {y}{z + kx}} + \\sqrt {\\frac {z}{x + ky}}\\geq \\frac {3}{\\sqrt {1 + k}}$\n\nholds if and only if $ 0\\leq k\\leq\\frac {49 - 9\\sqrt {17}}{32} = 0.371626542795\\cdots\\vee k\\geq\\frac {49 + 9\\sqrt {17}}{32} = 2.69087345720\\cdots$.[/quote][quote=\"Harry Potter\"]To Ji chen : \n\nCan you show your Comment about Value worth of $ k$ ? I think if it true, then it isn't actual easy.[/quote]Let $ \\frac {x}{y + kz} = \\frac {u^{2}}{1 + k},\\frac {y}{z + kx} = \\frac {v^{2}}{1 + k},\\frac {z}{x + ky} = \\frac {w^{2}}{1 + k},$ where $ u,v,w$ are positive numbers.\r\n\r\nHence, it remains to prove that $ u + v + w\\geq3,$ where $ (k^2 - k + 1)u^2v^2w^2 + k(v^2w^2 + w^2u^2 + u^2v^2) = (k + 1)^2.$\r\n\r\narqady's contradiction method works here: let $ u + v + w < 3,$ then\r\n\r\n$ \\left(k^2 - k + 1\\right)u^2v^2w^2 + k(v^2w^2 + w^2u^2 + u^2v^2)$\r\n\r\n$ < \\left(k^2 - k + 1\\right)u^2v^2w^2\\left(\\frac {3}{u + v + w}\\right)^6 + k\\left(v^2w^2 + w^2u^2 + u^2v^2\\right)\\left(\\frac {3}{u + v + w}\\right)^4$\r\n\r\n$ = (k + 1)^2 - \\sum{\\frac {(v - w)^2}{2(u + v + w)^6}\\Big\\{4\\left(16k^2 - 49k + 16\\right)u^3(v + w) + 2\\left(91k^2 - 142k + 91\\right)u^2vw + 6\\left(10k^2 - 7k + 10\\right)v^2w^2}$\r\n\r\n${ + (k + 1)^2\\left[58uvw(v + w) + v(v + 16w)(u - v)^2 + w(w + 16v)(u - w)^2\\right]\\Big\\}}$\r\n\r\n$ \\leq (k + 1)^2,$ which is a contradiction.\r\n\r\nSee also : http://www.mathlinks.ro/viewtopic.php?t=153451" } { "Tag": [ "MATHCOUNTS", "videos", "AMC 10", "\\/closed" ], "Problem": "At the end of next week, we'll be reorganizing the board. \r\n\r\nHere are the items we're considering; please add your thoughts and suggestions:\r\n\r\n1) Change the problem solving section to be 4 boards: Getting Started (MATHCOUNTS, AMC8/10 level), Intermediate Topics(AMC12/AIME), Advanced problems (Olympiad), Resources/Teaching (as it is now)\r\n\r\n2) Consolidate Books/Movies & Video Games\r\n\r\n3) Consolidate Other Programs & Other Contests\r\n\r\n4) Implement pruning on the Etc section - basically delete any thread that hasn't had a response in 10 days or so.", "Solution_1": "I agree with all the changes - particularly changing the problem solving section and pruning the Etc. forum...", "Solution_2": "Sounds good.", "Solution_3": "I vote \"Yes\" for these changes.\r\nAlthough I am surprised that \"prune\" was a verb...", "Solution_4": "We are still going to have the \"mathcounts\" and \"American Mathematics Competitions\" and so on sections within the Extracurricular Programs title right?\r\n\r\n-interesting_move", "Solution_5": "Yes; those will all still be there (though we'll probably fold the 'Other Programs' and 'Other Contests' into one board).", "Solution_6": "Pruning is a must. Except maybe in special situations, mainly the reference ones like \"what is your screen name?\" or \"how many digits of pi do you know\" (since these will probably come up over and over again if you delete them...)\r\n\r\nOne thing: if we delete posts in Etc., the authors of those posts will have their post count go down. This seems fair enough to me, since it's generally the posts in the other sections that are most relevant/helpful and therefore deserve to be credited. This would recognize big problem-solvers like TripleM and ComplexZeta for their hard work and many posts in math-related things. Thoughts on this?", "Solution_7": "[quote=\"mathfanatic\"]if we delete posts in Etc., the authors of those posts will have their post count go down. This seems fair enough to me, since it's generally the posts in the other sections that are most relevant/helpful and therefore deserve to be credited. This would recognize big problem-solvers like TripleM and ComplexZeta for their hard work and many posts in math-related things. Thoughts on this?[/quote]\r\nI agree. People who's posts are mostly useless (a.k.a. me:)) doesn't deserve high ranks/post numbers.", "Solution_8": "indeed. but at least tare is starting to solve in the amc folder now. thats a start at least.", "Solution_9": "I'm not at that level yet. I only got a measly 100... :) I do the best I can, but that's not a lot.", "Solution_10": "What grade are you in Tare?\r\n\r\n-interesting_move", "Solution_11": "[quote=\"interesting_move\"]What grade are you in Tare?[/quote]\r\nI took AMC 10 last year when I was in 7th. Now I'm about to be 8th.", "Solution_12": "dont get down. if youve been working hard, you should easaly make aime. not that many 7th graders can get a 100 on the amc 10. you should be proud. but just work hard, and you can do it.\r\n\r\n\r\nnow im done with my motivational speaking", "Solution_13": "[quote=\"Syntax Error\"]dont get down. if youve been working hard, you should easaly make aime. not that many 7th graders can get a 100 on the amc 10. you should be proud. but just work hard, and you can do it.\n\n\nnow im done with my motivational speaking[/quote]\r\n*claps* Thank you, HI...I only wish you'll stay that way :)", "Solution_14": "im always that way when someone doesnt believe in themselves." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let x1,x2,x3,x4>0,x1*x2+x2*x3+x3*x4+x4*x1\uff1d4\uff0cor x1*x2+x2*x3+x3*x4+x4*x1+x1*x3+x2*x4=6, k>=0, prove that\r\n\r\n1/x1+1/x2+1/x3+1/x4+k*(x1+x2+x3+x4)>=4*(1+k).", "Solution_1": "[size=100] Let $x_{1}$,$x_{2}$,$x_{3}$,$x_{4}>0$,\n$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{1}$ \uff1d4\uff0c or \n$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{1}+x_{1}x_{3}+x_{2}x_{4}=6$, $k \\ge 0$, prove that \n\n$\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\frac{1}{x_{3}}+\\frac{1}{x_{4}}+k(x_{1}+x_{2}+x_{3}+x_{4}) \\ge 4(1+k)$ .\n what is ' \uff1d4\uff0c '?[/size]", "Solution_2": "If $x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}=6$ then\r\n$(\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\frac{1}{x_{3}}+\\frac{1}{x_{4}})^{2}\\ge \\frac{8}{3}(\\frac{1}{x_{1}x_{2}}+\\frac{1}{x_{1}x_{3}}+\\frac{1}{x_{1}x_{4}}+\\frac{1}{x_{2}x_{3}}+\\frac{1}{x_{2}x_{4}}+\\frac{1}{x_{3}x_{4}}) \\ge \\frac{8}{3}\\frac{36}{x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}}=16$\r\nso $\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\frac{1}{x_{3}}+\\frac{1}{x_{4}}\\ge 4$ (*)\r\nalso $(x_{1}+x_{2}+x_{3}+x_{4})^{2}\\ge \\frac{8}{3}(x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})=16$\r\nso $k(x_{1}+x_{2}+x_{3}+x_{4}) \\ge 4k$ then add this with (*)" } { "Tag": [ "ratio", "AMC", "AIME", "percent" ], "Problem": "A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that $25\\%$ of these fish are no longer in the lake on September 1 (because of death and emigrations), that $40\\%$ of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1?", "Solution_1": "[hide]Because $25\\%$ of the tagged fish had died off or emigrated, there are only $60\\left(\\frac{3}{4}\\right)=45$ left in the lake. In the sample she collected on September 1, only $70\\left(\\frac{6}{10}\\right)=42$ were originally there on May 1. $3$ of these were tagged. Thus, the ratio of tagged to untagged fish on May 1 was $\\frac{3}{42}=\\frac{1}{14}$. Our answer is thus $14\\cdot \\text{number\\ of\\ tagged\\ fish}=14\\cdot 45=\\boxed{630}$.[/hide]", "Solution_2": "[hide=\"Answer\"]$\\frac{3}{70}$ fish in the lake at the later date are tagged. 40% were born since, so she is really looking at $\\frac{3}{42}$. 25% of the $60$ tagged fish have died, so there were only $45$ fish remaining, and she has caught $\\frac{1}{15}$ of them. Therefore, our answer is $15\\cdot 42=\\boxed{630}$.[/hide]", "Solution_3": "Bumped so that 4everwise can have the right solution in the solution's document. 4everwise, you multiplied $(14)(60)$ when it should be $(14)(45)$ :P", "Solution_4": "I think the answer actually would 14*60=840. We want the number of fish before the reduction of 25%.", "Solution_5": "Is the answer 840 or 630?", "Solution_6": "The answer should be 840. Here's azjps's solution from the wiki:\r\n\r\n[hide]\nOf the $ 70$ fish caught in September, $ 40\\%$ were not there in May, so $ 42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $ \\frac {3}{42} \\equal{} \\frac {60}{x} \\Longrightarrow \\boxed{x \\equal{} 840}$.\n\n(Note that the 25% death rate does not affect the answer.)[/hide]", "Solution_7": "I got 840 as well. Does anybody know the official AIME solution?\r\n\r\nI said that on may 1 there were x fish and on sept 1 there were y fish.\r\n\r\nSince 25 percent of x are gone, there are 3x/4 originals left in sept. Also, since 40% of fish in sept were not there in may, that means 60% of the fish in september WERE there in may, so the amount of original fish in september could also be written as 3y/5\r\n\r\n3y/5 = 3x/4 ---> y = 15x/12 = 1.25x\r\n\r\nSo in september there were 45 tagged fish (.75*60) and there were 1.25x fish in total. Since 3 of the 70 surveyed were tagged,\r\n\r\n3/70 = 45/1.25x ---> x = 840", "Solution_8": "Is the answer 630 or 840? Thanks.", "Solution_9": "840. \r\n\r\n45 of the initialled tagged fish are still there. Thus, 45 is 3/70 of the entire september population, so on 9/1, there were 1050 fish total.\r\n\r\nWhatever the number of fish were on may 1, after accounting for 25% deaths, experienced a 40% growth, to reach the 9/1 population. So, letting the number of fish on may 1 be x, we have\r\n((1-0.25)x)=((1-0.4))(1050),\r\n\r\nx = 840", "Solution_10": "[hide=Solution]After the $40$ percent reduction, $42$ of the $70$ fish caught on September 1 are still there, so $\\frac{3}{42}=\\frac{60}{x}$ where $x$ is the total. This means $x=\\boxed{840}$.[/hide]", "Solution_11": "[hide=Solution][i]m[/i] is the number of fish in the lake on May 1. Since 60 fish were tagged on May 1, but on September 1, 25% of all the fish were gone, that means 25% of the tagged fish were gone too, so 45 tagged fish were left in the lake. Since 40% of the fish in the lake on September 1 are new, that means [i]m[/i] is 60% of all the fish in the lake on September 1. 40 is 2/3 of 60, so the number of fish in the lake on September 1 is 5/3 * [i]m[/i]. The sample taken on September 1 tells us that 3/70 of the fish in the lake are tagged, but we just found out that 45 / (5/3 * m) is the ratio of the total tagged fish to the total number of fish in the lake! This means these ratios are equal: 3/70 = 45 / (5/3 * m) => 3/70 = 45 * 3 / 5m => we multiply both sides by (5m * 70): 5m = 45 * 70 => we divide both sides by 5 => m = 9 * 70 => we get [i]m[/i] = [b]630[/b].[/hide]\n\nThat's what I got. Is it 630 or 840?", "Solution_12": "According to the wiki, the answer is 840. Can someone edit the solution [url=https://artofproblemsolving.com/community/c4887_1990_aime_problems]here[/url]?" } { "Tag": [ "inequalities", "induction", "combinatorics solved", "combinatorics", "Enumeration" ], "Problem": "Three rods of length a,b,c can be made to form a triangle if and only if the three triangle inequalities are satisfied: a+b > c, b+c > a and c+a > b. Suppose one wants to make just one triangle and has at his disposal exactly one rod of each of lengths 1,2,3,...,n. He might decide on the (2,4,5) triangle, the (3,4,5), or the (3,5,7),... He can't make an isosceles triangle, of course, but there are still many possibilities to choose from. And we want to know how many ?", "Solution_1": "My guess is (if I understood correctly the problem) :\r\n\r\n. u(2n) = n(n-1)(4n-5)/6\r\n. u(2n+1) = (n-1)(4n-5n+6)/6 (= u(2n) + n-1)\r\n\r\n\r\nLet v(n) be the number of solution of :\r\n1 <= a < b < n, a+b > n\r\n\r\nBy induction, v(2n)=(n-1), v(2n+1)=n(n-1)\r\n\r\nThen sum v(k) from 1 to n to get u(n).", "Solution_2": "I have the same formula for even n but a different one for odd n. Maybe you can give your reasoning.\r\n\r\nu(n) = 1/24*n*(n-2)*(2n-5) with n being even \r\nu(n) = 1/24*(n-1)*(n-3)*(2n-1) with n being odd", "Solution_3": "It was a very last line calculation error (I'm prone that kind of silly mistake) :( :\r\n\r\n. u(2n+1) = u(2n) + n(n-1) (and not (n-1)) = n(4n+1)(n-1)/6 which is the same as you !" } { "Tag": [ "floor function", "AMC", "AIME" ], "Problem": "How many subsets of the set $\\{1,2,3,4,\\ldots,30\\}$ have the property that the sum of the elements of the subset is greater than 232?", "Solution_1": "[hide=\"As I recall...\"] This problem depends on the fact that $232$ is a special sum with relation to this set, but I don't remember how or why. [/hide]", "Solution_2": "[hide=\"hint\"]\n$232 = \\lfloor \\frac{465}{2}\\rfloor$\n[/hide]", "Solution_3": "Looks alot like aime 2006." } { "Tag": [], "Problem": "Show that the equation X\u00b2+2Y\u00b2=2p has integer solutions if and only if the equation U\u00b2+2V\u00b2=p has integer solutions.", "Solution_1": "[hide]\nSuppose there's a solution for X\u00b2+2Y\u00b2=2p. This implies that X is even. \nSo let X = 2 * Z: 4Z\u00b2 + 2Y\u00b2 = 2p.\nThen, Y\u00b2 + 2Z\u00b2 = p, which is a solution for U\u00b2+2V\u00b2=p when U = Y and V = Z.\n\nNow, suppose there's a solution for U\u00b2+2V\u00b2=p. Then (2V)\u00b2 + 2U\u00b2 = 2U\u00b2 + 4V\u00b2 = 2p, which is a solution for X\u00b2+2Y\u00b2=2p when X = 2V and Y = U.\n[/hide]" } { "Tag": [ "quadratics" ], "Problem": "$\\sqrt{8+4\\sqrt{3}}=\\sqrt{a}+\\sqrt{b}$, where $a$ and $b$ are positive integers and $a>b$. Find the ordered pair $(a, b)$.", "Solution_1": "[hide=\"Solution\"]$a+b+2\\sqrt{ab}=8+4\\sqrt{3}$\n\n$a+b=8$\n\n$\\sqrt{ab}=2\\sqrt{3}\\implies ab=12$\n\n$a\\left(8-a\\right)=12\\implies a^{2}-8a+12=0=\\left(a-6\\right)\\left(a-2\\right)=0$\n\nSo you have $\\left(a,b\\right)=\\boxed{\\left(6,2\\right)}$.[/hide]", "Solution_2": "[hide]Square both sides to get.\n\n$8+4\\sqrt{3}=a+2\\sqrt{ab}+b$.\n\nThen, this means that\n\n$8=a+b$, and $2\\sqrt{ab}=4\\sqrt{3}$.\n\n$a+b=8$\n$ab=12$.\n\nThen we can put this into a quadratic to get\n\n$x^{2}-8x+12=0$. So $x=6,2$.\n\nBut $a\\>b$, hence $a=6$ and $b=2$\n\n$(a,b)=(6,2)$[/hide]", "Solution_3": "[hide=\"solution\"]$\\text{We square both sides: }$\n$8+4\\sqrt3=a+b+2\\sqrt{ab}$\n$\\sqrt{ab}=2\\sqrt3=\\sqrt{12}$\n$a+b=8$\n$\\text{We use guess and check and see that: }$\n$\\boxed{(a,b)=((6,2),(2,6))}$[/hide]", "Solution_4": "[quote=\"nutz_for2.718281828\"][hide=\"solution\"]$\\text{We square both sides: }$\n$8+4\\sqrt3=a+b+2\\sqrt{ab}$\n$\\sqrt{ab}=2\\sqrt3=\\sqrt{12}$\n$a+b=8$\n$\\text{We use guess and check and see that: }$\n$\\boxed{(a,b)=((6,2),(2,6))}$[/hide][/quote]\r\n\r\nHe said $a>b$" } { "Tag": [ "algebra", "system of equations" ], "Problem": "If the following system of equations\r\n\r\n$x+y+z=3$\r\n\r\n$x^{3}+y^{3}+z^{3}=15$\r\n\r\n$x^{4}+y^{4}+z^{4}=35$\r\n\r\nhas a real solution for which $x^{2}+y^{2}+z^{2}>10$, find the value of $x^{5}+y^{5}+z^{5}$ for that solution.", "Solution_1": "Let $S_{n}=x^{n}+y^{n}+z^{n}(n=0,1,2,3,...)$ and $p=xy+yz+zx$, $q=xyz$.\r\nFrom $x+y+z=3$ follow: $S_{2}=9-2p$ and we have equality: $S_{4}-3S_{3}+pS_{2}-qS_{1}=0$ arriving $p^{2}=1$\r\nWith $p=1 S_{2}=7<10$(contradition)\r\nwith $p=-1\\Rightarrow S_{2}=11>10$ and $q=7$\r\nAlternatively, we have $S_{5}=3S_{4}-pS_{3}+qS_{2}=197$, done!", "Solution_2": "please tell me that how you eliminated q from equation\r\n$S_{4}-3*S_{3}....$.", "Solution_3": "We know that \\[\\begin{aligned}81 = (x+y+z)^{4}& = (x^{4}+y^{4}+z^{4})+4\\sum_{\\text{sym}}x^{3}y+6(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})+12xyz(x+y+z) \\\\ & = 35+4((x+y+z)(x^{3}+y^{3}+z^{3})-(x^{4}+y^{4}+z^{4}))+6(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})+36xyz \\\\ & = 75+6(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})+36xyz\\end{aligned}\\]So $1 = x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+6xyz = (xy+yz+zx)^{2}$. So $xy+yz+zx = \\pm 1$. But $10 < x^{2}+y^{2}+z^{2}= (x+y+z)^{2}-2(xy+yz+zx) = 9-2(xy+yz+zx)$, so $xy+yz+zx < \\frac{-1}{2}$. Therefore, $xy+yz+zx =-1$, i.e. $x^{2}+y^{2}+z^{2}= 11$. \r\n\r\nAlso, remark that \\[\\begin{aligned}27 & = (x+y+z)^{3}= x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(z+x) \\\\ &= 15+3(3-x)(3-y)(3-z) = 15+3(27-9(x+y+z)+3(xy+yz+zx)-xyz) = 15+3(-3-xyz)\\end{aligned}\\]Therefore, $xyz =-7$. \r\n\r\nLet $x^{5}+y^{5}+z^{5}= A$. Then \\[3^{5}= A+5\\sum_{\\text{sym}}x^{4}y+10\\sum_{\\text{sym}}x^{3}y^{2}+20xyz(x^{2}+y^{2}+z^{2})+30xyz(xy+yz+zx)\\]\r\nNow, \\[\\sum_{\\text{sym}}x^{4}y = (x^{4}+y^{4}+z^{4})(x+y+z)-(x^{5}+y^{5}+z^{5}) = 105-A\\] \\[\\sum_{\\text{sym}}x^{3}y^{2}= (x^{3}+y^{3}+z^{3})(x^{2}+z^{2}+y^{2})-(x^{5}+y^{5}+z^{5}) = 165-A\\]So \\[243 = A+5(105-A)+10(165-A)-1540+210 = 845-14A\\]And we obtain $A = 43$.", "Solution_4": "There is just one really weird thing about this... by Cauchy \r\n\r\n$x^{2}+y^{2}+z^{2}\\le \\sqrt{(x^{4}+y^{4}+z^{4})(1^{2}+1^{2}+1^{2})}= \\sqrt{105}$", "Solution_5": "[quote=\"Fibfreak\"]There is just one really weird thing about this... by Cauchy \n\n$x^{2}+y^{2}+z^{2}\\le \\sqrt{(x^{4}+y^{4}+z^{4})(1^{2}+1^{2}+1^{2})}= \\sqrt{105}$[/quote]\r\n\r\nHmm, you've got a point. The system doesn't have a real solution. (We get that the solutions are the roots of $x^{3}-3x^{2}-x+7 = 0$, which aren't all real.)", "Solution_6": "Correctly :[quote=\"Fibfreak\"]The following system of equations $\\{\\begin{array}{c}x+y+z=3\\\\\\ x^{3}+y^{3}+z^{3}=15\\\\\\ x^{4}+y^{4}+z^{4}=35\\end{array}$ has at least a real solution for which $\\boxed{x^{2}+y^{2}+z^{2}\\le10}$ and find the value of $x^{5}+y^{5}+z^{5}$ for this solution.[/quote]\r\nAnswer : $t^{3}-3t^{2}+t+1=0$ or $t^{3}-3t^{2}-t+7=0$." } { "Tag": [ "MATHCOUNTS" ], "Problem": "The numbers 1,2,3,...,9 are arranged, one per circle, in the triangle shown so that the sum, s, of the numbers on each side of the triangle is the same. The sum of the numbers in the three corner circles is 12. What is the value of s? Express your answer as a common fraction.", "Solution_1": "common fraction? i got 19", "Solution_2": "[hide]ok... you use the corners twice... and the others once...\n\n\n\nso \n\n\n\n3s = 1+2+3+4+5+6+7+8+9+12\n\n\n\n3s = 57\n\ns=19 [/hide]so you're right.", "Solution_3": "[hide]19/1[/hide]", "Solution_4": "Just as a note -- \"19\" is the proper way to express an integer as a \"common fraction,\" not \"19/1\" or anything else.", "Solution_5": "Yup, 19/1 is an improper fraction.", "Solution_6": "[color=cyan]and in mathcounts, \"improper\" fractions are common fractions, but 19/1 could be simplified, so yeah[/color]" } { "Tag": [ "Putnam", "induction", "college contests" ], "Problem": "Problem B2.\nLet n be a positive integer. Starting with the sequence \n1, 1/2, 1/3, ..., 1/n form a new sequence of n-1 entries \n3/4, 5/12, ..., (2n-1)/{2n(n-1)} by taking the averages of \ntwo consecutive entries in the first sequence. Repeat the \naveraging of neighbors on the second sequence to obtain a third \nsequence of n-2 entries and continue the final sequence produced \nconsists of a single number x_n. Show that x_n < {2/n} .\n----------------------------------------------------------------\n\nProve by induction that x_n= (1/2 n-1)* \\sum{k =0..n-1} (n-1)Ck/(k+1). (1)\nWe must prove (1)<2/n.\nAfter computations this comes to \\sum{k=1..n} nCk<2 n, or equivalently 2 n-1<2 n. This is obvious.", "Solution_1": "My solution is smaller that the problem text. :D", "Solution_2": "that's because you didn't make full-detailed redactation. :)\n\n[color=red][[b]Moderator edit:[/b] Now with 100% more solution at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=80&t=413607 .][/color]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "let a,b,c are non-negative such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$.prove that:\r\n$ \\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\plus{} \\frac {b^3}{b^5 \\plus{} b^2c \\plus{} a^3} \\plus{} \\frac {c^3}{c^5 \\plus{} c^2a \\plus{} b^3} \\leq 1$", "Solution_1": "[quote=\"tuantam1lan\"]let a,b,c are non-negative such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$.prove that:\n$ \\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\plus{} \\frac {b^3}{b^5 \\plus{} b^2c \\plus{} a^3} \\plus{} \\frac {c^3}{c^5 \\plus{} c^2a \\plus{} b^3} \\leq 1$[/quote]\r\ninequality problem have complicacy form.but it is an inequality very interesting and hard\r\nwe have:$ \\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\equal{} \\frac {a}{a^3 \\plus{} b \\plus{} \\frac {c^3}{a^2}}$\r\nto apply cauchuy-shawrs:\r\n$ a^3 \\plus{} b \\plus{} \\frac {c^3}{a^2} \\equal{} \\frac {a^4}{a} \\plus{} \\frac {b^4}{b^3} \\plus{} \\frac {c^4}{ca^2} \\geq \\frac {(a^2 \\plus{} b^2 \\plus{} c^2)^2}{a \\plus{} b^3 \\plus{} ca^2} \\equal{} \\frac {9}{a \\plus{} b^3 \\plus{} ca^2}$\r\n $ \\Leftrightarrow \\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\leq \\frac {a(a \\plus{} b^3 \\plus{} ca^2)}{9}$\r\n therefore\r\n $ \\sum_{cyc}\\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\leq \\frac {\\sum_{cyc} a^2 \\plus{} 2\\sum_{cyc} ab^3}{9} \\leq 1$\r\nequality is $ a \\equal{} b \\equal{} c \\equal{} 1$", "Solution_2": "I don't get the last inequality..\r\nCan you prove it clearly ?\r\nThanks before..", "Solution_3": "[quote=\"Ronald Widjojo\"]I don't get the last inequality..\nCan you prove it clearly ?\nThanks before..[/quote]\r\n if a,b,c are real numbers,then:\r\n$ (a^2 \\plus{} b^2 \\plus{} c^2)^2 \\geq 3(a^3b \\plus{} b^3c \\plus{} c^3a)$\r\n vasile cirtoaje created it in 1992\r\nit is hard inequality and very much deliberate\r\n inequality is equivalent:$ \\sum_{cyc}(a^2 \\minus{} 2ab \\plus{} bc \\minus{} c^2 \\plus{} ca)^2 \\geq 0$", "Solution_4": "[quote=\"tuantam1lan\"][quote=\"tuantam1lan\"]let a,b,c are non-negative such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$.prove that:\n$ \\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\plus{} \\frac {b^3}{b^5 \\plus{} b^2c \\plus{} a^3} \\plus{} \\frac {c^3}{c^5 \\plus{} c^2a \\plus{} b^3} \\leq 1$[/quote]\n\n $ \\Leftrightarrow \\frac {a^3}{a^5 \\plus{} a^2b \\plus{} c^3} \\geq \\frac {a(a \\plus{} b^3 \\plus{} ca^2)}{9}$\n [/quote]\r\nIt is a nice and really impressive proof ( :clap2: ) ; but this step has a little flaw - the inequality sign must be flipped. :blush:", "Solution_5": "@tuantam1lan: Nice hint to creat, my friend :)" } { "Tag": [ "geometry", "rectangle", "perimeter", "quadratics", "algebra" ], "Problem": "Hi, I need help with this ambiguous but otherwise fairly easy problem:\r\n\r\n[i]A rectangular piece of land of area 5000m^2 is to be enclosed by a wall, and then divided into three equal regions by partition walls parallel to one of its sides. If the total length of the wall is 445m, calculate the possible dimensions of the land.[/i]\r\n\r\nAttached are possible diagrams", "Solution_1": "If I understood the problem correctly:\r\n\r\nLet x be longer side of the rectangle and y its smaller side. Then the area is given by xy and the perimeter by 2(x+y). So we have to solve the equations $ 2(x+y) = 445$ and $ xy = 5000$. From the first we get\r\n$ y = 222.5-x$, and substituting in the second we get the quadratic equation $ x^{2}-222.5x+5000 = 0$, whose solution is $ x = 197.15\\, m$, and so $ y = 222.5-197.15 = 25.35\\, m$, and therefore these are the dimensions of the land. If then the land is to be divided in three equal regions, then one side of each will have 25.35 m, and the other side will have 197.15/3 = 65.72 m.", "Solution_2": "Thanks, I think I misread :roll:" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: R\\to R$ suct that $ f(f(x\\minus{}y)) \\equal{} f(x)f(y)\\minus{}f(x)\\plus{}f(y)\\minus{}xy$ for all real $ x,y$.", "Solution_1": "let $ f(0)\\equal{}c$\r\nput $ x\\equal{}y\\equal{}0$ we get that:\r\n$ f(c)\\equal{}c^{2}$\r\nnow put $ x\\equal{}y$ we get that:\r\n$ f(c)\\equal{}f(x)^{2}\\minus{}x^{2}\\equal{}c^{2}$\r\nfor all real numbers $ x$,now put $ x\\equal{}c$ we get that $ c^{2}\\equal{}f(c)^{2}\\minus{}c^{2}$ thus we get that $ c^{4}\\equal{}2c^{2}$ so we have $ c\\equal{}0$ or $ c\\equal{}\\pm\\sqrt{2}$\r\nso we get the following solutions:\r\n$ f(x)\\equal{}\\pm x$\r\n$ f(x)\\equal{}\\pm\\sqrt{x^{2}\\plus{}2}$", "Solution_2": "[quote=\"BaBaK Ghalebi\"]let $ f(0) \\equal{} c$\nput $ x \\equal{} y \\equal{} 0$ we get that:\n$ f(c) \\equal{} c^{2}$\nnow put $ x \\equal{} y$ we get that:\n$ f(c) \\equal{} f(x)^{2}\\minus{}x^{2}\\equal{} c^{2}$\nfor all real numbers $ x$,now put $ x \\equal{} c$ we get that $ c^{2}\\equal{} f(c)^{2}\\minus{}c^{2}$ thus we get that $ c^{4}\\equal{} 2c^{2}$ so we have $ c \\equal{} 0$ or $ c \\equal{}\\pm\\sqrt{2}$\nso we get the following solutions:\n$ f(x) \\equal{}\\pm x$\n$ f(x) \\equal{}\\pm\\sqrt{x^{2}\\plus{}2}$[/quote]\r\n$ f(x)\\equal{}\\pm\\sqrt{x^{2}\\plus{}2}$ is not solution.", "Solution_3": "neither is $ f(x)\\equal{}x$...\r\nso the only solution due to Rust is $ f(x)\\equal{}\\minus{}x$..." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Let $1=d_13$),\r\n$ p(p \\minus{} 7) \\equal{} 2(y^2 \\minus{} (2x)^2) \\equal{} 2(y \\minus{} 2x)(y \\plus{} 2x) , p > y \\minus{} 2x \\implies p |y \\plus{} 2x \\implies y < p,x < \\sqrt {p} \\implies p|y \\plus{} 2x < p \\plus{} 2\\sqrt {p} < 2p \\implies y \\plus{} 2x \\equal{} p$\r\nthe rest is trivial." } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem", "algebra unsolved" ], "Problem": "a is a real root of x^2 + 2x + 10x - 20; prove a^2 irrational.", "Solution_1": "hahhaha alex what's up.\r\n\r\nanyways, i assume it's $x^3+2x^2+10x-20=0$...\r\n\r\n$x^2(x+2)+10(x-2)=0$, so $x^2=-10(x-2)/(x+2)=-10(x-2)^2/(x^2-4)$. \r\n\r\nplug in $x=\\alpha$ and suppose $\\alpha^2$ is rational, then $\\alpha^4$ is also rational. Note that $\\alpha$ is irrational, so expanding that thing you have $\\alpha^4+6\\alpha^2+40=40\\alpha$, i.e the sum of 3 rational numbers is irrational so contradiction,", "Solution_2": "More naturally, write it as $x(x^2-10)=20-2x^2$. Thus if $k$ is a real root (there is at least one) then $k$ is rational iff $k^2$ is rational. But the rational root theorem shows that $k$ cannot be rational. The result follows.", "Solution_3": "You can generalise this result to any cubic quite easily by rational root thorem and Viete's formulawe.\r\nObviuosly, it must be real, not comlpex :) :)" } { "Tag": [ "Putnam", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Given a sequence ($x_{n}$) such that $\\lim_{n\\to\\infty}\\left(x_{n}-x_{n-2}\\right)=0$\r\nprove that\r\n\r\n$\\lim_{n\\to\\infty}\\frac{x_{n}-x_{n-1}}{n}=0$", "Solution_1": "Here's an ugly way: \r\n\r\n[hide]Choose $\\epsilon > 0$. Pick $N$ s.t. for $n \\geq N$ we have $|x_{n+2}-x_{n}| < \\frac{\\epsilon}{2}$. Let $|x_{N+1}-x_{N}| = d$.\n\nNow, choose $M \\geq \\max\\{\\frac{2d-N\\epsilon}{\\epsilon}, N\\}$. If $n \\geq M$ we have\n\n$\\frac{1}{n}|x_{n+1}-x_{n}| = \\frac{1}{n}|(x_{n+1}-x_{n-1})+(x_{n-1}-x_{n-3})+\\ldots-[(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\\ldots] \\pm (x_{N+1}-x_{N})| \\leq \\frac{(n-N)\\frac{\\epsilon}{2}+d}{n}= \\frac{\\epsilon}{2}+\\frac12 \\frac{2d-N}n \\leq \\epsilon$, and we're done.[/hide]", "Solution_2": "[quote=\"JBL\"]\n$\\frac{1}{n}|x_{n+1}-x_{n}| = \\frac{1}{n}|(x_{n+1}-x_{n-1})+(x_{n-1}-x_{n-3})+\\ldots-[(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\\ldots] \\pm (x_{N+1}-x_{N})|$ [/quote]\r\nwhy?I can't understand sorry :oops: \r\n\r\n(Edit:I know you are right now :) )\r\n\r\nHere is my proof:\r\nI just notice that the sub-sequences:\r\n$x_{1},x_{3},\\cdots,x_{2k+1},\\cdots$ and $x_{2},x_{4},\\cdots,x_{2k},\\cdots$ are independent.\r\nSo I want to prove:$\\lim_{n \\to \\infty}\\frac{x_{n}}{n}=0$\r\nbut this is not hard,because:\r\n$\\frac{x_{n}}{n}=\\frac{(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\\cdots}{n}$\r\nAnd use Cauchy's rule.\r\nthen is it easy to see:\r\n$\\lim_{n \\to \\infty}\\frac{x_{n}-x_{n-1}}{n}=\\lim_{n \\to \\infty}\\frac{x_{n}}{n}-\\lim_{n \\to \\infty}\\frac{x_{n-1}}{n}=0$:)", "Solution_3": "[quote=\"zhaobin\"][quote=\"JBL\"]\n$\\frac{1}{n}|x_{n+1}-x_{n}| = \\frac{1}{n}|(x_{n+1}-x_{n-1})+(x_{n-1}-x_{n-3})+\\ldots-[(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\\ldots] \\pm (x_{N+1}-x_{N})|$ [/quote]\nwhy?I can't understand sorry :oops: [/quote]\r\n\r\n$(x_{n+1}-x_{n}) = x_{n+1}-x_{n}+(x_{n-1}-x_{n-1})+(x_{n-2}-x_{n-2})+\\ldots+(x_{N+1}-x_{N+1})+(x_{N}-x_{N})$ and then reorder the sum. The $\\pm$ at the end is because we don't know which will be absorbed into our $\\epsilon$-pairs without breaking it into cases for the parity of $n-N$.\r\n\r\nYour way is nicer, though.", "Solution_4": "[quote=\"zhaobin\"]\n$\\frac{x_{n}}{n}=\\frac{(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\\cdots}{n}$\nAnd use Cauchy's rule.\nthen is it easy to see:\n[/quote]\nfirst question: what is Cauchy's rule?\n\n[quote=\"zhaobin\"]\n$\\lim_{n \\to \\infty}\\frac{x_{n}-x_{n-1}}{n}=\\lim_{n \\to \\infty}\\frac{x_{n}}{n}-\\lim_{n \\to \\infty}\\frac{x_{n-1}}{n}=0$:)[/quote]\r\nquestion 2: how do u get this since doesnt the previous expression come out to $x_{n}-x_{0}$ or $x_{n}-x_{1}$ depending on the parity of $n$?", "Solution_5": "The idea is that the condition shows that the sequences $x_{0}, x_{2}, ...$ and $x_{1}, x_{3}, ...$ (which are independent) both converge, say to $y_{1}$ and $y_{2}$. Then in the limit, the difference $|x_{n}-x_{n-1}|$ is about $|y_{1}-y_{2}|$, whence if you divide by $n$ it goes to 0.", "Solution_6": "[quote=\"Xevarion\"]the sequences $x_{0}, x_{2}, ...$ and $x_{1}, x_{3}, ...$ (which are independent) both converge[/quote]\r\nNo. For example, $x_{n}=n^{a}$, $02)2]2}2=144", "Solution_1": "Syntax Error wrote:Solve for x: {3+[2+(1+x2)2]2}2=144\n\n\n\n[hide]Work backwords. \n\n{3+[2+(1+x2)2]2}2=144 =>\n\n3+[2+(1+x2)2]2=12=>\n\n[2+(1+x2)2]2=9=>\n\n2+(1+x2)2]=3=>\n\n(1+x2)2]=1=>\n\nx :^2: =0. \n\n\n\nSo x =0[/hide]", "Solution_2": "Nice work. When I first did this problem, I plugged in [hide]0[/hide] the first time because I knew it wouldn't be too big and it worked.", "Solution_3": "[color=white]\nif {3+[2+(1+x :^2: ) :^2: ] :^2: } :^2: =144, then\n3+[2+(1+x :^2: ) :^2: ] :^2: =12\n[2+(1+x :^2: ) :^2: ] :^2: =9\n2+(1+x :^2: ) :^2: =3\n(1+x :^2: ) :^2: =1\n1+x :^2: =1\nx=0[/color]", "Solution_4": "yup yup", "Solution_5": "Can someone please tell me what every solution so far has left out (and then explain why it turned out okay anyhow)?", "Solution_6": "we could always have :pm: answers, but negetive answers would give us i's.", "Solution_7": "Rockin'.", "Solution_8": "Here is my hardcore algebra method .\n\n\n\n[hide](3+(2+(1+x :^2: ) :^2: ) :^2: ) :^2: =144\n\n(3+(2+(1+x :^2: ) :^2: ) :^2: )=12\n\n(2+(1+x :^2: ) :^2: ) :^2:=9\n\n(2+(1+x :^2: ) :^2: )=3\n\n(1+x :^2: ) :^2: )=1\n\n(1+x :^2: )=1\n\nx :^2: =0\n\nx=0 {0}\n\n[/hide]" } { "Tag": [ "calculus", "integration", "inequalities", "function", "real analysis", "real analysis unsolved" ], "Problem": "$ \\int_0^1f(x)dx \\int_0^1x^4f(x)dx \\le \\frac{4}{15} \\int_0^1f^2(x)dx$ , where $ f: [0,1]\\minus{}> R$ is continous .", "Solution_1": "Just use Cauchy-Schwartz for $ \\int_0^1\\left[\\frac 1{2\\sqrt 3}\\plus{}\\frac{\\sqrt 3}2x^4\\right]f(x)\\,dx$ and note that the left hand side is at least the square root of the product on the left hand side in the original inequality and that $ \\int_0^1\\left[\\frac 1{2\\sqrt 3}\\plus{}\\frac{\\sqrt 3}2x^4\\right]^2\\,dx\\equal{}\\frac 4{15}$. The equality is attained when both AM-GM and CS inequalities turn into equalities, i.e., for the function $ f(x)\\equal{}c(1\\plus{}3x^4)$ with $ c\\ge 0$." } { "Tag": [ "function", "trigonometry", "inequalities", "calculus", "derivative", "real analysis", "real analysis solved" ], "Problem": "Let $f:[0,1]\\to\\mathbb{R}$ a function which admits primitives on $[0,1]$ and $F$ one of them.\r\na) Prove that there exists $x\\in(0,1)$ s.t. $\\sqrt{2|f(x)F(x)|}<1+F^2(x)$\r\nb) Prove that for $f(x)=\\sin x$ the inequality from a) is true for all $x\\in[0,1]$.", "Solution_1": "For b)\r\nIt is a strange question, because $|f(x)|<1$, so $\\sqrt{2|fF|}\\leq \\sqrt{1+F^2}\\leq 1+F^2$, where at least one of \"\\leq\" is a strong sign.", "Solution_2": "Hmm... does anybody know exact constant in $\\sqrt{2|f(x)F(x)|}<1+F^2(x)$ instead of $2$? I obtained something like $\\frac{8}{\\pi}$... :?\r\n\r\n[P.S. It is a wrong observation :blush:]", "Solution_3": "and how did you find this constant Myth?", "Solution_4": "At first, I'd like to see your solution ;)", "Solution_5": "Sorry for delay\r\nhmm...\r\nI supposeed that for any $x\\in(0,1)$ the following inequality holds \r\n$\\sqrt{2|f(x)F(x)|}\\ge1+F^2(x)$ which is equivalent with\r\n$|\\left(\\frac1{F^2(x)+1}\\right)'|\\ge 1$ for any $x\\in R$\r\none ot the following cases must occur\r\n1.$\\left(\\frac1{F^2(x)+1}\\right)'\\ge 1$ \r\n2.$\\left(\\frac1{F^2(x)+1}\\right)'\\le -1$ \r\nfor 1. (it's analoguos for the other one)\r\nwe can integrate betweem 0 and $t$, $t\\in(0,1)$ and results that \r\n$\\frac1{1+F^2(t)}-\\frac1{1+F^2(0)}\\ge t$ but since $\\frac1{1+F^2(x)}$ is continuous results that we can make $t\\to 1$ an we have now that the difference between two number less than 1(and greater than0) is greater or equal to 1... which is a contradiction\r\n\r\nand the hypohesis is proved...\r\n\r\n\r\nI didn't think during the contest to find the biggest constant instead of $2$ but it would be an interesting problem to find it (I think ).", "Solution_6": "Ops... Sorry, I realized that I wrote a wrong primitive instead of $\\frac{1}{1+F^2}$, that is why I got a better constant. My solution was exactly the same.", "Solution_7": "I misunderstand one thing: how we can suppose that $|\\left(\\frac1{F^2(x)+1}\\right)'|\\ge 1$ or\r\n$|\\left(\\frac1{F^2(x)+1}\\right)'|\\le 1$ on the whole interval (0;1) (As far as i understand tending $t$ to 1 you supposed that one of the above inequality holds on the (0,1))", "Solution_8": "In my turn, I misunderstood what you mean.", "Solution_9": "Ok, i'll try to explain my problems:\r\n\r\n[quote=\"Kuba\"]Sorry for delay\n\n$|\\left(\\frac1{F^2(x)+1}\\right)'|\\ge 1$ for any $x\\in R$\none ot the following cases must occur\n1.$\\left(\\frac1{F^2(x)+1}\\right)'\\ge 1$ \n2.$\\left(\\frac1{F^2(x)+1}\\right)'\\le -11$ \nfor 1. (it's analoguos for the other one)\nwe can integrate betweem 0 and $t$, $t\\in(0,1)$ and results that \n$\\frac1{1+F^2(t)}-\\frac1{1+F^2(0)}\\ge t$ but since $\\frac1{1+F^2(x)}$ is continuous results that we can make $t\\to 1$ an we have now that the difference between two number less than 1(and greater than0) is greater or equal to 1... which is a contradiction\n\nand the hypohesis is proved...\n[/quote]\r\n\r\nHow we could integrate without knowing behavior of the function $\\left(\\frac1{F^2(x)+1}\\right)'$ near t=1:\r\nmaybe in some subinterval (0,a) the 1. holds and in the (a,1) the second holds (where a<1)", "Solution_10": "A derivative has a Darbou (intermediate) property, so it can't be $>1$ on $(0,a)$ and $<-1$ on $(a,1)$. Therefore, $\\left(\\frac1{F^2(x)+1}\\right)'\\ge 1$ OR $\\left(\\frac1{F^2(x)+1}\\right)'\\le -1$ on the whole segment.", "Solution_11": "Ok, thank you very much Myth, sorry for my stupidness-now i understand" } { "Tag": [ "function", "floor function", "algebra", "functional equation", "algebra proposed" ], "Problem": "Find all functions $f: \\mathbb R^\\plus{}\\to\\mathbb R^\\plus{}$ such that \\[ f(xf(y))\\equal{}f(x\\plus{}y)\\] for all positive reals $x$ and $y$.", "Solution_1": "[quote=\"pco\"]@angelstt : since you seem interested in this functional equation (4 different problems about it), here is a fifth one :\n\nFind all functions from $ \\mathbb R^ \\plus{} \\to\\mathbb R^ \\plus{}$ (the set of all positive real numbers) such that : $ f(xf(y)) \\equal{} f(x \\plus{} y)$ $ \\forall x,y > 0$[/quote]\r\nIs my solution right?\r\n$ f(x\\plus{}y)\\equal{}f(xf(y))\\equal{}f(yf(x))$\r\n[b]Case1[/b] if there aren't $ y_1 \\ne y_2$ such that $ f(y_1)\\equal{}f(y_2)$ then xf(y)=yf(x)\r\nThen, $ f(x)\\equal{}ax$ . Then check it which is not suitable.\r\n[b]Case2 [/b]if there are $ y_1 \\ne y_2$ such that $ f(y_1)\\equal{}f(y_2)$ \r\n then $ f(x\\plus{}y_1)\\equal{}f(xf(y_1))\\equal{}f(xf(y_2))\\equal{}f(x\\plus{}y_2)$ .\r\nThen, $ f(x)\\equal{}f(x\\plus{}c)$ .($ c\\geq 0$, c is the smallest frequency)\r\n [b] If[/b] $ c\\equal{}0$then $ f(x)\\equal{}const$.Check it suitable.\r\n [b] If [/b]$ c>0$then, there exits $ a$ that $ f(a) \\ne 1$ \r\n +$ f(a)>1$ \r\n $ f(xf(a))\\equal{}f(x\\plus{}a)$ so $ xf(a)\\equal{}x\\plus{}a\\plus{}nc$( $ n \\in N^*$ take x large enough)\r\nso $ x(f(a)\\minus{}1)\\equal{}a\\plus{}nc$\r\nThen , take $ x_0\\equal{}x\\plus{}\\frac{c}{2(f(a)\\minus{}1)}$we got $ x_0.(f(a)\\minus{}1)\\equal{} a\\plus{}nc\\plus{}c/2\\equal{}a\\plus{}mc$ ($ n,m \\in N^*$) Which can not true.\r\n Do the same for $ f(a)<1$\r\nThen we obtain $ f(x) \\equal{}const$", "Solution_2": "[quote=\"TRAN THAI HUNG\"]\n ...\n [b] If[/b] $ c \\equal{} 0$then $ f(x) \\equal{} const$.Check it suitable.\n ...\n[/quote]\r\n\r\nsorry, i don't see that. If $ c\\equal{}0$ we have $ f(x)\\equal{}f(x)$. Can you explain? Thanks", "Solution_3": "[quote=\"TRAN THAI HUNG\"][quote=\"pco\"]@angelstt : since you seem interested in this functional equation (4 different problems about it), here is a fifth one :\n\nFind all functions from $ \\mathbb R^ \\plus{} \\to\\mathbb R^ \\plus{}$ (the set of all positive real numbers) such that : $ f(xf(y)) \\equal{} f(x \\plus{} y)$ $ \\forall x,y > 0$[/quote]\nIs my solution right?\n$ f(x \\plus{} y) \\equal{} f(xf(y)) \\equal{} f(yf(x))$\n[b]Case1[/b] if there aren't $ y_1 \\ne y_2$ such that $ f(y_1) \\equal{} f(y_2)$ then xf(y)=yf(x)\nThen, $ f(x) \\equal{} ax$ . Then check it which is not suitable.\n[b]Case2 [/b]if there are $ y_1 \\ne y_2$ such that $ f(y_1) \\equal{} f(y_2)$ \n then $ f(x \\plus{} y_1) \\equal{} f(xf(y_1)) \\equal{} f(xf(y_2)) \\equal{} f(x \\plus{} y_2)$ .\nThen, $ f(x) \\equal{} f(x \\plus{} c)$ .($ c\\geq 0$, c is the smallest frequency)\n [b] If[/b] $ c \\equal{} 0$then $ f(x) \\equal{} const$.Check it suitable.\n [b] If [/b]$ c > 0$then, there exits $ a$ that $ f(a) \\ne 1$ \n +$ f(a) > 1$ \n $ f(xf(a)) \\equal{} f(x \\plus{} a)$ so $ xf(a) \\equal{} x \\plus{} a \\plus{} nc$( $ n \\in N^*$ take x large enough)\nso $ x(f(a) \\minus{} 1) \\equal{} a \\plus{} nc$\nThen , take $ x_0 \\equal{} x \\plus{} \\frac {c}{2(f(a) \\minus{} 1)}$we got $ x_0.(f(a) \\minus{} 1) \\equal{} a \\plus{} nc \\plus{} c/2 \\equal{} a \\plus{} mc$ ($ n,m \\in N^*$) Which can not true.\n Do the same for $ f(a) < 1$\nThen we obtain $ f(x) \\equal{} const$[/quote]\r\n\r\n1) As prester said, $ \\inf\\{c$ such that $ f(x\\plus{}c)\\equal{}f(x) \\forall x\\}\\equal{}0$ does not imply $ f(x)$ is constant (it would be true if $ f(x)$ was continuous, but this is not a constraint of the problem.\r\n\r\n2) If $ f(x)\\equal{}f(x\\plus{}c)$ $ \\forall x$, it does not allow you to conclude $ f(a)\\equal{}f(b)$ $ \\implies$ $ a\\equal{}b\\plus{}nc$ as you wrote (\"$ f(xf(a)) \\equal{} f(x \\plus{} a)$ so $ xf(a) \\equal{} x \\plus{} a \\plus{} nc$\")", "Solution_4": "1) I agree I forgot Dirichlete function also have that condition.(@to prester:when I write c=0, I mean there is no minimum of c). :) \r\n2) I agree If $ f(x) \\equal{} f(x \\plus{} c)$, it does not allow me to conclude $ f(a) \\equal{} f(b) \\implies a \\equal{} b \\plus{} nc$ but if we also have $ f(y_1) \\equal{} f(y_2)\\implies\\ y_1 \\minus{} y_2$ is a frequency then we can conclude that since c is the smallest frequency.\r\nAm I right? :) Thanks.\r\nHowever, I still fell in solve this problem. Can you show your proof, please?", "Solution_5": "[quote=\"TRAN THAI HUNG\"]1) I agree.\n2) I agree If $ f(x) \\equal{} f(x \\plus{} c)$, it does not allow me to conclude $ f(a) \\equal{} f(b) \\implies a \\equal{} b \\plus{} nc$ but if we also have $ f(y_1) \\equal{} f(y_2)\\implies\\ y_1 \\minus{} y_2$ is a frequency then we can conclude that since c is the smallest frequency.\nAm I right? :) Thanks.[/quote]\r\n\r\n :blush: You're right. $ f(a)\\equal{}f(b)$ $ \\implies$ $ |b\\minus{}a|$ is a period and so $ c$ divides $ |b\\minus{}a|$ if $ c\\ne 0$\r\nSorry.", "Solution_6": "Pco , can u please post your solution ?", "Solution_7": "[hide=\"Solution\"]\nWe claim that only constant functions work; it is clear that they do work. \n\n\n[b]Observation:[/b] If $ f(a) \\equal{} f(b)$ with $ b \\minus{} a \\equal{} s > 0$, then $ f(y \\plus{} a) \\equal{} f(yf(a)) \\equal{} f(yf(b)) \\equal{} f(y \\plus{} b)$ for positive $ y$. Set $ y \\equal{} x \\minus{} a$ to get that $ f(x) \\equal{} f(x \\plus{} s)$ for all $ x > a$. \n\n\nIf $ f$ is injective, then $ xf(y) \\equal{} x \\plus{} y$, so $ f(1) \\equal{} 1 \\plus{} \\frac {1}{x}$. However, this value is nonconstant, which is a contradiction, so $ f$ cannot be injective. By the observation, there are reals $ t$ and $ u$ so that for all $ x > u$, $ f(x) \\equal{} f(x \\plus{} t)$. Assume that $ f$ is nonconstant so that there is a $ z$ so that $ f(z) \\ne 1$. \n\nLet $ r$ be an arbitrary real greater than $ z$. If $ f(z) > 1$, then set $ x \\equal{} \\frac {z \\plus{} r}{f(z) \\minus{} 1}$. If $ f(z) < 1$, then set $ x \\equal{} \\frac {r \\minus{} z}{1 \\minus{} f(z)}$. In the first case, we get $ xf(z) \\equal{} x \\plus{} z \\plus{} r$, meaning that $ f(x \\plus{} z) \\equal{} f(x \\plus{} z \\plus{} r)$, and in the second case, we get that $ xf(z) \\equal{} x \\plus{} z \\minus{} r$, so $ f(x \\plus{} z \\minus{} r) \\equal{} f(x \\plus{} z)$. Either way, we have found a $ k$ and an $ l$ so that $ |k \\minus{} l| \\equal{} r$ and $ f(k) \\equal{} f(l)$. By the observation, there is a function $ g$ that maps reals to reals so that whenever $ x > g(r)$, $ f(x) \\equal{} f(x \\plus{} r)$.\n\nSet $ h(r) \\equal{} g(r) \\minus{} t\\lfloor {\\frac {g(r)}{t} }\\rfloor$ so that $ h(r) < t$ for all $ r$. Let $ \\max \\{t, u \\} \\equal{} v$; if $ x > v$, then $ f(x) \\equal{} f(x \\plus{} g(r) \\minus{} h(r)) \\equal{} f(x \\plus{} g(r) \\minus{} h(r) \\plus{} r) \\equal{} f(x \\plus{} r)$ for all $ r$. Then, there is a constant $ c$ so that $ f(x) \\equal{} c$ is constant whenever $ x > v$. For any $ w$, put in $ x \\equal{} \\frac {w}{c}$ and set $ y > v$. This gives that $ f(w) \\equal{} f(xf(y)) \\equal{} f(x \\plus{} y) \\equal{} c$, so $ f$ is constant everywhere. \n\nHence, only constant functions work.\n\n[/hide]", "Solution_8": "[u][b]@QuattroMaster [/b][/u]:\r\nI do agree with your solution with a little correction :\r\n[quote=\"The QuattoMaster 6000\"] ... then $ f(x) \\equal{} f(x \\plus{} g(r) \\minus{} h(r)) \\equal{} f(x \\plus{} g(r) \\minus{} h(r) \\plus{} r) \\equal{} f(x \\plus{} r)$ for all $ r$. [/quote]\r\nIn fact, this is not $ \\forall r$ : it is $ \\forall r > z$ but this does not destroy the proof.\r\n\r\nCongrats.\r\n\r\n[u][b]@Nixmtp[/b][/u] :\r\nHere is my solution, using the same idea than QuattroMaster, but with a little bit different end :\r\n\r\nLet $ P(x,y)$ be the equation $ f(xf(y)) \\equal{} f(x \\plus{} y)$\r\n\r\n$ f(x) \\equal{} 1$ $ \\forall x$ is a solution. So we'll from now consider $ \\exists u$ such that $ f(u)\\ne 1$\r\n\r\n1) If $ \\exists a,T > 0$ such that $ f(a) \\equal{} f(a \\plus{} T)$ then $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x\\ge a$\r\n=================================================================================\r\nLet $ x > a$ : \r\n$ P(x \\minus{} a,a)$ $ \\implies$ $ f((x \\minus{} a)f(a)) \\equal{} f(x)$\r\n$ P(x \\minus{} a,a \\plus{} T)$ $ \\implies$ $ f((x \\minus{} a)f(a \\plus{} T)) \\equal{} f(x \\plus{} T)$ and so $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x > a$\r\nAnd, since $ f(a) \\equal{} f(a \\plus{} T)$ : $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x\\ge a$\r\nQ.E.D.\r\n\r\n2) $ \\exists a,b > 0$ such that $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x > a,\\forall T > b$\r\n====================================================================\r\nLet $ u$ such that $ f(u)\\ne 1$\r\n2.1) if $ f(u) > 1$\r\n---------------\r\nLet $ T > 0$ : $ P(\\frac {u \\plus{} T}{f(u) \\minus{} 1},u)$ $ \\implies$ $ f(\\frac {uf(u) \\plus{} Tf(u)}{f(u) \\minus{} 1}) \\equal{} f(\\frac {u \\plus{} T}{f(u) \\minus{} 1} \\plus{} u)$ $ \\equal{} f(\\frac {uf(u) \\plus{} Tf(u)}{f(u) \\minus{} 1} \\minus{} T)$\r\nAnd so, using the point 1) above : $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x\\ge \\frac {uf(u) \\plus{} Tf(u)}{f(u) \\minus{} 1} \\minus{} T$\r\nQ.E.D.\r\n\r\n2.2) if $ f(u) < 1$\r\n----------------\r\nLet $ T > u$ : $ P(\\frac {u \\minus{} T}{f(u) \\minus{} 1},u)$ $ \\implies$ $ f(\\frac {uf(u) \\minus{} Tf(u)}{f(u) \\minus{} 1}) \\equal{} f(\\frac {u \\minus{} T}{f(u) \\minus{} 1} \\plus{} u)$ $ \\equal{} f(\\frac {uf(u) \\minus{} Tf(u)}{f(u) \\minus{} 1} \\plus{} T)$\r\nAnd so, using the point 1) above : $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x\\ge \\frac {uf(u) \\minus{} Tf(u)}{f(u) \\minus{} 1}$\r\nQ.E.D.\r\n\r\n3) The only solutions are constant functions\r\n============================================\r\nLet $ a,b > 0$ such that $ f(x) \\equal{} f(x \\plus{} T)$ $ \\forall x > a,\\forall T > b$ (according to point 2) above)\r\nLet $ x > y > a$ \r\n\r\nLet $ T_1 \\equal{} b \\plus{} 1 > b$. Using point 2, we get $ f(x) \\equal{} f(x \\plus{} b \\plus{} 1)$\r\nLet $ T_2 \\equal{} b \\plus{} 1 \\plus{} x \\minus{} y > b$. Using point 2, we get $ f(y) \\equal{} f(y \\plus{} b \\plus{} 1 \\plus{} x \\minus{} y) \\equal{} f(x \\plus{} b \\plus{} 1)$\r\nAnd so $ f(x) \\equal{} f(y) \\equal{} c$ $ \\forall x,y > a$ for some constant $ c$\r\n\r\nLet then $ u > 0$ $ P(\\frac u{f(a \\plus{} 1)},a \\plus{} 1)$ $ \\implies$ $ f(u) \\equal{} f(a \\plus{} 1 \\plus{} u{f(a \\plus{} 1)}$ and, since $ a \\plus{} 1 \\plus{} u{f(a \\plus{} 1)} > a$, we get $ f(u) \\equal{} c$ $ \\forall u > 0$\r\nQ.E.D\r\n\r\n\r\nAnd so (since the case $ f(x) \\equal{} 1$ $ \\forall x$ is also a constant solution and since obviously $ f(x) \\equal{} c$ fits the requested equation :\r\n\r\nThe only solutions to this equation are $ \\boxed{f(x) \\equal{} c}$ $ \\forall x > 0$ where $ c$ is a positive constant", "Solution_9": "Thank you very much , i really enjoyed your beautiful solutions.", "Solution_10": "my solution:\n$x=1$,we have $f(f(y))=f(y=1)$,so $f(x(f(y+1))=f(x+f(y))=f(x+y+1)$\nso $f(y)-y-1=T$ is period for any positive reals $y$\n$f(xf(y))-xf(y)-1=T$,$f(x+y)-xf(y)-1=T$,$f(x+y)-(x+y)-1=T$,\nso $T=x(1-f(y))+y$,if $f(x)$ is not constant,then we have $T$ can be all reals on interval $T>k$ or $Tk$ or $TN\\}$ .clearly $|xf(y)-x-y| \\in A$. let $t \\in A$. we have $f(y)^n t \\in A$ for every integer $n$ aswell. now if $u+v \\in A$ and also $v$ is not an element in $A$ we have :\n$$f((x+u)f(y+v))=f(xf(y))$$\nhence $x(f(y+v)-f(y))+uf(y+v)$ now since $f(v+y)$ and $f(y)$ are not equal we get that for some $ax_2$ such that $f(x_1)=f(x_2)$, then the equation implies, for all $x>x_2$, $f(x)=f(x+p)$ for some $p>0$. \n\nSince $f\\equiv 1$ works, let $f(n)\\neq 1$ for some $n$. If $f(n)>1$, then in the equation setting $x=\\tfrac{y+n}{f(n)-1}$ and $y=n$ yields $f(c)=f(c-y)$ for some $c$ and any $y>0$. If $f(n)<1$, then for any $y>n$, setting $x=\\tfrac{y-n}{1-f(n)}$ yields similar. And so from above, $f(x)=f(x+p)$ for all $x>c_1$ and for all $p>c_2$, where $c_1$, $c_2$ fixed. \n\nFor any $y>x$, setting $p=y-x+p$ gives $f(x)=f(y+p)=f(y)$, hence $f$ is constant over $(c_1,\\infty)$. Finally, original equation for fixed $y>c_1$ yields $f(xf(y))$ constant for all $x>0$. Hence $f$ is constant, which works." } { "Tag": [], "Problem": "Vote NOW!!!\r\nPlease state the month u were born =]\r\n(i couldnt do all 12 months cuz when i tried to post it said there was too many choices)", "Solution_1": "I may[size=0]ahem[/size] be born...", "Solution_2": "the 32nd day of the 13th month....", "Solution_3": "haha wait whered the poll go", "Solution_4": "month of luv....", "Solution_5": "October! Yay!", "Solution_6": "December huh", "Solution_7": "[quote=\".Ana.\"]December huh[/quote]\r\n\r\nME2!!!! Which day?\r\n\r\nMy birthday is posted on another thread, guess it!\r\n\r\nSomeone born February 29th?", "Solution_8": "I am born in October, like dynamo729!", "Solution_9": "10th and you???", "Solution_10": "October, like shentang and dynamo!!!", "Solution_11": "I've posted it in another thread. \r\n\r\n[hide=\"Hint\"][hide=\"Dont want to guess?\"][hide=\"Are you sure?\"]It is in my username[/hide][/hide][/hide]", "Solution_12": "Posted in another thread about this already.\r\nI was born in February." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Let $p(x)$ be a $1999$-degree polynomial with integer coeffficients that is equal to $\\pm 1$ for $1999$ different integer values of $x$. Show that $p(x)$ cannot be factored into the product of two plynomials with integer coefficients.", "Solution_1": "[hide]\nsuppose the polynomial can be factored into a product of polynomials with integer coefficients, $A(x)B(x)$\nwolog, let $A(x)$ have the lower degree. Then the degree of $A(x)$ is at most $999$. For each of the $1999$ values, $A(x)$ and $B(x)$ both give integer results, and the product of these results is $1$ or $-1$. This means that for each of the $1999$ values, $A(x)$ gives $1$ or $-1$. Then $A(x)$ gives $1$ at least $1000$ times, or $A(x)$ gives $-1$ at least $1000$ times, which is impossible for a polynomial of degree $999$.\n[/hide]" } { "Tag": [ "geometry", "parallelogram", "similar triangles", "geometry solved" ], "Problem": "let ABCD be convex quadrilateral .AC cuts BD at O .X,Y,Z,T are the projections of O on AB,BC,CD,DA.Prove that ABCD is parallelogram iff OX=OZ,OY=OT", "Solution_1": "If ABCD is a paralelogram, then obviously OX=OZ and OY=OT.\r\n\r\nThe converse: if OX=OZ and OY=OT.\r\nIf AB||CD (or AD||BC) then OX=OZ (or OY=OT) means that the similar triangles OAB and OCD (or OBC and OAD) have congruent corresponding heights, so they are congruent so ABCD is a paralelogram.\r\n\r\nSuppose no opposite sides of ABCD are paralel. Then there exists M=AB \\cap CD and N=AD \\cap BC.\r\nOX=OZ and OY=OT means MO and NO are bisectors for AMC and CNA.\r\n\r\nLets look at AMC and ANC. MO and No are bisectors of AMC and ANC, so CO/OA=CM/MA=CN/NA. Let O' be the point where the exterior bisector of AMC hits AC. We have O'C/O'A=OC/OA. It is easy to see that O' is also the point where the exterior bisector of ANC hits AC.\r\n\r\nIf we do the same looking at the triangles BMD and BND we get that the exterior bisectors of BMD and BND meet BD at a point O\".\r\n\r\nFrom these, we get O=O'=O\" which is impossible because for example O lies in [AC] while O' does not.\r\n\r\nSo ABCD has 2 paralel opposite sides, so from what we saw earlier, it is a paralelogram.\r\n----------------------------------------------------------------------------------\r\nSorry about my earlier mistake. I've put C instead of D someplaces.", "Solution_2": "NBC is triangle!!?? :? :?", "Solution_3": "I think it is a Television Network :D. \r\nReplace C by D where you see NBC or MBC." } { "Tag": [ "ratio", "LaTeX" ], "Problem": "Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball is fixed at the origin and the other is free. The balls are charged identically as a result of which the spring length increases n times. Determine the ratio of new frequency (when balls are charged) to old frequency (when balls were uncharged) when the free ball is displaced slightly from its mean position.\r\n\r\n\r\n[hide=\"Ans\"] $ \\sqrt{\\frac{3n\\minus{}2}{n}}$ [/hide]", "Solution_1": "$ \\minus{} kx \\equal{} ma$\r\n\r\n$ \\omega^2 \\equal{} \\frac {k}{m}$\r\n\r\nNow force at eq. pos is given by\r\n$ k(n \\minus{} 1)x \\equal{} \\frac {KQ^2}{(nx)^2}$ \r\n\r\nWhen the ball is displaced by dx , the net force is$ \\minus{} (k(n \\minus{} 1)(x \\plus{} dx) \\minus{} \\frac {KQ^2}{(nx \\plus{} dx)^2}) \\equal{} \\minus{} [\\frac {KQ^2}{(nx)^2} (1 \\minus{} \\frac {1}{(1 \\plus{} \\frac {2dx}{nx})}) \\plus{} k(n \\minus{} 1)dx] \\equal{} \\minus{} [ k(n \\minus{} 1)x(\\frac {2dx}{nx}) \\plus{} k(n \\minus{} 1)dx] \\equal{} \\minus{} k(n \\minus{} 1)dx(\\frac {2}{n} \\plus{} 1) \\equal{} ma$\r\n\r\nThis gives $ \\omega^2 \\equal{} k(n \\minus{} 1)(\\frac {2}{n} \\plus{} 1)/m$\r\n\r\nWrong again :( But where is the glitch", "Solution_2": "Cool work Conjurer , u made the same mistake i made initially but still appreciable effort :coolspeak: \r\n\r\nHere is my solution :wink: \r\n[hide][img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/c/d/2cd322fc424a562d303edf981271e041ce5424f2.gif[/img][/hide]", "Solution_3": "thank you very much :)", "Solution_4": "Yesterday I solved that problem, interesting, but I was waiting others tries.\r\n\r\nIn slogger's reply, when I click to reveal hidden content, I get only the word Image... Do you know how can I get his solution? Or, maybe you could \"transfer\" it to me using Latex?\r\n\r\nShould I put my solution?", "Solution_5": "Of course u can post ur solution\r\n\r\nGeniuses should not stop from posting a solution just because a lesser solution has been posted before :D\r\n\r\nAlso , u can right click on the place where Image is written and click view image", "Solution_6": "Slogger's solution, made easier to view :): [hide][img]http://www.mathlinks.ro/latexrender/pictures/2/c/d/2cd322fc424a562d303edf981271e041ce5424f2.gif[/img][/hide]" } { "Tag": [ "limit", "function", "complex numbers", "calculus", "calculus computations" ], "Problem": "$\\lim_{x\\to 0}(-1)^{x}=?$", "Solution_1": "You should first consider a more basic question: $(-1)^{x}=?$", "Solution_2": "I think sometimes it's not real, and when it's real, I think it's $-1$. So the answer is $-1$?", "Solution_3": "$-1 = e^{\\pi i}$.\r\n\r\nTherefore, $(-1)^{x}= (e^{\\pi i})^{x}= e^{\\pi i x}$.\r\n\r\nThus, $\\lim_{x\\to 0}(-1)^{x}= \\lim_{x\\to 0}e^{\\pi i x}= e^{\\pi i * 0}= e^{0}= 1$.", "Solution_4": "Suppose we agree that $(-1)^{x}$ means $e^{\\pi ix}$, where the exponential function is defined by its Taylor expansion $e^{z}=\\sum_{n=0}^{\\infty}\\frac{z^{n}}{n!}$. If so, then the answer is $1$, as PTynan89 wrote. \r\n\r\nBut do we all agree with this interpretation of $(-1)^{x}$? Some people think that $(-1)^{1/3}=-1$, which is not the same as $e^{\\pi i/3}=\\frac{\\sqrt{3}}{2}+\\frac{i}{2}$.", "Solution_5": "Any idea???", "Solution_6": "The point I tried to make is that one cannot define $(-1)^{x}$ for fractional values of $x$ in a way that is consistent with algebraic rules such as $(a^{x})^{y}=a^{xy}$. For instance, is $(-1)^{2/3}=((-1)^{2})^{1/3}=1^{1/3}=1$? Of course not, because $1^{3/2}$ is not equal to $-1$. And so on.", "Solution_7": "So $1$ is not the answer, right?\r\nWhat is it?", "Solution_8": "[quote=\"mlok\"]Suppose we agree that $(-1)^{x}$ means $e^{\\pi ix}$, where the exponential function is defined by its Taylor expansion $e^{z}=\\sum_{n=0}^{\\infty}\\frac{z^{n}}{n!}$. If so, then the answer is $1$, as PTynan89 wrote. \n\nBut do we all agree with this interpretation of $(-1)^{x}$? Some people think that $(-1)^{1/3}=-1$, which is not the same as $e^{\\pi i/3}=\\frac{\\sqrt{3}}{2}+\\frac{i}{2}$.[/quote]\r\n\r\nWell in that case we can define $-1$ as $e^{\\pi i+2\\pi i n}, \\ n\\in\\mathbb{Z}$, then we have $(-1)^{x}= e^{\\pi i x+2\\pi i n x}= e^{(\\pi i x)(1+2n)}$, and thus $\\lim_{x\\to 0}(-1)^{x}= \\lim_{x\\to 0}e^{(\\pi i x)(1+2n)}= e^{0}= 1$.", "Solution_9": "[quote=\"Jimmy\"][quote=\"mlok\"]Suppose we agree that $(-1)^{x}$ means $e^{\\pi ix}$, where the exponential function is defined by its Taylor expansion $e^{z}=\\sum_{n=0}^{\\infty}\\frac{z^{n}}{n!}$. If so, then the answer is $1$, as PTynan89 wrote. \n\nBut do we all agree with this interpretation of $(-1)^{x}$? Some people think that $(-1)^{1/3}=-1$, which is not the same as $e^{\\pi i/3}=\\frac{\\sqrt{3}}{2}+\\frac{i}{2}$.[/quote]\n\nWell in that case we can define $-1$ as $e^{\\pi i+2\\pi i n}, \\ n\\in\\mathbb{Z}$, then we have $(-1)^{x}= e^{\\pi i x+2\\pi i n x}= e^{(\\pi i x)(1+2n)}$, and thus $\\lim_{x\\to 0}(-1)^{x}= \\lim_{x\\to 0}e^{(\\pi i x)(1+2n)}= e^{0}= 1$.[/quote]\r\nThis is correct if n doesn't depend on x. But if we take n=[1/(2x)]", "Solution_10": "In my post $n$ was meant to be an arbitrary integer, although it could coincidentally be $\\frac{1}{2x}$.", "Solution_11": "So what's the conclusion?", "Solution_12": "[quote=\"OHO\"]So what's the conclusion?[/quote]\r\n\r\n$\\lim_{x\\to 0}(-1)^{x}= 1$", "Solution_13": "[quote=\"Jimmy\"]Well in that case we can define $-1$ as $e^{\\pi i+2\\pi i n}, \\ n \\in \\mathbb Z$, then we have $ (-1)^{x}= e^{\\pi i x+2\\pi i n x}= e^{(\\pi i x)(1+2n)}$, and thus $\\lim_{x \\to 0}(-1)^{x}= \\lim_{x \\to 0}e^{(\\pi i x)(1+2n)}= e^{0}= 1$.[/quote]\r\nYou're still not answering mlok's objection :!:\r\n\r\nFurthermore, if one assumes that \"the usual rules\" work with complex exponents, then one can find another proof of the fact that $0 = 1$. It's somewhere on the forum :huh:", "Solution_14": "[b]perfect_radio wrote:[/b][quote]\nYou're still not answering mlok's objection :!: \n\nFurthermore, if one assumes that \"the usual rules\" work with complex exponents, then one can find another proof of the fact that $0 = 1.$ It's somewhere on the forum :huh: [/quote]\r\nAs Jimmy wrote we can write -1 as $e^{\\pi i+2\\pi i n},\\n\\in \\mathbb Z$ \r\nso,in general whenever we write $(-1)^{1/3}$,we should express it in a more general form as $e^{(\\pi+2n\\pi)/3},$ which for different values of n will assume 3 values $-1,\\frac{\\sqrt{3}}{2}+\\frac{i}{2},\\frac{\\sqrt{3}}{2}-\\frac{i}{2}$", "Solution_15": "Well, is $(-1)^{x}$ is multivalued: $(-1)^{x}=\\{e^{\\pi i x(1+2n)}\\colon n\\in\\mathbb Z\\}$, then $\\lim_{x\\to 0}(-1)^{x}$ is multivalued too: it is the set $\\{z\\in\\mathbb C\\colon |z|=1\\}$. :ninja:\r\n\r\nAnd what is the definition of the limit of a multivalued function? :huh:", "Solution_16": "[b]mlok:3rd post[/b]\r\n[quote]\t\nThe point I tried to make is that one cannot define $(-1)^{x}$ for fractional values of $x$ in a way that is consistent with algebraic rules such as $(a^{x})^{y}=a^{xy}$. For instance, is $(-1)^{2/3}=((-1)^{2})^{1/3}=1^{1/3}=1?$ Of course not, because $1^{3/2}$ is not equal to $-1$. And so on.[/quote]\n\nIn previous post the point I wanted to clearify was that the above rule of algebra is valid even for complex numbers,\nand $1^{3/2}=e^{3n\\pi}=1 and-1$\n\n[b]mlok:5th post[/b]\n[quote]Well, is $(-1)^{x}$ is multivalued: $(-1)^{x}=\\{e^{\\pi i x(1+2n)}\\colon n\\in\\mathbb Z\\}$, then $\\lim_{x\\to 0}(-1)^{x}$ is multivalued too: it is the set $\\{z\\in\\mathbb C\\colon |z|=1\\}.$\n\nAnd what is the definition of the limit of a multivalued function? [/quote]\r\nI definitely agree with your above argument and would say that as you pointed out,that the given limit does not exit.But you should also note that as $x\\to 0$ $n$ can take infinite values,and choosing a large $n$ from the set will prove that [b]Jimmy's[/b] conclusion was wrong and the limit will not tend to 1.", "Solution_17": "[b]perfect_radio wrote:[/b]\r\n[quote]\nYou're still not answering mlok's objection :!: \n\nFurthermore, if one assumes that \"the usual rules\" work with complex exponents, then one can find another proof of the fact that $0=1$ . It's somewhere on the forum :huh: [/quote]\r\n\r\nI could not find it anywhere in the forum :maybe: can you help with it?\r\nAlso,do you agree with my above arguments?" } { "Tag": [ "inequalities", "geometry", "perimeter", "function", "inequalities proposed" ], "Problem": "Given an arbitrary triangle prove the inequality:\r\n\r\n$P \\geq \\frac{3\\sqrt{3}}{8} R^2$", "Solution_1": "i think its /4\r\n\r\nfix R, we want to maximize the area of the triangle. It can be written as $R^2/2(sinX + sinY + sinZ)$ where X,Y,Z are the angles OAB OBC OCA. The sinX + .. is well known, has maximum sqrt(3)/2. So minimum is R^2 3sqrt(3)/4\r\n\r\nps. if you mean P for perimeter, please state that.. you should state it if P is area anyway.", "Solution_2": "Singular,i think u found the maximum and not the minimum :\r\n\r\nP=(abc)/4R. a=2RsinA,etc so P=2R 2 sinAsinBsinC. The function ln(sinx) is concave and so sinAsinBsinC <= 3sqrt(3)/8 and thus P <= 3R 2 sqrt(3)/4\r\n\r\nAs far as the initial ineq is concerned, if P denotes the area, i think it is wrong : take A=2\u03c0/3, \u0392=C=\u03c0/6. Then sinAsinBsinC=sqrt(3)/8 an the whole expression is 2R 2 sqrt(3)/8.", "Solution_3": "You're right galeos, I've must have miscalculated when counting minimum.\r\n\r\nHowever $P\\leq \\frac{3\\sqrt{3}}{4}R^2$ is true, well-known and easy :D", "Solution_4": "This problem was very popular." } { "Tag": [ "calculus", "derivative", "parameterization", "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": "please if anyone has an idea how to solve p(x,t)=q(x,t)-q(x,t)q(x,t)'. here ' means the derivative with respect to t. I am sure that is no explicit solution but there must be some conditions such that I can write q in terms of p. One of the conditions is that p must be continuous, otherwise this has no sense. But can anyone give me an advise how to approach the problem? \r\n\r\nthanks a lot\r\nmikiduta", "Solution_1": "What is $ x$ doing there? :? Is it just a parameter and the equation you are interested in is merely $ q(t)(1\\minus{}q'(t))\\equal{}p(t)$ or, equivalently, $ q'(t)\\equal{}1\\minus{}\\frac{p(t)}{q(t)}$? \r\n\r\nAlso, what does it mean \"to write $ q$ in terms of $ p$\"? Are you just looking for the conditions for a solution to exist? One (rather trivial) statement of this sort is that if $ p$ is continuous and $ p\\le 0$, and $ q(t_0)>0$, then the solution exists for all $ t>t_0$. I doubt it is of much use for you but it is hard to tell more without having a more clear idea of what you are really after. :)", "Solution_2": "Thank you very much for looking at the problem that I posted.\r\nIndeed this equation has the form(after you rewrite it)[code]q_{'}(t)=1-frac{p(t)}{q(t)}. The x , indeed is not in the equation.\nI want to show that the equation is invertible (that I can take out q in terms of p), i.e has a solution, even though this solution is not an explicit one. The things that I didn't say is that I have no certain conditions about how p and g must be.I just need conditions (as generally as possible) such that this equation will have solutions.Can i have p in W^{1,1}? I found this problem while doing my research and no colleague was able to help me out with any idea.\nMay you please explain why it is trivial that the equation has solutions when [/code] p\\leq 0[code]. I am slow in understanding things but the nice thing is that once i get them I am fast:))?\r\n\r\nThank you soooooooo much for your help!\r\n\r\nSincerely,\r\nMikiduta", "Solution_3": "No mathematical help, but to use LaTeX on the forums, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 or [[LaTeX]].", "Solution_4": "Thank you very much for looking at the problem that I posted.\r\nIndeed this equation has the form(after you rewrite it) .\r\nCode:\r\n$ q_{'}(t)\\equal{}1\\minus{}\\frac{p(t)}{q(t)}$.\r\nThe Code: $ x$ , indeed is not in the equation.\r\nI want to show that the equation is invertible (that I can take out Code:$ q$ in terms of Code $ p$), i.e has a solution, even though this solution is not an explicit one. The things that I didn't say is that I have no certain conditions about how $ p$ and $ g$ must be.I just need conditions (as generally as possible) such that this equation will have solutions.Can i have $ p$ in $ W^{1,1}$?I found this problem while doing my research and none of my colleague was able to help me out with any idea.\r\nMay you please explain why it is trivial that the equation has solutions when Code: $ p\\leq 0$. I am slow in understanding things but the nice thing is that once i get them I am fast:))? \r\n\r\nThank you soooooooo much for your help!\r\n\r\nSincerely,\r\nMikiduta", "Solution_5": "Can any of you help me?\r\nI really need some ideas?\r\n\r\nThanks a lot!\r\nMikiduta", "Solution_6": "Hi I am new on this site.\r\nI saw your problem and as weird as it seams I am thinking at the same problem....I will suggest to use Peano theorem-theorem of existence of the solution. But I do not see why you need $ p\\leq 0$. The continuity is important because otherwise you cannot apply the theorem above. Do you need the solutions to be unique?", "Solution_7": "thanks a lot!\r\nI don not get either why the $ p$ must be less or equal then $ 0$....\r\nI have no other idea...I tried to expand in power series to see if I can work by hand this problem...but it is not working.\r\n\r\nI hope we will find an idea!\r\n\r\nThanks,\r\nmikiduta", "Solution_8": "Ah, of course, continuity of $ p$ is enough to ensure the [b]local[/b] existence and uniqueness. Just recall your classical existence and uniqueness theorem for general ODE. The problem is that if $ p$ and $ q$ are positive, the solution may easily develop a singularity in finite time when $ t$ goes up and $ q$ hits zero. The negativity of $ p$ together with the positivity of the initial data were there to prevent this and for nothing else.", "Solution_9": "thank you very much for your details.\r\nThis explains a lot.\r\n\r\nThanks,\r\nmikiduta", "Solution_10": "Do you think there can be any other general constraint that I can ask suck that this equation will have solution?for example if i have $ q_{'}$ bounded by a large constant, $ p$ continuous....etc\r\nThanks,\r\nmiki" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Let \r\n\\[ f_n(x)\\equal{}\\frac{x}{nx \\plus{} 1}\\]\r\nbe a sequence $ (f_n)_1^\\infty$. \r\nInvestigate uniform convergence on\r\n\r\n (i) $ [0,1]$\r\n\r\n\r\n (ii) $ [1,\\infty[$", "Solution_1": "we have uniform convergence on both intervals right :?:", "Solution_2": "For $ x\\ge 0,\\ 0 \\le \\frac {x}{nx \\plus{} 1} < \\frac {x \\plus{} \\frac1n}{nx \\plus{} 1} \\equal{} \\frac1n.$\r\n\r\nYes, that's uniform convergence." } { "Tag": [ "geometry", "Harvard", "college" ], "Problem": "what is it and how do i start it?\r\n\r\nseems like many acceptees to the so-called \"prestigious colleges\" have had some sort of experience with this.", "Solution_1": "[url=http://www.nmun.org/]http://www.nmun.org/[/url]\r\n\r\nThis looks like it, I could be wrong, though.\r\n\r\nI think I'm doing it now (sounds funny) for my AP World History class, but I don't know what about it goes on a resume or anything. All it means for me is more work for my already hw-flooded world history class.", "Solution_2": "Model UN could be run as a club (like in my school) or a class (as in paladin8's school- this is called Global Classrooms). it is run through the UN Assosiation of the USA (UNA-USA) and has more than one conference every year (look at [url=http://www.mun.org/]www.mun.org/[/url] for the dates and other info on the program)", "Solution_3": "it's a club in my school. most of the smart and top students in my school are in it. i'm not, i think it's boring.\r\nit's a serious club", "Solution_4": "Well, I don't know much about it. I worked a few years as High School teacher back in Costa Rica, and they were running that same project. It seems like a debate club or something like that, and somehow several of the kids on those highschools meet once a year. I think it is part of (or related to) the IB, because, at least in CR, just schools with the IB implemented participated in the project. Usually each \"team\" is assigned to a country, and they should read about the country and present debate as if they were defending such country.", "Solution_5": "I think it's easier to be enthusiastic about when run as a club. If you're not sure about to join, your school probably doesn't have a group. Therefore, I would suggest that you find at least 3 people who are interested, a conference somewhere in your area (many colleges will hold high school conferences, we have 3 major ones in SC that I know of), and a teacher who is willing to sponsor you. Usually history teachers or those interested in some form of debate are more than willing to coach. The focus of the club is primarily debate on international issues using a set of rules similar to that of the UN. Our team at Summerville is made up of 9 members, 3 people per member-state, and is run by an AP history teacher. We meet after school one day a week, usually for about an hour, though we typically are flexible and often cancel practice. It's a great way to show colleges a wider variety of talents, demonstrating your ability to deal with people and interest in world affairs. It especially helps in those that seek well-rounded students. If you have any more questions, talk to me and I'll try to answer.", "Solution_6": "my school's model un club is a complete joke. i've been a part of it this past year (9th grade, my first year in high school) and although there are a few smart people in it, the majority of people don't do any work. we go to the harvard model un conference and the ivy league model un concerence at upenn, and the kids just want a vacation. that's really all that anyone cares about the club...they just want the vacation from school where there's a dance. in the committee sessions, they go outside and talk with other kids and don't try to participate it all. it's really a shame, so i probably won't be doing it next year." } { "Tag": [ "probability", "ratio", "Putnam", "calculus", "real analysis", "function", "algebra" ], "Problem": "A fair die bearing number 1 through 6 on its faces is thrown repeatedly until the\r\nrunning total first exceeds 12. What is the most likely total that will be obtained?", "Solution_1": "[hide]If exceeding $ 12$ means at least $ 13$, then I think it's $ 13$.\n\nBefore first exceeding $ 12$, we may have a running total of $ 12,11,10,9,8,$ or $ 7$.\n\nFor each of the above running totals, there exists a number between $ 1$ and $ 6$ (inclusive) which , when added, can help it reach $ 13$.\n\nBut a running total of 7 before the last throw makes reaching $ 14$ impossible.\n\nSimilarly a running total of 7 or 8 before the last throw makes reaching $ 14$ and $ 15$ impossible.\n\nAnd a total of 18, the least likely of all, requires a running total of 12 before the last throw.[/hide]", "Solution_2": "You've neglected to consider the fact that arriving at those previous numbers might have different probabilities to begin with.", "Solution_3": "No: you have a weighted average of uniform distributions on $ \\{13\\}$, $ [13, 14]$, $ [13, 15]$, ..., $ [13, 18]$. This gives a greatest likelyhood of arriving at 13, regardless of the weights.\r\n\r\n(The actual probabilities are very ugly -- roughly speaking, they work out to 28%, 23.7%, 19.2%, 14.6%, 9.7% and 4.8%.", "Solution_4": "Notice that those probabilities are close to $ \\frac{6}{21}$, $ \\frac{5}{21}$, $ \\frac{4}{21}$, $ \\frac{3}{21}$, $ \\frac{2}{21}$, and $ \\frac{1}{21}$. Is that a lucky coincidence? :wink:", "Solution_5": "Makes sense, thanks.\r\n\r\nI take it that the probability of the most likely first roll exceeding $ n$ being $ n \\plus{} 1 \\ldots n \\plus{} 6$ would tend to $ \\frac {6}{21} \\ldots \\frac {1}{21}$, respectively, as $ n \\to \\infty$, or something like that?", "Solution_6": "Yes, that's right. Would anyone like to prove it?", "Solution_7": "See also this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=20848]old thread[/url] on a continuous version.\r\n\r\nThe nicest possible answer for the excess comes with a geometric distribution in the discrete case and an exponential distribution in the continuous case.", "Solution_8": "[quote=\"Ravi B\"]Notice that those probabilities are close to $ \\frac {6}{21}$, $ \\frac {5}{21}$, $ \\frac {4}{21}$, $ \\frac {3}{21}$, $ \\frac {2}{21}$, and $ \\frac {1}{21}$. Is that a lucky coincidence? :wink:[/quote]\r\n\r\nThis result is very intuitive.\r\nWe may take \"exceeding $ 1000$ as an example\r\n\r\nFor $ 1001$,\r\nobviously $ 995$, $ 996$, $ 997$, $ 998$ and $ 999$ ,$ 1000$ are possible running totals before the last throw.\r\n\r\n\r\nFor $ 1002$\r\n$ 996$, $ 997$, $ 998$, $ 999$ and $ 1000$ are possible running totals before the last throw.\r\n\r\nFor $ 1003$\r\n$ 997$, $ 998$, $ 999$ and $ 1000$ are possible running totals before the last throw.\r\n....\r\nFor 1006,\r\n$ 1000$ is the only possible running totals before the last throw.\r\nAnd the probabilities of arriving these running totals are very close. So the probabilities should be in the ratio of $ 6: 5: 4: 3: 2: 1$\r\n\r\nWe can expect the probability of a running total of $ n$ to be close to that of a running total of $ n \\plus{} 1$ for large $ n$, hence the result.\r\n\r\nNo idea how to prove it though :blush: .", "Solution_9": "Yes, you've got the basic idea. To make it rigorous, imagine that we keep rolling forever. It's well known that the probability of ever hitting a total of $ k$ approaches $ \\frac{2}{7}$ as $ k$ approaches infinity. (That was a Putnam problem from 1960.) Thus the probability of ever using a particular transition, say from $ k$ to $ k \\plus{} 2$, approaches $ \\frac{2}{7} \\cdot \\frac{1}{6}$, which is $ \\frac{1}{21}$.\r\n\r\nFor the first total exceeding $ n$ to be $ n \\plus{} 1$, we must have used one of the 6 transitions ending in $ n \\plus{} 1$. Hence the probability of doing so approaches $ \\frac{6}{21}$. In the same way, we get the other probabilities.", "Solution_10": "Now, actually proving that that probability of ever hitting $ k$ approaches $ \\frac27$ is harder. Several approaches follow:\r\n\r\n[hide=\"Special case, quantitative, minimal calculus\"]We think in terms of generating functions. Let $ f(n)$ be the probability of hitting $ n$ with one die roll, and $ F(n)\\equal{}\\frac16(x\\plus{}x^2\\plus{}x^3\\plus{}x^4\\plus{}x^5\\plus{}x^6)$ the corresponding generating function. Let $ h(n)$ be the expected number of times we hit $ n$ when rolling repeatedly and adding terms (which is also equal to the probability of ever hitting $ n$, since we can only hit once). We can easily see that $ h(0)\\equal{}1$ and $ h(n)\\equal{}\\sum_{m\\le n}f(m)h(n\\minus{}m)$; we reach $ n$ by first reaching $ n\\minus{}m$ and then rolling $ m$ with probability $ f(m)$.\nThat convolution is multiplication for generating functions, and the equations become $ H(x)\\equal{}1\\plus{}H(x)F(x)$. Solving for $ H$, $ H(x)\\equal{}\\frac1{1\\minus{}F(x)}\\equal{}\\frac{6}{6\\minus{}x\\minus{}x^2\\minus{}x^3\\minus{}x^4\\minus{}x^5\\minus{}x^6}$.\nNow, we decompose this with partial fractions. $ \\frac{6}{6\\minus{}x\\minus{}x^2\\minus{}x^3\\minus{}x^4\\minus{}x^5\\minus{}x^6}\\equal{}\\frac{2}{7}\\left(\\frac1{1\\minus{}x}\\plus{}\\frac{15\\plus{}10x\\plus{}6x^2\\plus{}3x^3\\plus{}x^4}{6\\plus{}5x\\plus{}4x^3\\plus{}3x^3\\plus{}2x^4\\plus{}x^5}\\right)$\n\nThat $ \\frac{2}{7}\\frac1{1\\minus{}x}$ is the limiting average of $ h(n)$; we want to estimate how fast the rest goes to zero.\nSince all zeros of $ 6\\plus{}5x\\plus{}4x^3\\plus{}3x^3\\plus{}2x^4\\plus{}x^5$ are strictly outside the unit circle, the coefficients of the power series of $ h$ go to zero geometrically fast. How fast?\nThe zeros of $ 6\\plus{}5x\\plus{}4x^3\\plus{}3x^3\\plus{}2x^4\\plus{}x^5$ are approximately $ \\minus{}1.4918$, $ \\minus{}0.8058\\pm 1.2229i$, and $ 0.5517\\pm 1.2533i$. The latter pair have the smallest absolute value, at approximately $ 1.369$. Expanding $ \\frac{15\\plus{}10x\\plus{}6x^2\\plus{}3x^3\\plus{}x^4}{6\\plus{}5x\\plus{}4x^3\\plus{}3x^3\\plus{}2x^4\\plus{}x^5}$ in complete partial fractions, we get the pair $ \\frac{\\minus{}0.2758\\mp 0.6267i}{x\\minus{}(0.5517\\pm 1.2533i)}$, the pair $ \\frac{0.4029\\mp 0.6115i}{x\\minus{}(\\minus{}0.8058\\pm 1.2229i)}$, and the singleton $ \\frac{0.7459}{x\\plus{}1.4918}$.\nEstimating by absolute values,\n$ \\left|\\frac27\\cdot\\frac{15\\plus{}10x\\plus{}6x^2\\plus{}3x^3\\plus{}x^4}{6\\plus{}5x\\plus{}4x^3\\plus{}3x^3\\plus{}2x^4\\plus{}x^5}\\right|<\\frac27\\left(\\frac{2\\cdot 0.7}{x\\minus{}\\frac43}\\plus{}\\frac{3\\cdot 0.75}{x\\minus{}\\sqrt{2}}\\right)$\n(termwise in the power series)\nWe then have that $ \\left|h(n)\\minus{}\\frac27\\right|<0.3\\cdot \\left(\\frac34\\right)^n\\plus{} \\frac12\\cdot 2^{\\minus{}\\frac n2}$.\n\nFrom here, the rest is as Ravi B wrote. The excess $ e_n(k)$ (the probability we get $ n\\plus{}k$ when we first get $ \\ge n$) has formula $ e_n(k)\\equal{}\\sum_{j>k}f(j)h(n\\plus{}k\\minus{}j)$ for $ k\\ge 0$, which converges to $ \\frac27\\sum_{j>k}f(j)$.[/hide]\n\n[hide=\"General case, includes complex analysis. Very long.\"]\nHere let $ f(n)$ be a nice enough probability distribution on the nonnegative integers with mean $ \\mu>0$ and variance $ \\sigma^2$. As before, let $ F(x)$ be the corresponding generating function, $ h(n)$ the expected number of times we hit $ n$, and $ H(x)$ the generating function for $ x$. We have $ H(x)\\equal{}\\frac1{1\\minus{}F(x)}$.\nNow, we want to estimate $ H(n)$ for large $ n$, but we can't just use partial fractions because we're not working with polynomials anymore. We use complex analysis instead. For $ 01$.\n\nFor $ r>1$, we must also account for poles; $ 1\\minus{}F(1)\\equal{}0$ since $ F$ comes from a probability distribution, and there may be other zeros of $ 1\\minus{}F$ or even barriers to analytically extending it. We draw on $ f$ being \"nice\" here; suppose that there is some $ R>1$ such that $ F(z)$ is analytic on the disk $ |z|k}f(j)h(n\\plus{}k\\minus{}j)\\equal{}\\sum_{j>k}f(j)\\left(\\frac1{\\mu}\\plus{}O(r^{\\minus{}n\\minus{}k\\plus{}j})\\right)$\n$ \\equal{}\\frac1{\\mu}\\sum_{j\\equal{}k\\plus{}1}^{n\\plus{}k}f(j)\\plus{}\\sum_{j\\equal{}k\\plus{}1}^{n\\plus{}k}O(r^{\\minus{}j})O(r^{\\minus{}n\\minus{}k\\plus{}j})\\equal{}O(nr^{\\minus{}n\\minus{}k})\\plus{}\\frac1{\\mu}\\sum_{j\\equal{}k\\plus{}1}^{n\\plus{}k}f(j)$\nAs $ n\\to\\infty$, this converges to the tail end distribution $ e(k)\\equal{}\\frac1{\\mu}P(X>k)$, where $ X$ is the base random variable with distribution $ f$. We can easily calculate its mean\n$ \\sum_k k\\cdot e_k\\equal{}\\frac1{\\mu}\\sum_k\\sum_{j>k}kf(j)\\equal{}\\sum_j\\sum_{k1$:\n$ g(n)\\equal{}\\frac1{2\\pi i}\\int_{|z|\\equal{}r}\\frac{G(z)}{z^{n\\plus{}1}}\\,dz \\minus{} \\text{Res}_{z\\equal{}1}\\frac{G(z)}{z^{n\\plus{}1}}$\n$ \\equal{}\\frac1{2\\pi}\\int_0^{2\\pi}\\frac{G(re^{i\\theta})}{r^n}e^{\\minus{}in\\theta}\\,d\\theta \\minus{} \\text{Res}_{z\\equal{}1}\\frac{1}{z^n(1\\minus{}z)(1\\minus{}F(z))}$\nSubstituting $ t\\equal{}z\\minus{}1$ and expanding $ F$ in its power series centered at $ 1$, that residue becomes $ \\text{Res}_{t\\equal{}0}\\frac{1}{(1\\plus{}t)^n(\\minus{}t)(\\minus{}\\mu t\\minus{}(\\mu^2\\minus{}\\mu\\plus{}\\sigma^2)\\frac{t^2}{2}\\plus{}O(t^3))}$\n$ \\equal{}\\text{Res}_{t\\equal{}0}\\frac1{\\mu t^2}(1\\plus{}t)^{\\minus{}n}\\left(1\\plus{}\\frac{\\mu^2\\minus{}\\mu\\plus{}\\sigma^2}{2\\mu}t\\plus{}O(t^2)\\right)^{\\minus{}1}$\n$ \\equal{}\\text{Res}_{t\\equal{}0}\\frac1{\\mu t^2}(1\\minus{}nt\\plus{}O(t^2))(1\\minus{}\\frac{\\mu^2\\minus{}\\mu\\plus{}\\sigma^2}{2\\mu}t\\plus{}O(t^2))$\n$ \\equal{}\\text{Res}_{t\\equal{}0}\\frac1{\\mu}t^{\\minus{}2}\\minus{}\\frac{n}{\\mu}t^{\\minus{}1}\\minus{}\\frac{\\mu^2\\minus{}\\mu\\plus{}\\sigma^2}{2\\mu^2}t^{\\minus{}1}\\plus{}O(1)$\n\nThe desired residue is the coefficient $ \\minus{}\\frac{n}{\\mu}\\minus{}\\frac{\\mu^2\\minus{}\\mu\\plus{}\\sigma^2}{2\\mu^2}$ of $ \\frac1t$, and thus $ g(n)\\approx \\frac{n}{\\mu}\\plus{}\\frac{\\mu^2\\minus{}\\mu\\plus{}\\sigma^2}{2\\mu^2}$ for large $ n$, with geometrically decaying error.\nThis method doesn't get the distribution of the excess directly, but it does get the expected value, and we can easily take differences of $ g$ to find $ h$ and get the distribution from there.\n\nThe whole argument here was constructed by analogy with Kent's argument in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=20848]old thread[/url] I linked; Laplace transforms in the continuous case become $ z$-transforms in the discrete case, and an inversion formula integrating over a line becomes an inversion formula integrating over a circle. It's quite likely that I assumed more niceness than necessary, but the residue arguments of I leave the disk are a lot nicer than anything I could do on the unit circle.[/hide]", "Solution_11": "In your first solution, just knowing that the zeros of $ 6 \\plus{} 5x \\plus{} 4x^2 \\plus{} 3x^3 \\plus{} 2x^4 \\plus{} x^5$ are outside the unit circle was enough to get the limit $ \\frac{2}{7}$, right? I guess your extra work was to get an explicit bound on the error.\r\n\r\nAnother solution is to apply the standard convergence result about finite Markov chains (stochastic matrices) to a particular 6-state diagram (matrix of order 6). But I guess that is relying on another result.\r\n\r\nThe Putnam book I own has an elementary proof of $ \\frac{2}{7}$. Using your notation, let $ h(n)$ be the probability that we ever hit $ n$. Let $ a(n)$ be the minimum of $ h(n \\minus{} 6)$, $ h(n \\minus{} 5)$, $ h(n \\minus{} 4)$, $ h(n \\minus{} 3)$, $ h(n \\minus{} 2)$, and $ h(n \\minus{} 1)$. Let $ b(n)$ be the corresponding maximum. Then we can show pretty easily that $ a$ is (weakly) increasing and $ b$ is (weakly) decreasing. With a little more work, we get the limiting probability squeezed in between.\r\n\r\nI know another nice proof of the $ \\frac{2}{7}$ limit using a coupling argument, but since that's going beyond \"Intermediate Topics\", I will omit it here. The idea is that if we start our running total artificially to be 1 with probability $ \\frac{6}{21}$, 2 with probability $ \\frac{5}{21}$, 3 with probability $ \\frac{4}{21}$, 4 with probability $ \\frac{3}{21}$, 5 with probability $ \\frac{2}{21}$, and 6 with probability $ \\frac{1}{21}$, then the probability of hitting $ n$ is [i]exactly[/i] $ \\frac{2}{7}$ for all $ n$. In fact, that same coupling argument can establish the results about Markov chains that I alluded to earlier.", "Solution_12": "Yes, the extra work in the first solution was to get an explicit bound.\r\n\r\nThat other argument looks like it still needs a version of \"the other roots are all strictly outside the unit disk\". We have some errors (which live in the invariant subspace of net probability zero), and we need to show that their contribution decays. I like the partial fractions for getting that out in the open.\r\nThe Markov chain version is annoyingly specific; it only works for bounded random variables. The elementary solution bounding between the last six values has the same restriction.\r\nAs for going far beyond \"Intermediate\"- that's why I put everything in spoiler tags.\r\n[Added in edit][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=185160]Here[/url] is another version of the argument for the discrete uniform case.[/edit]\r\n\r\nThe $ \\frac27$ is really $ \\frac {1}{\\text{mean}}$. If we don't worry about proving convergence, that's pretty easy to see intuitively.\r\nOne other note: my definition of excess is shifted by one from the original version in this thread, and starts at zero. It's an arbitrary choice, and easily adjusted between versions.", "Solution_13": "I see that you guys liked my problem huh?\r\n\r\nSay this was on the Putnam. Approximately where would it be?", "Solution_14": "Take a look at JBL's post #4, and re-consider whether it would be on the Putnam in the first place :P", "Solution_15": "That, and the Putnam has changed a lot in 50 years. The old Putnam question wasn't the original question of the thread, either.", "Solution_16": "Essentially this question (I think they sent you out to 700 instead of just 12) was also on the Princeton University Math Competition (they asked it as an expected value question, I think, but it ended up boiling down to the same type of reasoning) and it's appeared on the forum in at least one place, but I can't find it :(", "Solution_17": "I posted the general version [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242473]here[/url]. It's an extremely poor putnam problem because it's pretty straightforward if you know some markov chain theory, because the details are annoying to work out, and because there is almost nothing to do from a problemsolving perspective -- I figured out what the generalization was immediately upon reading the original PUMaC problem.\r\n\r\ne: this refers to my generalization. your problem is really just a textbook exercise and hence unsuitable for the putnam.\r\n\r\n[quote=\"Ravi B\"]I know another nice proof of the $ \\frac {2}{7}$ limit using a coupling argument, but since that's going beyond \"Intermediate Topics\", I will omit it here. The idea is that if we start our running total artificially to be 1 with probability $ \\frac {6}{21}$, 2 with probability $ \\frac {5}{21}$, 3 with probability $ \\frac {4}{21}$, 4 with probability $ \\frac {3}{21}$, 5 with probability $ \\frac {2}{21}$, and 6 with probability $ \\frac {1}{21}$, then the probability of hitting $ n$ is [i]exactly[/i] $ \\frac {2}{7}$ for all $ n$. In fact, that same coupling argument can establish the results about Markov chains that I alluded to earlier.[/quote]\r\n\r\nI would call this 'starting the chain at stationarity', but I guess calling it a coupling works just as well." } { "Tag": [ "logarithms" ], "Problem": "If $ \\log_2 \\Big(\\log_3 (\\log_2 x) \\Big) \\equal{} 0$, then $ x^{\\minus{}\\frac{1}{2}}$ equals", "Solution_1": "This means that $ \\log_{3}(\\log_{2}x)\\equal{}1$, which in turn means that $ \\log_{2}x\\equal{}3$, or that $ x\\equal{}8$. Then $ x^{\\minus{}1/2}\\equal{}\\frac{1}{2\\sqrt{2}}$" } { "Tag": [], "Problem": "This is a topic that has been on the news a lot in the recent years, illegal immigrants. What do you think we should do about it?", "Solution_1": "Go to war with Iran and make a draft. :wink:\r\n\r\ncome on.", "Solution_2": "[quote=\"1=2\"]Go to war with Iran and make a draft. :wink:\n\ncome on.[/quote]\r\n\r\nI was going more on the line of mexican immigrantion...because thats what the majority of illegal immigration is mexican...", "Solution_3": "Give them a chance to be a legal US citizen. I thought that's what the US was supposed to be about during the mid-1800s - a land of hope.", "Solution_4": "No but basically, in order to become an american citizen after you're 18, you need to pass a test, and most illegal immigrants would be unable to pass that test.\r\nAnd they also need funding.\r\nMost illegal immigrants are just people who were too poor to live in Mexico, which must be pretty poor seeing as Mexico is almost a third world country.", "Solution_5": "To possibly attract more responses, maybe this should go to The Round Table?", "Solution_6": "[quote=\"i_like_pie\"]To possibly attract more responses, maybe this should go to The Round Table?[/quote]\r\n\r\nIt might get chucked back here so this is fine. \r\n\r\n@ZhangPejin: Sure they need to pass a test, but that test is only a background check I believe. You can't have the entirety of illegal immigrants be registered as a terrorist...", "Solution_7": "Oh in order to get into the US, you just need a Visa.\r\nBut there's like a minor US history test for citizenship.", "Solution_8": "[quote=\"ZhangPeijin\"]Oh in order to get into the US, you just need a Visa.\nBut there's like a minor US history test for citizenship.[/quote]\r\n\r\nAAAAAHHHH! *runs away*\r\n\r\n[rhetorical]If this was a land of the free, why are people being shipped away?[/rhetorical]", "Solution_9": "[quote=\"ZhangPeijin\"]No but basically, in order to become an american citizen after you're 18, you need to pass a test, and most illegal immigrants would be unable to pass that test.\nAnd they also need funding.\nMost illegal immigrants are just people who were too poor to live in Mexico, which must be pretty poor seeing as Mexico is almost a third world country.[/quote]Actually, 60% of home-born Americans can't pass the test.", "Solution_10": "Oh.\r\nHoly ****.", "Solution_11": "It took me, an American-raised, legal immigrant, a guy who is more patriotic towards the US than many white Americans, someone who seeks to improve this country (and I think I can), 12 years to get my greencard, and I am not yet allowed to apply for citizenship. I've tutored American kids, gotten American peers off drugs, served the American elderly in nursing homes, and picked squash for American food kitchens.\r\n\r\nI don't think someone who is here illegally and who contributes less to the community than I do should have it easier. That's just not fair.\r\n\r\nAnd please, don't say it's that they have it much harder, I've lived without a bed...", "Solution_12": "You notice that new story awhile back?\r\nAbout the woman who had a son born in the US while she was here illegally?\r\nThey make it sound like the American Government is so cruel that they would separate the boy's mother from her etc, etc.\r\nThough I was pretty sure that the boy would've been able to went back to Mexico with her, but idk.", "Solution_13": "[quote=\"miyomiyo\"][quote=\"ZhangPeijin\"]No but basically, in order to become an american citizen after you're 18, you need to pass a test, and most illegal immigrants would be unable to pass that test.\nAnd they also need funding.\nMost illegal immigrants are just people who were too poor to live in Mexico, which must be pretty poor seeing as Mexico is almost a third world country.[/quote]Actually, 60% of home-born Americans can't pass the test.[/quote]\r\n\r\nwe took that test in History Class, and we got past the requirement... and i'm not a US citizen...\r\n\r\nI'm too lazy to take it again and become a US citizen." } { "Tag": [], "Problem": "What integer is closets to the fourth power of $ \\sqrt{2 \\plus{} \\sqrt{2 \\plus{} \\sqrt{2}}}$", "Solution_1": "We are given $ x=\\sqrt{2+\\sqrt{2+\\sqrt2}}\\implies x^2=2+\\sqrt{2+\\sqrt2}\\implies x^4=6+\\sqrt2+4\\sqrt{2+\\sqrt2}$. Since $ 1.4<\\sqrt2\\<1.5$ and thus\r\n\r\n$ \\sqrt{3.24}=1.8<\\sqrt{2+\\sqrt2}<\\sqrt{3.61}=1.9$, we have $ 6+1.4+4(1.8)=14.60$. Show that:\r\n\r\n\\[ \\frac{1}{a^2\\plus{}bc}\\plus{}\\frac{1}{b^2\\plus{}ca}\\plus{}\\frac{1}{c^2\\plus{}ab}\\le \\frac{a\\plus{}b\\plus{}c}{ab\\plus{}bc\\plus{}ca}\\left(\\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\right)\\]", "Solution_1": "[quote=\"Pain rinnegan\"]Let $ a,b,c > 0$. Show that:\n\\[ \\frac {1}{a^2 \\plus{} bc} \\plus{} \\frac {1}{b^2 \\plus{} ca} \\plus{} \\frac {1}{c^2 \\plus{} ab}\\le \\frac {a \\plus{} b \\plus{} c}{ab \\plus{} bc \\plus{} ca}\\left(\\frac {1}{a \\plus{} b} \\plus{} \\frac {1}{b \\plus{} c} \\plus{} \\frac {1}{c \\plus{} a}\\right)\\]\n[/quote]\r\nThe inequality is equivalent to\r\n\\[ \\sum \\frac{a\\plus{}b\\plus{}c}{b\\plus{}c}\\geq \\sum \\frac{ab\\plus{}bc\\plus{}ca}{a^{2}\\plus{}bc},\\]\r\nor\r\n\\[ \\sum \\frac{a}{b\\plus{}c}\\plus{}\\sum \\frac{a^{2}}{a^{2}\\plus{}bc}\\geq \\sum \\frac{a(b\\plus{}c)}{a^{2}\\plus{}bc}.\\]\r\nBy the Cauchy-Schwarz Inequality, we have\r\n\\[ \\begin{aligned}\r\n\\frac{1}{a^{2}\\plus{}bc} &\\equal{}\\frac{b\\plus{}c}{(a^{2}\\plus{}bc)(b\\plus{}c)}\\equal{}\\frac{b}{(a^{2}\\plus{}bc)(b\\plus{}c)}\\plus{}\\frac{c}{(a^{2}\\plus{}bc)(c\\plus{}b)} \\\\\r\n&\\leq \\frac{b}{\\left( a\\sqrt{b}\\plus{}\\sqrt{bc}\\sqrt{c}\\right) ^{2}}\\plus{}\\frac{c}{\\left( a\\sqrt{c}\\plus{}\\sqrt{bc}\\sqrt{b}\\right) ^{2}}\\equal{}\\frac{1}{(a\\plus{}c)^{2}}\\plus{}\\frac{1}{(a\\plus{}b)^{2}}.\r\n\\end{aligned}\\]\r\nTherefore,\r\n\\[ \\begin{aligned}\r\n\\sum \\frac{a(b\\plus{}c)}{a^{2}\\plus{}bc} &\\leq \\sum \\left[ \\frac{a(b\\plus{}c)}{(a\\plus{}b)^{2}}\\plus{}\\frac{a(b\\plus{}c)}{(c\\plus{}a)^{2}}\\right] \\equal{}\\sum \\left[ \\frac{b(c\\plus{}a)}{(b\\plus{}c)^{2}}\\plus{}\\frac{c(a\\plus{}b)}{(b\\plus{}c)^{2}}\\right] \\\\\r\n&\\equal{}\\sum \\frac{a}{b\\plus{}c}\\plus{}\\sum \\frac{2bc}{(b\\plus{}c)^{2}}.\r\n\\end{aligned}\\]\r\nIt suffices to prove that\r\n\\[ \\frac{a^{2}}{a^{2}\\plus{}bc}\\plus{}\\frac{b^{2}}{b^{2}\\plus{}ca}\\plus{}\\frac{c^{2}}{c^{2}\\plus{}ab}\\geq \\frac{2bc}{(b\\plus{}c)^{2}}\\plus{}\\frac{2ca}{(c\\plus{}a)^{2}}\\plus{}\\frac{2ab}{(a\\plus{}b)^{2}}.\\]\r\nSince $ (b\\plus{}c)^{2}<2(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}ab\\plus{}bc\\plus{}ca),$ we have\r\n\\[ \\frac{2bc}{(b\\plus{}c)^{2}}\\equal{}\\frac{1}{2}\\minus{}\\frac{(b\\minus{}c)^{2}}{2(b\\plus{}c)^{2}}\\leq \\frac{1}{2}\\minus{}\\frac{(b\\minus{}c)^{2}}{4(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}ab\\plus{}bc\\plus{}ca)}.\\]\r\nIt follows that\r\n\\[ \\begin{aligned}\r\n\\sum \\frac{2bc}{(b\\plus{}c)^{2}} &\\leq \\frac{3}{2}\\minus{}\\frac{(b\\minus{}c)^{2}\\plus{}(c\\minus{}a)^{2}\\plus{}(a\\minus{}b)^{2}}{4(a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}ab\\plus{}bc\\plus{}ca)} \\\\\r\n&\\equal{}\\frac{(a\\plus{}b\\plus{}c)^{2}}{a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}ab\\plus{}bc\\plus{}ca}.\r\n\\end{aligned}\\]\r\nTherefore, it is enough to check that\r\n\\[ \\frac{a^{2}}{a^{2}\\plus{}bc}\\plus{}\\frac{b^{2}}{b^{2}\\plus{}ca}\\plus{}\\frac{c^{2}}{c^{2}\\plus{}ab}\\geq \\frac{(a\\plus{}b\\plus{}c)^{2}}{a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}ab\\plus{}bc\\plus{}ca},\\]\r\nwhich is true according to the Cauchy-Schwarz Inequality. Equality holds if and only if $ a\\equal{}b\\equal{}c.$", "Solution_2": "The following is true too:\r\n\r\nLet $ a,b,c>0$. Prove that:\r\n\r\n\\[ \\frac{1}{a^2\\plus{}bc}\\plus{}\\frac{1}{b^2\\plus{}ca}\\plus{}\\frac{1}{c^2\\plus{}ab}\\le \\frac{1}{\\sqrt[3]{abc}}\\left(\\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\right)\\]", "Solution_3": "[quote=\"Pain rinnegan\"]The following is true too:\n\nLet $ a,b,c > 0$. Prove that:\n\\[ \\frac {1}{a^2 \\plus{} bc} \\plus{} \\frac {1}{b^2 \\plus{} ca} \\plus{} \\frac {1}{c^2 \\plus{} ab}\\le \\frac {1}{\\sqrt [3]{abc}}\\left(\\frac {1}{a \\plus{} b} \\plus{} \\frac {1}{b \\plus{} c} \\plus{} \\frac {1}{c \\plus{} a}\\right)\\]\n[/quote]\r\nYes, it's a hard problem :maybe: \r\nMy solution is based on the following ineq:\r\n$ 2\\sum\\ ab(a^4 \\plus{} b^4) \\plus{} 5\\sum\\ a^3b^3 \\plus{}$ $ 7abc\\sum\\ a^3 \\plus{} 12a^2b^2c^2 \\ge 2\\sum a^2b^2(a^2 \\plus{} b^2) \\plus{} 8abc \\sum ab(a \\plus{} b)$\r\nIt's obviously trues because \r\n$ 2\\sum ab(a^4 \\plus{} b^4) \\ge 2\\sum\\ a^2b^2(a^2 \\plus{} b^2)$\r\n$ 5\\sum\\ a^3b^3 \\ge \\frac {5}{2}abc\\sum ab(a \\plus{} b)$ (by AM-GM)\r\nand $ 7abc \\sum a^3 \\plus{} 12a^2b^2c^2 \\ge \\frac {11}{2}abc \\sum ab(a \\plus{} b)$\r\nIndeed, this proof is rather long. I'll post it soon. Wait for another proofs :)" } { "Tag": [ "calculus", "integration", "LaTeX" ], "Problem": "Find the largest integral value of $ x$ which solves: $ \\frac{1}{3}<{\\frac{x}{5}}<{\\frac{5}{8}}$", "Solution_1": "1/3=0.3333333333etc\r\n\r\n5/8=.625\r\n1/5= 0.2, 2/5=0.4 3/5=0.6 4/5=0.8 the greatest is 3", "Solution_2": "In reality only an upper bound is necessary:\r\nx/5<5/8 => x<25/8=3.125, so x=3 is the largest integer.", "Solution_3": "Please use LaTeX!!!\r\n\r\nBecause each $ \\frac{1}{5}$ is $ 0.2$, We find multiples of $ 0.2$ less than $ \\frac{5}{8}\\equal{}0.625$, finding that $ \\frac{3}{5}\\equal{}0.6$ is the largest, since another $ 0.2$ added would bring the solution above the decimal value of $ \\frac{5}{8}$. In general, it is much easier to compare fractions when converted to a decimal.", "Solution_4": "If $ \\frac{x}{5} < \\frac{5}{8}$, then $ 8x < 25$. Therefore, $ x < \\frac{25}{8}$, so the answer is 3." } { "Tag": [ "Harvard", "college", "Princeton", "MIT" ], "Problem": "I was on Emory University's website checking out their early admission acceptance rate\r\n\r\nhttp://www.emory.edu/ADMISSIONS/admission-aid/apply.htm\r\n\r\nand the website showed that 32% acceptance rate for Regular, and 34% for early decision.\r\n\r\nThat's almost like no differences. could anyone explain this?\r\n\r\nthanks", "Solution_1": "Why should applying ED make you more likely to get in? There's really only one factor, which is the commitment factor. Applying ED suggests that the school is your top choice, and that makes you slightly more desirable. But otherwise, there's nothing. When a school has a significantly higher ED acceptance rate, it's probably because the ED pool was self-selected into being higher quality. \r\n\r\nPersonally, I don't see the point of ED and I'm glad that Harvard precipitated its decline in the Ivy League (well I don't know if any other top university has followed yet besides Princeton, but presumably after those two did, other schools are considering it).", "Solution_2": "Virginia has also followed the lead of Harvard and Princeton in going to a single-deadline system, if I remember correctly. \r\n\r\nhttp://www.virginia.edu/undergradadmission/admission.html", "Solution_3": "MIT uses [b]non-binding[/b] early action :D ." } { "Tag": [ "trigonometry" ], "Problem": "Hi everyone,\r\ndoes anyone know how to do these two problems??? i have been working on them for hours and am unable to get an answer. please if you could help me i would appreciate it!!! :)\r\n\r\nsin[Arccos(-square root 2/2)-pie/4]\r\n\r\n my attempt: with using the unit circle i got 90 degrees or 1 or 0 or 180 degrees. i dont know if the anwswer is to be in degrees or not as the instructions for both of these problems was just to find each value.\r\ncos(2Arctan squareroot 3/3)\r\n my attempt was getting cos(2 times the square root of 12) by using the unit circle\r\n\r\nPLEASE if you could help me work out these problems i would very much appreciate it!\r\n\r\n\r\nThank you very much", "Solution_1": "[quote=\"chocolatelover\"]Hi everyone,\ndoes anyone know how to do these two problems??? \ncos(2Arctan squareroot 3/3)...\n[/quote]\r\n\r\nStep by step evaluation... That's the Key:\r\nFirst off,\r\n$ \\arctan \\frac {\\sqrt {3}}{3} \\equal{} \\arctan \\frac {1}{\\sqrt {3}}.$\r\n\r\nDraw an equilateral triangle of side 2. It is straightforward now that\r\n\r\n$ \\arctan \\frac {1}{\\sqrt {3}} \\equal{} \\frac {\\pi}{6}.$\r\n\r\nHence, $ \\cos \\left(2 \\arctan \\frac {\\sqrt {3}}{3}\\right) \\equal{} \\cos \\left(\\frac {\\pi}{3}\\right) \\equal{} \\frac {1}{2}.$\r\n\r\nHope that helps...", "Solution_2": "thank you so much for taking your time to help me!!\r\ni have one question though...what happend to the 2???\r\nthanks", "Solution_3": "Thought there were 2 problems there. All I did was sketch the resolution process of the second one [8th line of your original post].", "Solution_4": "no you were right there were two problems... but i ment in the problem that you did it stated 2Arctan what happened to the 2 in the process that you did? i did not follow what happened to the 2", "Solution_5": "[quote=\"chocolatelover\"]thank you so much for taking your time to help me!!\ni have one question though...what happend to the 2???\nthanks[/quote]\r\n\r\nJust edited that part, man... Sorry.", "Solution_6": "its ok thanks for the help...i thought that was what probably happened just forgot the 2...but the answer still works out the same so its ok. but thank you very much for the help", "Solution_7": "[quote=\"chocolatelover\"]\n\nsin[Arccos(-square root 2/2)-pie/4]...\n\n[/quote]\r\n\r\nReal problem is the precomputation of $ \\arccos \\left( \\minus{} \\frac {\\sqrt {2}}{2}\\right)$. The rest would follow from the [b]sin-of-a-sum[/b] identity. Gimme a sec to work out the sol. in some detail.\r\n\r\nYour idea of introducing the [color=red]unit circle[/color] is OK. Though, you have to be careful of the way you pick the value for $ \\arccos \\left( \\minus{} \\frac {\\sqrt {2}}{2}\\right)$. It turns out to be $ \\frac {3\\pi}{4}$ in this case. Therefore,\r\n\r\n$ \\arccos \\left( \\minus{} \\frac {\\sqrt {2}}{2}\\right) \\minus{} \\frac {\\pi}{4} \\equal{} \\frac {\\pi}{2},$\r\n\r\nand the mentioned identity would'nt be necessary after all. You know, $ \\frac {\\pi}{2}$ is one of them notable angles in Trigs.", "Solution_8": "ok thanks i would really appreciate that :)", "Solution_9": "ok i understand that but then dont you have to take the sin of pie/2??", "Solution_10": "[quote=\"coquitao\"][quote=\"chocolatelover\"]\n\nsin[Arccos(-square root 2/2)-pie/4]...\n\n[/quote]\n\n\nYour idea of introducing the [color=red]unit circle[/color] is OK. Though, you have to be careful of the way you pick the value for $ \\arccos \\left( \\minus{} \\frac {\\sqrt {2}}{2}\\right)$. It turns out to be $ \\frac {3\\pi}{4}$ in this case. [/quote]\n\nWhy? The n\u00e4ive approach dictates to our minds that there are 2 possibilities for an x to satisfy $ \\cos x \\equal{} \\minus{} \\frac {\\sqrt {2}}{2}.$ Yet, we must always pick the value for x that belongs to $ [0,\\pi]$.\n\n[quote=\"chocolatelover\"]ok i understand that but then dont you have to take the sin of pie/2??[/quote]\r\n\r\nYou are right... Didn't write it, though. In any case, here it is:\r\n\r\n$ \\sin \\frac{\\pi}{2} \\equal{} 1$.\r\n\r\nSee you!" } { "Tag": [ "geometry", "ratio", "geometric transformation", "reflection", "parallelogram" ], "Problem": "1. Does there exist a positive integer $N$ such that the number formed by the last two digits of the sum \r\n$1 + 2 + 3 + \\cdot\\cdot\\cdot + N$ is $98$?\r\n\r\n2. Let $ABCD$ be a rectangle. If $P$ and $Q$ are points respectively on $AD$ and $DC$ such that the areas of the triangles $BAP, PDQ$ and $QBC$ are all equal, find the ratios $DP/PA$ and $DQ/QC$.\r\n\r\n3. Let $P$ be an interior point of an equilateral triangle $ABC$ such that $AP^2 = BP^2 + CP^2$.\r\nProve that $\\angle BPC = 150^o$.\r\n\r\nPS: I didn't solve the last. Any idea for it?", "Solution_1": "[quote=\"Andreas\"]1. Does there exist a positive integer $N$ such that the number formed by the last two digits of the sum \n$1 + 2 + 3 + \\cdot\\cdot\\cdot + N$ is $98$?\n\n2. Let $ABCD$ be a rectangle. If $P$ and $Q$ are points respectively on $AD$ and $DC$ such that the areas of the triangles $BAP, PDQ$ and $QBC$ are all equal, find the ratios $DP/PA$ and $DQ/QC$.\n\n3. Let $P$ be an interior point of an equilateral triangle $ABC$ such that $AP^2 = BP^2 + CP^2$.\nProve that $\\angle BPC = 150^o$.\n\nPS: I didn't solve the last. Any idea for it?[/quote]\r\n\r\n[hide=\"1.\"]1. Consider $n(n+1) \\mod 10$. In order for $n(n+1)$ to end in $6$, and $\\frac{n(n+1)}{2}$ to possibly end in $8$, $n \\equiv 7 \\mod 10$ \nSo let $n$ be of the form $7 + 10a_1 + 100a_2 + ..... 10^{n}a_n$. \nSo $n+1$ is in the form $8 + 10a_1 + 100a_2 + ..... 10^{n}a_n$ \n\nTo see if $\\frac{n(n+1)}{2}$ ends in $98$, it suffices to consider multiplying the last two digits of $n$ and $n+1$, then dividing by $2$. \n\n$\\frac{(7 + 10a_1)(8+10a_1)}{2} = \\frac{56 + 150a_1 +100{a_1}^2}{2} = 28 + 75a_1 + 50{a_1}^2.$\n\nSo for $\\frac{n(n+1)}{2}$ to end in $98$, $28 + 75a_1$ must end in $98$. \n\nBut $28 + 75a_1$ ends in only $28, 03, 53, 78$.\n\nSo $\\frac{n(n+1)}{2}$ never ends in $98$.[/hide]", "Solution_2": "Let $R$ be the reflection of the point $P$ w.r.t. the midpoint $M$ of the side $[BC].$\r\n\r\n$PA^2=PB^2+PC^2,\\ 4AM^2=3a^2,\\ 4AM^2=2(AR^2+PA^2)-4MP^2$ $,\\ 4MP^2=2(PB^2+PC^2)-a^2;$\r\n\r\n$2AR^2=4AM^2-2PA^2+4MP^2=3a^2-2(PB^2+PC^2)$ $+[2(PB^2+PC^2)-a^2]=2a^2\\Longrightarrow AR=a.$\r\n\r\nTherefore $R\\in C(A,a)$, the quadrilateral $BPCR$ is a parallelogram and $A=60^{\\circ}.$\r\n\r\nThus, $m(\\widehat {BPC})=m(\\widehat {BRC})=\\frac 12(360^{\\circ} -A)=150^{\\circ}.$", "Solution_3": "Last one is cute...\r\n\r\nLet Q be the point such that $\\angle QBP = 60$ and QB = PB. Since we then have QBP equilateral and QBA and PBC congruent, $PB^2 + PC^2 = PQ^2 + QA^2$ which equals $AP^2$ only when $\\angle PQA = 90 \\implies \\angle BQA = 150 \\implies \\angle BPC = 150$.", "Solution_4": "For 1, if $ \\frac {n(n \\plus{} 1)}{2} \\equal{} 98 \\plus{} 100k$, then $ (2n \\plus{} 1)^2 \\equal{} 100q \\plus{} 85$, but a square can't be 10 mod 25.\r\n\r\nFor 2, WLOG let AB=1, AP=m, PD=n, DQ=p, then $ m\\equal{}np\\equal{}(m\\plus{}n)(1\\minus{}p)\\equal{}n\\minus{}pm$\r\nThus $ np\\equal{}n\\minus{}p(pn)$, $ p\\equal{}\\frac{\\sqrt{5}\\minus{}1}{2}$, the rest follows easily." } { "Tag": [ "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ p$ be a prime and let $ F$ be a field. Let $ K$ be a Galois extension of $ F$ whose Galois group is a p-group.\r\nSuppoe that $ L$ is a Galois extension of $ K$ whose Galois group is a p-group.\r\nProve that the Galois closure of $ L$ over $ F$ is a Galois extension of $ F$ whose Galois group is a p-group.", "Solution_1": "[hide=\"Hint\"]Every $ \\sigma\\in Gal(L/F)$ induces an automorphism $ \\sigma'\\in Gal(K/F)$ by restriction to $ K$.[/hide]", "Solution_2": "[quote=\"Third Edition\"]Every $ \\sigma\\in Gal(L/F)$ induces an automorphism $ \\sigma'\\in Gal(K/F)$ by restriction to $ K$.[/quote]\r\n$ L/F$ may not be Galois extension.\r\nWhat is $ Gal(L/F)$?", "Solution_3": "It is notation for the automorphisms of $ L$ that fix $ F$. The extension doesn't have to be Galois to talk about that (maybe some places use $ Aut(L/F)$ when the extension is not Galois?)." } { "Tag": [ "limit", "calculus", "symmetry", "function", "calculus computations" ], "Problem": "What is the maximum value of (1/2)^x + (1/2)^(1/x) for x > 0", "Solution_1": "hello, we have $ f(x) \\le 1$ for all $ x \\in \\mathbb{R}$ and $ x>0$. Further we have $ \\lim_{x \\to \\infty}f(x)\\equal{}1$.\r\nSonnhard.", "Solution_2": "Ugh we can't even apply Jensen's to this. I can't think of anything but calculus here.\r\n\r\nIf it does end up going to calculus, I would suggest the substitution $ x \\equal{} 2^y \\implies \\frac{1}{x} \\equal{} 2^{\\minus{}y}$ to gain some degree of symmetry: \\[ f(y) \\equal{} \\left( \\frac{1}{2} \\right) ^{2^y} \\plus{} \\left( \\frac{1}{2} \\right) ^{2^{\\minus{}y}}\\; ,\\; \\; y \\in \\mathbb{R} \\]\r\n[hide=\"Calculus\"]Differentiate, set to 0, and rearrange to get $ \\frac{2^y \\minus{} 2^{\\minus{}y}}{2} \\equal{} y$. We know a lot about this function because it is closely related to hyperbolic sin! In fact, we can easily show that there must be exactly 3 solutions to this: $ y\\equal{}0$ and $ y\\equal{} \\pm a$ for some $ a$.\n\nFrom there, we can use the fact that $ f$ is eventually increasing and approaches $ 1$ as $ y$ increases without bound, to argue that the the graph must have one of two possible shapes:\n1. A relative minimum at $ y\\equal{}0$, non-extremal stationary points at $ \\pm a$, and increases asymptotically to $ 1$ (symmetric for negative $ y$)\n2. A relative maximum at $ y\\equal{}0$, relative minima at $ \\pm a$, and then increases asymptotically to $ 1$ (symmetric for negative $ y$)\n\nCase (1), we have a global minimum at $ y\\equal{}0$, but we know that $ f(0)\\equal{}1 \\equal{} \\lim_{y \\to \\infty} f(y)$, which is invalid in our situation. This means our graph assumes the shape outlined in case (2). Now it is obvious that the maximum must either be the attainable value at $ y\\equal{}0$ or the unattainable limit as $ y$ increases (or decreases) without bound. We find that the maximum is the attainable value $ f(0)\\equal{}1$.[/hide]" } { "Tag": [], "Problem": "The number 3003 is a palindrome because its digits are the same forward or backward. The number 1 is also a palindrome. How many whole numbers between 1 and 3003 are palindromes?", "Solution_1": "[hide] Lets find all the plaindromes including 1 and 3003.\n\nx= 9\n\nxx=9\n\nxxx=90\n\nxxxx (maximum of 2)=20\n\n3003= 1\n\n9 + 9 + 90 + 20 + 1 = 129, 129-2=[b]127[/b] [/hide]", "Solution_2": "[hide=\"Can\"] [hide=\"any\"] [hide=\"of\"] [/hide] [/hide] [/hide] It requires some math.....", "Solution_3": "delete this", "Solution_4": "1 digits:\r\n9\r\n\r\n2 digits:\r\n9\r\n\r\n3 digits:\r\n\r\n9*10 = 90\r\n\r\n4 digits:\r\n\r\n2*10 + 1\r\n\r\nTotal: 129?", "Solution_5": "G-UNIT is correct, because it asks for palindromes [i]between[/i] 1 adn 3003.", "Solution_6": "oh oops :blush:", "Solution_7": "[quote=\"G-UNIT\"]delete this[/quote]\r\nwhat was the point of that?", "Solution_8": "[quote=\"math92\"][quote=\"G-UNIT\"]delete this[/quote]\nwhat was the point of that?[/quote]\r\n\r\nMy guess is that G-UNIT posted something, then edited it. You wouldn't see the \"edited\" line then. Although, then G-UNIT could just delete it then.", "Solution_9": "To delete your own post click on the x next to the quote button." } { "Tag": [ "ratio", "geometry", "symmetry", "geometric transformation", "reflection" ], "Problem": "Consider $ \\triangle ABC$ with vertices $ A(4,8),B( \\minus{} 1,2),$ and $ C(0, \\minus{} 3).$ Find the point $ D$ such that $ \\triangle ABD,\\triangle ACD$ and $ \\triangle BCD$ all have the same area.", "Solution_1": "The centroid is such a point...", "Solution_2": "How would you prove it?", "Solution_3": "Let $ G$ be the centroid; let $ A'\\in AG\\cap BC$. Then $ A'$ is the midpoint of $ BC$. It is well known that $ \\frac{AA'}{GA'}\\equal{}\\frac{1}{3}\\equal{}\\frac{[ABC]}{[BCG]}$ since we can drop altitudes from $ A$; $ G$ to $ BC$ the the ratio of the heights is $ 1: 3$ and the bases are the same so the area ratio is also $ 1: 3$. By symmetry of argument, all these triangles have an area that is a third of the original triangle so they are equal.", "Solution_4": "I didn't know that, I used matrices instead. But I didn't know whether the absolute values of the expressions were positive or negative so I had to try out possibilities.", "Solution_5": "I didn't prove that I found all solutions. In fact, there are three other solutions!\r\n\r\nIn fact, one can show that the triple ratio $ u: v: w\\equal{}[PBC]: [PCA]: [PAB]$ uniquely defines a point $ P$ in the plane of $ ABC$. (The areas are signed, if $ P$ is in the same half plane as $ A$ wrt line $ BC$, $ [PBC]$ is positive. If $ P$ and $ A$ are on opposite sides of $ BC$, then $ [PBC]$ is negative.)\r\n\r\nNoting that $ [PBC]\\plus{}[PCA]\\plus{}[PAB]\\equal{}[ABC]$ (using directed areas), the problem postulates that\r\n\r\n$ |[PBC]|\\equal{}|[PCA]|\\equal{}|[PAB]|$\r\n\r\nWe can normalize to $ u\\plus{}v\\plus{}w\\equal{}1$. So we either have the solutions $ (u,v,w)\\equal{}\\left(\\frac{1}{3},\\frac{1}{3},\\frac{1}{3}\\right),(1,1,\\minus{}1)$ and permutations.\r\n\r\nThe first solution corresponds to the centroid. The second corresponds to the reflection of a vertex in the midpoint of the opposite side." } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Given a non negative integer $ n$\r\nProof: there [b]exists[/b] a [b]unique[/b] polynomial $ p(x)$ which satisfies\r\n(1)$ P(0)\\equal{}1$\r\n(2)$ degP(x)\\equal{}n$\r\n(3)$ (1\\plus{}x)P(x)\\plus{}(1\\minus{}x)P(\\minus{}x)\\equal{}2$", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=248878" } { "Tag": [ "inequalities" ], "Problem": "$x,y,z$ are positive numbers. Given that $x^{2}+y^{2}+z^{2}=1$. Find the minimum value of $S=\\frac{xy}{z}+\\frac{yz}{x}+\\frac{zx}{y}$.\r\n\r\n[hide=\"Spoiler\"]Consider $S^{2}$[/hide]", "Solution_1": "[quote=\"BanishedTraitor\"]$x,y,z$ are positive numbers. Given that $x^{2}+y^{2}+z^{2}=1$. Find the minimum value of $S=\\frac{xy}{z}+\\frac{yz}{x}+\\frac{zx}{y}$.\n\n[hide=\"Spoiler\"]Consider $S^{2}$[/hide][/quote]\r\nWe have $S^{2}\\geq3 ({x}^{2}+{y}^{2}+{z}^{2})=3$ by using AM-GM inequality\r\nSo $minS=\\sqrt{3}$ when $x=y=z=\\frac{1}{\\sqrt{3}}$" } { "Tag": [ "articles", "summer program", "Mathcamp", "ARML", "videos", "Ross Mathematics Program", "calculus" ], "Problem": "This article was published this summer, but only recently came to my attention. What do you all think of it?\r\n[url=http://www.theglobeandmail.com/servlet/ArticleNews/TPStory/LAC/20030816/MATHMAIN16/TPFocus/]Strength in numbers[/url]\r\n\r\nI'm somewhat annoyed. Granted, it's not a major newspaper, but they could have done much better. For instance, they call it an \"arithmetic program\"; they make fun of the campers (\"Those campers who have made it out of bed shovel scrambled eggs down the hatch -- one, seeking optimal efficiency, lowers his mouth to the same horizontal plane as his plate -- and another day at Mathcamp has begun.\"), they ridicule the idea of spending time doing math instead of 'normal' things (they quote someone, probably very loosely: \"there's a lot of freedom and no curfew . . . because if we had to go to bed at a certain time, that might interrupt some important mathematical ideas.\"), etc.", "Solution_1": "And don't forget all the bad puns. \"Strength in numbers,\" indeed. This is a fact of being a math person -- the New York City Math Team makes in into the New York Post from time to time when we win ARML or NYSML, but it's always in this tiny column at the bottom of the page that usually has pedophilia-type stories above it. (I'm actually serious -- in 2002, we were on a page with a story about R. Kelly's sex video with underage girls and a story about an overly-active priest.) Anyhow, this guy seems particularly set on degrading the mathcampers, which is a real shame. But there will always be the point that they \"could be running around playing sports,\" or some other nonsense like that. It's life as a math person to be so regarded.", "Solution_2": "while at the ROSS program, we were interviewed by a newspaper, and in it they referred to \"number theory\" as \"numbers theory\" and got mod incorrectly (i think they said somehting like 25 is 3 mod 24) or something like that", "Solution_3": "Oddly enough, Mathcamp's website, http://www.mathcamp.org, links to the article. I suppose comparatively ignorant press is better than no press at all. Some people might recall an anti-drug ad that had someone say \"I'd rather go to math camp than smoke a joint,\" which rather annoyed us, but became a joke (indeed, I learned about it through such jocular references).\r\n\r\nI am myself a huge fan of Mathcamp, having gone in 2001 and 2002 (and I'm intending to return again this year). It is definitely much more than what any such article could make it to be.\r\n\r\nOn the subject of bad math puns:\r\nDecimals have their point.\r\nCalculus has its limits.\r\nAlgebra is full of unknowns.\r\nTrig lets you see things from many different angles.\r\nAnd many more...", "Solution_4": "[quote]Oddly enough, Mathcamp's website, http://www.mathcamp.org, links to the article.[/quote]\r\n\r\nThat's how I found the article. I'm somewhat surprised that they linked to it (and said it \"gives a good overall feel for life at Mathcamp\"), but I suppose anyone who is serious about math would just laugh at that article and ignore the insults anyway.", "Solution_5": "I think an online friend (it may even have been Alison's mom, although I'm not remembering for sure) first told me about the link through an email message to an email discussion list. I agree that the article conveys an attitude of the reporter that it's weird to like math, but that's what made the story newsworthy: finding a whole camp for unexpected people, namely likers of math. I printed out the whole article to show my son someday, but I haven't actually shown it to him yet (again, if I remember right). I think math-likers do need to meet one another in person to get over the many occasions on which they will be regarded as weird. \r\n\r\nAll the rest that I have heard about MathCamp, and I definitely have heard reports that Alison's mom has passed on from other people, has been uniformly favorable. It's on our list here as a program to apply to when my son is the right age.", "Solution_6": "\"Mathcamp is the negative reciprocal of all that.\"\r\n\r\n???\r\n\r\n\"Mathcamp was not created to provide a summer wonderland in the woods, but to be an incubator where gifted students aged 15 to 18 could pursue their beloved science beyond the bounds of outdated and unimaginative curricula, not to mention burnt-out teachers.\"\r\n\r\nOutdated curricula? I agree that the math curricula in school isn't the best, but I don't like the sound of \"retro-math\" at all.\r\n\r\n\"He has four adult daughters in his native England, and on this side of the Channel, two teenaged boys as well as a two-year old son he watches two days a week, each grouping from a subsequent marriage.)\"\r\n\r\nWouldn't \"this side of the English Channel\" be referring to France, since the implication is that it's on the side of the English channel opposite of England?\r\n\r\nBut I like the second half of the article better. I was amused at the fact that there is a cheerleader camp at the same place...", "Solution_7": "[quote=\"mathfanatic\"]\"Mathcamp is the negative reciprocal of all that.\"\n\n\"He has four adult daughters in his native England, and on this side of the Channel, two teenaged boys as well as a two-year old son he watches two days a week, each grouping from a subsequent marriage.)\"[/quote]\r\n\r\nIt looks like you could teach the editor of that newspaper a thing or two about English, as well as about math.", "Solution_8": "Gollly Gee TokenAdult you stay up late I think its 12 olock rightnow.", "Solution_9": "My son has attended Mathcamp for each of the past 3 years, and hopes to go again. I'm a big Mathcamp fan. But I wasn't offended at all by the article. If was written by a non-math person as entertainment for an audience of non-math persons. It was meant to be amusing, not repectful. Taken in that light, I thought the description of events gave a pretty positive picture of why someone who loves math might want to go to Mathcamp.\r\n\r\n(some of the language peculiarities like \"arithmetic camp\" and \"other side of the channel\" may have to do with the author being Canadian. For instance, maybe they use the word arithmetic in a more general sense than we do, to mean mathematics. Canadians, what do you think?)", "Solution_10": "A lot of mathematicians say arithmetic to mean number theory.", "Solution_11": "I didn't really like the way the writer did this article....." } { "Tag": [ "geometry" ], "Problem": "Let AB be circle O's diameter, CD is a chord, AE is perpendicular to CD, AE intersects CD at E, BF is perpendicular to CD, BF intersects CD at F. Prove EC=DF. :ninja:", "Solution_1": "[hide]If $M$ is the midpoint of $EF$ , then in trapez $AEFB$ we have that $OM // AE //BF$. But then $OM \\perp EF$ , so $M$ is also the midpoint of $CD$. Finally\n \n $CE = ME - MC= MF - MD = DF$[/hide]\r\n Babis", "Solution_2": "Anyone else with a different solution. :D ;)", "Solution_3": "I don't understand your last step, it seems like you are assuming it, can you explain why?", "Solution_4": "Anyone with a solution :(" } { "Tag": [], "Problem": "Simplify the following:\r\n\r\n$\\sqrt[4]{284+48\\sqrt{35}}$", "Solution_1": "Anyone?\r\n\r\nbump back to first page", "Solution_2": "[quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\r\n[hide] $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}=\\sqrt{5}+\\sqrt{7}$, which is our answer. [/hide]", "Solution_3": "[quote=\"The QuattoMaster 6000\"][quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\n[hide] $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}=\\sqrt{5}+\\sqrt{7}$, which is our answer. [/hide][/quote]\r\nI don't get your steps. Can you expand your solution a bit longer?", "Solution_4": "[quote=\"ckck\"][quote=\"The QuattoMaster 6000\"][quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\n[hide] $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}=\\sqrt{5}+\\sqrt{7}$, which is our answer. [/hide][/quote]\nI don't get your steps. Can you expand your solution a bit longer?[/quote]\r\n\r\nQuatto means that he found the fourth root of 284 and 84, simplifing it to \r\n$\\sqrt{12+2\\sqrt{35}}$\r\n\r\nwait noo that doesn't make sense....quatto explain urself :huh:", "Solution_5": "[quote=\"7h3.D3m0n.117\"][/quote][quote=\"ckck\"][quote=\"The QuattoMaster 6000\"][quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\n[hide] $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}=\\sqrt{5}+\\sqrt{7}$, which is our answer. [/hide][/quote]\nI don't get your steps. Can you expand your solution a bit longer?[/quote][quote=\"7h3.D3m0n.117\"]\n\nQuatto means that he found the fourth root of 284 and 84, simplifing it to \n$\\sqrt{12+2\\sqrt{35}}$\n\nwait noo that doesn't make sense....quatto explain urself :huh:[/quote]\r\n\r\nUsually with problems like these, you can assume a clean and tidy denesting, i.e. that $\\sqrt{a \\pm b \\sqrt{c}}=m \\pm n \\sqrt c$. Solve by squaring, then setting rational and irrational parts equal. If this doesn't work, you can try solving other systems in a similar manner by assuming it denests into the form $\\sqrt{m}\\pm \\sqrt{n}$, or other combinations of integers and radicals.\r\n\r\nJust remember that the $\\pm$ should be determined by what your nested expression looks like; and keep all radicals and denested radicals as real numbers (i.e. don't accidentally make something negative when it cannot be.)", "Solution_6": "[quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\r\n\r\n[hide=\"solution/2\"]Let $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{a+b\\sqrt{c}}$\n\n$\\sqrt{284+48\\sqrt{35}}=a+b\\sqrt{c}$\n\n$284+48\\sqrt{35}=a^{2}+2ab\\sqrt{c}+b^{2}c$\n\nWe know that $c=35$\n\n$2ab=48\\Rightarrow ab=24$\n\n$a^{2}+35b^{2}=284$\n\nSolving the 2 equasions we get $a=12$ and $b=2$.\n\nSo $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}$.\n\nTo simplify this, do something similar. Im tired :| [/hide]", "Solution_7": "[quote=\"jli\"][quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\n\n[hide=\"solution/2\"]Let $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{a+b\\sqrt{c}}$\n\n$\\sqrt{284+48\\sqrt{35}}=a+b\\sqrt{c}$\n\n$284+48\\sqrt{35}=a^{2}+2ab\\sqrt{c}+b^{2}c$\n\nWe know that $c=35$\n\n$2ab=48\\Rightarrow ab=24$\n\n$a^{2}+35b^{2}=284$\n\nSolving the 2 equasions we get $a=12$ and $b=2$.\n\nSo $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}$.\n\nTo simplify this, do something similar. Im tired :| [/hide][/quote]\r\nSo then $\\sqrt{12+2\\sqrt{35}}$ simplified would be in the form of $a+b\\sqrt{c}$ right?", "Solution_8": "[quote=\"ckck\"][quote=\"jli\"][quote=\"b-flat\"]Simplify the following:\n\n$\\sqrt[4]{284+48\\sqrt{35}}$[/quote]\n\n[hide=\"solution/2\"]Let $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{a+b\\sqrt{c}}$\n\n$\\sqrt{284+48\\sqrt{35}}=a+b\\sqrt{c}$\n\n$284+48\\sqrt{35}=a^{2}+2ab\\sqrt{c}+b^{2}c$\n\nWe know that $c=35$\n\n$2ab=48\\Rightarrow ab=24$\n\n$a^{2}+35b^{2}=284$\n\nSolving the 2 equasions we get $a=12$ and $b=2$.\n\nSo $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}$.\n\nTo simplify this, do something similar. Im tired :| [/hide][/quote]\nSo then $\\sqrt{12+2\\sqrt{35}}$ simplified would be in the form of $a+b\\sqrt{c}$ right?[/quote]\r\n\r\n[b]IF[/b] it leads to a more simple expression, then yes. But if it does not, or if the system is inconsistent, then try other types of expressions, e.g. $\\sqrt{m}+\\sqrt{n}$ or others.", "Solution_9": "Well, actually, the way I did it way guessing and checking, kind of. \r\n[hide] I assumed that the equation could be simplified. Now, say that $\\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{a+b\\sqrt{35}}\\Rightarrow a^{2}=35b^{2}+2ab\\sqrt{35}=284+48\\sqrt{35}.$ This means that $a^{2}+35b^{2}=284$ and $2ab=48\\sqrt{35}\\Rightarrow ab=24\\sqrt{35}$. Guessing and checking from here gives that $a=12, b=2\\Rightarrow \\sqrt[4]{284+48\\sqrt{35}}=\\sqrt{12+2\\sqrt{35}}$. Now do a similar method for our new squareroot. [/hide]", "Solution_10": "\\begin{eqnarray*}\\sqrt{12+2\\sqrt{35}}&=& \\sqrt{5+2\\sqrt{35}+7}\\hfill \\\\ &=& \\sqrt{(\\sqrt 5+\\sqrt 7 )^{2}}\\hfill \\\\ &=& \\sqrt 5+\\sqrt 7 \\hfill \\\\ \\end{eqnarray*}" } { "Tag": [ "ratio", "geometric series" ], "Problem": "Let $s$ be the limiting sum of the geometric series\r\n\r\n\\[ 4 - \\frac{8}{3} + \\frac{16}{9} - \\cdots, \\]\r\nas the number of terms increases without bound. Then $s$ equals:\r\n\r\n(A) a number between 0 and 1\r\n(B) 2.4\r\n(C) 2.5\r\n(D) 3.6\r\n(E) 12", "Solution_1": "Hello, MCrawford!\r\n\r\n[quote]Let $S$ be the limiting sum of the geometric series: $ 4 - \\frac{8}{3} + \\frac{16}{9} - \\cdots, $\nas the number of terms increases without bound. Then $S$ equals:\n\n(A) a number between 0 and 1\n(B) 2.4\n(C) 2.5\n(D) 3.6\n(E) 12[/quote]\r\n\r\nThe sum of a convergent infinite geometric series is: $S \\;= \\;\\frac{a}{1 - r}$\r\nfor first term $a$ and common ratio $r$.\r\n\r\nWe have: $a = 4,\\;r = -\\frac{2}{3}.$\r\n\r\nTherefore: $\\displaystyle{S \\;= \\;\\frac{4}{1 - (-\\frac{2}{3})} \\;= \\;\\frac{12}{5} \\;= \\;2.4}$ . . . answer choice [b](B)[/b]", "Solution_2": "hi\r\nit is a geometic series\r\nwith a=4 & r=-2/3:\r\n on calculating using infinite geometric series (S =a/(1-r))\r\nit is 12/5", "Solution_3": "hi\r\nit is a geometic series\r\nwith a=4 & r=-2/3:\r\n on calculating using infinite geometric series (S =a/(1-r))\r\nit is 12/5" } { "Tag": [], "Problem": "[hide=\"click to tell me your secrets\"]1st - Brendan Fletcher & Bryce Taylor - 72\n3rd - Christine Hong - 62\n4th - David Lucia - 60\n5th - Elizabeth someone (8th grade girl)\n6th / 7th - Kevin Lang & someone else\n8th - Shannon Taylor - 58\n9th - 4? people tied with 51[/hide]\r\n\r\nThere were 18 questions worth 4 points each (1 point for a blank) on the contest.\r\n\r\nOverall, the contest did a fairly good job of separating the state Qualifiers from the non-qualifiers.\r\n\r\nAs a related note, Forsyth Country Day school won Algebra II and I don't remember any other results, but I'll post them once they are on the web site.", "Solution_1": "[quote=\"b-flat\"]\n\nI don't remember any other results, but I'll post them once they are on the web site.[/quote]\r\n\r\nXD\r\n\r\nI don't think any words are necessary...", "Solution_2": "[url=http://www.math.uncc.edu/~hbreiter/problems/Writeup08.htm]Here are the results.[/url]\r\n\r\n[hide=\"just comprehensive\"]\n[b]Individual Results[/b]\n1st - Brendan Fletcher & Bryce Taylor - 72\n3rd - Christine Hong - 62\n4th - David Lucia - 60\n5th - Alexander Mauney(8th grade girl)\n6th - Elizabeth Shen\n7th - Kevin Lang\n8th - Shannon Taylor - 58\n9th - 4? people tied with 51\n\n[b]in-county schools:[/b]\n1st - Providence Day School\n2nd - Myers Park High School\n3rd - Charlotte Latin School\n\n[b]out of county schools:[/b]\n1st - C. E. Jordan High School\n2nd - Piedmont High School\n3rd - North Lincoln High School[/hide]" } { "Tag": [], "Problem": "Find all positive integer solutions to $abc-2=a+b+c$.", "Solution_1": "[hide] WLOG, assume $a \\ge b \\ge c$. We have $a = \\frac{b+c+2}{bc-1}$ and $a \\ge 1$, so $bc-1 \\le b+c+2 \\Rightarrow (b-1)(c-1) \\le 4$. This gives us several cases:\n\n$c = 1$: $ab-2 = a+b+1 \\Rightarrow (a-1)(b-1) = 4$, which gives us solutions $(5, 2, 1); (3, 3, 1)$.\n\n$c = 2$: $b = 2, 3, 4, 5$ which gives us the solution $(2, 2, 2)$.\n\n$c = 3$: $b = 3$, no solutions.\n\nSo our solutions are $(2,2,2); (3,3,1); (5,2,1)$ and all permutations. [/hide]", "Solution_2": "[hide=\"easy\"]If all of $a,b,c \\ge 2$ then $abc\\ge 4c$ Permutation and addition gives $abc \\ge \\frac{4}{3}(a+b+c)$ and so from the equation we obtain $a+b+c \\le 6$. This means that $a=b=c=2$. Let $c=1$ then ...\n:oops_sign: - it's nearly the same with [b]paladin8[/b][/hide]\r\nP.S. I offen observe \"WLOG\" here. Does it mean \"easy\", \"brutal force\" or somewhat else?", "Solution_3": "It means \"Without loss of generality\"", "Solution_4": "O, it was so easy to guess but I failed!" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "prove the following", "Solution_1": "It's the same argument as proving that the product of limits is the limit of the products. If you make both terms close enough to the bound, you can make the product as close as you want to the product of the bounds.\r\nThere's also the step of showing that the product is an upper bound; that's what the positivity conditions are needed for." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Simplify:\r\n\r\n$\\frac{(11+6\\sqrt{2})\\sqrt{11-6\\sqrt{2}}-(11-6\\sqrt{2})(\\sqrt{11+6\\sqrt{2}})}{(\\sqrt{\\sqrt{5}+2}+\\sqrt{\\sqrt{5}-2})-(\\sqrt{\\sqrt{5}+1})}$", "Solution_1": "[hide]\nAll I know is that the top reduces down to $14\\sqrt{2}$. \n[/hide]", "Solution_2": "And I can find nothing better than $7(2+\\sqrt{2})\\sqrt{\\sqrt{5}-1}$", "Solution_3": "I second Farenhajt's answer." } { "Tag": [ "analytic geometry", "trigonometry", "vector", "trig identities", "Law of Cosines", "calculus", "calculus computations" ], "Problem": "Find the shortest distance between two points in polar coordinates. \r\n(hint: use $ ds^2 \\equal{} dr^2 \\plus{} r^2d\\theta^2$)", "Solution_1": "There's no need for differential notation. The distance formula in polar coordinates is the Law of Cosines:\r\n\r\nThe distance from $ (r_1,\\theta_1)$ to $ (r_2,\\theta_2)$ is\r\n\\[ \\sqrt{r_1^2\\plus{}r_2^2\\minus{}2r_1r_2\\cos(\\theta_1\\minus{}\\theta_2)}\\]\r\n\r\n(I think I've assumed that $ r_1$ and $ r_2$ are nonnegative; I'll let someone else worry about what happens if you allow negative $ r.$)", "Solution_2": "Note that if our points are written as vectors $ \\mathbf{v}_1, \\mathbf{v}_2$, then $ |\\mathbf{v}_1 \\minus{} \\mathbf{v}_2| \\equal{} \\sqrt { |\\mathbf{v}_1|^2 \\plus{} | \\mathbf{v}_2|^2 \\minus{} 2 \\mathbf{v}_1 \\cdot \\mathbf{v}_2 }$, which is coordinate-free (and depends only on the inner product we specify) and fairly easy to compute in polar coordinates besides. As I understand it, the form of the differential is only important if we specify a different metric." } { "Tag": [ "reflection" ], "Problem": "It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress. What do you think?", "Solution_1": "Well Christmas is more of a greed holiday. I feel bad that people die close to Christmas and life goes on.", "Solution_2": "[quote=\"kstan013\"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress. What do you think?[/quote]\r\n\r\nI really don't care, but imo people are overprotective of their religion and its public image.", "Solution_3": "Holiday controversy need not always be about religion.\r\nFor instance, in Belgium there is a \"Dynasty Day\" on the fifteenth of november. Anti-monarchists don't like this holiday.\r\nAnd then there is a national Belgian holiday on the 21st of July, while the Flemish celebrate their own on the 11st of July. Lots of Flemings don't get a day off then.\r\n\r\nThere must be many other examples of controversies , like \"Columbus Day\".", "Solution_4": "[quote=\"fredbel6\"]Holiday controversy need not always be about religion.\nFor instance, in Belgium there is a \"Dynasty Day\" on the fifteenth of november. Anti-monarchists don't like this holiday.\nAnd then there is a national Belgian holiday on the 21st of July, while the Flemish celebrate their own on the 11st of July. Lots of Flemings don't get a day off then.\n\nThere must be many other examples of controversies , like \"Columbus Day\".[/quote]\r\n\r\nI don't relaly care...but my Jewish friend celebrates Hannakuh (god my spelling sucks thats the closest i can get lol) and Christmas....got like $\\$ 500$ worth of presents...", "Solution_5": "[quote=\"paladin8\"][quote=\"kstan013\"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress. What do you think?[/quote]\n\nI really don't care, but imo people are overprotective of their religion and its public image.[/quote]I very much agree with that statement.", "Solution_6": "somebody's always going to feel discriminated against no matter what holiday it is. It is impossible for all humankind to reach a consensus, therefore discrimination exists.", "Solution_7": "Why should I feel discriminated against? A holiday is, after all, a holiday, and the more of them the better. We have a long holiday during durga pujas. We have a holiday on the day of each of the smaller pujas. We have christmas holidays and easter holidays and eid holidays (no halloween though). Then there is muharram and guru nanak's birthday. But we don't have jewish holidays, as there aren't many jews here.\r\n\r\nHow about a random allocation of holidays in the beginning of each year? :P", "Solution_8": "Ours is a church affiliated christian school...80% are hindus,even then we have 3 weeks for christmas and a week for easter and only 2 days for diwali and 10 days for Dussehra! :( :( Truly unfair...but...\r\n[b] As long as we GET holidays at SOME point of the year..i'm not complaining! :lol: [/b]\r\nThankfully there are so many religions....each one's festivals SHOULD be honoured...(We get so many holidays!!)", "Solution_9": "Ours was a church affiliated christian school too.\r\nThe christmas songs are sweet, though the rest is rubbish.", "Solution_10": "The carols are amazing and so are the hymns...Don't forget the christmas plum-cakes they hand out each year!!\r\nI totally agree...the rest is rubbish!", "Solution_11": "a holiday is just that, each person celebrates what they want, what's everyone else complaining about? people who get too upset over a holiday ruin what the season is about; isn't it supposed to be a time of peace, goodwill, personal reflection regardless what religion you're from? why get mad over a bunch of holidays about peace and happiness?", "Solution_12": "[quote=\"kstan013\"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress.[/quote]\r\n\r\nSpelling aside, the very way in which you word the question is biased (e.g., you ascribe a victim's mentality to those who may be (legitimately) troubled by how major (essentially Christian) holidays are celebrated).", "Solution_13": "Everybody says \"holiday\" but we all know it's christmas they're talking about.\r\nlike my \"unprejudiced\" school played christmas music the whole day between classes.", "Solution_14": "[quote=\"blahblahblah\"][quote=\"kstan013\"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress.[/quote]\n\nSpelling aside, the very way in which you word the question is biased (e.g., you ascribe a victim's mentality to those who may be (legitimately) troubled by how major (essentially Christian) holidays are celebrated).[/quote]\n\nIs it only me or can no one else spot the 'spelling' mistake either?\n\n[quote=\"Padmavathi\"]The carols are amazing and so are the hymns...Don't forget the christmas plum-cakes they hand out each year!! \nI totally agree...the rest is rubbish!\n[/quote]\r\n\r\nThey didn't hand out christian plumcakes in our school. :( \r\nAnd I do not (rather cannot) distinguish between carols and hymns.", "Solution_15": "[quote=\"bubka\"][quote=\"blahblahblah\"][quote=\"kstan013\"]It seems like through the holidays some feel discriminated by other religions' holidays and are always quick to distress.[/quote]\n\nSpelling aside, the very way in which you word the question is biased (e.g., you ascribe a victim's mentality to those who may be (legitimately) troubled by how major (essentially Christian) holidays are celebrated).[/quote]\n\nIs it only me or can no one else spot the 'spelling' mistake either?[/quote]\r\n\r\nYou're right, I'm wrong. My mistake.", "Solution_16": "I feel that a lot of people (my mom included) are WAY to oversensitive about the whole Christmas thing in the United States. I think that people are stupid to assume that everyone celebrates Christmas, but their stupidity doesn't really hurt anyone. Schools are now so busy being politically correct, that it ends up just annoying everyone. I would personally rather have a Christmas party than have a regular day of school, even though I am Jewish. I have a Jewish friend who goes caroling just because he likes singing and spreading holiday cheer.", "Solution_17": "it's more about many people using others to get what they want as in gifts.", "Solution_18": "yeah about holidays it has gotten a much more diverse meaning but holiday= holy day and has a christian influence\r\n\r\nbut truely I believe that some people are overprotective of their religon they really might have some cause for that but it is not as strong as they believe\r\n\r\nalso most holidays have been overy commercilized making them bigger then what they used to be", "Solution_19": "As someone who celebrates Hanukka and not Christmas, I really don't like it when we do Christmas things in school or in public places.\r\n\r\nNow, it's not really about \"holidays;\" it's about religion. You see, in America at least, although it may not be so in other countries, you can't promote one particular religion in public schools; it comes from the doctrine of \"Separation Between Church and State\" in our laws. So, I feel uncomfortable when we're doing Christian religious things in a public school! I don't believe in those thing and they're against my beliefs, why don't we do Hanukka stuff too? You just really can't mix religion in with school or politics or anything public...\r\n\r\nI mean wouldn't you (assuming you're Christian) feel uncomfortable if your school was promoting Jewish religious viewpoints? :huh: \r\n\r\n500TH POST!!! :D :D :lol: :lol:", "Solution_20": "The only solution is to ban Christmas.\r\n\r\nAs an added bonus, it will diminish the influence of the Pope on our beloved country.", "Solution_21": "Christmas is awesome, the movie theaters are mostly empty so no having to get there really early and we always get Chinese.", "Solution_22": "[quote=\"Dark_Shadow\"]The only solution is to ban Christmas.\n\nAs an added bonus, it will diminish the influence of the Pope on our beloved country.[/quote]\r\n\r\n\r\nChristmas is still a valued tradition for many people. Other people should be able to enjoy it as well. You can just not celebrate Xmas if you don't want, no one's forcing you against your will. Religion is like a decision. If you want to, you can do it. It's just like you can believe in god if you want. You can study the Bible and attend church if you would like. Yeah, you get the point. \r\n\r\n\r\n :lol:", "Solution_23": "I think Christmas has been commercialized enough that even though someone may mention baby Jesus here or there, the Christian doctrine isn't inherent in the celebration of Christmas so it shouldn't discriminate against many. Sure, Hannukah may not get as much press, but in reality it's not that important of a holiday, anyways (please excuse my ignorance if it is, but I'm going from what I know of the holiday and what others have told me). I'd consider Passover an 'important' holiday just for reference.\r\n\r\nDoes anyone else think it's odd that there's no poll option of \"I do not feel discriminated by the holidays\". Very biased, meh.\r\n\r\n\"The only solution is to ban Christmas. \"\r\n\r\nChristmas is not solely religious but is inherently cultural nowadays. Banning it would create much uproar and animosity towards whoever proposed the bill. Christmas is a time of happiness (I mean, you're off from school!) so enjoy it! It's also good for the economy.", "Solution_24": "I was in a christian school, and it was petty of them to make sing hymns and all. But I have seen Hindu schools and heard of Muslims schools, and they are just as petty.\r\n\r\nReligion makes men petty, god makes men petty, hence my war against religion, against god.\r\n\r\nBut holidays of course make people feel happy, so I like them. :)", "Solution_25": "[quote=\"bubka\"]I was in a christian school, and it was petty of them to make sing hymns and all. But I have seen Hindu schools and heard of Muslims schools, and they are just as petty.\n\nReligion makes men petty, god makes men petty, hence my war against religion, against god.\n\nBut holidays of course make people feel happy, so I like them. :)[/quote]\r\n\r\nReligion makes people happy, why don't you like them?", "Solution_26": "Holidays make people happy without making them petty.\r\nReligion may or may not make people happy, but it makes them petty.\r\n\r\nSo I like holidays but denounce religion.", "Solution_27": "Just felt like posting this too. :wink: \r\n[hide][quote]Please accept with no obligation, implied or implicit, my best wishes for an environmentally conscious, socially responsible, non-addictive, gender-neutral celebration of the winter solstice holiday, practiced within the traditions of the religious persuasion of your choice, or secular practices of your choice (with respect for the religious/secular persuasions and/or traditions of others or their choice not to practice such traditions at all).*\n\nFurthermore, I offer my (non-binding) best wishes for the onset of the generally accepted calendar year of 2006, but not without due respect for the calendars of choice of other cultures whose contributions to society have helped make America great. (Which is not to imply that America is any greater than any other country or is the only \u201cAmerica\u201d in the western hemisphere.) These wishes are offered without regard to the race, creed, color, age, physical ability, choice of computer platform or sexual preference of the wishee.\n\nHappy (if happiness is in your belief system)\n\nHolidays (if you recognize them)\n\n*This greeting is subject to clarification or withdrawal. It implies no promises by the wisher to actually implement any of the wishes for himself or others, is void where prohibited by law and is revocable at the sole discretion of the wisher. This wish expires within one year or until the issuance of a subsequent holiday greeting, whichever comes first. [/quote][/hide]", "Solution_28": "As long as we get to stay at home and somewhere else...someone is celebrating...thats fine with me!!", "Solution_29": "[quote=\"WindSlicer\"]Does anyone else think it's odd that there's no poll option of \"I do not feel discriminated by the holidays\". Very biased, meh.[/quote]That would be \"I really don't care\"[quote=\"calcrulz\"]I mean wouldn't you (assuming you're Christian) feel uncomfortable if your school was promoting Jewish religious viewpoints?[/quote]I would be more confused than anything. You have to realize how nondiverse my community is. There are no Jews in my entire school and the nearest synagogue is 20 miles away, with the next closest being God knows where. Probably in Green Bay. Even non-religious people in my community celebrate Christmas and Easter. It's pretty much a given where I live.", "Solution_30": "i am crying because of christmas it is such a bad holiday :(", "Solution_31": "im hindu, and i still kind of celebrate christmas. hey, its a reason to have parties and make cookies and get gifts.\r\n\r\ni really don't care that christmas gets so much attention. its just that I think that people should have more awareness about different cultural holidays.\r\n\r\nrandom, but: i was reading something interesting in the newspaper about how there are shows on christmas that are sometimes called 'kosher comedy' where people who arent really celebrating christmas come for a fun time. they have chinese takeout because its the only thing open,", "Solution_32": "[quote=\"kstan013\"][quote=\"WindSlicer\"]Does anyone else think it's odd that there's no poll option of \"I do not feel discriminated by the holidays\". Very biased, meh.[/quote]That would be \"I really don't care\"[/quote]\r\n\r\nBut I do care, I have opinions on the matter and am not apathetic.", "Solution_33": "I feel that the attempts of schools to become less focused on Christmas isn't making them 'multicultural,' it is making them 'noncultural.' I am not religious at all, but we still celebrate Christmas and Easter as a family (my mother's upbringing), and celebrate Hanukkah and others with our Jewish neighbors. \r\n\r\nI would much rather hear Christmas songs on the radio than whatever is on the rest of the year, and am not against making Christmas tree ornaments in chemistry, instead of a metal plating lab. \r\n\r\nI guess what it comes down to is that I would rather have culture, even if I don't agree and follow it, than plain nothingness.", "Solution_34": "Christmas is so commercialized these days that in some aspects, it seems secular. Even the atheist's association celebrates it for the \"sharing and happiness\" aspect.\r\n\r\nChristmas, in America, has been rooted so much in the base of culture that it is closeminded (and impossible) to avoid it completely. Thus we get the problems mentioned above with the separation of Church and State.\r\n\r\nNot being Christian, I view Christmas as a cultural icon and more of an American tradition than a religious holiday.\r\n\r\nIf you think about it, in all countries with one religious majority, how do religious minorities avoid discrimination (even in countries that observe separation of Church and State)?", "Solution_35": "[quote=\"fredbel6\"][quote]Please accept with no obligation, implied or implicit, my best wishes for an environmentally conscious, socially responsible, non-addictive, gender-neutral celebration of the winter solstice holiday, practiced within the traditions of the religious persuasion of your choice, or secular practices of your choice (with respect for the religious/secular persuasions and/or traditions of others or their choice not to practice such traditions at all).*\n\nFurthermore, I offer my (non-binding) best wishes for the onset of the generally accepted calendar year of 2006, but not without due respect for the calendars of choice of other cultures whose contributions to society have helped make America great. (Which is not to imply that America is any greater than any other country or is the only \u201cAmerica\u201d in the western hemisphere.) These wishes are offered without regard to the race, creed, color, age, physical ability, choice of computer platform or sexual preference of the wishee.\n\nHappy (if happiness is in your belief system)\n\nHolidays (if you recognize them)\n\n*This greeting is subject to clarification or withdrawal. It implies no promises by the wisher to actually implement any of the wishes for himself or others, is void where prohibited by law and is revocable at the sole discretion of the wisher. This wish expires within one year or until the issuance of a subsequent holiday greeting, whichever comes first. [/quote][/quote]\r\n\r\nRotfl, saw that on GCFL.net\r\n\r\nI believe that people have the right to not celebrate Christmas (sorry for the split infinitive :D) but shouldn't keep others from doing so if they want to.\r\n\r\nWow, that grammar was pretty bad.", "Solution_36": "You could avoid that dubious construction saying 'right to forego Christmas celebration'. :) \r\n\r\nChristmas is in Europe as Durga Puja is in India. It is primarily a Hindu festival, but celebrated over five days by people of all religions, including atheists like me. I do not object to celebration or holidaying, I only object to attaching any supernatural or divine meaning to it.", "Solution_37": "What I really hate is that when people know you're a Christian they automatically think you celebrate Christmas. Christmas isn't even a Christian holiday. It is a pagan holiday.", "Solution_38": "[quote=Irshad]What I really hate is that when people know you're a Christian they automatically think you celebrate Christmas. Christmas isn't even a Christian holiday. It is a pagan holiday.[/quote]\n\nYeah as if the pagans believed in Jesus. Just because some of the traditions have pagan roots doesn't mean it's a pagan holiday. Also I don't think the holiday celebrated by pagans on December 25th was called Christmas or had anything to do with Christ.", "Solution_39": "Yes, Christians renamed it and added that \"Jesus was born on this day\". When you insert pagan things in worshiping God it's called sin. Take the case of the golden calf, for example, during the times of Moses. In fact, Aaron who was organizing it said that it would be a holiday to the Lord. People indeed worshiped God, but they introduced pagan ideologies, in this case the golden statue of a calf, pagan music and dancing, etc. That was a sin. God wanted to destroy the entire nation right there and then, but Moses \"persuaded\" God to not do it. Easter, Christmas, icons in church, all these things are the same as what happened with the golden calf. We should not insert anything pagan in worshiping God.", "Solution_40": "[quote=Dark_Shadow]The only solution is to ban Christmas.\n\nAs an added bonus, it will diminish the influence of the Pope on our beloved country.[/quote]\n\nHmmm. You seem to dislike catholicism. [quote=Irshad]What I really hate is that when people know you're a Christian they automatically think you celebrate Christmas. Christmas isn't even a Christian holiday. It is a pagan holiday.[/quote]\n\nIt is meant to be a christian holiday. people turn it into \"Santa's Holiday\" though sadly.", "Solution_41": "[quote=bubka]Holidays make people happy without making them petty.\nReligion may or may not make people happy, but it makes them petty.\n\nSo I like holidays but denounce religion.[/quote]\n\nIt's Satan who makes men \"petty\". It's this funny thing nowadays, people talk about being \"happy\". A holiday cannot make you happy. Without a God you are empty. With God there is true happiness. I, as a person who has experienced God in a real way, and seen him work in people around me, can say that what I have just said is true. ", "Solution_42": "Being an agnostic, I would prefer having the whole week off for someone's religious holiday then just one day. If I had a week whole off, I could go on vacation somewhere or drive to my country house. If I just had one day off, there really would not be that much to do.", "Solution_43": "If we were to have one holiday each year on the winter solstice exactly, and call it Jeffrey, would that solve everyone's problems? ", "Solution_44": "@above that's my name...\nSo then people would be jealous of me :P" } { "Tag": [], "Problem": "What is the least positive multiple of 72 that has exactly 16 positive factors?", "Solution_1": "Haven't checked, but 216?", "Solution_2": "Yes, it's $ \\boxed{216}$.\r\n\r\nApart from guess-and-check, I'm not really sure how to solve it.", "Solution_3": "YAY! GO GUESSTIMATION! :D" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "I want to prove by the $ \\epsilon-\\delta$ definition that $ \\underset{(x,y)\\to (1,2)}{\\mathop{\\lim }}\\,\\frac{x^{2}}{x+2y}=\\frac{1}{5}.$\r\n\r\nThen, I started by making the following:\r\n\r\n\\begin{eqnarray*}\r\n \\left| \\frac{x^{2}}{x+2y}-\\frac{1}{5} \\right|&=&\\left| \\frac{5x^{2}-x-2y}{5(x+2y)} \\right| \\\\ \r\n &=&\\left| \\frac{5x^{2}-5x+4x-2y+4-4}{5(x+2y)} \\right| \\\\ \r\n & =&\\left| \\frac{5x(x-1)+4x-2(y-2)-4}{5(x+2y)} \\right| \\\\ \r\n & \\le& \\frac{5\\left| x \\right|\\left| x-1 \\right|+4\\left| x \\right|+2\\left| y-2 \\right|+4}{5(x+2y)},\r\n\\end{eqnarray*}\r\n\r\nnow I can suppose that $ \\delta=1$ to bound the other expressions.\r\n\r\nAnother way of doing it is that given $ |x-1|<\\delta$ and $ |y-2|<\\delta$ and with $ \\delta=1$ we get $ 02$ and $ \\left| 5x^{2}-x-4 \\right|<14\\delta,$ hence by taking $ \\delta =\\min \\left\\{ 1,\\frac{5\\epsilon }{8} \\right\\}$ we get\r\n\r\n$ \\left| \\frac{x^{2}}{x+2y}-\\frac{1}{5} \\right|=\\left| \\frac{5x^{2}-x-4-2y+4}{5(x+2y)} \\right|<\\frac{14\\delta +2\\delta }{10}=\\frac{8}{5}\\delta =\\epsilon .$\r\n\r\nWhich one of the proceduresis the best? I like the second one.", "Solution_1": "The second one is the method I'm more used to seeing and using when proving a limit via $ \\varepsilon$-$ \\delta$.", "Solution_2": "Excellent, I just wanted some clarification.\r\n\r\nThanks for your answer." } { "Tag": [ "vector", "quadratics", "linear algebra" ], "Problem": "Hi all! I am struggling with the following problem:\r\n\r\nLet E be a finite dimensional Euclidean space and let [b]T: E -> E[/b] be a symmetric operator, which additionally satisfies the relation [b](T(x), T(x)) = (x,x)[/b] for every vector [b]x [/b]in [b]E[/b].\r\nProve that there exist subspaces [b]M =< E[/b] and [b]N =< E[/b] such that [b]E=M+N[/b] (direct sum), and [b]T(a) = a[/b] for each [b]a [/b]in [b]M [/b]and [b]T(b)= -b[/b] for each [b]b [/b]in [b]N[/b]. Check that [b]N [/b]is orthogonal complement of [b]M[/b].\r\n\r\nAnyone got any idea???", "Solution_1": "The theorems:\r\n\r\nA symmetric/Hermitian operator is orthogonally/unitarily diagonalizable with real eigenvalues.\r\nAn orthogonal/unitary operator has $|\\lambda|=1$ for all eigenvalues $\\lambda$.\r\nYour $T$ satisfies both conditions. Diagonalize, and see what happens.", "Solution_2": "an orthogonal/unitary operator is defined as (T(x), T(y)) = (x,y). But does (T(x), T(x)) = (x,x) still mean that T is orthogonal/unitary????", "Solution_3": "Yes. You can derive the inner product from the quadratic form." } { "Tag": [ "geometry" ], "Problem": "dayereye C va W yekdigar ro to P,Q ghat mikonan\r\ndayereii be markaze P ,C ro dar S,R va W ro dar M,N ghat mikone\r\n\r\nagar mahale barkhorde MS , RN ro K benamin :\r\nsabet konid PQ bar KQ amoode [/list]", "Solution_1": "nohteye $MN\\cap RS=Q'$roye ghotbiye $K$ gharar darad .agar az $Q'$bar $PK$ amood konim ta $PK$ ra dar $K'$ ghat konad \r\n\r\n$K'$ varoone $K$ khahd bood.\r\n\r\naz tarafi $\\angle PCA=\\angle PCQ'=\\angle PQC$ dar natige$PQ'\\times PQ=PK\\times PK'=r^2$($r$ shoae dayere be \r\nmarkaze $P$)\r\n\r\npas chehar zeleiye $KK'QQ'$ mohati ast dar natige $\\angle PQK=90^{\\circ}$" } { "Tag": [ "Putnam", "function", "integration", "logarithms", "algebra", "functional equation", "absolute value" ], "Problem": "Find all differentiable functions $f: (0,\\infty)\\mapsto (0,\\infty)$ for which there is a positive real number $a$ such that\r\n\\[ f'\\left(\\frac ax\\right)=\\frac x{f(x)} \\]\r\nfor all $x>0.$", "Solution_1": "Let\r\n$g(x)=f\\left(\\frac ax\\right)f(x)$\r\nso\r\n$g'(x)=f\\left(\\frac ax\\right)f'(x)-\\frac{a}{x^2}f'\\left(\\frac ax\\right)f(x)=f\\left(\\frac ax\\right)f'(x)-\\frac{a}{x}$\r\n$g'(\\frac{a}{x})=f\\left(x)f'(\\frac ax\\right)-x=0$\r\n\r\nso $g(x)$ is constant.\r\n\r\nSuppose\r\n$f\\left(\\frac ax\\right)f(x)=c$\r\n$f\\left(\\frac ax\\right)=\\frac{c}{f(x)}$\r\n$f'\\left(\\frac ax\\right)=\\frac{-f'(x)c}{f(x)^2}$\r\n\r\nSo the initial condition becomes\r\n\r\n$\\frac{f'(x)}{f(x)}=\\frac{a}{c}$\r\nso\r\n$f(x)=Bx^r$, $r=\\frac{a}{c}$, $B>0$\r\n\r\nand these give all the solutions, with the only exception that $B=1$ when $r=1$ (you get this by checking if this actually is a solution).", "Solution_2": "I got precisely this solution: $f(x) = Bx^{a/C}$, where $B > 0$ and $C$ is some constant. Then, in the end I bungled things by showing that $f(x) = x$ is the only solution that satisfies the conditions of the problem for all $a > 0$. I missed the part that I only need to find solutions that satisfy the given conditions for some $a > 0$ :(\r\n\r\nDoes anyone think I will get credit for my solution? Help!!!", "Solution_3": "Here's my solution.\r\n\r\nWe have $f''(t)=\\frac{d}{dt}\\frac{a/t}{f(a/t)}=\\frac{d}{dt}\\frac{a}{tf(a/t)}= \\frac{-a}{t^2f(a/t)}-\\frac{af'(a/t)}{t(f(a/t))^2}\\cdot\\frac{-a}{t^2}$\r\n$f''(t)=-\\frac1tf'(t)+\\frac{a^2}{t^4f(a/t)}\\cdot\\frac{a}{\\frac at f(t)}= -\\frac1tf'(t)+\\frac{(f'(t))^2}{tf(t)}.$\r\n\r\nThis is a second order differential equation with no singular points in the specified range, so by the uniqueness half of the existence-uniqueness theorem, $f$ is completely determined by $f$ and $f'$ at a single point. Since $f'(\\sqrt{a})=\\frac{a}{\\sqrt{a}}$, $f$ is completely determined by $f(\\sqrt{a})$. A little experimentation (how about powers of $x$?) gives us that for positive $c$, $f_c(x)=\\frac{x^c}{\\sqrt{ca^{c-1}}}$ works, with $f_c(\\sqrt{a})=\\sqrt{c}$. The previous argument shows that these are all solutions, and we are done.\r\n\r\nThis didn't take very much insight.", "Solution_4": "The only problem I see with that is that the statement of the problem only said that $f'$ exists; it didn't say anything about $f''.$", "Solution_5": "That was a proof that $f''$ exists; since $f'$ depends algebraically on something differentiable, $f'$ is differentiable.", "Solution_6": "Here is what I did in the exam and it worked ofcourse. \r\n( infact I was really happy to see a functional equation :D )\r\n\r\nDefine g(x) = e^f(ln x)\r\n\r\nthen \r\n\r\ng'(x) = a /e^[g(x) + g(lna - x)]\r\n\r\nthe right hand side is symmetric with respect to x and lna -x. so if b=lna\r\nthen g'(x) = g'(b-x) so g(x) + g(b-x) = const so g'(x) = const and therefore f(x) = a x^n\r\n\r\n-Ali", "Solution_7": "Here is my solution. It is somewhat similar to dystopia's though the technique is slightly different. Anyways, here's what I did.\r\n\r\n$f'(\\frac ax) = \\frac x{f(x)}$ $... (1)$\r\n\r\nReplace $x$ by $\\frac ax$ in both sides of the equation above. We obtain \r\n$f'(x) = \\frac a{xf(\\frac ax)}$ $... (2)$\r\n\r\nDividing $(1)$ by $(2)$ and rearranging terms, we get\r\n$\\frac{af'(\\frac ax)}{x^2f(\\frac ax)} = \\frac{f'(x)}{f(x)}$\r\n\r\nIntegrating both sides with respect to $x$, we get\r\n\r\n$- \\int \\frac{d(f(\\frac ax))}{f(\\frac ax)} = \\int \\frac{d(f(x))}{f(x)}$\r\n\r\nOr, $-ln|f(\\frac ax)| = ln|(f(x))| + lnC_1$, where $C_1 > 0$\r\n\r\nSince $f(x) > 0$, we can drop the absolute value sign, $| . |$. We then get\r\n$f(\\frac ax) = \\frac{C_2}{f(x)}$, where $C_2 > 0$\r\n\r\nDifferentiating both sides above with respect to $x$ and combining again with $(1)$ while noting that $x > 0$, we get\r\n$\\frac{f'(x)}{f(x)} = \\frac{aC_3}x$, where $C_3 > 0$\r\n\r\nIntegrating further both sides above with respect to $x$, we finally obtain\r\n$f(x) = Bx^{\\frac aC}$ as the required solution. Here, $B > 0$ and $C$ is some real constant.\r\n\r\nI skipped a few steps above though the original solution I wrote was quite detailed. Well, in the end I messed up by showing that $f(x) = x$ is the required solution for all $a > 0$, forgetting the fact that equation $(1)$ holds true for some $a > 0$.\r\n\r\nCould anyone tell me how many points I might get for this?", "Solution_8": "i got a differnt answer and was wondering if it is right or not.\r\n\r\n[hide=\"here is my solution\"]\n$f'(x)=\\frac{a}{xf\\left(\\frac{a}{x}\\right)}$\n\n$g(x)=f(x)f\\left(\\frac{a}{x}\\right)$\n\n$g'(x)=f'(x)f\\left(\\frac{a}{x}\\right)+f(x)\\left(\\frac{-a}{x^{2}}\\right)f'\\left(\\frac{a}{x}\\right)$\n\nso $g'(x)=0$ and $g(x)=k$ for some constant.\n\n$f(x)f(a/x)=k$\n\n$f(a/x)=\\frac{k}{f(x)}$\n\n$f'(a/x)=\\frac{x}{f(x)}=\\frac{-kf'(x)}{f(x)^{2}}$\n\n$xf(x)=-kf'(x)\\longrightarrow\\frac{-f'(x)}{f(x)}=\\frac{x}{k}$\n\nintegrating both sides yields\n\n$-\\ln(f(x))=\\ln\\left(\\frac{1}{f(x)}\\right)=\\frac{x^{2}}{2k}+C$ so we get\n\n$f(x)=\\frac{A}{e^{Bx^{2}}}$ where $B=\\frac{1}{2k}$\n\ni am wondering if i made a mistake or if this solution is correct.\n[/hide]", "Solution_9": "Well, your answer does not survive being substituted into the original equation. It also fails to satisfy your functional equation\r\n\r\n$f(x)f\\left(\\frac{a}{x}\\right)=k.$\r\n\r\nSo where does it go wrong? I'd suggest looking at this step:\r\n\r\n[quote]$f(a/x)=\\frac{k}{f(x)}$\n\n$f'(a/x)=\\frac{x}{f(x)}=\\frac{-kf'(x)}{f(x)^{2}}$[/quote]\r\nThe problem with that is that $f'(a/x)$ is not the same thing as $\\frac{d}{dx}f\\left(\\frac{a}{x}\\right).$", "Solution_10": "ooh i see now thanks.", "Solution_11": "Can someone tell me how FieryHydra exactly got from step 1 to step 2.\n[quote=\"FieryHydra\"]\nDividing $(1)$ by $(2)$ and rearranging terms, we get\n\n[u]Step 1:[/u] $\\frac{af'(\\frac ax)}{x^2f(\\frac ax)} = \\frac{f'(x)}{f(x)}$\n\n\n[u]Step 2:[/u] Integrating both sides with respect to $x$, we get\n\n\n$- \\int \\frac{d(f(\\frac ax))}{f(\\frac ax)} = \\int \\frac{d(f(x))}{f(x)}$\n[/quote]\n\nPeace.\nFaustus\n :yinyang:", "Solution_12": "Obviously, $\\dfrac{d}{dx}\\left[f(x)\\right] = f'(x)$. So $\\dfrac{f'(x)}{f(x)}\\,dx = \\dfrac{d(f(x))}{f(x)}$\n\nUsing the Chain Rule: $\\dfrac{d}{dx}\\left[f\\left(\\dfrac{a}{x}\\right)\\right]$ $= \\dfrac{d}{dx}\\left[\\dfrac{a}{x}\\right]f'\\left(\\dfrac{x}{a}\\right)$ $= -\\dfrac{a}{x^2}f'\\left(\\dfrac{x}{a}\\right)$. \n\nThus, $\\frac{af'(\\frac{a}{x})}{x^{2}f(\\frac{a}{x})}\\,dx = -\\dfrac{d(f(\\frac{a}{x}))}{f(\\frac{a}{x})}$. \n\nTherefore, $ \\frac{af'(\\frac{a}{x})}{x^{2}f(\\frac{a}{x})}=\\frac{f'(x)}{f(x)} $ is equivalent to $ -\\frac{d(f(\\frac{a}{x}))}{f(\\frac{a}{x})}= \\frac{d(f(x))}{f(x)} $.", "Solution_13": "We did this problem in Wednesday Putnam Seminar today, and I think have a solution which is different from all the other ones posted here. (EDIT: After reading the other solutions, I've realized that mine is essentially equivalent to many of the ones written here, just written in a different form.)\n\n[hide=\"Solution\"]\nRewrite the original FE as $f(x)f'\\left(\\frac ax\\right)=x$; under the substitution $x\\mapsto \\frac ax$ we get $f\\left(\\frac ax\\right)f'(x)=\\frac ax$. Now integrate both sides with respect to $x$. The RHS is just $a\\ln x + C_1$ for some $C_1\\in\\mathbb{R}$. To compute the LHS, we use integration by parts. Specifically, let $u=f\\left(\\frac ax\\right)$ and $dv=f'(x)\\, dx$. Then $du=-\\frac{a}{x^2}f'\\left(\\frac ax\\right)\\, dx$ and $v=f(x)$. As a result, \\begin{align*}\\int f\\left(\\frac{a}{x}\\right)f'(x)\\, dx&=f\\left(\\frac ax\\right)f(x)-\\int f(x)f'\\left(\\frac ax\\right)\\left(-\\frac{a}{x^2}\\right)\\,dx\\\\&=f\\left(\\frac ax\\right)f(x)+\\int x\\cdot\\frac{a}{x^2}\\,dx\\\\&=f\\left(\\frac ax\\right)f(x)+a\\ln x+C_2.\\end{align*} Thus $f(\\tfrac ax)f(x)=b$ for some constant $b$. (The reason we use $b$ instead of $C$ will become apparent later.)\n\nWe're almost done with the problem now. Remark that solving for $f\\left(\\frac ax\\right)$ and plugging into $f\\left(\\frac ax\\right)f'(x)=\\frac ax$ gives $\\frac{b f'(x)}{f(x)}=\\frac ax$. Integrating and utilizing the fact that $f>0$ for all $x$ gives \\[\\ln f(x)=\\frac ab\\ln x + C\\implies f(x)=Cx^{a/b}.\\] Now it just suffices to plug into the original FE for additional constraints. Specifically, we see that \\[f(x)f'\\left(\\frac ax\\right) = Cx^{a/b}\\cdot \\left(\\frac ax\\right)^{a/b-1}=Cxa^{a/b-1}.\\] Setting this equal to $x$ yields $Ca^{a/b-1}=1\\implies C=a^{1-a/b}$. As a result, $f(x)=\\boxed{Cx^{a/b},\\,C=a^{1-a/b}}$ works.\n\nThis looks slightly different than jmerry's parametrization, but I think by and large they're the same.[/hide]", "Solution_14": "Let $g(x) = f(a/x)$ then $[g(x)f(x)]' = g'(x) f(x) + f'(x) g(x) = -a/x^2 f'(a/x)f(x) + f'(x) f(a/x) = 0$ so $g(x) f(x)$ is constant. The rest of the solution is the same.", "Solution_15": "Observe:\n$x\\to a/x \\implies f'(x)=\\frac{a}{xf(a/x)} \\implies f''=-\\frac{a}{x^2f(a/x)}+\\frac{a^2f'(a/x)}{x^3f(a/x)^2}.$\n$f(a/x)\\to a/(xf'(x)) \\implies f''(x)=\\frac{(f'(x))^2}{f(x)}-\\frac{f'(x)}{x}.$\nLet $g(x)=\\log f(x).$ It follows $g'(x)=f'(x)/f(x),$ obviously $g'(x)$ is positive.\n$g''(x)=\\frac{f(x)f''(x)-f'(x)^2}{(f(x))^2}=\\frac{-f(x)f'(x)/x}{(f(x))^2}=\\frac{-f'(x)}{xf(x)}=\\frac{-g'(x)}{x}.$\nLet $\\phi(x)=\\log g'(x),$ so $\\phi'(x)=g''(x)/g'(x)=-1/x.$ \nIntegrating yields $\\phi(x)=c-\\log x,$ where $c$ is any real. \nExponentiation yields $g'(x)=k/x,$ where $k=\\exp c.$\nIntegrate again $g(x)=d+k\\log x.$\nExponentiate again $\\boxed{f(x)=yx^k},$ where $y=\\exp k.$ \nEasy to see this works.\nIf $k=1,$ easy to get $\\boxed{f(x)=x},$ which works too.\n" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ p_n$ a sequence of positive integers such that :\r\n\r\nif $ n \\neq m$ then $ gcd(p_n,p_m)\\equal{}1$ and $ p_n$ is composite for all $ n$.\r\n\r\nDo we have $ \\sum_{n \\equal{}1}^{\\infty} \\frac{1}{p_n} < \\infty$ ?\r\n\r\nIt is maybe false, but i did not succeed in finding a counterexample, could you help me?", "Solution_1": "Disclaimer: It is now 4:30 in the morning. No guarantees :maybe: \r\n\r\nLet $ q_n$ be the smallest prime dividing $ p_n$. Since $ p_n$ is composite we have $ p_n \\geq {(q_n)}^2$. Since $ \\gcd(p_n,p_m)\\equal{}1$ we have $ q_n \\neq q_m$ when $ m \\neq n$. This means that the sequence $ \\frac{1}{{(q_n)}^2}$ is a subsequence of $ \\frac{1}{m^2}$ in some order. So\r\n\r\n$ \\sum_{n\\equal{}0}^\\infty \\frac{1}{p_n} \\leq \\sum_{n\\equal{}0}^\\infty \\frac{1}{{(q_n)}^2} \\leq \\sum_{m\\equal{}0}^\\infty \\frac{1}{m^2} < \\infty$.", "Solution_2": "thank you :)" } { "Tag": [ "geometry", "incenter", "blogs", "analytic geometry", "geometry proposed" ], "Problem": "3. The incircle of triangle $ABC$ touches $AB$ at $D$. $DE^{'}$ is a diameter of the incircle. Tangent to the incircle in point $E^{'}$ intersects $AC$ at $A^{'}$ and $BC$ at $B^{'}$. If $M$ is the midpoint of $AB$, Prove that:\r\n\r\n$a)$ $A^{'}B,AB^{'}$ and $CM$ reach at one point.\r\n\r\n$b)$ $M,$ the incenter of $ABC$ and the midpoint of $CD$ are on a straight line. \r\n\r\nhere's it's link in my blog:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/weblog_entry.php?p=749422#749422", "Solution_1": "For a), $AB$//$A'B'$ -> obvious!\r\nb), barycentric coordinate calculation!" } { "Tag": [ "geometry", "analytic geometry", "geometry open" ], "Problem": "hi! view the problem in gif!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_1": "[quote=\"Alisher\"]hi! view the problem in gif!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]sorry", "Solution_2": "try coordinate geometry" } { "Tag": [ "logarithms", "trigonometry" ], "Problem": "$ find$ $ 2^{i}$", "Solution_1": "hmm not sure if this is right but here's an attempt\r\n\r\n[hide]\n[b]Case I:[/b] Let $ 2^{i}=re^{2\\pi k}$, where $ k\\in \\mathbb{Z}$. Then we can find that $ r=\\frac{1}{2}$ so $ \\boxed{2^{i}=\\frac{1}{2}e^{2\\pi k}}$.\n\n[b]Case II[/b]: Let $ 2^{i}=re^{\\pi+2\\pi k}$, where $ k\\in \\mathbb{Z}$. Then we can find that $ r=-\\frac{1}{2}$ so $ \\boxed{2^{i}=-\\frac{1}{2}e^{\\pi+2\\pi k}}$. \n\n\n[/hide]", "Solution_2": "I'm not sure how you're getting your results.\r\n\r\n[hide=\"Solution\"] $ 2^{i}= e^{i \\ln 2}= \\cos \\ln 2+i \\sin \\ln 2$ [/hide]" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "[hide][img]http://img410.imageshack.us/img410/9100/orhanng1.jpg[/img][/hide]", "Solution_1": "Please , \r\n say if $AC \\perp EB$\r\n\r\nor not. \r\n\r\n Babis", "Solution_2": "no it is not right." } { "Tag": [ "geometry", "geometric transformation", "rotation", "function", "analytic geometry", "circumcircle", "trigonometry" ], "Problem": "A triangle is inscribed in a circle. The length of the sides of the triangle are 7, 8, 9 inches. Find the radius of the triangle.\r\n\r\nIs there a special formula?\r\n\r\n\r\nLet x^2 - sqrt(3)*x*y - 1 = 0. What is the angle of rotation from it's parent function?\r\n\r\nHow do you figure out angle of rotations?\r\n\r\n\r\nThe graph of x^2 + 6x +y^2 - 6y +15 = 0 lies in which quadrants?\r\n\r\nHow would I do this without a graphing calculator?\r\n\r\n\r\nThe point (-3,3) is rotated 30 deg. clockwise about the origin. The coordinates of the point after the rotation is?\r\n\r\n\r\nThe altitude of PQR forms 2 right triangles,PQX and RQX.\r\nPQ=11, RX = 10, angle RQX = 48 deg.\r\nWhat is PX?", "Solution_1": "[quote]the radius of the triangle. [/quote] I think you means the radius of $ R$:circumradius of $ \\triangle ABC$. You can use Heron's formula to find the area of $ \\triangle ABC$ and Using $ Area\\equal{}\\frac{abc}{4R}$ you'll get R. Or you can use law of cosines to find 1 angle, and use law of sines to find R.", "Solution_2": "[hide=\"3rd question\"]\nFor this question, you simply \"complete the square\" and change into the standard form of a circle as follows:\n$ x^2 \\plus{} 6x \\plus{} y^2 \\minus{} 6y \\plus{} 15 \\equal{} 0 \\implies (x^2 \\plus{} 6x \\plus{} 9) \\plus{} (y^2 \\minus{} 6y \\plus{} 9) \\minus{} 3 \\equal{} 0 \\implies (x \\plus{} 3)^2 \\plus{} (y \\minus{} 3)^2 \\equal{} 3 \\implies (x \\plus{} 3)^2 \\plus{} (y \\minus{} 3)^2 \\equal{} \\left(\\sqrt {3}\\right)^2$\nThis equation's graph is the circle with center $ ( \\minus{} 3,3)$ and radius $ \\sqrt3$, which lies only in the $ \\boxed{2^{nd} \\text{ quadrant}}$.\nYou should look at example 3 on [url=http://www.analyzemath.com/CircleEq/Tutorials.html]this website[/url] if you did not understand my method\n[/hide]\n[hide=\"4th question\"]\nThe easiest thing to do is to change this point into [url=http://en.wikipedia.org/wiki/Polar_coordinates]polar form[/url]. Since the point $ (\\minus{}3,3)$ is $ 3\\sqrt2$ units away from the origin, and this point has an angular value of $ 135^{\\circ}$, the point $ (\\minus{}3,3)$ is also equivalent to the point $ (3\\sqrt2,135^{\\circ})$ in oplar form.\nIn order to rotate this point 30 degrees clockwise about the origin, you simply have to decrease the angle by 30 degrees, resulting in the point $ (3\\sqrt2,105^{\\circ})$.\nFinally, we reverse the process to turn this point in polar form to its equivalent point in rectangular ((x,y) form). Since this point has a angular value of $ 105^{\\circ}$, it is in the form $ (r\\cos(105^{\\circ}),r\\sin(105^{\\circ}))$, where r is the radius. The final answer is $ \\boxed{(3\\sqrt2 \\cos(105^{\\circ}),3\\sqrt2 \\sin(105^{\\circ}))}$\n[/hide]" } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "Find the limit :\r\n$ lim (\\sqrt{2\\minus{}\\sqrt{2}}.\\sqrt{2\\minus{}\\sqrt{2\\plus{}\\sqrt{2}}}...\\sqrt{2\\minus{}\\sqrt{2\\plus{}\\sqrt{2\\plus{}\\sqrt{2\\plus{}....\\plus{}\\sqrt{2}}}}})$", "Solution_1": "Call that sequence is $ a_n$ (has n product)\r\n$ b_{n} = \\sqrt {2 + \\sqrt {{2}... + {\\sqrt{2}}}}$\r\nThen \r\n$ a_n.b_n=\\sqrt {2}$\r\nbut $ \\lim_{n\\to\\infty} b_n = 2$\r\nSo $ \\lim_{n\\to\\infty} a_n = \\frac {\\sqrt {2}}{2}$" } { "Tag": [ "inequalities", "number theory", "inequalities unsolved" ], "Problem": "If $a,b,c$ are positive reals number. Find all positive real numbers $x,y,z$ such that:\r\n\r\n$x+y+z=a+b+c$\r\n\r\n$4xyz-(a^{2}x+b^{2}y+c^{2}z)=abc$\r\n :wink:", "Solution_1": "I bet it is not a number theory problem. Don't know what mods do on this forum...\r\n\r\nI can analyze it for case $x,y,z\\leq \\frac{a+b+c}{2}$...", "Solution_2": "Yeah, looks suspiciousally cauchy-swartzy...", "Solution_3": "[quote=\"Myth\"]I bet it is not a number theory problem. Don't know what mods do on this forum...[/quote]\r\n :( \r\nI can't even remember having seen this.\r\nPS: you are mod here ;)", "Solution_4": "Yes, it was already posted and Myth and I gave different solutions, but it was on the inequalities forum." } { "Tag": [], "Problem": "Find all positive integer solutions of\r\n\r\n$ (1+\\frac{1}{a})(1+\\frac{1}{b})(1+\\frac{1}{c})=2$.", "Solution_1": "[hide]\n\nWe'll first write the equation in form: $ (a+1)(b+1)(c+1)=2abc$. Let's now make a substitution $ a=u+4, b=v+4, c=w+4 (-3 \\le u,v,w)$.\n\nThe equation becomes: $ (u+5)(v+5)(w+5) = 2(u+4)(v+4)(w+4)$. When we expand that we get $ uvw+3uv+3vw+3wu+7u+7v+7w+3=0$.\n\nNow note that at least one of $ u,v,w$ need to be negative because we want the LHS to equal 0. Now we just do a little case work, symmetrically we can just check the negative values for u:\n1) $ u=-3$\n$ 2v+2w =-18$, this is impossible, because it would require $ u \\mathrm{or}v <-3$.\n2) $ u=-2$\n$ vw+v+w=11$\nWe get the following (w,v) pairs: $ (11,0), (5,1), (3,2)$.\n3) $ u=-1$\n$ vw+2v+2w=2$ results $ (1,0), (-1,4)$.\n\nSo the solutions $ (a,b,c)$ are: $ (2,15,4), (2,9,5), (2,7,6), (3,5,4), (3,3,8)$ and all their permutations.\n[/hide]", "Solution_2": "i think it s not work", "Solution_3": "im not sure im mad or this pro's mad :D\r\nWLOG $ \\frac{1}{a}\\geq \\frac{1}{b}\\geq \\frac{1}{c}$\r\nthen we have $ (1+\\frac{1}{a})^{3}\\geq (1+\\frac{1}{a})(1+\\frac{1}{b})(1+\\frac{1}{c})=2$\r\nequivalents to $ 1 \\geq (\\sqrt[3]{2}-1)a$ finally, we obtain $ a \\leq 3$\r\nsimilar way to the other variables" } { "Tag": [ "algebra", "polynomial", "analytic geometry" ], "Problem": "Find all points of intersection of the two graphs. One intersection occurs at x=1.\r\n\r\n(Problem has the picture of these graphs below)\r\n\r\nY = x^4 + 2x^3 - 9x^2 +20 and Y = -2x^2 + 7x - 2\r\n\r\nThis is from a math book. I already know the answers. By looking at the graph I can easily tell what the answers are. Also by plugging and plotting my visual guesses from the looks of the picture of the two graphs, I can be sure of those guesses. Even though I have the answers, I want to know the \"correct\" method of solving polynomial interestions. I want to know of an exact and specific method to solving this problem. Unless the method I did was the way to do it? All I did was look at the two graphs, estimate where they would intersect and plugged in the the x or y coordinates to get the other and do the same for both equations. But I know this is not the \"correct way\" of solving this problem.\r\n\r\nPlease describe to me how to solve this problem and the general method of solving intersections of polynomials. Thank you very much for reading my post.", "Solution_1": "Intersections are where the equations equal each other.\r\nBut at x=1, these two equations don't equal each other.\r\n1+2-9+20=14 doesn't equal -2+7-2=3.\r\nAre you sure you typed the problem correctly?", "Solution_2": "In general, you solve the two equations (kind of like solving linear simultaneous equations.\r\n\r\n$y=x^4+2x^3-9x^2+20$\r\n$y=-2x^2+7x-2$.\r\n\r\nSubtract the two equations to get something in terms of $x$ only, then solve that equation for $x$.\r\n\r\nThen plug in those values of $x$ into the original equation to get the $y$ values.", "Solution_3": "Am sry, the second equation is -2x^2 + 8x +8. Very sorry. :(", "Solution_4": "So we have two polynomials:\r\n$y=x^4+2x^3-9x^2+20$\r\n$y=-2x^2+8x+8$\r\nif they intersect somewhere they attain the same value for the same $x$ so:\r\n$x^4+2x^3-9x^2+20=-2x^2+8x+8$\r\n$x^4+2x^3-7x^2-8x+12=0$\r\nWe find roots among divisors of $12$ and we get :$-3,-2,1,2$", "Solution_5": "Math is one of my hardest subjects to understand, still I enjoy it a lot. If you don't mind, can you describe how you got -3, -2, 1 and 2 from the equation x^4 + 2x^3 - 7x^2 - 8x + 12 = 0? I realize that the equation factors into (x+3)(x+2)(x-1)(x-2) from those 4 numbers you had told me. Am basically asking if you may describe how you factored the equation (if there's a method / formula)? If I were to look at that equation I would have never simplified it (x+3)(x+2)(x-1)(x-2) and gotta -3, -2, 1 and 2 for the \"X's\". The book only goes are far as giving me a few detailed examples, but not enough for me to have a complete understanding. Please help me, and I thank you very much for helping me already with the equation.", "Solution_6": "[b]Theorem[/b]: if polynomial has integer coefficients:\r\n$P(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$\r\nAnd this polynomial has integer root $a$, then $a$ is a divisor of $a_0$. \r\nIn our case $a_0=12$ so we check whether $-1, -2, -3, -4, -6, -12, 1, 2, 3, 4, 6, 12$ are roots of this polynomial.\r\n\r\nThe other method is factoring\r\n$x^4+2x^3-7x^2-8x+12=0$\r\n$x^4+2x^3-6x^2-x^2-6x-2x+12=0$\r\n$x^4+2x^3-x^2-2x-6x^2-6x+12=0$\r\n$x(x^3+2x^2-x-2)-6(x^2+x-2)=0$\r\n$x(x^2(x+2)-(x+2))-6(x^2+2x-x-2)=0$\r\n$x((x+2)(x^2-1))-6(x(x+2)-(x+2))=0$\r\n$x(x+2)(x+1)(x-1)-6(x+2)(x-1)=0$\r\n$(x+2)(x-1)((x+1)x-6)=0$\r\n$(x+2)(x-1)(x^2+x-6)=0$\r\n$(x+2)(x-1)(x^2+3x-2x-6)=0$\r\n$(x+2)(x-1)(x(x+3)-2(x+3))=0$\r\n$(x+2)(x-1)(x+3)(x-2)=0$", "Solution_7": "Thank you very much dondigo for the explanation. It helped me solve the other problems. One more question though. :) This will be the last I promise. After you find the divisors, is there a definte way of checking to see if they are the roots of the equation? How did you weed out -1,-4,-6,-12,3,4,6, and 12 from the divisors of 12, and figure out that -3, -2, 1 and 2 would be the roots of the equation? Am sorry if am giving you a hard time asking so many questions. If you don't want to answer I understand and I am sorry. Thank you.", "Solution_8": "First of all. If you have some problem don't hesitate to ask. This is a Mathforum, so other people will be pleased to help you. You can ask as many question as you can. If there is something easy, that you don't understand post the topic in Getting Started, if you have more difficult problem-ask in Intermediate just like you have done it. \r\nThe answer to your question:\r\nFor example we have a polynomial:\r\n$W(x)=x^2+2x+1$\r\nWe check whether for example $2$ is the root:\r\n$W(2)=2^2+2\\cdot2+1=4+4+1=9$\r\n$2$ is not the root. \r\nWe check $-1$\r\n$W(-1)=(-1)^2+2\\cdot(-1)+1=0$\r\n$W(-1)=0$, so $-1$ is the root of this polynomial", "Solution_9": "Thank you Dondigo, I understand completely now. This is my first post on MathLinks. I'll be sure to post more problems that I may have trouble with in the future. I'll be sure to browse and help others that with theirs too. :)" } { "Tag": [ "number theory" ], "Problem": "excuse me can anyone fill me in,\r\nwith some number theory links please..........", "Solution_1": "There is avaliable a special NT forum here in mathlinks [url]http://www.mathlinks.ro/index.php?f=6[/url].Enjoy it !", "Solution_2": "thanks friend but i think it is too assorted \r\nwhat about some pdf link if anyone has??\r\nthanks for replying", "Solution_3": "What kind of number theory?\r\nConsidering you're posting in the Pre-Olympiad forum it is sort of implied that you want the olympiad level stuff (what enndb0x linked to).\r\n\r\nIf you're looking for something introductory, AoPS has an introduction to number theory book." } { "Tag": [ "ratio", "rotation" ], "Problem": "As illustrated below, we can dissect every triangle $ABC$ into four pieces so that piece 1 is a triangle similar to the original triangle, while the other three pieces can be assembled into a triangle also similar to the original triangle. Determine the ratios of the sizes of the three triangles and verify that the construction works.\n\n[asy]\nimport olympiad;size(350);defaultpen(linewidth(0.7)+fontsize(10));\npath p=origin--(13,0)--(9,8)--cycle;\npath p2=rotate(180)*p,\np3=shift(-26,0)*scale(2)*p,\np4=shift(-27,-24)*scale(3)*p,\np1=shift(-53,-24)*scale(4)*p;\npair A=(-53,-24), B=(-8,16), C=(12,-24), D=(-17,8), E=(-1,-24), F=origin, G=(-13,0), H=(-9,-8);\nlabel(\"1\", centroid(A,D,E));\nlabel(\"2\", centroid(F,G,H));\nlabel(\"3\", (-10,6));\nlabel(\"4\", (0,-15));\ndraw(p2^^p3^^p4);\nfilldraw(p1, white, black);\npair point = centroid(F,G,H);\nlabel(\"$\\mathbf{A}$\", A, dir(point--A));\nlabel(\"$\\mathbf{B}$\", B, dir(point--B));\nlabel(\"$\\mathbf{C}$\", C, dir(point--C));\nlabel(\"$\\mathbf{D}$\", D, dir(point--D));\nlabel(\"$\\mathbf{E}$\", E, dir(point--E));\nlabel(\"$\\mathbf{F}$\", F, dir(point--F));\nlabel(\"$\\mathbf{G}$\", G, dir(point--G));\nlabel(\"$\\mathbf{H}$\", H, dir(point--H));\nreal x=90;\ndraw(shift(x)*p1);\nlabel(\"1\", centroid(shift(x)*A,shift(x)*D,shift(x)*E));\ndraw(shift(130,0)*p4);\ndraw(shift(130,0)*shift(-27,-24)*p);\ndraw(shift(130,0)*shift(-1,-24)*p3);\nlabel(\"2\", shift(130,0)*shift(-27,-24)*centroid(F,(9,8),(13,0)));\nlabel(\"3\", shift(130,0)*shift(-1,-24)*(-10,6));\nlabel(\"4\", shift(130,0)*(0,-15));\nlabel(\"Piece 2 rotated $180^\\circ$\", (130,10));[/asy]", "Solution_1": "This problem is very nicely. \r\nI think that the answer is $\\frac{GH}{1}=\\frac{DE}{4}=\\frac{BC}{5}\\ \\left(=\\frac{HE}{2}=\\frac{GE}{3}\\right).$" } { "Tag": [], "Problem": "A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring\r\nat a distance much greater than R and allowed to fall towards the ring under the influence of the\r\nring\u2019s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same\r\n(linear) mass density but radius 2R, and the experiment is repeated. What is the new maximum\r\nspeed of the particle?\r\n\r\n$ \\textbf{(A)}\\ \\frac{1}{2}v\\qquad \\textbf{(B)}\\ \\frac{1}{\\sqrt{2}}v\\qquad \\textbf{(C)}\\ v\\qquad \\textbf{(D)}\\ \\sqrt{2}v\\qquad \\textbf{(E)}\\ 2v$", "Solution_1": "Conservation of energy - work done by the ring on the particle = change in kinetic energy of the particle. \r\ntherefore, find the gravitational potential energy of the ring. However, this does not change when the ring is replaced, and so the velocity stays the same.\r\nAnother topic about this question can be found at: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190314[/url]", "Solution_2": "[quote=\"xsk13\"]Another topic about this question can be found at: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190314[/url][/quote]\r\n\r\nThanks! Sorry, I searched for \"2008 F=ma #18\" and didn't find anything, so I posted this up.", "Solution_3": "Doesn't the gravitational potential energy change though? The mass of the ring change because at a radius of 2R; it is larger and will have two times as much mass as it did before. The problem is not saying that mass is remaining the same. It is saying that mass density is remaining the same.", "Solution_4": "Qualitatively, the larger radius means a greater distance between the mass and the $ 2R$ ring than the distance between the mass and the $ R$ ring, at every point along the x-axis. Thus, increase in mass and increase in distance \"balances\" out to produce the same net work.\r\n\r\nQuantitatively, you can form an expression for the force, integrate it along the x-axis to find work. Since the starting distance $ x$ is much greater than $ R$, you can treat $ R \\equal{} 0$ when you have something along the lines of $ \\frac {something}{x \\plus{} R}$. Conclusion: $ R$ doesn't matter for a sufficiently large $ x$.", "Solution_5": "Oh wow...I was considering all mass at the Center of Mass and ignored the increased distance because the center of mass didn't change.", "Solution_6": "I don't understand... Since gravitational energy can be described as $ G\\frac{Mm}{r}$, if the new ring is twice as massive, wouldn't the speed then be $ \\sqrt{2}$ times as much?", "Solution_7": "[quote=\"pascal12\"]I don't understand... Since gravitational energy can be described as $ G\\frac {Mm}{r}$, if the new ring is twice as massive, wouldn't the speed then be $ \\sqrt {2}$ times as much?[/quote]\r\nGravitational Potential Energy at Centre of the ring is $ U_{centre} \\equal{} \\frac {GMm}{R}$. \r\nChange in K.E in the process of the falling of the particle from $ \\infty$ to the centre of the ring(where the particle has max. K.E) is equal to the change in the P.E.\r\nSo $ (K.E)_{centre} \\equal{} U_{centre}$\r\nBut $ U_{centre}$ doesn't change as \"M\"(mass of ring) is also doubled along with the radius \"R\".", "Solution_8": "But if it's falling from infinity, wouldn't the radius of the ring be negligible? After all, isn't $ r$ the distance from the object to the center of mass?", "Solution_9": "By telling you that it starts from infinity, it is basically telling you that the original potential gravitational energy is zero.\r\n\r\nWhen the particle gets closer to the ring, the distance is no longer negligible." } { "Tag": [], "Problem": "\u039d\u03b1 \u03bb\u03c5\u03b8\u03b5\u03b9 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7:\r\n$ 1! \\plus{} 2! \\plus{} .....x! \\equal{} y^2$", "Solution_1": "[quote=\"\u0391\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7\u03c2 6\"]\u039d\u03b1 \u03bb\u03c5\u03b8\u03b5\u03b9 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7:\n$ 1! \\plus{} 2! \\plus{} .....x! \\equal{} y^2$[/quote]\r\n\r\n\u0393\u03b9\u03b1 $ x>5$ \u03ad\u03c7\u03c9 $ LHS\\equal{}3(mod5)$ \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b7 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b9\u03ba\u03cc \u03ba\u03b1\u03c4\u03ac\u03bb\u03bf\u03b9\u03c0\u03bf $ mod 5$. \u039c\u03ad\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03bb\u03ad\u03b3\u03be\u03bf\u03c5\u03bc\u03b5 \u03b3\u03b9\u03b1 $ x\\equal{}1,2,3,4$\r\n\r\n$ qed$", "Solution_2": "\u03a3\u03c9\u03c3\u03c4\u03bf\u03c2 \u039d\u03b9\u03ba\u03bf\u03bb\u03b1\u03ba\u03b9..... :lol: \r\n\u0391\u03bb\u03bb\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7\u03bd \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5....\u03ba\u03c1\u03b9\u03bc\u03b1 \u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9....\r\n\u039d\u03b1 \u03bb\u03c5\u03b8\u03b5\u03b9 \u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7:\r\n$ 1! \\plus{} 2! \\plus{} ...x! \\equal{} y^n$ (\u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 $ n$ \u03c6\u03c5\u03c3\u03b9\u03ba\u03bf \u03bc\u03b5\u03b3\u03b1\u03bb\u03c5\u03c4\u03b5\u03c1\u03bf \u03c4\u03bf\u03c5 $ 2$).\u039d\u03b9\u03ba\u03bf good luck WAGM!\u0396\u0397\u03a4\u0391\u03a9 \u039a\u0391\u0399 \u03a4\u0397\u039d \u0395\u039d\u0391\u03a3\u03a7\u039f\u039b\u0397\u03a3\u0397 \u03a4\u039f\u03a5 \u0391\u039d\u03a4\u039f \u039a\u0391\u0399 \u03a4\u039f\u03a5 CRETANMAN \u0393\u0399\u0391 \u03a4\u039f \u03a0\u03a1\u039f\u0392\u039b\u0397\u039c\u0391....KAI O\u039b\u03a9\u039d \u03a4\u03a9\u039d \u0391\u03a1\u0399\u0398\u039c\u039f\u0398\u0395\u03a9\u03a1\u0397\u03a4\u0399\u039a\u03a9\u039d \u03a6\u03a5\u03a3\u0399\u039a\u0391.... :wink:", "Solution_3": "T\u03b5\u03bb\u03b9\u03ba\u03ac \u03c0\u03bf\u03b9\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1?" } { "Tag": [ "MATHCOUNTS", "ARML" ], "Problem": "Prove that ${ \\sqrt[3] {\\sqrt [3]{2} - 1} = \\sqrt [3]{\\frac{1}{9}} - \\sqrt [3]{\\frac{2}{9}} + \\sqrt [3]{\\frac{4}{9}}}.$", "Solution_1": "It's too hard!!!", "Solution_2": "I don't think proofs are MC level...", "Solution_3": "This is not the appropriate forum to discuss proofs. Please keep the problems Mathcounts level. Examples of some relatively easy mathcounts problems are [url]http://www.mnspe.org/mathcounts/08_state_test/2008_sprint.pdf[/url].\r\n\r\nThis should be moved to an appropriate forum. I'd suggest Intermediate topics, since I've seen this before in there (I think). This was an ARML problem, except it was find $ \\sqrt[3]{\\sqrt[3]{2}\\minus{}1}\\equal{}\\sqrt[3]{a}\\minus{}\\sqrt[3]{b}\\plus{}\\sqrt[3]{c}$, and we wanted $ a\\plus{}b\\plus{}c$.", "Solution_4": "[quote=Ji Chen]Prove that ${ \\sqrt[3] {\\sqrt [3]{2} - 1} = \\sqrt [3]{\\frac{1}{9}} - \\sqrt [3]{\\frac{2}{9}} + \\sqrt [3]{\\frac{4}{9}}}.$[/quote]\n[url]https://artofproblemsolving.com/community/c7h2598821p22420229[/url]\n", "Solution_5": "\n[hide=Post #5 by sqing][url=aops.com/community/user/148231][b]sqing[/b][/url] \u00b7 Jun 23, 2021, 1:41 PM [url=aops.com/community/p22425693](view)[/url][color=transparent]helo[/color]\n[quote=Ji Chen]Prove that ${ \\sqrt[3] {\\sqrt [3]{2} - 1} = \\sqrt [3]{\\frac{1}{9}} - \\sqrt [3]{\\frac{2}{9}} + \\sqrt [3]{\\frac{4}{9}}}.$[/quote]\n[url]https://artofproblemsolving.com/community/c7h2598821p22420229[/url]\n\n-----------\n[color=#5b7083][aops]x[/aops] 5[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]\n\n[tip=@sqing][img]https://avatar.artofproblemsolving.com/avatar_148231.jpg[/img] [center]AoPS User[/center][/tip] why the 12-year bump?", "Solution_6": "Just use x^3+y^3=(x-y)(x^2+y^2-xy) first,notice the r.h.s has some match with this eqn simplify then use x^3-y^3=(x-y)(x^2+x-y+y^2).\nAnyway as far I know these formulas are taught in class 9 in my country,I think it's taught before in some other countries.", "Solution_7": "I think some people cannot see the dates and are commenting/replying stuff randomly /shrug", "Solution_8": "@above what I told is a truth,and it doesn't matter what the date is,people learn everyday.\nFor your information aops is not like any social media,even 15 year old problems are also solved hear.\nP.S.--I reckon you are new here so surf a little bit more." } { "Tag": [ "analytic geometry", "topology", "complex analysis" ], "Problem": "Prove that every discrete set in $ \\mathbb{C}$ is countable", "Solution_1": "See [url]http://mathworld.wolfram.com/DiscreteSet.html[/url]\r\n\r\nThere is a standard argument.\r\nBy the definition of a discrete set $ S$, for any $ x \\in S$ there exists a neighborhood $ U_x$ of $ x$, such that $ U_x \\cap S \\equal{} \\{x\\}$. Then there is a ball $ B(x, \\rho_x) \\subseteq U_x$, and a point $ P_x(a_x, b_x) \\in B(x, \\rho_x)$ of rational coordinates. But $ \\mathbb{Q} \\times \\mathbb{Q}$ is countable.", "Solution_2": "How does one prove that every open set in $ \\mathbb{R}^{n}$ has atleast one point with rational coordinates", "Solution_3": "First, put the exponent $ ^n$ outside the brackets of $ \\mathbb{R}$, in order to get $ \\mathbb{R}^n$, and not $ \\mathbb{R^n}$.\r\n\r\nNext, any open set contains a ball, and $ \\mathbb{Q}^n$ is dense in $ \\mathbb{R}^n$.", "Solution_4": "rite thanks!" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "if $x_{1}=3$ and $ 1+x_{n}x_{n+1}=2n^{2} (x_{n+1}-x_{n})$, then compute the limit of $x_{n}$\r\n\r\nhave fun! nice and easy! :cool:", "Solution_1": "I discover that the sequence {x_n} is 3, 5, 7, 9, 11, 13, ...", "Solution_2": "Please check your computation! It's not good!\r\n\r\nyou rewrite the sequence as:\r\n(x_(n+1)-x_n)/(1+x_n*x_(n+1))=1/(2n^2) now you apply arctan to this equality\r\nand we know that:\r\n\r\nacrtan ((x_(n+1)-x_n)/(1+x_n*x_(n+1)))=arctan(x_(n+1))-arctan(x_n)\r\nand arctan(1/(2n^2))=arctan(n/(n+1))-arctan((n-1)/n) and you sum it up and get:\r\narctan (X_(n+1))-arctan(x_1)=arctan(n/(n+1)) and then you get that:\r\nx_(n+1)=(4n+1)/(-2n+1) so lim(x_n)=-2\r\nI hope i didn't make any mistakes! \r\n\r\ncheers!~ :D :D :cool:", "Solution_3": "oops...i made a mistake during the computation... :( \r\n\r\nYour answer is right, Lagrangia." } { "Tag": [ "trigonometry", "vector", "complex numbers" ], "Problem": "I took some time to study complex numbers today with AoPS Vol 2. I understood all the concepts (except for roots of unity) but I failed to put them to use. I encountered a few problems. One of them is that I don't know how to express general numbers in the exponential form. For example, $-64$ can be written as $64e^{i\\pi}$ and $1 = e^{2ki\\pi}$\r\n\r\nThe following probles are taken from the AoPS textbook.\r\n\r\n1. Find $(1+i)^{4}(2-2i)^{3}$ - Try using polar form to complete the problem.\r\n\r\n2. Find the product of the $n$ $n$th roots of $1$ in terms of $n$.\r\n\r\n3. Find $Im(\\cos12^\\circ+i\\sin12^\\circ+\\cos48^\\circ+i\\sin48^\\circ)^{6}$ (Try not to use a geometric method.)\r\n\r\nIf anyone has additional resources (webstes) for complex numbers, the help would be greatly appreciated. \r\n\r\nThanks for looking.", "Solution_1": "[hide=\"Problem 1\"] \\begin{eqnarray*}(1+i)^{4}(2-2i)^{3}& = & \\left(\\sqrt{2}\\left(\\cos{\\frac{\\pi}{4}}+i \\sin{\\frac{\\pi}{4}}\\right)\\right)^{4}\\left(2\\sqrt{2}\\left(\\cos{\\frac{7\\pi}{4}}+i \\sin{\\frac{7\\pi}{4}}\\right)\\right)^{3}\\\\ & = & \\left(\\sqrt{2}\\right)^{4}\\left(2\\sqrt{2}\\right)^{3}\\left(\\cos{\\frac{\\pi}{4}}+i \\sin{\\frac{\\pi}{4}}\\right)^{4}\\left(\\cos{\\frac{7\\pi}{4}}+i \\sin{\\frac{7\\pi}{4}}\\right)^{3}\\\\ & = & (4)\\left(16\\sqrt{2}\\right) \\left(\\cos{\\pi}+i \\sin{\\pi}\\right)\\left(\\cos{\\frac{21\\pi}{4}}+i \\sin{\\frac{21\\pi}{4}}\\right) \\\\ & = & 64\\sqrt{2}\\left(\\cos{\\frac{\\pi}{4}}+i \\sin{\\frac{\\pi}{4}}\\right) \\\\ & = & \\boxed{64+64i}\\end{eqnarray*} Note that DeMoivre's Theorem was used to find $\\left(\\cos{\\frac{\\pi}{4}}+i \\sin{\\frac{\\pi}{4}}\\right)^{4}= \\left(\\cos{\\pi}+i \\sin{\\pi}\\right)$, etc.[/hide]\n\n[hide=\"Problem 2\"]Do you mean the nth roots of 1?\n\nThe nth roots of 1 are of the form $\\cos{\\frac{2k\\pi}{n}}+i \\sin{\\frac{2k\\pi}{n}}= \\mbox{cis}\\frac{2k\\pi}{n}$, where $k = 0,1,2,...,n$. By DeMoivre's Theorem, their product is: \\begin{eqnarray*}\\mbox{cis}\\frac{0\\pi+2\\pi+4\\pi+...+2n\\pi}{n}& = & \\mbox{cis}\\frac{2\\pi(1+2+3+...+n)}{n}\\\\ & = & \\mbox{cis}\\frac{2\\pi n(n+1)}{2n}\\\\ & = & \\boxed{\\mbox{cis}\\left(\\pi(n+1)\\right)}\\end{eqnarray*} [/hide]\r\n\r\nAnd what do you mean by Im(...)?", "Solution_2": "[hide=\" :D$_{1}$\"]\n\n\\begin{eqnarray*}(1+i)^{4}(2-2i)^{3}& = & \\left(2^{3}\\right)\\left(1+i\\right)\\left(1+i\\right)^{3}\\left(1-i\\right)^{3}\\\\ & = & \\left(2^{3}\\right)\\left(1+i\\right)\\left(2\\right)^{3}\\\\ & = & 64+64i\\end{eqnarray*}\n[/hide]\n[hide=\" :D$_{2}$\"] \\[\\text{The n n-th roots of 1 satisfy the equation: }x^{n}-1=0\\] \\[\\text{Let this equation be F(x,n). We factor it into n roots: }\\prod_{i=1}^{n}(x-r_{i}0\\] \\[\\text{The constant term of F(x,n) is the only term without a power of x. Therefore it is the product of the roots: }\\] \\[\\prod_{i=1}^{n}(x-r_{i}) \\ \\ \\text{mod x}\\equiv \\prod_{i=1}^{n}(-r_{i})=(-1)^{n}\\left(\\prod_{i=1}^{n}r_{i}\\right) \\text{(which we know equals }-1 \\text{)}\\] $\\therefore \\prod_{i=1}^{n}r_{i}=\\frac{(-1)}{(-1)^{n}}\\implies \\boxed{\\prod_{i=1}^{n}r_{i}=(-1)^{n+1}}$\nNotice that: $\\mbox{cis}\\left(\\pi(n+1)\\right)=\\mbox{cis}(n\\pi)\\mbox{cis}(\\pi)=(-1)^{n+1}$\n[/hide]\n[hide=\" :D$_{3}$\"]\n\\begin{eqnarray*}(\\cos12^\\circ+i\\sin12^\\circ+\\cos48^\\circ+i\\sin48^\\circ)^{6}& = & \\left(\\mbox{cis}(30^\\circ)\\mbox{cis}(18^\\circ)+\\frac{\\mbox{cis}(30^\\circ)}{\\mbox{cis}(18^\\circ)}\\right)^{6}\\\\ & = & \\left(\\mbox{cis}(18^\\circ)+\\frac{1}{\\mbox{cis}(18^\\circ)}\\right)^{6}\\\\ & = & \\left(\\cos(18^\\circ)+i\\sin(18^\\circ)+\\cos(-18^\\circ)+i\\sin(-18^\\circ)\\right)^{6}\\\\ & = & (2\\cos(18^\\circ)^{6}\\end{eqnarray*}\nAnd that is precisely the answer, so the imaginary part ($\\text{Im(...)}$) is 0. The geometric method uses vectors, but it is really the same as the algebra :) \n[/hide]", "Solution_3": "[quote=\"xsquaredster\"]And what do you mean by Im(...)?[/quote]\r\n\r\nIt refers to the imaginary part. That is, given $a+bi \\in \\mathbb{C}$ with $a, b \\in \\mathbb{R}$ we define\r\n\r\n$\\Re(a+bi) = a$\r\n$\\Im(a+bi) = bi$", "Solution_4": "[quote=\"t0rajir0u\"][quote=\"xsquaredster\"]And what do you mean by Im(...)?[/quote]\n\nIt refers to the imaginary part. That is, given $a+bi \\in \\mathbb{C}$ with $a, b \\in \\mathbb{R}$ we define\n\n$\\Re(a+bi) = a$\n$\\Im(a+bi) = bi$[/quote]\r\nIsn't $\\Im(a+bi) = b$? :maybe:", "Solution_5": "Hmm. I'm not so sure. It doesn't make too much of a difference, since it's only ever used in isolation :P" } { "Tag": [ "inequalities", "trigonometry", "geometry", "circumcircle", "IMO Shortlist" ], "Problem": "Let $ ABCD$ be a convex quadrilateral, and let $ R_A, R_B, R_C, R_D$ denote the circumradii of the triangles $ DAB, ABC, BCD, CDA,$ respectively. Prove that $ R_A \\plus{} R_C > R_B \\plus{} R_D$ if and only if $ \\angle A \\plus{} \\angle C > \\angle B \\plus{} \\angle D.$", "Solution_1": "$ R_A = \\frac {AD}{2sin\\angle ABD}$\r\n$ R_B = \\frac {BC}{2sin\\angle CAB}$\r\n$ R_C = \\frac {BC}{2sin\\angle BDC}$\r\n$ R_D = \\frac {AD}{2sin\\angle DCA}$\r\n\r\nWLOG $ \\angle DAB$ + $ \\angle BCD$ $ > =$ $ \\angle CDA$ + $ \\angle ABC$\r\nThe sum of all angles is $ 360$ and that implies $ S =$ $ \\angle DAB$ + $ \\angle BCD$ $ > =$ $ 180$\r\n\r\n1)If $ S = 180$ then A,B,C and D lie on the same circle, so $ R_A + R_C = R_B + R_D$ \r\n\r\n2)If $ S > 180$ then $ \\angle DCA > \\angle ABD$ and $ \\angle CAB > \\angle BDC$ which implies \r\n$ R_A > R_D$ and $ R_C > R_B$ \r\nSum these two inequalities and you get the result", "Solution_2": "@Bugi: $\\angle DCA > \\angle ABD$ does not imply $\\sin \\angle DCA > \\sin \\angle ABD$ or the opposite inequality, since some of these angles may be obtuse.\n\n[hide]\nBy capital letters $A, B, C, D$, I mean those angles of the quadrilateral $ABCD$ with the specified point as vertex. Also, let $R(XYZ)$ denote the circumradius of triangle $XYZ$.\n\nClearly the if and only if statements are equivalent. Assume then that $A + C > B + D$ First we dispose of most quadrilaterals with one method, and then we look for those for which this method is invalid and prove the problem for them separately.\n\nAssume WLOG $\\sin C < \\sin A$ so that $R_C > R_A$ by the law of sines. Extend $AC$ to meet the circumcircle of $ABD$ at point $C'$. We have $R_A + R_C > R_A + R(BC'D)$ since $R(BC'D) = R_A$, and $R_A + R(BC'D) = R(ABC') + R(ADC')$. Next, if $BC' > BC$ and $CD' > CD$, then we would have $R(ABC') = \\frac{BC'}{\\sin A} > \\frac{BC}{\\sin A} > R(ABC)$, and similarly $R(ADC') > R(ADC)$, so we would be done.\n\nHence it remains to consider $ABCD$ for which $BC' > BC$ and $CD' > CD$ are not both true. Clearly at least one of $\\angle ACB$, $\\angle ACD$ is acute, so at most one of these inequalities is false. WLOG, assume $BC' < BC$. Since the other inequality $CD' > CD$ holds we can show as before that $R_A > R_D$. We are left with proving $R_C > R_B$.\n\nFrom $BC' < BC$ it follows that $\\angle BCA$ is obtuse. Next, we show that $\\angle BDC < \\angle BAC$ and both are acute. The acuteness follows from the obtusenesses of $\\angle BCA$ and $C$. The inequality is obvious from the fact that the ray $DC$ meets the circumcircle of $ABD$ at a point on minor arc $BC'$. Hence $\\sin \\angle BDC < \\sin \\angle BAC$ so $\\frac{BC}{\\sin \\angle BDC} > \\frac{BC}{\\sin \\angle BAC}$ and $R_C > R_B$.\n[/hide]" } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities proposed" ], "Problem": "[b]By: Pham Dinh Khanh -10A2t-khtn[/b]\r\nLet $ x,y,z > 0$ Prove that \r\n$ \\frac {x^2}{\\sqrt {x^2 \\plus{} xy \\plus{} y^2}} \\plus{} \\frac {y^2}{\\sqrt {y^2 \\plus{} yz \\plus{} z^2}} \\plus{} \\frac {z^2}{\\sqrt {z^2 \\plus{} zx \\plus{} x^2}} \\geq \\frac {x\\plus{} y \\plus{} z}{\\sqrt {3}}$", "Solution_1": "This problem can be written as follows\r\n$ \\sum_{cyc}\\sqrt{x^2\\plus{}xy\\plus{}y^2} \\ge \\sum_{cyc}\\frac{y(x\\plus{}y)}{\\sqrt{x^2\\plus{}xy\\plus{}y^2}} \\plus{}\\frac{x\\plus{}y\\plus{}z}{\\sqrt{3}}$\r\nAfter that, using Cauchy inequality, we derive\r\n$ x^2\\plus{}y^2\\plus{}xy \\ge \\frac{3}{4} (x\\plus{}y)^2$\r\nWe are done!", "Solution_2": "Could you show me how do you use Cauchy?", "Solution_3": "Just rewrite $ x^2\\plus{}y^2\\plus{}xy \\ge \\frac{3}{4} (x\\plus{}y)^2$:\r\n\r\n$ 4(x^2\\plus{}y^2\\plus{}xy) \\ge 3(x\\plus{}y)^2 \\equal{} 3x^2\\plus{}6xy\\plus{}3y^2$\r\n$ x^2 \\plus{} y^2 \\minus{} 2xy \\ge 0$\r\n$ (x\\minus{}y)^2 \\ge 0$.", "Solution_4": "I'm so sorry. This $ x^2 \\plus{} y^2 \\plus{} xy \\ge \\frac {3}{4}(x \\plus{} y)^2$ is obvious, but I thought about something else. Now I understand the whole proof, but previously I didn't, but I don't know why...\r\n\r\nSorry for creating confusion :blush:", "Solution_5": "[quote=\"dragon1\"][b]By: Pham Dinh Khanh -10A2t-khtn[/b]\nLet $ x,y,z > 0$ Prove that \n$ \\frac {x^2}{\\sqrt {x^2 \\plus{} xy \\plus{} y^2}} \\plus{} \\frac {y^2}{\\sqrt {y^2 \\plus{} yz \\plus{} z^2}} \\plus{} \\frac {z^2}{\\sqrt {z^2 \\plus{} zx \\plus{} x^2}} \\geq \\frac {x \\plus{} y \\plus{} z}{\\sqrt {3}}$[/quote]\r\nIt's weaker than the well-known:\r\n$ \\frac{a}{\\sqrt{a\\plus{}b}} \\plus{} \\frac{b}{\\sqrt{b\\plus{}c}} \\plus{} \\frac{c}{\\sqrt{c\\plus{}a}} \\geq\\ \\frac{\\sqrt{a} \\plus{} \\sqrt{b} \\plus{} \\sqrt{c}}{\\sqrt{2}}$\r\nWith this ineq, I remember that hxy09 and me have found a nice proof for it :lol:", "Solution_6": "Guys,\r\n\r\nUnfortunately this solution is wrong, because the inequality is reversed: $ x^2 \\plus{} xy \\plus{} y^2 \\geq \\frac34(x \\plus{} y)^2 \\iff \\frac 1{\\sqrt {x^2 \\plus{} xy \\plus{} y^2}} \\leq \\frac 2{\\sqrt3(x \\plus{} y)}$.\r\n\r\nAlso unfortunately, I couldn't come up with a solution too. But I'll keep trying!\r\n\r\nEDIT: I was referring to the post above nguoivn's. By the way, I tried to use the Fermat point of a triangle with sides $ \\sqrt{x^2\\plus{}xy\\plus{}y^2}$, $ \\sqrt{y^2\\plus{}yz\\plus{}z^2}$ and $ \\sqrt{z^2\\plus{}zx\\plus{}x^2}$ (so $ x\\plus{}y\\plus{}z$ is a natural minimum), but I still haven't found anything nice about the LHS.", "Solution_7": "[quote=\"cyshine\"]Guys,\n\nUnfortunately this solution is wrong, because the inequality is reversed: $ x^2 \\plus{} xy \\plus{} y^2 \\geq \\frac34(x \\plus{} y)^2 \\iff \\frac 1{\\sqrt {x^2 \\plus{} xy \\plus{} y^2}} \\leq \\frac 2{\\sqrt3(x \\plus{} y)}$.\n\n[/quote]\nNo! nguoivn means that $ x^2 \\plus{} xy \\plus{} y^2\\leq\\frac {3(x^2 \\plus{} y^2)}{2}.$ :wink:\n[quote=\"nguoivn\"]\nIt's weaker than the well-known:\n$ \\frac {a}{\\sqrt {a \\plus{} b}} \\plus{} \\frac {b}{\\sqrt {b \\plus{} c}} \\plus{} \\frac {c}{\\sqrt {c \\plus{} a}} \\geq\\ \\frac {\\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c}}{\\sqrt {2}}$\nWith this ineq, I remember that hxy09 and me have found a nice proof for it :lol:[/quote]\r\nWhat is your proof for this one, nguoivn? Thank you!\r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071", "Solution_8": "Here is our poof (hxy09 and me) :) \r\nWe can write it into: $ \\sum\\ \\frac {a^2}{\\sqrt {a^2 \\plus{} b^2}} \\geq\\ \\frac {a \\plus{} b \\plus{} c}{\\sqrt {2}}$\r\nAfter square and use Cauchy Schwarts and Am-Gm, we only need to prove:\r\n$ \\sum\\ \\frac {a^4 \\plus{} b^4}{a^2 \\plus{} b^2} \\plus{} \\frac {2(ab \\plus{} bc \\plus{} ca)^2}{a^2 \\plus{} b^2 \\plus{} c^2} \\geq\\ (a \\plus{} b \\plus{} c)^2$\r\nAnd the last step is a nice hint of hxy09 :lol: \r\n[quote=\"hxy09\"][b]AND[/b] it is [b]EQUIVALENT TO[/b] \n$ \\sum \\frac {(a^2 \\plus{} b^2 \\minus{} ca \\minus{} cb)^2(a \\minus{} b)^2}{a^2 \\plus{} b^2} \\ge 0$\nWhich is obviously true.:) [/b][/quote]", "Solution_9": "I have checked it. It's very interesting proof. My congratulations!" } { "Tag": [ "calculus", "derivative", "function", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "prove that for all $x\\in \\mathbb{R}$ function $f(x)=\\sum_{i=1}^{\\infty}\\frac 1{2^{i}}.\\cos (3^{i}x)$ is continious and isn't derivative.", "Solution_1": "${f(x)=\\sum_{i=1}^{\\infty }\\frac{\\cos \\left(3^{i}x\\right)}{2^{i}}}$\r\n\r\n${f'(x)=\\frac{\\partial \\left(\\sum_{i=1}^{\\infty }\\frac{\\cos \\left(3^{i}x\\right)}{2^{i}}\\right)}{\\partial x}}$\r\n${f'(x)=\\sum_{i=1}^{\\infty }\\frac{\\partial \\frac{\\cos \\left(3^{i}x\\right)}{2^{i}}}{\\partial x}}$\r\n${f'(x)=-\\sum_{i=1}^{\\infty }\\left(\\frac{3}{2}\\right)^{i}\\sin \\left(3^{i}x\\right)}$\r\n\r\nbut $1\\geq \\sin \\left(3^{i}x\\right)\\geq$ \r\n\r\nAnd\r\n\r\n${\\sum_{i=1}^{\\infty }\\left(\\frac{3}{2}\\right)^{i}}$\r\n\r\nDiverge\r\n\r\n\r\ni dont think if this is a proof.... but, the derivative diverges ..." } { "Tag": [ "geometry", "trigonometry", "function" ], "Problem": "VOne side of a triangle has length 8 and a second side has length 5. Which of the following could be the area of the triangle?\r\n\r\nI 24\r\nII 20\r\nIII 5", "Solution_1": "[hide]\nAssume the triangle is right and the given sides of the legs, then $A = \\frac{ab}{2}= \\frac{5 \\cdot 8}{2}= \\frac{40}{2}= 20$\nSo $II$\n[/hide]", "Solution_2": "I forgot to mention that there are 2 correct answers. What is the 2nd 1? and why is it.", "Solution_3": "[hide=\"Furthered\"]\nThen from the area formula of any triangle $A = \\frac{a b \\sin \\theta}{2}$.\nAgain let $a=5$ , $b=8$ and then let $A=24$.\nRearranging $\\theta = \\sin^{-1}\\left( \\frac{24}{20}\\right)$, this then eliminates this as a solution because the $\\sin^{-1}$ function is not defined at numbers greater than one.\nSo the two correct ones must be $II \\ , \\ III$.\n[/hide]", "Solution_4": "What if you didn't know there are 2 answers? \r\n\r\nand it was either \r\n\r\n5\r\n20\r\nor both. How would you know its 5.", "Solution_5": "[hide=\"Furthering the Previously Furthered\"]\nUsing the same method, just let $A=5$\nThen $\\theta = \\sin^{-1}\\left( \\frac{5}{20}\\right)$, since $\\frac{5}{20}= \\frac{1}{4}< 1$ where the $\\sin^{-1}$ is defined.\nThe $sin^{-1}$ being defined there implies that there exists such a triangle.\n[/hide]", "Solution_6": "What was the first formula that was used in the beginning of the thread? I've never seen it before, A=ab/2", "Solution_7": ".. You have never seen the forumla for the area of a right trangle, but you know your trig? :rotfl: \r\n\r\nSry, just had to say.", "Solution_8": "[quote=\"rofler\"].. You have never seen the forumla for the area of a right trangle, but you know your trig? :rotfl: \n\nSry, just had to say.[/quote]\r\nHe actually never demonstrated 'knowing his trig'.\r\nThat was another user...", "Solution_9": "Uber-Furthered:\r\n\r\n[hide]\nThe generic triangle ABC has area $\\frac{1}{2}ab\\sin{C}$ for two sides and their included angle.\n\nThus all possible areas of this triangle have area $\\frac{1}{2}(5)(8)\\sin{x}$ where x is the angle between the sides of length 5 and 8. We don't really care about the length of the third side, but clearly the triangle exists for all values of x degrees, 0=2 (except for the case p=2 and n=2), find an explicit monic polynomial g(x) of degree strictly less than p^n such that for every y in Z_(p^n), g(y) = 0 (mod p^n).", "Solution_1": "I'm not sure why you gave an expression for $g(x)$ and then ask for an expression for $g(x)$, but if I understand your question correctly, this problem is very easy by induction as long as the formal derivative of $f(x)$ is not $0 \\bmod p$." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Evaluate\r\n\r\n\\[\\int^\\pi_0 xe^x\\sin x dx\\]", "Solution_1": "[hide=\"Hint\"]$e^x\\sin x=\\mathrm{Im}\\left(e^{(1+i)x}\\right).$[/hide]", "Solution_2": "You may use Integral by parts.", "Solution_3": "I think here is what Kent wrote :D\r\n[hide=\"HINT\"]$e^xsin x=\\mathrm{Im}\\left(e^{(1+i)x}\n\\right)$[/hide]", "Solution_4": "In Japan,ordinary high school student don't study the integral in using Euler's formula.\r\nThis problem is solved by those who are 17 or 18.(Pre college). That's why Integral by parts will do or be sufficient.\r\nBut I have another solution more basically in only one minute. :D", "Solution_5": "Even with the complex exponential, there will be one integration by parts; it's just that without that simplifying notation, the organization of the several necessary integrations by parts gets fairly messy. Do-able, certainly, but long-winded.", "Solution_6": "Here is my solution.\r\n\r\n$(xe^x\\sin x)'=e^x\\sin x+xe^x\\sin x+xe^x\\cos x$\r\n$(xe^x\\cos x)'=e^x\\cos x+xe^x\\cos x-xe^x\\sin x$\r\n\r\nSubtracting of both sides we obtain $(xe^x\\sin x)'-(xe^x\\cos x)'=2xe^x\\sin x-(e^x\\cos x)'$.\r\n\r\nTherefore $\\int xe^x\\sin xdx=\\frac{1}{2}e^x\\{x{(\\sin x-\\cos x)+\\cos x\\}+Const.}$\r\n\r\nAs you see, Integral by parts is not necesary except special case." } { "Tag": [ "geometry", "trigonometry", "circumcircle", "ratio", "projective geometry", "geometry proposed" ], "Problem": "Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \\equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \\equal{} FG$.", "Solution_1": "Assume WLOG that $ \\angle {B} \\leq \\angle{C}$ (in this case $ F$ will be on the segment $ AE$, as in the statement). Denote by $ Y$, $ Z$ the tangency points of the incircle with the sides $ CA$, and $ AB$. Let $ T$ be the intersection point of the lines $ YZ$ and $ BC$. Since $ DZEY$ is an harmonic quadrilateral (because of the concurrency of the tangects at $ Y$, $ Z$ on the line $ DE$), the line $ TD$ is tangent to the incircle at $ D$, and thus $ \\angle{TED} \\equal{} \\angle{TDE}$. But $ \\angle{TED} \\equal{} \\angle{TFE}$, and so the lines $ TD$ and $ CF$ are parallel. Now, since the quadruple $ (B, E, C, T)$ is harmonic, the pencil $ D(B, E, C, T)$ is harmonic, and by intersecting it with the line $ CF$, we conclude that $ F$ is the midpoint of $ CG$ (according to the parallelism of $ DT$ and $ CF$).", "Solution_2": "If we denote $ K\\in\\omega(I)\\cap BD$,$ L\\in\\omega(I)\\cap CD,N\\in\\omega(I)\\cap AB,M\\in\\omega(I)\\cap AC$,then problem follows by the fact that the following fours are harmonic:$ D(DMEN)\\cap l,D(DNKE)\\cap l,D(DMLE)\\cap l$,where $ l\\equal{}CF$.", "Solution_3": "Let the incircle interescts $ BD$ and $ CD$ at $ K,L$ respectively\r\n\r\nThen $ \\frac{GF}{CF}\\equal{}\\frac{DG \\times EK}{DC \\times EL}\\equal{}\\frac{DL \\times EK}{DK \\times EL}$\r\n\r\nI think I have proved that $ DL \\times EK\\equal{}DK \\times EL$,\r\n\r\nbut the proof is long and complicated.\r\n\r\nDoes anyone have a nice proof for this?", "Solution_4": "See here: [url]http://www.mathlinks.ro/viewtopic.php?t=2619[/url] (Read especially the 5th post). \r\nFor some extensions, see also here: [url]http://www.mathlinks.ro/viewtopic.php?p=887161#887161[/url].", "Solution_5": "Let tangent to incircle in a point $ D$ intersect line $ BC$ in $ K$. $ DK\\parallel CF$. Line parallel to $ DK$ through $ B$ intersect line $ AE$ in $ M$ Let $ L$ be a 'point' respective to a set of lines parallel to $ DK$.\r\nThen $ (GFCL)=(DFEM)=(KCEB)=-1$ So $ GF=FC$", "Solution_6": ":oops: \r\nI have made one unnecessary turn.\r\nLet tangent to incircle in a point $ D$ intersect line $ BC$ in $ K$. $ DK\\parallel CF$. Let $ L$ be a 'point' respective to a set of lines parallel to DK. \r\nThen $ (GFCL)\\equal{}(KCEB)\\equal{}\\minus{}1$. So $ GF\\equal{}FC$.", "Solution_7": "I will use [b]pohoata[/b]'s notation.\r\n\r\nSince $ T$ lies on polar of $ A$, $ A$ lies on polar of $ T$. But $ TE$ is tangent to given circle so $ AE$ is polar of $ T$ and we already have that $ D$ is on both $ AE$ and incircle, so $ TD$ is tangent to incircle.\r\n\r\nNow it is easy to conclude that $ CF$ and $ TD$ are parallel and by Menelaus theorem:\r\n\r\n$ \\frac{GF}{FC}\\equal{}\\frac{EB}{EC}\\frac{DG}{BD}\\equal{}\\frac{BZ}{ZY}\\frac{CT}{TB}\\equal{}\\frac{BZ}{TB}\\frac{CT}{CY}$ $ \\equal{}\\frac{\\sin \\angle BTZ}{\\sin \\angle AZY}\\frac{\\sin \\angle AYZ}{\\sin \\angle CTY}\\equal{}1$ which implies $ CF\\equal{}FG$.", "Solution_8": "See from the post #2 that $DT\\parallel CF$ and $T$ is the harmonic conjugate of $E$ w.r.t. $BC$, then apply Menelaos to $\\triangle BCG$ and the transversal $EFD$, tol get $\\frac{CE}{BE}\\cdot \\frac{BD}{DG}\\cdot \\frac{GF}{FC}=1$, but $\\frac{BD}{DG}=\\frac{BT}{CT}$ from parallel lines and $\\frac{BT}{CT}=\\frac{BE}{CE}$ from the harmonic division and remains $\\frac{GF}{FC}=1$.\n\nBest regards,\nsunken rock", "Solution_9": "I will use pohoatza's notation. In addition, let $BD$ intersect the incircle again at $Q$. Let $I$ be the incenter. \n\nWe have $\\angle EQD = \\angle CEF = \\angle CFE = \\angle DFG$, thus, $GQFE$ is cyclic. \n\nFurthermore, note that $C$ is the circumcenter of $\\triangle EFY$, so $\\angle EFY = 180 - \\frac {C} {2} $, and $\\angle DFY = \\frac {C} {2}$. Furthermore, $\\angle FDY = 90 - \\frac {C} {2}$, so $\\angle DYF = 90$. Thus, $\\triangle DYE \\sim \\triangle IYC$. \n\nNow I will calculate $GF$. We have $\\frac {GF} {QE} = \\frac {DF} {DQ}$ from $GQFE$ cyclic, so $GF = QE \\cdot \\frac {DF} {DQ}$. Now using $DF = \\frac {DY} {\\sin {\\frac {C} {2}}}$, we get $GF = \\frac {QE} {DQ} \\cdot \\frac {DY} {\\sin {\\frac {C} {2}}}$. \n\n$ZE$ is the symmedian of $QEDZ$, so $\\frac {QE} {DQ} = \\frac {ZE} {2ZD}$. But we also have $\\frac {ZE} {ZD} = \\frac {AE} {AZ} = \\frac {AE} {AY} = \\frac {YE} {DY}$. Thus, we can multiply out to get $GF = \\frac {YE} {2 \\cdot \\sin {\\frac {C} {2}}} = CE$, as desired.", "Solution_10": "I found a solution without using harmonic properties, though admittedly quite a bit more computational.\n\n[hide=\"hidden for length\"]Let $ A' $ be the foot of the altitude from $ A $ to $ BC $. Firstly, using the usual notation, \\[ 4s(s-c)-2ab=(a+b+c)(a+b-c)-2ab=(a+b)^2-c^2-2ab=a^2+b^2-c^2, \\] so \\[ \\frac{2s(s-c)}{ab}-1=\\frac{4s(s-c)-2ab}{2ab}=\\frac{a^2+b^2-c^2}{2ab}=\\cos C. \\] We also have $ \\cos C=\\frac{CA'}{b}=\\frac{s-c-A'E}{b} $, so \\[ \\frac{2s(s-c)}{ab}-1=\\frac{s-c-A'E}{b}\\implies a\\cdot A'E=sa-ac-2s(s-c)+ab=(s-a)(c-b). \\]\n\nFor brevity, let $ \\angle AEC=\\theta $ and $ A=[ABC] $. Then, from $ \\triangle AA'E $, we have\n\n\\[ \\tan\\theta=\\frac{a\\cdot AA'}{a\\cdot A'E}=\\frac{2A}{(s-a)(c-b)}. \\]\n\nNow let $ C' $ be the projection of $ C $ onto $ AE $, and let $ E' $ be the point on the incircle diametrically opposite $ E $. Then, since $ \\angle DE'E=\\angle CEC' $ and $ \\angle CC'E=\\angle DE'E=90^\\circ $, so $ \\triangle CC'E\\sim\\triangle EE'D $. Then $ \\frac{DE}{CC'}=\\frac{EE'}{CE}=\\frac{2r}{s-c} $. Also, $ \\frac{CC'}{C'E}=\\tan\\theta $, so multiplying these two ratios, we find that\n\n\\[ \\frac{DE}{C'E}=\\frac{2r\\tan\\theta}{s-c}=\\frac{2A\\tan\\theta}{s(s-c)}=\\frac{4A^2}{(s)(s-a)(s-c)(c-b)}=\\frac{4(s-b)}{c-b}. \\]\n\nSince $ FE=2\\cdot C'E $, then $ \\frac{DE}{FE}=\\left(\\frac{1}{2}\\right)\\left(\\frac{4(s-b)}{c-b}\\right)=\\frac{2(s-b)}{c-b} $. Therefore, $ \\frac{DF}{FE}=\\frac{DE}{FE}-1=\\frac{2(s-b)-(c-b)}{c-b}=\\frac{a}{c-b} $.\n\nTherefore, $ (BC)(FE)=(DF)(c-b)=(DF)((s-b)-(s-c))=(DF)(BE-CE) $, and $ DF\\cdot CE+FE\\cdot BC=BE\\cdot DF $.\n\nRearranging,\n\n\\[ 1+\\frac{FE\\cdot BC}{DF\\cdot CE}=\\frac{BE}{CE}. \\]\n\nNow, consider $ \\triangle BDE $ and the line $ GFC $ passing through it. By Menelaus, $ \\frac{DF\\cdot CE\\cdot BG}{FE\\cdot BC\\cdot GD}=1 $, so $ \\frac{FE\\cdot BC}{DF\\cdot CE}+1=\\frac{BG}{GD}+1=\\frac{BD}{GD} $, and therefore,\n\n\\[ \\frac{BE}{CE}=\\frac{BD}{GD}. \\]\n\nA mass point argument then finishes off the problem with $ \\triangle CDB $ and cevians $ DE $ and $ CG $.[/hide]", "Solution_11": "Let the ray $DN$ intersect the circle $\\omega$ with center in point $C$ and has radius $CE$, the second time in point $P$. \\\\ 1)$\\angle CNP=\\angle DNA=\\angle DEN=\\frac{\\angle FCN}{2} $ so $\\angle FNP=\\angle FNC+\\angle CNP=90^{\\circ}-\\frac{\\angle FCN}{2}+\\frac{\\angle FCN}{2}= 90^{\\circ}$\nbecause $EC=FC=NC$. And since point $C$ is the center of $\\omega$, we must have points $F,C,P$ collinear. \\\\ 2)$\\frac{2FC}{FD}=\\frac{FP}{FD}=\\frac{\\sin\\angle FDP}{\\sin \\angle FPD}=\\frac{\\sin\\angle FDP}{\\sin \\angle DEN}=\\frac{EN}{DN}$\\\\ 3) Let the incircle touch $AB$ and $AC$ at points $Z,N$ respectively. Let point $M$ be the middle of $ZE$. Then since $BD$ is the symmedian of $\\angle ZDE$, we have $\\angle ZDM=\\angle GDF=90^{\\circ}-\\angle MDE-\\angle \\frac{ABC}{2} $.Also we find that $\\angle EZD=\\angle DNA+\\angle ENC=90^{\\circ}-\\angle \\frac{FCE}{2}$.\\\\ 4)$\\angle ZMD =180^{\\circ}-\\angle ZDM-\\angle DZM=\\angle \\frac{ABC}{2}+\\angle \\frac{FCE}{2}+\\angle MDE$.\\\\$\n\\angle DZA=\\angle ZED =\\angle ZMD -\\angle MDE$. \\\\ 5)\n$\\angle DGF= \\angle FCE+\\angle GBC=\\angle FCE+\\angle ABC-\\angle ZBD=\\angle FCE+\\angle ABC-\\angle AZD+\\angle ZDB=\\angle FCE+\\angle ABC-\\angle ZMD +\\angle MDE+\\angle ZDB=\\angle \\frac{ABC}{2}+\\angle \\frac{FCE}{2}+\\angle ZDB . $. So we found that $\\angle DGF=\\angle ZMD$ and $\\angle ZDM=\\angle GDF$ . Which gives us that $\\triangle GDF\\sim\\triangle MDZ$.\\\\ 6)So $\\frac{2GF}{DF}=\\frac{2ZM}{ZD}=\\frac {ZE}{ZD}=\\frac{EN}{DN}=\\frac{2FC}{FD}$, because $AE$ is the symmedian of $\\angle ZEN$.\\\\ From (6) we find $GF=FC$ and we are done.", "Solution_12": "Projective geometry yields the following quick solution.\nLet the incircle touch $AB$ and $AC$ at $F$ and $G$ respectively. Also, let $BC \\cap GF \\equiv X$ and $AE \\cap GF \\equiv Y$.\nBy the Symmedian Lemma, $DFEG$ is harmonic. This implies that $XD=XE$ from tangency, thus $XD$ is parallel to $CF$. Therefore we can write\n\\[ -1 = (X, Y; G, F) = (X, E; C, B) = (\\infty, F; C, G) \\]\nand the conclusion follows.\n\n", "Solution_13": "[hide=Solution][quote=China TST 2008 P1]Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \\equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \\equal{} FG$.[/quote]\n[b][color=#000]Solution:[/color][/b] Let $(I)$ be tangent to $AB,AC$ at $I_B,I_C$. Let the tangent at $D$ to $(I)$ intersect $BC$ at $X$. Since, $DI_BEI_C$ is a Harmonic Quadrilateral, hence, $X$ $\\in$ $I_BI_C$. From the Statement, $\\angle CEF$ $=$ $\\angle XDE$ $=$ $\\angle CFE$ $\\implies$ $XD$ $||$ $CF$.\n\\begin{align*} -1= (I_C, BI_C \\cap (I) ; E, I_B) \\overset{I_C}{=} (C, B ; E, X) \\overset{D}{=} (C, G; F, \\infty_{CF}) \\end{align*}\nThis implies, $CF=FG$ $\\qquad \\blacksquare$\n[/hide]\n\n", "Solution_14": "Hmm easy even for a P1\n[quote=Fang-jh]Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \\equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \\equal{} FG$.[/quote]\nLet the incircle touch $AC,AB$ at $G,H$ respectively.Let $T=GH\\cap BC$.Then since $GHDE$ is harmonic hence $TD$ is tangent to the incircle.Hence $TD=TE\\implies TD\\parallel CF$.Now \\[-1=(T,E,B,C)\\overset{D}{=}(\\infty_{CF},F,G,C) \\implies CF=FG\\square\\]" } { "Tag": [ "search", "geometry", "rhombus" ], "Problem": "Prove that a line through M to bisect any side of the quadrilateral is perpendicular to the opposite side.(M is not hte centre of the circle.)", "Solution_1": "This is the Brahmagupta theorem, you can search for it.", "Solution_2": "[quote=\"Arne\"]This is the Brahmagupta theorem, you can search for it.[/quote]\r\nAre you sure I think I can do it without it.\r\n[hide=\"Solution\"]\nAs the diagonals of the quadrilateral are perpendicular to each other then the quadrialteral is a rhombus or a square.\nIt is a square in this case because in a cylic quadrilateral the sum of the opposite sides is $180^\\circ$ So, then a perpendicular to one side will be parallel to the adjacent side and hence it would be perpendicular to side opposite the bisected side.\n[/hide]", "Solution_3": "[quote=\"riddler\"]\nAre you sure I think I can do it without it.[/quote]\n\nRead my post carefully -\nI said that it IS the Brahmagupta theorem, not that you have to use it...\n\n[quote=\"riddler\"]As the diagonals of the quadrilateral are perpendicular to each other then the quadrialteral is a rhombus or a square.\n[/quote]\r\n\r\nAnd that is false.", "Solution_4": "[quote=\"Arne\"][quote=\"riddler\"]\nAre you sure I think I can do it without it.[/quote]\n\nRead my post carefully -\nI said that it IS the Brahmagupta theorem, not that you have to use it...\n\n[quote=\"riddler\"]As the diagonals of the quadrilateral are perpendicular to each other then the quadrialteral is a rhombus or a square.\n[/quote]\n\nAnd that is false.[/quote]\r\nAs for the $1$st thing sorry :blush: but as for the $2$nd thing I fail to understand why it wouldnt be a rhombus or a square.", "Solution_5": "Take the points (1, 0), (-3,0), (0, 5) and (0, -2). Do these points form a quadrilateral? Are the diagonals perpendicular? Is the quadrilateral a rhombus?", "Solution_6": "[quote=\"riddler\"]...I fail to understand why it wouldnt be a rhombus or a square.[/quote]\r\n\r\nIf you think it's true, then you have to prove it. Otherwise it remains a conjecture.", "Solution_7": "[quote=\"Arne\"]Take the points (1, 0), (-3,0), (0, 5) and (0, -2). Do these points form a quadrilateral? Are the diagonals perpendicular? Is the quadrilateral a rhombus?[/quote]\r\n :oops: :oops: :oops: :oops:", "Solution_8": "SO this question is the brhmagupta theorem....it's in our 10th grade math school text book as a question...thnx for tellin me 'll search :lol:" } { "Tag": [ "geometry", "trapezoid", "cyclic quadrilateral" ], "Problem": "1. An isosceles trapezoid has lengths of 5,5,5 and 8. What is the sum of the lengths of its diagonals? \n\n\n\nI got an answer of [hide]2 :sqrt: 65[/hide]. I believe that this is a cyclic quadrilateral with can then be used with the theorem: ab+cd=ef, (Ptolemy's?) Can anyone shed light on this?", "Solution_1": "Your answer is right, it is $2 \\sqrt{65}$. Actually, what you used is\r\n\r\n[b](a)[/b] the fact that every isosceles trapezoid is cyclic;\r\n[b](b)[/b] the Ptolemy theorem for cyclic quadrilaterals;\r\n[b](c)[/b] the fact that the diagonals of an isosceles trapezoid are equal in length.\r\n\r\nBut note that with the usual notations (ABCD cyclic quadrilateral, AB = a, BC = b, CD = c, DA = d, AC = e, BD = f), the Ptolemy theorem is ac + bd = ef, and not ab + cd = ef as you wrote.\r\n\r\n Darij", "Solution_2": "Oh, I do apologize for writing the wrong thing for Ptolemy. I did mean that the a and the b were supposed to be opposite sides, same for the c and d. Thank you for verifying the answer though." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Let $ c_{n}$ be the number of permutations $ \\sigma$ of the set $ \\{1, 2, \\ldots, n\\}$ which have the property that there do not exist $ i < j < k$ such that $ \\sigma(i) < \\sigma(k) < \\sigma(j)$. Thus, for example, $ \\sigma = 32541$ is not such a permutation because $ 2 = \\sigma(2) < 4 = \\sigma(4) < 5 = \\sigma(3)$, but $ \\pi = 34521$ is. Compute $ c_{n}$. (These are called $ 132$-avoiding permutations.)\r\n\r\nSimilarly, let $ d_{n}$ be the number of permutations $ \\sigma$ of $ \\{1, 2, \\ldots, n\\}$ which have the property that there do not exist $ i < j < k$ such that $ \\sigma(i) < \\sigma(j) < \\sigma(k)$. Thus, $ \\sigma = 32541$ is such a permutation but $ \\pi = 34521$ is not because $ 3 = \\sigma(1) < 4=\\sigma(2) < 5 = \\sigma(3)$. Compute $ d_{n}$. (These are called $ 123$-avoiding permutations.)", "Solution_1": "Maybe it helps:\r\n[url]http://www.combinatorics.org/Volume_13/PDF/v13i1r51.pdf[/url]" } { "Tag": [], "Problem": "Un tetraedru $ ABCD$ este inscris intr-o sfera de raza $ R$ si are baza $ ABC$ un triunghi echilateral.Se stie ca sfera de raza minima care contine $ ABC$ si care admite pe $ D$ ca punct interior are raza egala cu $ m\\equal{}\\frac{R}{2}$ sa se arate ca V =$ m(6m^2\\minus{}DA^2\\minus{}BD^2\\minus{}CD^2)$", "Solution_1": "Ceva nu-i in regula cu enuntul! Sfera cea mai mica care contine atat triunghiul ABC cat si punctul D,\r\ncontine tetraedrul ABCD. Insa cea mai mica sfera care contine tetraedrul ABCD este sfera circumscrisa!\r\n\r\n$ \\mbox{In tetraedrul }A_1A_2A_3A_4, \\mbox{ notand cu } a_{ij} \\equal{} |A_iA_j| \\mbox{ si cu } \\\\\r\nt \\equal{} \\frac {a_{12}a_{34} \\plus{} a_{13}a_{24} \\plus{} a_{14}a_{23}}{2}\\mbox{; avem: } \\\\\r\n6.R.V \\equal{} \\sqrt {t(t \\minus{} a_{12}a_{34})(t \\minus{} a_{13}a_{24})(t \\minus{} a_{14}a_{23})}.\\ \\ (1) \\\\\r\n\\mbox{In cazul in care triunghiul }A_1A_2A_3 \\mbox{ este echilateral, avem: }\\\\\r\na_{12} \\equal{} a_{13} \\equal{} a_{23} \\equal{} m \\mbox{, iar relatia }(1) \\mbox{ devine: } \\\\\r\n24.R.V \\equal{} m^2.\\sqrt {(a_{14} \\plus{} a_{24} \\plus{} a_{34})(a_{24} \\plus{} a_{34} \\minus{} a_{14})(a_{14} \\plus{} a_{34} \\minus{} a_{24})(a_{14} \\plus{} a_{24} \\minus{} a_{34})} \\\\\r\n\\mbox{sau: } \r\n24.R.V \\equal{} m^2.\\sqrt {2(a_{14}^2a_{24}^2 \\plus{} a_{14}^2a_{34}^2 \\plus{} a_{24}^2a_{34}^2) \\minus{} (a_{14}^4 \\plus{} a_{24}^4 \\plus{} a_{34}^4)}.$", "Solution_2": "Problema apare sub aceasta forma la ONM 1991.Intr-adevar ,la rezolvari, se specifica faptul ca o astfel de situatie e practic imposibila,insa exista o rezolvare (destul de neclara).", "Solution_3": "[quote=\"mihai miculita\"] Insa cea mai mica sfera care contine tetraedrul ABCD este sfera circumscrisa!\n\n[/quote]\r\nNu neap\u0103rat. De exemplu, cel mai mic cerc care con\u0163ine un triunghi obtuzunghic nu e cercul circumscris.", "Solution_4": "[quote=\"Adriana N.\"]Intr-adevar ,la rezolvari, se specifica faptul ca o astfel de situatie e practic imposibila,insa exista o rezolvare (destul de neclara).[/quote]\r\nDaca te referi la culegerea de la editura B\u00eerchi, afirmatia \"sfera de raza minima in general nu exista\" cred ca e gresita. Solutia data acolo pare OK." } { "Tag": [ "USAMTS", "email", "\\/closed" ], "Problem": "I faxed off my solutions in time, and it says that the results are available, but I don't have a score. Why not? How can I see what I got? I've gone to the \"My Usamts\" but it shows the - of it not being graded yet. :(", "Solution_1": "Please send an email with your full name, username, and USAMTS Id to usamts@usamts.org; if you didn't put your username or USAMTS Id on your paper, then we probably have the paper but it's in our stack of papers we can't map to id numbers." } { "Tag": [], "Problem": "994 cages each contain 2 chickens. Each day we rearrange the chickens so that the same pair of chickens are never together twice. What is the maximum number of days we can do this?", "Solution_1": "do you take guesses?\r\n\r\nthere are 994 cages, and 994 pairs of chiken. to count the maximum number of days, we need to move only two chickens a day. we also only need to move one chicken from each cage and fix the other chicken.\r\n\r\nso on the first day, we move the chicken in cage 1 to 2 and 2 to 1. on day two, we move the chicken in box 2 to 3, and 3 to 2. keeping this up, on the 993rd day, the first chicken will be at the 994th cage, and every other chicken displaced one cage down.\r\n\r\nthen we count the chicken that is now in the 1 box, moving it up to the 993th box, we have 992 days to count.\r\n\r\nthen we count the third chicken, which is in the first box, we have 991 days.\r\n\r\ndoing this on to the 994th chicken, we should have 993+992+...+2+1 days, or $d=\\frac{993(1+993)}{2}$.\r\n\r\nEDIT: oops. i think the answer should be $d=993+991+989+987+...+5+3+1=\\frac{497(1+993)}{2}=247,009.$\r\n\r\nEDIT: actually i'm not sure any more.\r\n\r\nplease tell me if i reasoned incorrectly.", "Solution_2": "[quote=\"arsemesomething\"]do you take guesses?\n\n[hide=\"be warned\"]there are 994 cages, and 994 pairs of chiken. to count the maximum number of days, we need to move only two chickens a day. we also only need to move one chicken from each cage and fix the other chicken.\n\nso on the first day, we move the chicken in cage 1 to 2 and 2 to 1. on day two, we move the chicken in box 2 to 3, and 3 to 2. keeping this up, on the 993rd day, the first chicken will be at the 994th cage, and every other chicken displaced one cage down.\n\nthen we count the chicken that is now in the 1 box, moving it up to the 993th box, we have 992 days to count.\n\nthen we count the third chicken, which is in the first box, we have 991 days.\n\ndoing this on to the 994th chicken, we should have 993+992+...+2+1 days, or $d=\\frac{993(1+993)}{2}$.[/hide]\n\nplease tell me if i reasoned incorrectly.[/quote]\r\n\r\nWell considering you only have 1988 chickens, if you consider the possible pairs the first chicken could have and there are always two chicken in a cage, the maximum possible number of days is only 1987.. :maybe: maybe I have the question wrong\r\nBut if your method did work, then I think you'd have to prove that by moving the first chicken you can achieve the maximum..so moving chickens might not be the best idea :|" } { "Tag": [ "search", "geometry", "area of a triangle" ], "Problem": "From 2001 National Sprint Round, # 25:\r\n\r\nThe lengths of the perpendiculars drawn to the sides of a regular hexagon from an interior point are 2, 4, 5, 9, 10, and 12 centimeters. What is the number of centimeters in the length of a side of this hexagon? Express your answer as a common in simplest radical form.", "Solution_1": "This problem has been posted before, I suggest you go search for it (you'll probably find a better explanation than mine), but if you're too lazy (like me), you can read my solution:\r\n\r\n[hide]Use the area of a triangle 6 times ($\\frac{bh}{2}$), so you get the area of the hexagon to be $\\frac{(2+4+5+9+10+12)s}{2}=21s$, where s is the length of a side. Then you find it using equilateral triangles (since it is a REGULAR hexagon), $\\frac{s^2\\sqrt{3}}{4}\\cdot6=\\frac{3s^2\\sqrt{3}}{2}$. Then you do \n$21s=\\frac{3s^2\\sqrt{3}}{2}$\n$s=\\frac{14\\sqrt{3}}{3}$\n[/hide]" } { "Tag": [ "Olimpiada de matematicas" ], "Problem": "Sea $T$ un conjunto de $2007$ puntos en el plano donde no hay 3 puntos colineales. Sea $P$ un punto cualquiera de $T$. Pruebe que el numero de triangulos que contienen a $P$ en su interior y cuyos vertices estan en $T$ es par.", "Solution_1": "[color=darkblue]\nSea $\\Delta_{P}$ el numero de tri\u00e1ngulos que contienen a $P$ en su interior. Ahora veamos cada grupo de cuatro puntos cualesquiera de $T$ distintos de $P$. Si un tri\u00e1ngulo tiene en su interior el punto $P$ este triangulo se encontrara en un total de 2003 conjuntos. Asi sumados de cada conjunto obtendriamos en total $2003 \\Delta_{P}$.\n\nAhora por otro lado en cada conjunto de cuatro puntos tenemos que estos formaran o bien un cuadrilatero convexo, o bien un cuadrilatero no convexo que lo podemos ver como un tri\u00e1ngulo sub dividido en tres triangulitos internos, es facil ver que cada conjunto de cuatro puntos tiene o bien 0 tri\u00e1ngulos que contienen a $P$ en su interion o 2 tri\u00e1ngulos que contienen a $P$ en su interior. Si sumamos el total de tr\u00e1ingulos en cada conjunto de cuatro puntos obtenemos un numero par, digamos $2k$. \n\nTenemos por las dos formas de contar que $2k = 2003 \\Delta_{P}$.\n\nDe donde se concluye que $\\Delta_{P}$ es par.[/color]\r\n\r\n :ninja: :pilot:" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ \\triangle ABC$ and $ \\triangle DEF$ be inversely congruent.Show that midpoint of $ AD,BE,CF$ are collinear", "Solution_1": "Very nice fact and Lemma 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=564145#564145]http://www.mathlinks.ro/Forum/viewtopic.php?t=82897 post #7[/url].\r\n\r\n Darij" } { "Tag": [ "geometry", "rhombus" ], "Problem": "what is the greatest area for a square that can fit in a rhombus with side 20\r\n\r\nexpress you answer to the nearest whole number", "Solution_1": "[hide]Area (square) =< Area (rhombus)\n\nsay an angle of the rhombus is x, the area is 400*sin(x)\n\nif x=90, rhombus is at a max, \n\nArea(square) =< 400\n\nequality does occur since if Area (rhombus)=400, x=90, the rhombus is the square... $\\boxed{400}$[/hide]", "Solution_2": "haha, that's clever :D", "Solution_3": "oops...sorry\r\n\r\ni forgot to mention that one of the angles of the rhoumbus is 50 degrees\r\n\r\n :blush:" } { "Tag": [ "calculus", "integration", "ratio", "articles" ], "Problem": "0.000100010001.... repeating \r\nis written in base sqrt(2).\r\n\r\nWhat is this number in base 10?", "Solution_1": "[hide]the number is equivalent to $\\frac{1}{(\\sqrt{2})^4}+\\frac{1}{(\\sqrt{2})^8}+\\frac{1}{(\\sqrt{2})^{12}}+\\ldots=\\frac{1}{4}+\\frac{1}{16}+\\frac{1}{64}+\\ldots=\\frac{1}{3}$[/hide]", "Solution_2": "of course drunner is correct, but the problem statement is confusing because i don't see how you can have digits for base $\\sqrt{2}$", "Solution_3": "Since 1 is less than sqrt(2), using the digit 1 seems legit, but yes weird.\r\n\r\nIn what base b>0, does 1000 = 110?\r\n\r\n(Note for bases b less than your answer, 1000<110.)", "Solution_4": "Um, how can a number equal to another number in any base? Am I missing something here?", "Solution_5": "[quote=\"mathgeniuse^ln(x)\"]Um, how can a number equal to another number in any base? Am I missing something here?[/quote]\r\n\r\nuh, $10_2=2_{10}$. A number in base 2 is equal to another number in base 10. \r\n\r\nIs that what you want?", "Solution_6": "we have $b^3 = b^2 + b$, so rewrite it to be $b(b^2 - b - 1) = 0$, since base 0 and base negative is absurd, I think the answer is $\\frac{1+\\sqrt{5}}{2}$.", "Solution_7": "Good job. BTW, yes both numbers were in the same base, b. Weird things happen for non-integral bases. :D \r\nNote that the solution for b is the Golden Ratio :) \r\n\r\nI did specify b>0, Pekerichang, but you shouldn't have mentioned the 'absurdity' of negative bases :lol: \r\n\r\nA few more challenges!\r\nUsing only 0 and 1 (and no negative signs), in base negative 2:\r\n\r\n1. Count from 1 to 10\r\n2. Count from -1 to -10\r\n (Note: in negative bases, all integers can be written WITHOUT the negative symbol!)\r\n3. Convert the base ten numbers 100 and -100 to base negative 2.\r\n4. Write one-third in base negative two.\r\n5. Write negative one-third in base negative two.\r\n\r\nHere's a blurb from Keith Devlin, a British author, about bases in general. http://www.maa.org/devlin/devlinfeb.html\r\n\r\nBTW: Keith presents an additional 'base' problem at the end of his article. No solution given and I can't solve it. Anyone?", "Solution_8": "I'll do #1:\r\n\r\n[hide]1, 110, 111, 100, 101, 11010, 11011, 11000, 11001, 11110.[/hide]\r\n\r\nWouldn't it be fun to always use negative bases? They're more versatile since you no longer need negative signs.", "Solution_9": "The notion of \"base\" is being stretched very badly here. Any reasonable definition of the word base requires that there be [i]unique[/i] representations in that base. If you want to talk about base-like representations, go ahead, but these are not bases in any conventional sense of the term.", "Solution_10": "Yes, 'useful' bases probably would require the 'extra' property of unique representation. But who would have appreciated the need for the extra property without playing around with the base-like b's? Ergo, 'weird bases' challenge us to think more carefully about exactly what a base is (or should be).\r\n\r\nAs far as I can tell, representations in negative integer bases are unique. In addition, the Devlin article mentions that some 1950's UMC-1 computers were developed and sold that were based on base negative 2. Actually, I've liked them since a high school friend of mine did a neat class presentation on them. \r\n\r\nBTW, any takers on 1/3 or -1/3? Both answers are pretty nice.", "Solution_11": "With two's complement you can do any integer without negative signs (and i think you can do fractions?).", "Solution_12": "My guess is that programmers quickly realized that twos complement was a much more practical way to \"avoid the negative sign\" than base negative 2.", "Solution_13": "I believe one of the USACO online competition program is to convert a number from base 10 to base negative 2", "Solution_14": "For -1/3 in base -2, it's $.\\overline{1}$ since $-\\frac12+\\frac14-\\frac18+\\cdots = -\\frac13$. \r\n\r\nFor 1/3 in base -2, it's $.\\overline{01}$ since $\\frac14+\\frac{1}{16}+\\cdots = \\frac13$.\r\n\r\nAlso, $100_{10} = 110100100_{-2}$ and $-100_{10} = 11101100_{-2}$." } { "Tag": [ "probability" ], "Problem": "We throw a dice. What is the probability of seem each number after $ n$ rolling?", "Solution_1": "I think this is what you're asking: \"We throw a dice. What is the probability of seeing a given number after $ n$ rolls?\r\n\r\n[hide]The probability of seeing a given number on the first roll is $ \\frac{1}{6}$, and the probability of not seeing it is $ \\frac{5}{6}$. This probability remains the same for each subsequent roll (if the number hasn't already been rolled). The answer to your question (I think) is $ 1\\minus{}(\\frac{5}{6})^n$.[/hide]", "Solution_2": "I do not mean that $ cf249$. I want to see all of 1,2,3,4,5,6 after rolling dice $ n$ times. Of course $ n>6$.", "Solution_3": "OK, then, I think this is the same as the first question under [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186425]this topic.[/url]", "Solution_4": "It's a different question as I see it; we want to know the probability that it works within the first $ n$ steps, which is essentially the cumulative distribution.\r\n\r\nIt will be complicated.\r\n\r\nThe probability of not seeing the number $ 1$ is $ \\left(\\frac56\\right)^n$. The probability of not seeing $ 1$ or $ 2$ is $ \\left(\\frac46\\right)^n$, and so on. Build an inclusion/exclusion sum; the probability of getting everything is $ \\sum_{S\\subset \\{1,\\dots,6\\}} (\\minus{}1)^{|S|}\\left(\\frac{6\\minus{}|S|}{6}\\right)^n$\r\n$ \\equal{} 1\\minus{}6\\left(\\frac56\\right)^n\\plus{}15\\left(\\frac46\\right)^n\\minus{}20\\left(\\frac36\\right)^n\\plus{}15 \\left(\\frac26\\right)^n\\minus{}6\\left(\\frac16\\right)^n\\plus{}\\left(\\frac06\\right)^n$\r\n\r\nThis works for all $ n$, as long as we interpret the last term as $ 1$ when $ n$ is zero.", "Solution_5": "Ah. I see. Nice solution, jmerry! :)" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "angle bisector", "perpendicular bisector", "geometry unsolved" ], "Problem": "Let $\\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\\Gamma$ such that $AB < \\sqrt{3}r$. Let the circle with centre $B$ and radius $AB$ meet $\\Gamma$ again at $C$. Let $P$ be the point inside $\\Gamma$ such that triangle $ABP$ is equilateral. Finally, let the line $CP$ meet $\\Gamma$ again at $Q$. \r\n\r\nProve that $PQ = r$.", "Solution_1": "Let $\\triangle A_0BP_0$ be an equilateral triangle inscribed in the circle $\\Gamma$ with the center O. Let the point A move on the minor arc $A_0B$. The equilateral triangles $\\triangle ABP \\sim \\triangle A_0BP_0$ remain similar, the vertex B is fixed and the vertex A moves on the circle $\\Gamma$. It follows that the vertex P also moves on a circle $\\Gamma'$. When $A \\longrightarrow A_0,\\ P \\longrightarrow P_0$, when $A \\longrightarrow B,\\ P \\longrightarrow B$ and when $AB \\longrightarrow r,\\ P \\longrightarrow O$. Therefore, the locus circle $\\Gamma'$ is the circumcircle of the isosceles triangle $\\triangle BOP_0$ with the angle $\\angle BOP_0 = 120^\\circ$. It follows that the circles $\\Gamma' \\cong \\Gamma$ are congruent, the center O' of $\\Gamma'$ lies on $\\Gamma$, and the angle $\\angle BOO' = 60^\\circ$. Hence, the lines $OO' \\perp BP_0$ are perpendicular and the line OO' meets the circle $\\Gamma$ again at $A_0$ and $OA_0 = r$.\r\n\r\nSince the angle $\\angle BAP = 60^\\circ$, the line AP passes through $P_0$. The angles $\\angle CAP_0 = \\angle CBP_0$ spanning the same arc $CP_0$ of $\\Gamma$ are equal. On the other hand, $\\angle CAP_0 \\equiv \\angle CAP = \\frac{\\angle CBP}{2}$, because $\\angle CBP$ is the central angle of the major arc CP of the circle (B) with radius AB and the angle $\\angle CAP$ spans this arc. Consequently, $BP_0$ bisects the angle $\\angle CBP$. Since the triangle $\\triangle CBP$ is isosceles with CB = PB, the angle bisector $BP_0 \\perp CP$ is perpendicular to its base CP. As a result, the lines $OA_0 \\equiv OO' \\parallel CP$ are parallel, both being perpendicular to $BP_0$. But then form the congruence of the circles $\\Gamma' \\cong \\Gamma$ with their centers separated by $OO' = r$, we get $PQ = OA_0 = r$.", "Solution_2": "This problem is simply and nicely. [hide=\"Here is a short proof.\"][color=darkred]Denote $2x=m(\\widehat {CBP})$ and calculate the values of the all angles from this figure. Suppose w.l.o.g. $CB=BA=AP=PB=1$. Thus, $m(\\widehat {BPC})=m(\\widehat {BCP})=90^{\\circ}-x$, $m(\\widehat {APQ})=30^{\\circ}+x$, $m(\\widehat {BAC})=m(\\widehat {BCA})=60^{\\circ}-x$, $m(\\widehat {ACQ})=m(\\widehat {PBQ})=\\boxed {m(\\widehat {ABQ})=30^{\\circ}}$, $BQ\\perp AP$, $AQ=PQ$ $\\Longrightarrow$\n$m(\\widehat {CAP})=x$, $m(\\widehat {AQB})=m(\\widehat {BQP})=60^{\\circ}-x$.\nTherefore, $1=AB=2r\\sin \\widehat {BCA}=2r\\sin (60^{\\circ}-x)\\Longrightarrow \\boxed {r=\\frac{1}{2\\cos (30^{\\circ}+x)}}\\ .$\n$\\frac{PQ}{\\sin \\widehat {PBQ}}=\\frac{PB}{\\sin \\widehat {BQP}}\\Longrightarrow \\frac{PQ}{\\sin 30^{\\circ}}=\\frac{1}{sin (60^{\\circ}-x)}\\Longrightarrow \\boxed {PQ=\\frac{1}{2\\cos (30^{\\circ}+x)}}\\ .$\nAt last, we draw the conclusion $\\boxed {\\ PQ=r\\ }\\ .$[/color][/hide]", "Solution_3": "Since $ABP$ is equilateral, $P$ lies on the circle with centre $B$ and radius $AB$.\nBy the Inscribed Angle Theorem, $\\angle ABQ=\\angle ACP=\\frac{\\angle ABP}{2}=30^{\\circ}$, which means $BQ$ is the perpendicular bisector of $AP$.\nBy the Sine's Law, $PQ=AQ=2r\\sin\\angle ABQ=2r\\sin30^{\\circ}=r$, QED." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all integers satisfying $y(x-y)(x+2y)\\neq0$ and $(x-y)^{4}+4y^{2}(x+2y)^{2}=z^{2}$.", "Solution_1": "You should be able to get all solutions by writing $z=r^{2}+s^{2}$, $r^{2}-s^{2}= (x-y)^{2}$, and $rs = y(x+2y)$ for some positive integers $r\\ne s$." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $ u_1,u_2,...,u_n(n>2)$ be a sequence of positive real numbers such that\r\n\r\n1)$ \\frac{1004}{k}\\equal{}u_1\\ge u_2\\ge ....\\ge u_n$ for some positive integer $ k$\r\n2) $ u_1\\plus{}u_2\\plus{}...\\plus{}u_n\\equal{}2008$\r\nProve that it is possible to select $ k$ elements from the set $ (u_n)$ such that in this collection of $ k$ numbers, the smallest one is at least half of the largest", "Solution_1": "Not really hard,if I got correctly.\r\nassume the contrary at first,afterwards note that the sum $ 1004(1\\plus{}\\frac{1}{2}\\plus{}\\frac{1}{4}\\plus{}\\dots)$ is less than $ 2008$.\r\nthe last step is to divide all these $ n$ numbers into the groups,each with $ k\\minus{}1$ elements,(the last one may contain less number of elements).\r\nI can elaborate it,if necessary." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Show that :\r\n$ V$ is simple left $ R$-modul if and only if exist $ M$ maximal left ideal of $ R$ and $ V\\simeq R\\backslash M$", "Solution_1": "Which direction are you having trouble with?", "Solution_2": "I have poblem with this $ \\Rightarrow$", "Solution_3": "Construct a nonzero map $ R\\to V$. Why is it surjective? If $ V\\cong R/I$, why must $ I$ be maximal?" } { "Tag": [ "geometry", "incenter", "geometry solved" ], "Problem": "Let $BC$ meet perpendicular line from $A$ on $BC$ at $D$ in triangle $ABC$. Let $E$ and $F$ be incenters of triangles $ABD$ and $ACD$ respectively. Let $EF$ meet $AB$ and $AC$ at $K$ and $L$ respectively. Prove that $AK=AL$ iff $AB=AC$ or $\\angle BAC = 90$.", "Solution_1": "see http://www.mathlinks.ro/Forum/viewtopic.php?t=78401 .\r\n\r\n darij" } { "Tag": [ "inequalities", "factorial" ], "Problem": "So I got inspired from Poland Inequality problem so I decided to write one up. It follows from exactly same method as the original problem so it's probably very easy but I think proofs shouldn't be at Intermediate so I'm posting here. :D \r\n\r\nFor real numbers $ x,y,z,w$, prove that:\r\n\r\n\\[ (x\\plus{}y\\plus{}z\\plus{}w)^{2!} \\plus{} 2! w^{\\frac{3!}{2!}} \\leq 2! \\left(x^{2!} \\plus{} y^{2!} \\plus{} z^{2!} \\plus{} w^{2!}\\right) \\plus{} \\frac{4!}{3!} \\left((x\\plus{}1)(y\\plus{}1)(z\\plus{}1) \\minus{} (x\\plus{}y\\plus{}z\\plus{}xyz\\plus{}1!) \\plus{} \\frac{w^{\\frac{3!}{2!}}}{2!}\\right)\\]", "Solution_1": "What about $ x\\equal{}y\\equal{}z\\equal{}0$, $ w \\rightarrow \\infty$?\r\n\r\nThen it reduces to $ w^2 \\plus{} 2w^3 \\le 2w^2 \\plus{} \\frac{2w^3}{3}$, and $ w^3$ clearly grows faster...\r\n\r\n(also, what is up with the factorials?)", "Solution_2": "Yeah obviously I haven't realized that condition (this is my FIRST one) so ignore that...\r\n\r\nAnd there shouldn't be $ \\frac{2w^3}{3}$ on RHS. $ 4! \\equal{} 4*3*2*1$ and $ 3! \\equal{} 3*2*1$ so $ 4!/3! \\equal{} 4$ and you should end up with $ 2w^3$ on RHS as well (which cancels LHS's one). \r\n\r\nFactorials are nothing. I just added them at the last minute to make problem more fancy. :rotfl:", "Solution_3": "[quote=\"Silverfalcon\"]Factorials are nothing. I just added them at the last minute to make problem more fancy. :rotfl:[/quote]\r\nYeah, it looks quite silly with all the factorials :D\r\n\r\n[hide=\"As for the problem,\"]after simple expanding it reduces to $ (x\\plus{}y\\plus{}z\\minus{}w)^2\\geq 0$.[/hide]", "Solution_4": ":D haha.\r\n\r\n[hide]\nMy intended solution was (by Cauchy)\n\n$ ((x \\plus{} y \\plus{} z) \\plus{} w)^2 \\leq 2((x \\plus{} y \\plus{} z)^2 \\plus{} w^2)$ [/hide]" } { "Tag": [ "pigeonhole principle", "algebra open", "algebra" ], "Problem": "guys try to solve\r\n\r\nIf n is odd and a_1; a_2;.......... ; a_n are the numbers 1; \r\n2;...............; n arranged in some order, prove that the product \r\n(a_1 - 1)(a_2 - 2)...........(a_n - n) must always be even. :?:", "Solution_1": "Assume that all $ a_i\\minus{}i$ are odd numbers. We have $ \\sum(a_i\\minus{}i)\\equal{}0$. But $ \\sum(a_i\\minus{}i)$ is odd (it follows from n-odd). Contradiction.", "Solution_2": "use the pigeonhole principle.\r\n\r\nin the numbers 1,..., n, if n is odd, there will be (n+1)/2 odd numbers and (n-1)/2 even numbers.\r\n\r\namong the 2n numbers a1, a2, ...., an and 1, 2, 3, ..., n, there are n+1 odd numbers. but there are only n places in which to fill them. therefore, one of the factors will contain two odd numbers, whose difference is even.\r\n\r\ntherefore, the product is even." } { "Tag": [], "Problem": "In the diagram, if points $ A$, $ B$ and $ C$ are points of tangency, then $ x$ equals:\r\n[asy]unitsize(5cm);\ndefaultpen(linewidth(.8pt)+fontsize(8pt));\ndotfactor=3;\n\npair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);\npair O=(0,3/8);\n\ndraw((-2/3,9/16)--(2/3,9/16));\ndraw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));\ndraw(Circle(O,3/16));\ndraw((-2/3,0)--(2/3,0));\n\nlabel(\"$A$\",A,SW);\nlabel(\"$B$\",B,SE);\nlabel(\"$C$\",C,N);\nlabel(\"$\\frac{3}{8}$\",O);\ndraw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));\ndraw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));\nlabel(\"$\\frac{1}{2}$\",(.5,.25));\ndraw((.5,.33)--(.5,.5),EndArrow(3));\ndraw((.5,.17)--(.5,0),EndArrow(3));\nlabel(\"$x$\",midpoint((.5,.5)--(.5,9/16)));\ndraw((.5,5/8)--(.5,9/16),EndArrow(3));\nlabel(\"$60^{\\circ}$\",(0.01,0.12));\ndot(A);\ndot(B);\ndot(C);[/asy]$ \\textbf{(A)}\\ \\frac {3}{16}\" \\qquad \\textbf{(B)}\\ \\frac {1}{8}\" \\qquad \\textbf{(C)}\\ \\frac {1}{32}\" \\qquad \\textbf{(D)}\\ \\frac {3}{32}\" \\qquad \\textbf{(E)}\\ \\frac {1}{16}\"$", "Solution_1": "[hide=\"Click for solution\"]\nThe distance from the center of the circle to the intersection point of the tangents is $ \\frac{3}{8}$, so $ CD\\equal{}\\frac{3}{8}\\plus{}\\frac{3}{16}\\equal{}\\frac{9}{16}\\equal{}x\\plus{}\\frac{1}{2} \\implies x\\equal{}\\frac{1}{16}$ or $ \\boxed{\\textbf{(E)}}$.\n[/hide]" } { "Tag": [ "ARML", "analytic geometry", "HCSSiM" ], "Problem": "So who will be there?\r\n\r\nI'll be there for the South Carolina team.", "Solution_1": "I will be there as a captain of New York City A team.", "Solution_2": "I will be there with the TJ A team.", "Solution_3": "In looking to see what ARML's up to this year, I discovered this little gem in Penn State's application. Great idea, not so great implementation.\r\n\r\nHint: Any time someone feels like defending themselves after one sentence into the description of the event, there's a problem. \r\n\r\nConvoluted description of how a song ends up winning this competition? CHECK. \r\n\r\nBizarre, inexplicable 30-second song length requirement? CHECK. \r\n\r\nNo prizes at all to get people to actually show up to this thing? CHECK. \r\n\r\nI could launch into a diatribe about how this is the type of thing you get when schools simply participate in the same competitions year after year [i]without even thinking about it[/i], or taking time to ask \"is this something we really ought to be spending money on?\" But I'll save that for another day...ARML needs to stop reading from Mu Alpha Theta's [u]How Not To Run A National Convention[/u] playbook, though. \r\n\r\n\r\n[b]ARML Song Contest [/b]\r\nAt Penn State University \r\n\r\nTeams are encouraged to write a song about the ARML competition, their trip to Penn \r\nState, their ARML team, or math in general. This is a great team building activity for \r\nteams to work on during their trip to Penn State or during the day on Friday. \r\nA sign up sheet will be posted outside the registration booth in Findlay Commons and \r\nteams are encouraged to register for the competition upon arrival. All teams wishing to participate must register by 7:00 pm Friday evening. \r\n\r\nThe preliminaries will be held on Friday evening in Findlay Commons at 9:35 pm. The \r\naudience, along with ARML judges, will select five teams who will perform at the main \r\ncompetition on Saturday after the Super Relay. The winning song will be selected by \r\naudience applause. \r\n\r\nTeams are allowed to have as few, or as many performers as they want (1 or 2 up to 15), but each team is limited to one entry. Alternates may compete on the team that they came to Penn State with. \r\nSongs will be judged based on originality, creativeness, and overall presentation. Songs must be limited to thirty seconds and any entries with inappropriate content or dress will be disqualified.", "Solution_4": "I don't know if you're aware, but the song contest is about 4 years (maybe 5?) old, and is solely for the purpose of filling time to allow them to finish grading power, coordinate among the various sites, get awards ready, etc. The final voting is audience applause. As there are no prizes, there is in fact 0 cost, and yet people do participate. Your message is the equivalent of writing a nasty letter about the ultimate frisbee games that take place, or something -- it's just there for fun to fill up some time.", "Solution_5": "If this is a tradition, as well as something that serves a functional purpose within their tournament schedule, so be it.\r\n\r\nI don't have any problem with people having fun, either. \r\n\r\n\r\n\r\n[quote]it's just there for fun to fill up some time.[/quote]\r\n\r\n\r\nInstead of just \"filling time,\" and occupying a captive audience, I wish math competitions (not just ARML, by any means, as its a great event) would MARKET themselves toward new groups of students and new schools, not just the same ones year after year. [b]Putting a little bit more thought - [u]and a lot more incentive[/u] - behind participation in the recreational events that go on at math competitions would be a tremendous step forward in this regard.[/b]", "Solution_6": "... :? \r\n\r\nbut anyway, getting away from this diatribe against (of all things) the song contest, North Carolina has not chosen it's ARML team yet (the contest will be this thursday), but hopefully I will be there.", "Solution_7": "I'll be there for South Carolina.", "Solution_8": "I'll also be there for south carolina. but after my bad performance at practice, it looks like i'll be on the B team. :?", "Solution_9": "I'll be there for Vermont. :D \r\n\r\nMaybe I can go for a 7 or 8 this year-that would be nice. question-if you're on a team that qualifies for the A division, how many years do you have to stay in that division?", "Solution_10": "I'll be there for Georgia, definitely on the B team.\r\n\r\nEDIT: or alternate ... forgot they did those :(", "Solution_11": "mkkool: 2 years i think, but i could be making that up.", "Solution_12": "North Carolina A team... I wonder how many other 8th graders are going? And yes, mysticterminator is on there again (getting top two as usual).", "Solution_13": "Im going to be there representing the NYC A Team...\r\n Stan", "Solution_14": "I'll be there, giving the Friday night talk...", "Solution_15": "are we meeting or something?", "Solution_16": "Alabama team represent", "Solution_17": "NYC A\r\n\r\n[quote=\"Bictor717\"]If you expect me to be at all alive, you'll be sorely disappointed because my prom is on Thursday night-Friday morning.[/quote]\r\n\r\ndo you really want to miss out on the manhattan bus antics? :D", "Solution_18": "Hi. I'm coming, as a coach. For NYC C. That's right, I found this place too.\r\n\r\nThere is too much to miss on the Manhattan bus. You know that I will try to get so many games running...\r\n\r\nHmm...Sarah-Marie is running the Friday talk this year...I doubt if she remembers me though. HCSSiM 2001...and I was too little then.", "Solution_19": "So much to do! I can't not sleep for 2 days and expect to do well on the third, after roughly 5-6 hours of sleep. I love Manhattan bus antics, but I also don't want to miss Sarah-marie's talk. I was in HCSSiM 2003. Will there be a training session as well?\r\nAnd we'll meet fellow AoPSers, right? Just to say hello.\r\nI am so looking forward to this. Must go pack now.", "Solution_20": "i don't know if i want to meet any of you. i might have to put a mask on to conceal my identity. :ninja:", "Solution_21": "Bictor717 and other HCSSiM alumns, you've likely seen this talk before: it's \"Seventeen plus Seven Flavors of Wallpaper.\" (I think we've done it as a prime time almost every year since 2000.) You may recall it as the one involving some jumping around.\r\n\r\n[quote=\"Midnight\"]\nHmm...Sarah-Marie is running the Friday talk this year...I doubt if she remembers me though. HCSSiM 2001...and I was too little then.[/quote]\r\n\r\nI remember everyone from 2001 (though not all the last names). Given the number of 2001 alumns who are not in NYC, you're at most one of 11 people. If by \"little\" you meant that you were one of the younger students (rather than one of the shorter students) then you're likely to be one of three people. Of course, you might look different now---most people change quite a bit in four years of teenage-ness---so maybe I won't recognize you! Be sure to say hi to me.", "Solution_22": "quick question:\r\ndoes anyone know the answer to number 8 team round? (the parametric equations one)\r\ni got 0 as an answer 5 seconds after time was up.\r\nIs that right?", "Solution_23": "[quote=\"qweretyq\"]quick question:\ndoes anyone know the answer to number 8 team round? (the parametric equations one)\ni got 0 as an answer 5 seconds after time was up.\nIs that right?[/quote] No- it's 100. Look at a=b.", "Solution_24": "hey ARML was pretty fun", "Solution_25": "Congratulations Arnav!\r\n\r\nGreat job.\r\n\r\nSame to all the top scorers!\r\n\r\nEdit: And to all the people on our team, including hmmm, who scored really high!\r\n\r\nGo Lehigh Fire!\r\n\r\nEdit 2: That guy at the podium! OUCH. He screwed up everybody's name - Wow...", "Solution_26": "yea I know that was funny I was like hahaha that's funny\r\n\r\nw00t I actually know some people on lehigh fire go lehigh fire!!!", "Solution_27": "Boom,\n19 year old bump.\nTAKE THAT!", "Solution_28": "LOL WHAT\nNAH NO WAY\nAnd im doing arml next year btw", "Solution_29": "yes way, it happened.\nI did the impossible\n19 year, 6 months, and 12 day bump.\nI think... if im doing the math right.\n" } { "Tag": [ "ratio" ], "Problem": "The inside wheels of a car traveling on a circular path are rotating half as fast as the outside wheels. The front two wheels are six feet apart. What is the number of feet in the path traced by the inside front wheel in one trip around the circle?", "Solution_1": "[hide=\"long way\"]If $x$ is the distance of the inner wheel from the center of the circle, then $x+6$ is the distance of the outer wheel from the center. For the inner wheel to be traveling at $\\frac{1}{2}$ the speed of the outer wheel, the circumferences need to be in a ratio of $1: 2$ (inner to outer). Then you set $2*\\pi*(x+6)$ equal to $2*(2*\\pi*x)$\n$2\\pi(x+6)=4x\\pi$\n$x\\pi+6\\pi=2x\\pi$\nDivide by pi: $x+6=2x$\n$x=6$ and the circumference of the inner cirle is $12\\pi$[/hide]\n[hide=\"short way\"]The distances from the center of the circle are $x$ and $2x$ or $(x+6)$ so then $2x=x+6$ and $x=6$\ncircumference is $12\\pi$[/hide]", "Solution_2": "wait, since when is there an, \"Inside,\" wheel for a car?\r\n\r\n-jorian", "Solution_3": "[quote=\"jhredsox\"]wait, since when is there an, \"Inside,\" wheel for a car?\n\n-jorian[/quote]\r\nInside is referring to the wheel that is closer to the center of the circular path.", "Solution_4": "Oh, I think I get it now!\r\n\r\nI stink at figuring out stuff like this, and have attempted the problem, but I got it wrong.\r\n\r\n-jorian" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Calculate \r\n\\[ \\exp{\r\n\\begin{bmatrix}\r\nt&1&\\cdots&0\\\\ \r\n&t&1&0\\\\ \r\n&0 & \\ddots &1\\\\ \r\n& & &t\\\\ \r\n\\end{bmatrix}}, ( t\\in R)\\]\r\nHelp me!", "Solution_1": "Let's start by letting $ B\\equal{}\r\n\\begin{bmatrix} 0 & 1 & \\cdots & 0 \\\\\r\n& 0 & 1 & 0 \\\\\r\n& 0 & \\ddots & 1 \\\\\r\n& & &0 \\\\\r\n\\end{bmatrix}.$\r\n\r\nIts powers can be explicitly computed (it is nilpotent) and put into the definition of the exponential. The result is\r\n\r\n$ e^B\\equal{}\\begin{bmatrix} 1 & 1 &\\frac12&\\cdots & \\frac1{(n\\minus{}1)!} \\\\\r\n0 & 1 & 1 &\\cdots &\\frac1{(n\\minus{}2)!} \\\\\r\n0 & 0 &1 &\\cdots & \\frac1{(n\\minus{}3)!} \\\\\r\n\\vdots &\\vdots&\\vdots&\\ddots&\\vdots\\\\\r\n0&0 & 0&\\cdots&1 \\\\\r\n\\end{bmatrix}.$\r\n\r\nThe matrix whose exponential we want is $ A\\equal{}tI\\plus{}B.$ Since $ tI$ and $ B$ commute, the result is $ e^A\\equal{}e^{tI}e^B\\equal{}e^{t}Ie^B\\equal{}e^te^B$ where $ e^B$ is computed as above." } { "Tag": [ "inequalities", "algebra" ], "Problem": "Let $a,b,c$ be positive numbers such that $a+b+c \\geq \\dfrac 1a + \\dfrac 1b + \\dfrac 1c$. Prove that\r\n\\[ a+b+c \\geq \\frac 3{abc}. \\]", "Solution_1": "[hide]\nAM-HM:\n$\\sqrt[3]{abc} \\geq \\frac{3}{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}} $\n\n$\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq \\frac{3}{\\sqrt[3]{abc}} \\geq \\frac{3}{abc}$\n\nFrom the question:\n\n$a+b+c \\geq \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq \\frac{3}{abc}$\n[/hide]", "Solution_2": "[hide=\"Not complete\"]\nWhy should $\\frac{3}{\\sqrt[3]{abc}} \\geq \\frac{3}{abc}$ ? \n[/hide]", "Solution_3": "[hide]\nIf a,b,c are positive, then:\n\n$abc \\geq \\sqrt[3]{abc}$\n\n$1 \\geq \\frac{\\sqrt[3]{abc}}{abc}$\n\n$\\frac{1}{\\sqrt[3]{abc}} \\geq \\frac{1}{abc}$\n\nIs that correct?\n[/hide]", "Solution_4": "[quote=\"devious_\"][hide]\nIf a,b,c are positive, then:\n\n$abc \\geq \\sqrt[3]{abc}$\n\n$1 \\geq \\frac{\\sqrt[3]{abc}}{abc}$\n\n$\\frac{1}{\\sqrt[3]{abc}} \\geq \\frac{1}{abc}$\n\nIs that correct?\n[/hide][/quote]No. Same mistake.", "Solution_5": "\\[\r\nabc(a + b + c) \\geqslant bc + ca + ab \\geqslant \\sqrt {3abc(a + b + c)} \r\n\\]\r\nso,\r\n\\[\r\nabc(a + b + c) \\geqslant 3\r\n\\]", "Solution_6": "siuhochung, can you please explain how you arrived at $\\sqrt {3abc(a + b + c)}$?", "Solution_7": "[quote=\"devious_\"]siuhochung, can you please explain how you arrived at $\\sqrt {3abc(a + b + c)}$?[/quote]\r\nIt is just $(x+y+z)^2\\geq 3(xy+yz+zx)$.", "Solution_8": "I was just coming to edit my post and say that I figured it out, then I noticed two replies. :D", "Solution_9": "I don't know whether it has a name, but you see that it is $x^2+y^2+z^2\\geq xy+yz+zx$, which is very well known and widely used.", "Solution_10": "Its most ancient ancestor is called the trivial inequality. :lol:", "Solution_11": "[quote=Valentin Vornicu]Let $a,b,c$ be positive numbers such that $a+b+c \\geq \\dfrac 1a + \\dfrac 1b + \\dfrac 1c$. Prove that\n\\[ a+b+c \\geq \\frac 3{abc}. \\][/quote]\n\neasy!\n\nmy solution = \n\nthe given condition is $abc(a+b+c)\\ge ab+bc+ca$\n\nbut $(ab+bc+ca)^2\\ge 3abc(a+b+c)\\ge 3(ab+bc+ca)\\Longrightarrow ab+bc+ca\\ge 3\\Longrightarrow abc(a+b+c)\\ge 3$\n\nso we are done :D", "Solution_12": "Let $a,b,c$ be positive real numbers such that: $abc(a+b+c) =3.$ Prove that\n$$a+b+c \\geq \\frac{3}{ab+bc+ca}+\\frac{2}{abc}$$\n$$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{3}{a+b+c}+\\frac{2}{abc}$$\nLet $a,b,c$ be positive real numbers, such that: $a+b+c \\geq \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}.$ [url=https://artofproblemsolving.com/community/c6h148342p838560]Prove that[/url]\n$$a+b+c \\geq \\frac{3}{a+b+c}+\\frac{2}{abc}$$\n[url]https://artofproblemsolving.com/community/c6h1490860p8749593[/url]\n[quote=Valentin Vornicu]Let $a,b,c$ be positive numbers such that $a+b+c \\geq \\dfrac 1a + \\dfrac 1b + \\dfrac 1c$. Prove that\n\\[ a+b+c \\geq \\frac 3{abc}. \\][/quote]\n[url]https://artofproblemsolving.com/community/c6h1184601p5748463[/url]" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Does anybody know where to find results from USAMOs before 2002? The 2002-2005 results are on the AMC website.", "Solution_1": "Here is to 98: http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/ .", "Solution_2": "I meant statistics (_% of people got _, ...); I guess I wasn't very clear." } { "Tag": [], "Problem": "Solve for $x$: $\\sqrt{x+\\sqrt{x+74}}+\\sqrt{x-\\sqrt{x+74}}=10$.", "Solution_1": "[hide]$\\sqrt {x+\\sqrt {x + 74}} + \\sqrt {x - \\sqrt {x + 74}} = 10$\n$2x + 2 \\sqrt {x^2 - x - 74} = 100$\n$x + \\sqrt {x^2 - x - 74} = 50$\n$x^2 - x - 74 = 2500 - 100x + x^2$\n$99x = 2574$\n$x = 26$\n[/hide]", "Solution_2": "[hide]$\\sqrt{x+\\sqrt{x+74}}+\\sqrt{x-\\sqrt{x+74}}=10$\nSquaring both sides, we get $x+\\sqrt{x+74}+2\\sqrt{(x+\\sqrt{x+74})(x-\\sqrt{x+74})}+x-\\sqrt{x+74}=100$. Simplifying, we get $x+\\sqrt{x^{2}-x-74}=50$. Rearranging and then squaring both sides, we see $x^{2}-x-74=2500-100x+x^{2}$, so $99x=2574$ and $x=26$.[/hide]" } { "Tag": [], "Problem": "A student on a frictionless skateboard at rest has a box of 20 baseballs. Each ball has a mass of 175 grams. She has a mass of 49 kg. The box is 100 grams and the skateboard is 900 grams. \r\n\r\n1. first she threw all of the baseballs horizontally at once away from her from the box. She held on to the box. They were going 33m/s with respect to her, not the ground. How fast was her recoil speed across the ground? How fast were the baseballs going relative to the ground? \r\n\r\n2. Next, she threw the 20 baseballs one at a time. Each time they traveled away from her horizontally 33m/s with respect to her. How fast was her recoil speed across the ground after the last ball was thrown?", "Solution_1": "It's conservation of momentum in inertial and non-inertial frames.", "Solution_2": "[quote=\"climskywell\"]A student on a frictionless skateboard at rest has a box of 20 baseballs. Each ball has a mass of 175 grams. She has a mass of 49 kg. The box is 100 grams and the skateboard is 900 grams. \n\n1. first she threw all of the baseballs horizontally at once away from her from the box. She held on to the box. They were going 33m/s with respect to her, not the ground. How fast was her recoil speed across the ground? How fast were the baseballs going relative to the ground? \n\n2. Next, she threw the 20 baseballs one at a time. Each time they traveled away from her horizontally 33m/s with respect to her. How fast was her recoil speed across the ground after the last ball was thrown?[/quote]\r\n\r\nMomentum conservation as ^ said.\r\n\r\nLet the girl's absolute velocity be 'v'. abs velocity of baseball is 33-v (moving in opp dir)\r\n(20x175)x(33-v) = (49000+100+900)xv\r\n\r\nsolve for v.\r\n\r\n2. first ball:\r\n\r\n(175)x(33-v)=(m-175)xv\r\nv1=(175)x33\r\n -----------\r\n m\r\n\r\nagain:\r\n\r\n(m-175)xv1=(m-350)xv2 -175(33-v2)\r\n\r\nsolve. It follows a geometric progression. get the general form and substitute for n=20/\r\n\r\n \r\n \r\n\r\nP.S: Does this even belong in advanced physics?", "Solution_3": "Probably not." } { "Tag": [ "LaTeX" ], "Problem": "I just recently downloaded both Latex programs and I'm trying to type something for my math assignment but I seem to be caught in a lot of problems. Can someone help me take a look at this? \r\n\r\nThanks\r\n\r\n[code]2. Determine the length of each side, ie. AB , BC , AC \n\nAB = $\\sqrt{{(2-8)}^2+{(5+1)}^2}=\\sqrt{36+36}=\\sqrt{72}=6\\sqrt{2}\n\nBC = $\\sqrt{{(-4-8)}^2+{(3+1)}^2}=\\sqrt{144+16}=\\sqrt{160}=4\\sqrt{10}\n\nAC = $\\sqrt{{(2+4)}^2+{(5-3)}^2}=\\sqrt{36+4}=\\sqrt{40}=2\\sqrt{10}\n\n3a. The triangle is scalene because all the lengths are different\nb. The triangle is obtus because it satisfies ${c^2+b^24$ , we have\r\n\r\n$5^4(4+5^{n-4}$\r\n\r\nLet $4+5^{n-4}=z^2$\r\n\r\n$\\implies 5^{n-4}=(z+2)(z-2)$\r\n\r\nSince they diff by 4 , it must be $n-4=1$ or $n=5$ . \r\n\r\nFor $n\\leq 4$ , it should be easy to check that there are no solution ." } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $ABC$ be a triangle and $w$ it's circumcircle. Denote $AD$ the bisector of $\\angle{BAC}$, where $D \\in (BC)$, and consider the circles $k_{1}$ tangent to $w, BD,AD$ and $k_{2}$ tangent to $w,CD,AD$. Prove that $k_{1}$ and $k_{2}$ are tangent in $I$, where $I$ is the incircle of $\\triangle{ABC}$.\r\n\r\n[hide=\"Remark\"]See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=146656[/url][/hide]", "Solution_1": "Let $E$ the intersection of $AD$ and $w$. We know that $EB = EC = EI$.\r\nDo one inversion $(E, EB^{2}= EC^{2}= EI^{2})$.\r\nThe circle w is mapped to the line $BC$, because $B = B'$ and $C = C'$ (and, obviously, $BC$ is mapped to $w$)\r\nSee that the line $AD$ is fixed, because $E\\in AD$. So, $k_{1}'$ will be tangent to $B'C'$, $w'$ and $(AD)'$, and this implies that $k_{1}' = k_{1}$. \r\nSo, The intersection $X$ of $k_{1}$ and $AD$ will be fixed, and $EX^{2}= EI^{2}\\Longrightarrow X = I$.\r\n\r\n(The same for $k_{2}$)", "Solution_2": "See, for example, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=307]here[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46342]here[/url]. Check Darij's post!", "Solution_3": "Oh, thanks. I knew that it is a nice particulary case of the Thebault theorem, but I didn't know that something like that was given in a contest before; I was just generalizing the problem proposed by stergiu here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=146656[/url]" } { "Tag": [ "probability" ], "Problem": "What is the probability that a positive integer less than or equal to 24 is a factor of 24? Express your answer as a common fraction.", "Solution_1": "The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24, making a total of 8 out of 24 numbers, or a probability of 1/3." } { "Tag": [ "calculus", "inequalities", "derivative", "algebra", "function", "domain", "parameterization" ], "Problem": "Find the maximum value of the expression \r\n$A = 13\\sqrt{x^2-x^4}+9\\sqrt{x^2+x^4}$, where $0\\leq x\\leq 1$.", "Solution_1": "this is a hard one....i dunno i think it might be 162, ie when $x=1$ ...lol...like i simplified the expresion to:\r\n\r\n$A^2 = 250x^2 - 88x^4 + 234\\sqrt{x^4 - x^16}$", "Solution_2": "Using Calculus, I have found that [i][b]maxiumum of A = 16[/b][/i] when $x = \\frac 4 5$.\r\nBut I want to know if there is any other way to find the maximum without using Calculus.\r\n\r\nEdit: maximum of A is attained when $x^2=\\frac 4 5$.", "Solution_3": "I think it can be solved by BCS .See here in posts #52 and #54 for a very similar problem\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=352548#p352548", "Solution_4": "Sorry, I don't understand the language in the link provided.\r\nDo you mind explaining the detail in English? :lol:", "Solution_5": "In the post #54 it says <>\r\n\r\nAt the post #52 the problem tells <>", "Solution_6": "I suppose BSC = Cauchy-Schwarz inequality ?\r\nI tried it but the result doesn't follow. :P", "Solution_7": "You should find numbers $k,m$ such that the root of the equation \r\n\r\n$\\frac{m}{\\sqrt{x^2-x^4}}=\\frac{k}{\\sqrt{x^2+x^4}}$ has root $x=\\frac{4}{5}$", "Solution_8": "Sorry ! Why root $x=\\frac 45$ ? At first you find this root with calculus ... and then You painted proof !", "Solution_9": "luimichael can you post your solution using calculus?\r\nhow did you find the roots of the derivative?", "Solution_10": "First of all, I have to apologize for telling an incorrect value of x for the maximum of A.\r\nIn fact, A is maxiumum when $x^2=\\frac 4 5$.\r\nHere is my calculations.\r\nLet $t=x^2$.\r\nThen $A=13\\sqrt{t-t^2}+9\\sqrt{t+t^2}$, where $0\\leq t\\leq 1$.\r\n$\\frac{dA}{dt}=\\frac{13(1-2t)}{2\\sqrt{t-t^2}}+\\frac{9(1+2t)}{2\\sqrt{t+t^2}}$.\r\n$\\frac{dA}{dt}=0$ if and only if $\\frac{13(2t-1)}{2\\sqrt{t-t^2}}=\\frac{9(1+2t)}{2\\sqrt{t+t^2}}$----(*)\r\n.......\r\n$2t(500t^3-375t+44)=0$\r\n$t=0, \\frac 4 5, -\\frac{3\\sqrt{3}+4}{10}, \\frac{3\\sqrt{3}-4}{10}$.\r\nt = 0 corresponds to A = 0 , which is clearly the absolute minimum of A.\r\nThe third root is out of range.\r\nWhile the fouth root does not satisfy (*).\r\nChecking second derivative, $t=\\frac 4 5$ corresponds to a maxiumum.\r\nAs there is only one turning point in the given domain and this value corresponds to a maxiumum, $t=\\frac 4 5$ corresponds to the absolute maximum of A.\r\nWhen $t= \\frac 4 5$, $A = 16$.\r\n[b]\nAnd, Virgil, I agree with your comment[/b] :!:", "Solution_11": "Aren't you missing $\\sqrt{t-t^2}+\\sqrt{t+t^2}$ in the derivative of A?\r\nI mean by the product rule, the derivative of $13\\sqrt{t-t^2}$ is $\\sqrt{t-t^2}+\\frac{13(1-2t)}{2\\sqrt{t-t^2}}$\r\nAm I wrong?", "Solution_12": "[quote=\"manuel\"]Am I wrong?[/quote]\r\n\r\nYes. You're taking the derivative of $t\\sqrt{t - t^2}$ or something.", "Solution_13": "oops, true. the derivative of 13 is 0 not 1. sorry :oops:", "Solution_14": "This problem reminds me of why in the United States graphing calculators are recommended. With a graphing calculator, you can find the maximum in one minute.", "Solution_15": "Cautiong: Just an outline.\r\nI don't like using x as a parameter, and since the problem seeks the max of A and does not care for x, we change:\r\n$0 \\le x \\le 1, x^2 = t \\Rightarrow 0 \\le t \\le 1$\r\n$A = 13\\sqrt{t - t^2} + 9\\sqrt{t + t^2}$\r\nWe have x and y parametrically defined:\r\n$x = \\sqrt{t - t^2}$ and $y = \\sqrt{t + t^2}$\r\nThese give two equations (both of which must be simult. true):\r\n$x^2 + y^2 = 2t$ and $y^2 - x^2 = 2t^2$\r\n$\\Rightarrow 0 = 2(0) \\le x^2 + y^2 \\le 2(1) = 2$\r\n\r\nWe also have a family of vertical hyperbolas. We can find our maximum from the points that lie in the circle with radius $\\sqrt{2}$ and on on of the hyperbolas. We have the line $13x + 9y = k$ for some $k$. The greater the y-intercept, the greater the $k$. So, we know if the line is tangent to the outer circle, we will have our max, except the tangency point may not lie on one of the hyperbolas. If it doesn't, we must find the closest points that do. Gotta run." } { "Tag": [ "inequalities", "integration", "function", "limit", "logarithms", "geometry", "real analysis" ], "Problem": "For natural number $ n$, prove that\r\n\r\n$ \\sum_{i\\equal{}1}^{n}\\frac{1}{\\sqrt{n^2\\plus{}i^2}} < \\ln(1\\plus{}\\sqrt{2})\\minus{}\\frac{1}{2(2\\plus{}\\sqrt{2})n}.$", "Solution_1": "We have to show that $ \\frac {d_{n} \\plus{} D_{n}}{2} < \\int^{1}_{0} \\frac {1}{\\sqrt {1 \\plus{} x^{2}}}$ where $ d_{n}$ is the lower and $ D_{n}$ the upper Darboux sum of the function $ f(x) \\equal{} \\frac {1}{\\sqrt {1 \\plus{} x^{2}}}$ on $ [0,1]$. However, since the function is convex, I would say that we have $ \\frac {d_{n} \\plus{} D_{n}}{2} > \\int^{1}_{0}\\frac {1}{\\sqrt {1 \\plus{} x^{2}}}$ for all $ n$-s?", "Solution_2": "I didn`t have time before to explain my thoughts, so I will be clearer now.\r\n\r\n$ \\sum_{i \\equal{} 1}^{n}\\frac {1}{\\sqrt {n^2 \\plus{} i^2}} \\equal{} \\sum_{i \\equal{} 1}^{n}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}}$, which is the Riemann sum of the function $ f(x) \\equal{} \\frac {1}{\\sqrt {1 \\plus{} x^{2}}}$ for the divison of the interval $ [0,1]$ on $ n$ equal segments and choosing the right most point on each of this segments.\r\n\r\nNow we know that $ \\lim_{n \\to \\infty}\\sum_{i \\equal{} 1}^{n}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}} \\equal{} \\int^{1}_{0} \\frac {1}{\\sqrt {1 \\plus{} x^{2}}} \\equal{} \\ln{(1 \\plus{} \\sqrt {2})}.$\r\n\r\nNow we have to show that $ \\sum_{i \\equal{} 1}^{n}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}} < \\ln(1 \\plus{} \\sqrt {2}) \\minus{} \\frac {1}{2(2 \\plus{} \\sqrt {2})n} \\qquad (\\ast)$.\r\n\r\nWe have to somehow find $ A, B$, such that:\r\n\\[ \\ln(1 \\plus{} \\sqrt {2}) \\minus{} \\frac {1}{2(2 \\plus{} \\sqrt {2})n} < A\\leq \\sum_{i \\equal{} 1}^{n}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}}\\leq B < \\ln(1 \\plus{} \\sqrt {2})\r\n\\]\r\n.\r\nTherefore we can study the upper and the lower Darboux sum of the function $ f$ for the same division.\r\n\r\nSince $ f^{\\prime}(x) \\equal{} \\minus{} \\frac {x}{\\sqrt {(x^{2} \\plus{} 1)^{3}}} \\Rightarrow f^{\\prime}(x) < 0$ on $ (0,1]$, the function $ f$ is strictly decreasing on $ [0,1]$. Therefore we take the right most point on each interval of our division to get the lower Darboux sum $ d_{n}$ and the left most point on each interval to get the upper Darboux sum $ D_{n}$.\r\nTherefore\r\n\\[ d_{n} \\equal{} \\sum_{i \\equal{} 1}^{n}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}} \\quad,\\quad D_{n} \\equal{} \\sum_{i \\equal{} 0}^{n \\minus{} 1}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}}\r\n\\]\r\nWe can see that\r\n\\[ D_{n} \\minus{} d_{n} \\equal{} \\frac {2}{2n(2 \\plus{} \\sqrt {2})}\r\n\\]\r\n.\r\nUsing this in $ (\\ast)$, we have to show\r\n\\[ \\frac {D_{n} \\plus{} d_{n}}{2} < \\ln{(1 \\plus{} \\sqrt {2})}.\r\n\\]\r\nNow we know, that on the intervals, where $ f$ is convex, the mean value of the upper and the lower Darboux sum is greater than the area below $ f$ and vice versa for concave parts of $ f$. Hence $ \\frac {D_{n} \\plus{} d_{n}}{2}$ partitions into $ \\frac {D_{n} \\plus{} d_{n}}{2} \\equal{} \\frac {D^{\\prime}_{n} \\plus{} d^{\\prime}_{n}}{2} \\plus{} \\frac {D^{\\circ}_{n} \\plus{} d^{\\circ}_{n}}{2}$, where $ D^{\\prime}, d^{\\prime}$ are from convex parts of $ f$ and $ D^{\\circ}, d^{\\circ}$ are from concave parts of $ f$. If there wouldn`t be any concave parts of $ f$ on $ [0,1]$, we would have $ \\frac {D_{n} \\plus{} d_{n}}{2} > \\ln{(1 \\plus{} \\sqrt {2})}$, but if there wouldn`t be any convex parts of $ f$ on $ [0,1]$, we would have $ \\frac {D_{n} \\plus{} d_{n}}{2} < \\ln{(1 \\plus{} \\sqrt {2})}$.\r\n\r\nBut the problem appears, since $ f^{\\prime\\prime}(x)$ gives us, that $ f$ is concave on $ [0,\\frac {\\sqrt {2}}{2}]$ and convex on $ [\\frac {\\sqrt {2}}{2},1]$. Now it remains to decide which part prevails. I am stuck here.", "Solution_3": "I have made a mistake in the upper proof in the line:\r\n\\[ \\ln(1 \\plus{} \\sqrt {2}) \\minus{} \\frac {1}{2(2 \\plus{} \\sqrt {2})n} < A\\leq \\sum_{i \\equal{} 1}^{n}\\frac {1}{n}\\frac {1}{\\sqrt {1 \\plus{} \\left(\\frac {i}{n}\\right)^2}}\\leq B < \\ln(1 \\plus{} \\sqrt {2}).\\]\r\nIt is not needed, so just ignore it.\r\n\r\n\r\nHowever, does anybody have any hint on how to show that the concave part of $ f(x)$ prevails over convex in the upper reasoning?" } { "Tag": [ "trigonometry", "geometry", "circumcircle", "incenter", "geometry unsolved" ], "Problem": "Help me to solve this problems(can you solve them using trigonometry).\r\n\r\n[b]1.[/b] On the sides $AB$, $BC$, $CD$, and $DA$ of quadrilatera $ABCD$ are given points $M$, $N$, $P$ and $Q$ and for them is $AM=BN=CP=DQ$. If $MNPQ$ is square then $ABCD$ is also square.\r\n\r\n[b]2.[/b] Let $k$ be circumcircle of triangle $ABC$ and $D$ be an arbitrary point on side $AB$. Circle $k_1$ touches Circle $k$ inside, $AD$ and $CD$, and circle $k_2$ touches $k$ inside, $BD$ and $DC$. Prove that $k_1$ touches $k_2$ iff $CD$ is bisector of $\\angle ACB$.\r\n\r\n[b]3.[/b] The incircle $s$ of triangle $ABC$ touches $BC$ in $K$. Let $M$ be the midpoint of the altitude $AD$. Let $N$ be the second intersection point line $KM$ with circle $s$. Prove that $N$ is the common point of circle $s$ and circumcircle of triangle $BCN$.", "Solution_1": "[quote=\"X-man\"][b]3.[/b] The incircle $s$ of triangle $ABC$ touches $BC$ in $K$. Let $M$ be the midpoint of the altitude $AD$. Let $N$ be the second intersection point line $KM$ with circle $s$. Prove that $N$ is the common point of circle $s$ and circumcircle of triangle $BCN$.[/quote]\n\nThe point N is the tangency point of the incircle of $\\triangle ABC$ and circumcircle of $\\triangle BCN$. For a proof, see [url=http://www.mathlinks.ro/Forum/post-141487.html]posted before? [a circumcircle tangent to the incircle][/url], with a significant contribution from pestich = armpist:\n\n[quote=\"pestich\"]:?[/quote]\n(no kidding). \n\n[quote=\"X-man\"][b]2.[/b] Let $k$ be circumcircle of triangle $ABC$ and $D$ be an arbitrary point on side $AB$. Circle $k_{1}$ touches Circle $k$ inside, $AD$ and $CD$, and circle $k_{2}$ touches $k$ inside, $BD$ and $DC$. Prove that $k_{1}$ touches $k_{2}$ iff $CD$ is bisector of $\\angle ACB$.[/quote]\r\n\r\nTo use such a valuable tool as trigonometry for a problem that can be solved synthetically at the first glance would be a waste.\r\n\r\nAssume that CD bisects the angle $\\angle C$. Then it passes through the midpoint M of the circumcircle arc AB opposite to the vertex C. Invert the figure in a circle (M) with radius MA = MB. The line AB is carried into the circumcircle k, the circumcircle k into the line AB, and the line CD passing through the inversion center M is carried into itself. This means that the circles $k_{1}, k_{2}$ are carried into themselves, which is possible only if they are perpendicular to the inversion circle. But the line CD is perpendicular to the inversion circle (M), hence, the circles $k_{1}, k_{2}$ have to touch this line at the same point, i.e., they have to touch each other.\r\n\r\nConversely, assume that the circles $k_{1}, k_{2}$ touch each other. Let o, m be tangents to the circumcircle k and to the circle (M) at their intersection A. We have $\\angle (AB, o) = 180^\\circ-\\angle C$ and $\\angle (AB, m) = \\frac{\\angle AMB}{2}= \\frac{180-\\angle C}{2}$. This means that the circle (M) bisects the angle formed by the line AB and the circumcircle k. Hence, any circle $k_{i}$ centered on the same side of AB as the vertex C, touching the line AB, and touching the circumcircle k internally is perpendicular to the circle (M). Consequently, any 2 such circles $k_{1}, k_{2}$ touching each other have their tangency point on the circle (M), which means that their single common internal tangent must pass through the center M of this circle. If this tangent also passes through the vertex C, it is the bisector of the angle $\\angle C$.", "Solution_2": "[quote=\"X-man\"][b]1.[/b] On the sides $AB$, $BC$, $CD$, and $DA$ of quadrilateral $ABCD$ are given points $M$, $N$, $P$ and $Q$ and for them is $AM=BN=CP=DQ$. If $MNPQ$ is square then $ABCD$ is also square.[/quote]\r\n\r\nAssuming that the points $M, N, P, Q$ lie on the segments $AB, BC, CD, DA.$ The vertices $A, B, C, D$ lie on 4 congruent circles $(M), (N), (P), (Q)$ with radius $r = AM = BN = CP = DQ$, respectively. The line $AM$ is either tangent to the circle $(N)$ at a point $B_0$ or it meets it at 2 points $B_1, B_2$ on different sides of the line $B_0N$, such that $AB_1 < AB_2$. The line $B_0N$ is tangent to the circle $(P)$ and the line $B_1N$ does not intersect it. Hence, the quadrilateral vertex $B$ is identical either with the tangency point $B_0$ or with the more distant point $B_2$. Assume that $B \\equiv B_2$ is different from $B_0$ and let $F$ be the midpoint of the chord $B_1B_2$. Since the angles $\\angle MNP = \\angle AFM = 90^\\circ$ are both right, it follows that the angles $\\angle(NP, FN) = \\angle BMN$ are equal and $\\angle CNP \\equiv \\angle(NP, BN) < \\angle(NP, FN) = \\angle BMN$. In exactly the same way, we can show that $\\angle DPQ < \\angle CNP, \\angle AQM < \\angle DPQ, \\angle BMN < \\angle AQM,$ etc. Combining,\r\n\r\n$\\angle BMN > \\angle CNP > \\angle DPQ > \\angle AQM > \\angle BMN$\r\n\r\nwhich is a contradiction. Hence, the quadrilateral vertex $B$ must be identical with the tangency point $B_0$. But the line $BN$ is then tangent to the circle $(P)$ and the single tangency point must be identical with the quadrilateral vertex $C$. Likewise, the line $CP$ is then tangent to the circle $(Q),$ the tangency point identical with the quadrilateral vertex $D,$ the line $DQ$ is tangent to the circle $(M)$, the tangency point identical with the quadrilateral vertex $A$, and finally, the line $AM$ is tangent to the circle $(N)$, the tangency point identical with the quadrilateral vertex $B$, thus closing the square $ABCD$.", "Solution_3": "[color=darkred][b]The proof of the problem II.[/b] The circles $k_{1}$ and $k_{2}$ are tangent (in the point $T\\in CD$) $\\Longleftrightarrow$ the rays $[AT$, $[BT$ are the bisectors of the angles $\\widehat{BAC}$, $\\widehat{ABC}$ respectively $\\Longleftrightarrow$ the point $T$ is incentre of the triangle $ABC$ $\\Longleftrightarrow$ the ray $[CD$ is the bisector of the angle $\\widehat{ACB}\\ .$[/color]\r\n\r\n[b]EDIT:[/b] This is wrong, I am sorry ! (It doesn't follow that the rays $[AT$, $[BT$ are the bisectors of the angles $\\widehat{BAC}$, $\\widehat{ABC}$ respectively!) I wanted to apply the following lemma, but I made a great mistake confusing the letters. I shall come back soon for correcting of proof.\r\n\r\n[color=darkred][b]Lemma.[/b] The circle $k_{1}$ is interior tangent to the circle $k$ in the point $T$. The tangent to the circle $k_{1}$ in a point $M\\in k_{1}$ meet the circle $w$ in the points $N$ and $P$. Then the ray $[TM$ is the bisector of the angle $\\widehat{NTP}$.[/color]\r\n\r\nTry to apply this lemma to the circles $k_{1}$ and $k_{2}$ for the their common tangents $AB$ and $CD$ !" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Hi!\r\n\r\nSorry for my non-mathematic description of my problem. But here we go...\r\n\r\nLet's say I have an ordered sequence of objects. Let's take N letters for examples.\r\nHow many ways can I create k groups of letters without changing the orders.\r\n\r\nExample : ABCDE\r\n\r\nWith k=4 (I want 4 groups), I can divide it 4 different ways :\r\n[list][*]{AB},{C},{D},{E}\n[*]{A},{BC},{D},{E}\n[*]{A},{B},{CD},{E}\n[*]{A},{B},{C},{DE}[/list]With k=3 (I want 3 groups), I can divide it 6 different ways :\n[list][*]{ABC},{D},{E}\n[*]{A},{BCD},{E}\n[*]{A},{B},{CDE}\n[*]{AB},{CD},{E}\n[*]{AB},{C},{DE}\n[*]{A},{BC},{DE}[/list]With k=2, I can only divide is two ways :\n[list][*]{ABCD},{E}\n[*]{A},{BCDE}\n[*]{AB},{CDE} (Edit)\n[*]{ABC},{DE} (Edit)[/list]Is there a simple formula/theorem that can give me the total number of partition I can create for a give N and k ?\r\n\r\nMat", "Solution_1": "[quote=\"BleuBleu\"]With k=2, I can only divide is two ways :\n[list][*]{ABCD},{E}\n[*]{A},{BCDE}[/list][/quote] What about AB-CDE and ABC-DE? Assuming you just overlooked these (and so there aren't any other restrictions you haven't mentioned), the answer is $ \\binom{n \\minus{} 1}{k \\minus{} 1} \\equal{} \\frac {(n \\minus{} 1)!}{(k \\minus{} 1)!(n \\minus{} k)!}$ -- from among $ n \\minus{} 1$ gaps between the letters you must choose $ k \\minus{} 1$ to be the dividers between the different groups.", "Solution_2": "Cool, I never tough about looking at the problem in terms of \"dividers\".\r\nThat makes perfect sense!\r\nI just edited my original post to include those 2 missing combinations.\r\n\r\n[code]Binomial[4, 3] = 4\nBinomial[4, 2] = 6\nBinomial[4, 1] = 4[/code]\r\n\r\nIt works! Thank you sooo much!\r\n\r\nMat" } { "Tag": [ "algebra", "polynomial", "calculus", "derivative", "induction", "number theory", "relatively prime" ], "Problem": "Prove that the remainder of the polynomial \r\np(x)=x^2007+2x^2006+3x^2005........+2006x^2+2007x+2008 \r\n\r\nis the same upon division by x(x+1) as upon division by x(x+1)^2", "Solution_1": "It's shorter than it looks at first glance, but it still bothers me that I ended up using calculus.\r\n\r\n[hide=\"Solution\"]There exists a unique combination $ P(x)$, $ C_1$, $ C_2$, $ C_3$, such that: \\[ x^{2007} + \\ldots + 2007x+2008 = P(x) \\cdot x (x+1)^2 + C_1 x(x+1) + C_2 x + C_3 \\] We must show that $ C_1 = 0$ in the above formulation.\n\n[b]1.[/b] The instance $ x=0$ gives $ C_3 = 2008$. Subtract that out and divide by $ x$ to get: \\[ x^{2006} + \\ldots + 2006x + 2007 = P(x) \\cdot (x+1)^2 + C_1 (x+1) + C_2 \\]\n[b]2.[/b] The instance $ x=-1$ gives $ C_2 = 1004$. Subtract that out to get: \\[ x^{2006} + \\ldots + 2006x + 1003 = P(x) \\cdot (x+1)^2 + C_1 (x+1) \\]\n[b]3.[/b] We want to show that $ (x+1)^2 \\; | \\; x^{2006} + \\ldots + 2006x + 1003$, or equivalently, [i]both[/i] of the following: \\[ x+1 \\; | \\; x^{2006} + \\ldots + 2006x + 1003 \\] \\[ x+1 \\; | \\; \\frac{\\partial}{\\partial x} ( x^{2006} + \\ldots + 2006x + 1003 ) \\]However, the first is already asserted by the right-hand side of step 2, so we must only prove the second.\n\n[b]4.[/b] Note that all non-constant terms of the polynomial on the right-hand side take the form $ n x^{2007 - n}$, for $ n \\in \\{ 1 \\ldots 2006\\}$. We can sum the terms as pairs $ n x^{2007 - n} + m x^{2007 - m}$ such that $ n+m = 2007$. Observe that one of $ n,m$ is even and the other is odd.\n\\[ \\frac{\\partial}{\\partial x} n x^{2007 - n} = n(2007-n) x^{2006-n} = nm x^{m-1} \\] \\[ \\frac{\\partial}{\\partial x} m x^{2007 - m} = n(2007-n) x^{2006-n} = nm x^{n-1}\\] \nTesting divisibility by $ x+1$ is equivalent to testing if the instance $ x=-1$ results in a root of the polynomial. Plug in $ x=-1$ to find that: \\[ \\; \\frac {\\partial}{\\partial x} | _{x=-1} ( n x^{2007 - n} + m x^{2007 - m} ) \\; = nm (-1)^{m-1}+nm (-1)^{n-1} = 0 \\]\nThus, the sum of the derivatives at each pair $ n,m$ evaluates to $ 0$. Similarly, the derivative of the constant term evaluates to zero, and thus at $ x=-1$, the entire derivative is $ 0$, establishing divisibility as desired, and we are done.[/hide]", "Solution_2": "[hide]You can prove using induction that \\[ \\frac{p(x)}{x(x\\plus{}1)} \\equal{} x^{2005} \\plus{} x^{2004} \\plus{} 2x^{2003} \\plus{} 2x^{2002} \\plus{} \\ldots \\plus{} 1003x \\plus{} 1003 \\plus{} \\frac{1004x}{x(x\\plus{}1)}.\\] The quotient is divisible by $ x\\plus{}1$ and the result follows.[/hide]", "Solution_3": "[quote=\"TZF\"]it still bothers me that I ended up using calculus.[/quote]\r\nIt shouldn't; the formal derivative is a powerful and perfectly reasonable way to tease out the behavior of a polynomial modulo the power of an irreducible polynomial. One of the things I didn't elaborate on in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]this post[/url] is a certain asymmetry between how Lagrange interpolation and CRT are usually presented: whereas CRT is stated to apply to any set of pairwise relatively prime ideals of $ \\mathbb{Z}$, Lagrange interpolation tends to be restricted to the prime ideals of $ \\mathbb{C}[x]$. On the other hand, it makes just as much sense to replace questions modulo linear factors with questions modulo \"powers of primes,\" that is, questions such as\r\n\r\n$ P(x) \\equiv a \\bmod x$\r\n$ P(x) \\equiv b \\plus{} c(x \\plus{} 1) \\bmod (x \\plus{} 1)^2$\r\n\r\nwhich is equivalent to asking\r\n\r\n$ P(x) \\equiv a \\bmod x$\r\n$ P(x) \\equiv b \\bmod (x \\plus{} 1)$\r\n$ P'(x) \\equiv c \\bmod (x \\plus{} 1)$\r\n\r\nwhich is an equally sensible interpolation problem; asking to specify the behavior of a polynomial $ \\bmod (x \\plus{} 1)^2$ specifies both its value at $ x \\equal{} \\minus{} 1$ and its slope. This is a special case of the more general problem of defining [url=http://en.wikipedia.org/wiki/Intersection_number]intersection multiplicity[/url]; in my opinion, the generic methods of defining intersection multiplicity are actually more difficult to understand than the use of the derivative in this special case.", "Solution_4": "Thank's everyone for your solutions :)" } { "Tag": [ "inequalities", "geometry", "circumcircle", "inequalities unsolved" ], "Problem": "I had a proof for this problem , it's very nice :\r\nLet ABC is a triangle .Prove that \r\n$R\\leq \\frac{4abc\\sqrt{3}.\\sum a^2b^2}{18a^2b^2c^2-\\sum a^2.\\sum a^2b^2}$.", "Solution_1": "hello, is R the radius of the circumcircle?\r\nSincerely,Sonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, is R the radius of the circumcircle?\nSincerely,Sonnhard.[/quote]\r\nYes .This inequality equivalent with :\r\n$ \\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2} \\geq \\frac{18}{a^2+b^2+c^2+4S_{ABC} \\sqrt{3}}$. :D", "Solution_3": "nice inequality.For all positive real numbers x,y,:\r\n [tex]\\frac{1}{x^2+y^2+xy}+\\frac{1}{y^2+z^2+yz}+\\frac{1}{z^2+x^2+zx}\\geq\\frac{9}{(x+y+z)^2}[/tex]\r\nIn triangle ABC,let M be toroxelli point,apply this inequality for [tex]MA=x;MB=y;Mc=z..[/tex],we obtain reyes\"s inequality", "Solution_4": "[quote=\"dhthstn\"]nice inequality.For all positive real numbers x,y,:\n [tex]\\frac{1}{x^2+y^2+xy}+\\frac{1}{y^2+z^2+yz}+\\frac{1}{z^2+x^2+zx}\\geq\\frac{9}{(x+y+z)^2}[/tex]\nIn triangle ABC,let M be toroxelli point,apply this inequality for [tex]MA=x;MB=y;Mc=z..[/tex],we obtain reyes\"s inequality[/quote]\r\nOk. :) .This's my friend's solution.I have a question for you .Ok?How do you find this solution ?" } { "Tag": [ "function" ], "Problem": "Given the numbers $ 1,2,2^2, \\ldots ,2^{n\\minus{}1}$, for a specific permutation $ \\sigma \\equal{} x_1,x_2, \\ldots, x_n$ of these numbers we define $ S_1(\\sigma) \\equal{} x_1$, $ S_2(\\sigma)\\equal{}x_1\\plus{}x_2$, $ \\ldots$ and $ Q(\\sigma)\\equal{}S_1(\\sigma)S_2(\\sigma)\\cdot \\cdot \\cdot S_n(\\sigma)$. Evaluate $ \\sum 1/Q(\\sigma)$, where the sum is taken over all possible permutations.", "Solution_1": "Cute. Here's an inductive solution:\r\n\r\n[hide]Forget for a moment that we know some values for the $ x_i$. I claim that the given expression is equal to $ \\frac {1}{x_1\\cdot x_2 \\cdots x_n}$. The proof is inductive: it's clearly true for $ n = 1$. Let $ S(k)$ be the set of permutations of $ \\{x_1, \\ldots, x_n\\}$ that end in $ k$ and let $ T(k)$ be the set of permutations of $ \\{x_1, \\ldots, x_n\\} \\setminus \\{x_k\\}$. Then for $ n \\geq 2$,\n\\begin{eqnarray*} \\sum_\\sigma 1/Q(\\sigma) & =& \\sum_k \\sum_{\\sigma \\in S(k)} 1/Q(\\sigma) \\\\\n& = &\\sum_k \\sum_{\\sigma \\in T(k)} \\frac {1}{Q(\\sigma)(x_1 + \\ldots + x_n)} \\\\\n& = &\\frac {1}{x_1 + \\ldots + x_n} \\sum_{k} \\sum_{\\sigma \\in T(k)} 1/Q(\\sigma) \\\\\n& = &\\frac {1}{x_1 + \\ldots + x_n} \\sum_{k} \\frac {x_k}{x_1 \\cdots x_n} \\\\\n& = &\\frac {1}{x_1 \\cdots x_n} \\end{eqnarray*}\nwhere we apply the inductive hypothesis going from the third to the fourth line.\n[/hide]", "Solution_2": "That's nice. Of course, the final expression can be simplified to $ 2^{\\minus{}{n\\choose 2}}$.", "Solution_3": "Yes. I wonder if there is a more direct solution, or something interesting with generating functions? (I suppose there could also be something clever for the specific case of powers of 2, but my guess is that anything nice will apply in the general case.)", "Solution_4": "The general/normal solution is by induction. As far as I know, this problem was posted before and solved by towerfreak with the use of induction.", "Solution_5": "Indeed, Towersfreak2006's solution at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125603 is essentially the same as mine." } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "Factor completely:\r\n\r\n$ 6x^2\\plus{}2x\\minus{}6y^2\\minus{}5xy\\plus{}23y\\minus{}20$", "Solution_1": "[quote=\"Smartguy\"]Factor completely:\n\n$ 6x^2 \\plus{} 2x \\minus{} 6y^2 \\minus{} 5xy \\plus{} 23y \\minus{} 20$[/quote]\r\n\r\n$ 6x^2 \\plus{} 2x \\minus{} 6y^2 \\minus{} 5xy \\plus{} 23y \\minus{} 20\\equal{}6x^2\\plus{}(2\\minus{}5y)x\\minus{}6y^2\\plus{}23\\minus{}20\\equal{}0$,\r\n\r\nby solving this for $ x$ we have\r\n\r\n$ x\\equal{}\\frac{\\minus{}(2\\minus{}5y)\\pm \\sqrt{(2\\minus{}5y)^2\\minus{}4\\cdot 6(\\minus{}6y^2\\plus{}23y\\minus{}20)}}{12}$\r\n\r\n$ \\equal{}\\frac{5y\\minus{}2\\pm (13y\\minus{}22)}{12}\\equal{}\\frac{3y\\minus{}4}{2},\\frac{\\minus{}2y\\plus{}5}{3}$,\r\n\r\nso finally we have $ 6x^2 \\plus{} 2x \\minus{} 6y^2 \\minus{} 5xy \\plus{} 23y \\minus{} 20\\equal{}6(x\\minus{}\\frac{3y\\minus{}4}{2})(x\\minus{}\\frac{\\minus{}2y\\plus{}5}{3})$\r\n$ \\equal{}\\boxed{(2x\\minus{}3y\\plus{}4)(3x\\plus{}2y\\minus{}5)}$", "Solution_2": "using the quadratic formula would make it very messy--> easy to make a mistake\r\n\r\ncan you do guess and check? \r\n\r\nFirst $ 6x^2\\minus{}5xy\\minus{}6y^2\\plus{}2x\\plus{}23y\\minus{}20$\r\nFactor--> $ (3x\\plus{}2y)(2x\\minus{}3y)\\plus{}2x\\plus{}23y\\minus{}20$\r\n\r\nThen look at the -20 and find factors to fill the following spaces:\r\n\r\n$ (3x\\plus{}2y\\pm$___$ )$$ (2x\\minus{}3y\\pm$___$ )$\r\n\r\n\r\nComments?", "Solution_3": "This solution is easier, and finding factors is not to cumbersome here. But how did (or would) you find (3x+2y)(2x-3y) without looking at the previous post?", "Solution_4": "you can factor the $ 6x^2\\minus{}5xy\\minus{}6y^2$\r\nright?", "Solution_5": "Okay, but I guess I'm just wondering if there was a reason that you should factor that part first... :|", "Solution_6": "it does make the problem simpler, because once you factor that part, you can look at the other part of the equation and try to factor that, less terms are easier to factor unless its 2x+23y-20 where trial and error is necessary with the exception of the quad. for.", "Solution_7": "Yes,in this problem,I believe your solution is the best.\r\n\r\nWith quadratic formula way we can reach the answers automatically.If the constant term has many dividers,it could be onerous to find the pairs.", "Solution_8": "Ok. I just didn't know the general steps you should do when facing a quadratic in 2 variables like this.\r\n\r\nNow I get it." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "A sequence $ (a_n)$ satisfies $ a_0\\equal{}0,0\\leq a_{k\\plus{}1}\\minus{}a_k\\leq 1$. Prove that\r\n\r\n$ \\sum_{k\\equal{}0}^na_k^3\\leq (\\sum_{k\\equal{}0})^na_k)^2$", "Solution_1": "Correct version:\r\n\r\nA sequence $ (a_n)$ satisfies $ a_0 \\equal{} 0,0\\leq a_{k \\plus{} 1} \\minus{} a_k\\leq 1$. Prove that\r\n\r\n$ \\sum_{k \\equal{} 0}^na_k^3\\leq (\\sum_{k \\equal{} 0}^na_k)^2$", "Solution_2": "Nobody? ...", "Solution_3": "Still?.....", "Solution_4": "[quote=\"Carlez Tevos\"]\nA sequence $ (a_n)$ satisfies $ a_0 \\equal{} 0,0\\leq a_{k \\plus{} 1} \\minus{} a_k\\leq 1$. Prove that\n\n$ \\sum_{k \\equal{} 0}^na_k^3\\leq (\\sum_{k \\equal{} 0}^na_k)^2$[/quote]\r\nFor $ n \\equal{} 0$ it's true.\r\nIf $ \\sum_{k \\equal{} 0}^{n}a_k^3\\leq \\left(\\sum_{k \\equal{} 0}^{n}a_k\\right)^2$ then $ \\sum_{k \\equal{} 0}^{n \\plus{} 1}a_k^3\\leq \\left(\\sum_{k \\equal{} 0}^{n}a_k\\right)^2 \\plus{} a_{n \\plus{} 1}^3.$\r\nHence, it remains to prove that $ \\left(\\sum_{k \\equal{} 0}^{n}a_k\\right)^2 \\plus{} a_{n \\plus{} 1}^3\\leq\\left(\\sum_{k \\equal{} 0}^{n \\plus{} 1}a_k\\right)^2,$ \r\nwhich is equivalent to $ a_{n \\plus{} 1}^2\\leq2\\sum_{k \\equal{} 0}^{n}a_k \\plus{} a_{n \\plus{} 1},$ which is still true for $ n \\equal{} 0.$\r\nIf $ a_{n \\plus{} 1}^2\\leq2\\sum_{k \\equal{} 0}^{n}a_k \\plus{} a_{n \\plus{} 1}$ then $ a_{n \\plus{} 2}^2\\leq a_{n \\plus{} 2}\\left(a_{n \\plus{} 1} \\plus{} 1\\right) \\equal{} a_{n \\plus{} 2}a_{n \\plus{} 1} \\plus{} a_{n \\plus{} 2}\\leq$\r\n$ \\leq \\left(a_{n \\plus{} 1} \\plus{} 1\\right)a_{n \\plus{} 1} \\plus{} a_{n \\plus{} 2}\\equal{} a_{n \\plus{} 1}^2 \\plus{} a_{n \\plus{} 1} \\plus{} a_{n \\plus{} 2}\\leq2\\sum_{k \\equal{} 0}^{n \\plus{} 1}a_k \\plus{} a_{n \\plus{} 2}.$\r\nThus, $ a_{n \\plus{} 1}^2\\leq2\\sum_{k \\equal{} 0}^{n}a_k \\plus{} a_{n \\plus{} 1}$ is true for all $ n\\geq0.$\r\nId est, $ \\sum_{k \\equal{} 0}^{n}a_k^3\\leq \\left(\\sum_{k \\equal{} 0}^{n}a_k\\right)^2$ is true for all $ n\\geq0.$", "Solution_5": "Nice...\r\n\r\nWhat about if $ a_0\\equal{}0$ and $ a_{k\\plus{}1}\\geq a_k\\plus{}1$. Prove that \r\n\r\n$ \\sum_{k\\equal{}1}^n a_k^3\\geq (\\sum_{k\\equal{}1}^na_k)^2$.", "Solution_6": "[quote=\"Carlez Tevos\"]\n\nWhat about if $ a_0 \\equal{} 0$ and $ a_{k \\plus{} 1}\\geq a_k \\plus{} 1$. Prove that \n\n$ \\sum_{k \\equal{} 1}^n a_k^3\\geq (\\sum_{k \\equal{} 1}^na_k)^2$.[/quote]\r\nI think, this is the same." } { "Tag": [], "Problem": "Given S = ( 1/2 , 1/3 , 2/3, 3/4 , 5/8 , 5/16 , 15/32 )\r\n\r\nWhat is the least positive difference between two distinct numbers from set S?\r\n\r\nA. 1/48 B. 1/32 C. 1/24 D. 1/12 \tE. 1/8", "Solution_1": "[hide=\"my solution\"]Ok. So you have to pick the 2 lowest numbers...which would give you the answer as E..I don't quite know if this is right [/hide]", "Solution_2": "What about $ \\frac12\\minus{}\\frac{15}{32}\\equal{}\\frac{1}{32}$? :wink:" } { "Tag": [ "geometry", "rectangle", "modular arithmetic" ], "Problem": "A $9 \\times 9$ square consists of $81$ unit squares. Some of these unit squares are painted black, and the others are painted white, such that each $2 \\times 3$ rectangle and each $3 \\times 2$ rectangle contain exactly 2 black unit squares and 4 white unit squares. Determine the number of black unit squares.", "Solution_1": "Hmm... what if we just triple-cover every square, then we can find a perfect covering, and have $3B = 3\\frac{2}{6}81 = 81 \\Rightarrow B=27$ so there are 27 black squares. Did I miss something? :?", "Solution_2": "The answer is correct, but I don't have a clue what you mean. Please try to write a clear solution. For example, what do you mean by \"triple-cover\"? And \"perfect covering\"?", "Solution_3": "Ah well, won't make a computer drawing for this. :)\r\n\r\nBut even easier: use the rectangles to cover all but the middle 3x3 square. We have covered 24 so far.\r\n\r\nSo the 3x3 must either have 3 black squares, or 4 in the pattern\r\nb w b\r\nw w w\r\nb w b\r\n\r\nIf 4, taking the rectangle $A_{5-7,4-5}$ tells that either $A_{7,4}$ or $A_{7,5}$ is black, which gives 3 blacks in $A_{6-7,4-6}$, contradiction. So we have 3. And 24+3=27.\r\n\r\nAnd such a coloring exists: put $A_{i,j}$ black iff $i=j \\pmod 3$." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Let $a \\in R$ s.t. $0\\leq a\\leq \\frac{1}{n}\\forall n \\in Z^{+}$ then $a=0$, please explain the proof.", "Solution_1": "The Archimedian axiom tells us that for every $x>0$, there exists an integer $n$ with $n>x$. Taking reciprocals, we see that for every $y>0$, there exists $n$ such that $1/nn$. Symmetry with respect to the line that separates $n$th row and $n+1$th row.", "Solution_5": "We remove the corresponding number of unit squares from every line\r\nor just from one?", "Solution_6": "Sorry for taking back this topic !\nBut if anyone has a solution please post it here.\nThanks...", "Solution_7": "[quote=\"shobber\"]Consider a $2n \\times 2n$ board. From the $i$th line we remove the central $2(i-1)$ unit squares. What is the maximal number of rectangles $2 \\times 1$ and $1 \\times 2$ that can be placed on the obtained figure without overlapping or getting outside the board?[/quote]\nI have tried to solve this problem for a long time.\nI want to ask you if the solutions are :if $n$ is odd we have $\\frac{(n-1)^2}{2}$\nif $n$ is even then $\\frac{(n-1)^2+1}{2}$", "Solution_8": "[quote=\"Arberi1\"][quote=\"shobber\"]Consider a $2n \\times 2n$ board. From the $i$th line we remove the central $2(i-1)$ unit squares. What is the maximal number of rectangles $2 \\times 1$ and $1 \\times 2$ that can be placed on the obtained figure without overlapping or getting outside the board?[/quote]\nI have tried to solve this problem for a long time.\nI want to ask you if the solutions are :if $n$ is odd we have $\\frac{(n-1)^2}{2}$\nif $n$ is even then $\\frac{(n-1)^2+1}{2}$[/quote]\n\nThe solution is certain higher:\n\nwe find for $n=1,2,3,4$ resp. $2,6,12,20$ $(=n(n+1))$\n, because here we can full in all squares (there are there for each $n$ $2n(n+1)$ when we calculate that.\n\n\nThe way to the solution:\n\nDivide the $2n*2n-square$ in $4$ $n*n-$suares.\nIn this squares, we have $\\frac{n(n+1)}{2}$ litle squares, when we colour this in the chesscoloring (black, white,...),\nwe have ${\\frac{(n+1)}{2}^2}$ black (the corner is black) and $\\frac{n^2-1}{4}$ white squares.\nThis mean we have $\\frac{n+1}{2}$ black ones more, we can only use $2$ with an other litle square of the $n*n-squares.$ \n\nSo we have $2(n-3)$ litle squares we can't fill in.\nWe get $n^2+3$ rectangles we can set in the picture.\nOne remark; this formula is other for only $3.$\n\n\nUse we the same strategy, we find if $n$ is even, we can fill in the whole picture\nand then the answer is $n^2+4.$\nHere the formula doesn't work for $2$\n\n\n\nEdit: We can write the answer as $n(n+1)-2\\lfloor{\\frac{n-3}{2}}\\rfloor$ where the floorfunction is 0 if the number is negative\n(wording of Alberi1)", "Solution_9": "Problem assumes that we remove $2(i-1)$ squares if $i\\leq n$, and $2(2n-i)$ squares if $i>n$.\n\nDivide the entire board into 4 quadrants each containing $n^2$ unit squares. \n\nFirst we note that the $2$ squares on the center on each of the $4$ bordering lines of the board can always be completely covered by a single tile, so we can count in the first and last unit squares (which are diagonally opposite) of each quadrant as being filled in completely by a tile.\n\nSo in each quadrant we have:\n\nif $n$ is even, there are exactly $(n-4)/2$ unit squares which cannot be filled by the tiles and if $n$ is odd, there are exactly $(n-3)/2$ unit squares which cannot be filled by the tiles.\n\nAbove can be seen by drawing a diagram and noticing that alternate columns have even and odd number of unit squares (leaving a unit square uncovered by tiles in odd numbered blocks of columns).\n\nAlso, note that the total number of unit squares which were removed from each quadrant = $(1 + 2+ 3 + ... n-1) = n(n-1)/2$\n\nLet us consider the 2 cases for parity of $n$:\n\n\n$Case 1: n$ is even\n\nfor $n = 2$, it can be seen easily that we can use a maximum of $6$ tiles.\n\nfor $n \\ge 4:$\nTotal number of squares that cannot be filled in each quadrant is:\n$n(n-1)/2 + (n-4)/2 = (n^2 - 4)/2$\nSo total number of squares that cannot be filled on the entire board = $2(n^2 - 4)$\nSo total number of squares that CAN be filled completely by the tiles = $4n^2 - 2(n^2 - 4) = 2n^2 + 8$\nSo the maximum number of tiles that can be used = $(2n^2 + 8)/2 = n^2 + 4$\n\n\n$Case 2: n$ is odd\n\nfor $n = 1$, it can be seen easily that we can use a maximum of $2$ tiles.\n\nfor $n \\ge 3:$\nTotal number of squares that cannot be filled in each quadrant is:\n$n(n-1)/2 + (n-3)/2 = (n^2 - 3)/2$\nSo total number of squares that cannot be filled on the entire board = $2(n^2 - 3)$\nSo total number of squares that CAN be filled completely by the tiles = $4n^2 - 2(n^2 - 3) = 2n^2 + 6$\nSo the maximum number of tiles that can be used = $(2n^2 + 6)/2 = n^2 + 3$" } { "Tag": [], "Problem": "1!+1+2!+2+3!+3.....8!+8+9!+9 what is the units digit? this is a countdown question so its not exactly effective if i write it all out is it?\r\nthanks", "Solution_1": "Pick at the little pieces you know. First, break in into pieces you like. Then, think about what you know about last digits of factorials.", "Solution_2": "[hide]8\n\nSince 5! on ends in a 0, then the units digit for 1! + 2! + 3! + 4! is 3\n\n1+...+9's unit digit is 5... so 8[/hide]", "Solution_3": "Is that supposed to be 1!+1+2!+2+..., or should it be 1!*1+2!*2+3!*3+...? The latter occurs quite frequently on math contests." } { "Tag": [ "LaTeX" ], "Problem": "Thanks. =]", "Solution_1": "[hide=\"1\"]$3^{105} \\equiv 3 \\bmod 5$, $4^{105} \\equiv -1 \\bmod 5$.[/hide]", "Solution_2": "Thank you, Andreas.\r\nHowever, can I have \r\n3^105 + 4^105 = 5Q + R, where R is a product of prime factors except 5,\r\nas the answer?\r\nAlso, I don't know what is \"mod\". Could you tell me what it is? Thanks. =]\r\n\r\nBesides, how can I put the LATEX in my posts instead of in the form of an attachment?\r\n\r\nSorry for being troublesome. =]", "Solution_3": "It's clear that $p+q\\geq 0$. So $p^3+q^3=(p+q)(p^2-pq+q^2)=(p+q)((p+q)^2-3pq)\\geq \\frac{1}{4}(p+q)^3$ and the rest is trivial.", "Solution_4": "In number 2, you just have to use the power mean inequality. :P", "Solution_5": "[quote=\"M4RI0\"]In number 2, you just have to use the power mean inequality. :P[/quote]\r\n\r\nWhat if $p$ and $q$ are not both positive? Then it requires some more work. My method avoids that difficulty." } { "Tag": [ "geometry", "3D geometry", "rational numbers" ], "Problem": "Prove that every positive rational number can be represented in the form \\[\\frac{a^{3}+b^{3}}{c^{3}+d^{3}}\\] for some positive integers $a, b, c$, and $d$.", "Solution_1": "For a positive rational $\\frac{m}{n}$, find another positive rational $\\frac{s}{t}$ such that\r\n\r\n$\\frac{t^3m}{s^3n}=\\frac{p}{q}$ lies between $\\frac12$ and $2$. Such a rational clearly exists.\r\n\r\nThen $\\frac{p}{q}=\\frac{(p+q)^3+(2p-q)^3}{(p+q)^3+(2q-p)^3}$, and all four cubes are positive.\r\n\r\nThen just multiply back, giving $\\frac{m}{n}=\\frac{(s(p+q))^3+(s(2p-q))^3}{(t(p+q))^3+(t(2q-p))^3}$." } { "Tag": [], "Problem": "I have students that are preparing for the Pythagoras Contest which is a nation wide Grade 6 math challenge, Do you have any Past tests?", "Solution_1": "There is some prepatory material on the sponsor's site:\r\n\r\n[url]http://www.mathematica.ca/eng/prep.htm[/url]" } { "Tag": [ "geometry" ], "Problem": "Hi!\r\n\r\nI need help to solve this problem:\r\n\r\nGiven: Two Circles k1 and k2, that meet in A and B. The tangent to k1 in A meets k2 in C2. The tangent to k2 in A meets k1 in C1. The line C1C2 meets k1 in D. \r\n\r\nProof that the line BD bisects the chord AC2.\r\n\r\n\r\nGreetings, \r\nUlysses", "Solution_1": "STOP !\r\n\r\nThe problem is exactly problem 3 of the ONGOING 2nd Round of the BWM 2004 (BWM = Bundeswettbewerb Mathematik = German national mathematical competition).\r\n\r\nDON'T POST SOLUTIONS !\r\n\r\n Darij", "Solution_2": "Well, cool, i am from italy. I got this problem from an old book called \"mathematical problems\". But in the book, there're no solutions for the problems.", "Solution_3": "Could you give me the name, authors and publishing house of the book, please?\r\n\r\n[b]PS.[/b] Your first posting (before you edited it) contained a link to a figure for the problem on an Austrian provider (I'm sad I didn't save the link); so I may assume that you are an Italian with some close relationship to German-speaking countries. Maybe you can even read German? If so, I am delighted to announce you that on 2/9/2004 a detailed solution of the problem will appear on [url=http://de.geocities.com/darij_grinberg/Dreigeom/Inhalt.html#bwm]my website[/url]. I am almost sure your patience will suffice for this one week.\r\n\r\n Darij", "Solution_4": "Sure, i can: \"problemi matematici\" by \"Damiano Scarponi\" published 1951 in italy.\r\n\r\nYou're right... i live close to austria, but my german is quite bad, it is as bad as my english :-)\r\nIt would be great if you offered a detailed solution on your website :-)\r\nThank you,\r\n\r\nUlysses\r\n\r\nPS: you're quite good in math, aren't you?\r\n[quote]\nA brief note about me: I am a school student (gymnasiast) in Karlsruhe, Germany and participant of several mathematical olympiads like the Bundeswettbewerb Mathematik, the Landeswettbewerb Mathematik Baden-Wrttemberg and the Deutsche Mathematik-Olympiade. Have won first prizes in the 1st Rounds of the Bundeswettbewerb Mathematik 2001, 2002, 2003 and 2004, as well as in the 2nd Round of the Bundeswettbewerb Mathematik 2003 (where I gained third prizes in 2000 and 2002), and became Bundessieger 2003. Participated in the \"Deutsche Mathematik-Olympiade\" since 2002, and gained a 1st prize in the 41th olympiad 2002 (Hamburg), an honorable mention in the 42th olympiad 2003 (Bremen), and a 2nd prize in the 43th olympiad 2004 (Essen). Silver medal in the 45th IMO 2004.[/quote]\r\ncongratulations! :-)", "Solution_5": "[quote=\"Ulysses\"]It would be great if you offered a detailed solution on your website :-)[/quote]\r\n\r\nIt will be detailed, actually it will be the solution I submitted to the olympiad. I will also post my solution on the forum (in English). But I'll do this all on 2 September, when the deadline will be over.\r\n\r\nWhat is really strange is that this problem is really identic with the 3rd problem of the BWM 2004, 2nd round. The wording is almost equal, and the notations are completely the same. Maybe the BWM guys have their secret sources...\r\n\r\nThanks for the congratulations ;) \r\n\r\n Darij", "Solution_6": "I'm going to lock this thread; you can remind a mod to unlock it once the contest is over.", "Solution_7": "I think it would be best to lock this thread until the contest is over (just in case someone solves the problem without looking at Darij's post, thinking it would contain a solution). Concur?", "Solution_8": "Okay, now the contest is over, let me now announce that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16201]the \"right\" thread contains some solutions to this problem[/url].\r\n\r\n Darij" } { "Tag": [ "modular arithmetic", "number theory", "prime numbers", "number theory unsolved" ], "Problem": "6. Find all prime numbers p,q for which pq divides (5^p-2^p)(5^q-2^q).\r\n\r\nA solution is given to this problem, but there is a part of it that I don't understand:\r\n\r\n\"If p|5^p-2^p, then p|5-2 by Fermat's theorem, so p=3. Suppose p,q!=3; then p|5^q-2^q and q|5^p-2^p. Without loss of generality, assume p>q, so that (p,q-1)=1.\" Continuing, \"Then if a is an integer such that 2a==5(mod q), then the order of a mod q divides p as well as q-1, a contradiction.\"\r\n\r\nThe proof continues to give the solutions. But I do not understand this statement: how can you justify that the order of a mod q divides p and q-1?\r\n\r\nBy the way, the symbol \"!=\" means \"not equal to\" and the symbol \"==\" means \"is congruent to.\"\r\n\r\nCould anyone please tell me what kind of theorem about orders justifies this statement?", "Solution_1": "The order of $a \\mod q$ must divide $q-1$ since $a^{q-1} \\equiv 1 \\pmod{q}$ by Fermat's Little Theorem. By our definition of $a$ as $2a \\equiv 5 \\pmod{q}$, raising both sides of the congruence by a power of $p$, we get $(2^p)(a^p) \\equiv 5^p \\pmod{q}$, but we earlier have $q \\mid 5^p-2^p \\Leftrightarrow 5^p \\equiv 2^p \\pmod{q}$, thus we have $a^p \\equiv 1 \\pmod{q}$, thus the order of $a \\mod q$ must divide $p$, and consequently we have the contradiction as desired, since $a \\neq 1$." } { "Tag": [ "geometry", "3D geometry", "email", "search" ], "Problem": "Hi Guy's ... Did anybody have a 4x4x4 Rubik's Revenge Cube solution ? ( In PDF file or u can email me at kofxi123@gmail.com ) Thank's a million.... :blush: :P", "Solution_1": "You could always search for it. Try this: \r\n\r\nhttp://www.jaapsch.net/puzzles/cube4.htm" } { "Tag": [ "ratio" ], "Problem": "Two regular polygons have sides of 6 inches and 8 inches respectively. Find the ratio of their perimeters.\r\n\r\n\r\nI am totally confused on how to answer this. :(", "Solution_1": "[quote=\"ReyBoB\"]Two regular polygons have sides of 6 inches and 8 inches respectively. Find the ratio of their perimeters.\n\n\nI am totally confused on how to answer this. :([/quote]\r\n\r\nIs there some info missing? i don't think you can get a definite answer with that much info. unless, of course, you want to take the number of sides as $x$ and $y$.", "Solution_2": "I guess the polygons have the same number of sides?", "Solution_3": "could always assume one has n sides and the other m sides and then multiply by the 6 and 8 respectively and divide. If they have the same number of sides then its trivial" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $ ABC$ be a triangle such that $ \\angle ABC \\equal{}120$. \r\nThe foot of the bisector of $ \\angle ABC$ is $ L$.\r\n$ K$ is a point on $ BC$ such that $ \\angle KAC \\equal{} 30$.\r\nProve that $ AK$ is a bisector of $ \\angle BKL$.", "Solution_1": "Let $ BM$ be $ \\angle LBC$ bisector and $ P$ the intersection point of $ BM$ and $ AK$. We have $ \\angle PBL\\equal{}\\angle PAL$ hence points $ B,P,L,A$ lie on a circle and $ \\angle KPL\\equal{}120^{\\circ}$. We also see that $ BKMA$ is inscribed quadrilateral hence $ \\angle KMC\\equal{}120^{\\circ}$ which means that $ PKML$ is inscribed => $ \\angle BKA\\equal{}\\angle BMA\\equal{}\\angle PKL$ QED", "Solution_2": "[quote=\"Nak\"]Let $ ABC$ be a triangle such that $ \\angle ABC \\equal{} 120$. \nThe foot of the bisector of $ \\angle ABC$ is $ L$.\n$ K$ is a point on $ BC$ such that $ \\angle KAC \\equal{} 30$.\nProve that $ AK$ is a bisector of $ \\angle BKL$.[/quote]\r\n\r\n___________\r\nCan I add the following question (NAK agrees?): Prove or disprove that $ CK \\equal{} CL$", "Solution_3": "[quote=\"Ligouras\"][quote=\"Nak\"]Let $ ABC$ be a triangle such that $ \\angle ABC \\equal{} 120$. \nThe foot of the bisector of $ \\angle ABC$ is $ L$.\n$ K$ is a point on $ BC$ such that $ \\angle KAC \\equal{} 30$.\nProve that $ AK$ is a bisector of $ \\angle BKL$.[/quote]\n\n___________\nCan I add the following question (NAK agrees?): Prove or disprove that $ CK \\equal{} CL$[/quote]\r\n\r\n$ CK \\neq CL$ becouse if $ AB < BC$ then there is a point $ T$ on $ [AB]$ such $ AL\\equal{}AT$ but $ \\angle TCA < 30$.", "Solution_4": "Next problem (harder than first)\r\nLet $ ABC$ be an equilateral triangle with center $ O$.\r\nLine $ l$ passes throught $ O$ intersects $ AB$ and $ AC$ at $ P$ and $ Q$ respectively.\r\n$ P'$, $ Q'$ are points on $ BC$ such that $ BP'\\equal{}BP$, $ CQ'\\equal{}CQ$.\r\nProve that there is a point $ K$ on $ l$ such that $ P'K\\equal{}P'B$, $ Q'K\\equal{}Q'C$.", "Solution_5": "this problem is solving with harmonic conjugate. D is intersection of BL and AK. E is intersection of KL and AB. F is intersection ED and AC. therefore A,F,L,C is harmonic conjugate. if CB is exterior bisector, BF is right to BC. and BDFA is cyclic quadrilateral. $ \\frac {x^2 \\minus{} 1}{(x^2 \\plus{} 7)^{2/3} \\plus{} 2(x^2 \\plus{} 7)^{1/2} \\plus{} 4} \\plus{} \\frac {6(x \\minus{} 1)}{(2x \\minus{} 1)^{1/2} \\plus{} 1} \\plus{} \\frac {x \\minus{} 1}{x^{1/2} \\plus{} 1} \\equal{} 0$\r\n\r\n<=> $ (x \\minus{} 1)[\\frac {x \\plus{} 1}{(x^2 \\plus{} 7)^{2/3} \\plus{} 2(x^2 \\plus{} 7)^{1/2} \\plus{} 4} \\plus{} \\frac {6}{(2x \\minus{} 1)^{1/2} \\plus{} 1} \\plus{} \\frac {1}{x^{1/2} \\plus{} 1}] \\equal{} 0$\r\n\r\n$ \\frac {x \\plus{} 1}{(x^2 \\plus{} 7)^{2/3} \\plus{} 2(x^2 \\plus{} 7)^{1/2} \\plus{} 4} \\plus{} \\frac {6}{(2x \\minus{} 1)^{1/2} \\plus{} 1} \\plus{} \\frac {1}{x^{1/2} \\plus{} 1} > 0$ with $ x\\ge \\frac {1}{2}$\r\n\r\nthus, $ x \\minus{} 1 \\equal{} 0$ equivalence to $ x \\equal{} 1$", "Solution_2": "[quote=\"bunhiacovski\"][quote=\"thanhnam2902\"][color=darkblue]Slove this equation: $ \\sqrt [3]{x^2 \\plus{} 7} \\plus{} 3\\sqrt {2x \\minus{} 1} \\plus{} \\sqrt {x} \\equal{} 6$[/color][/quote]\ndetermination $ x$ be $ x\\ge \\frac {1}{2}$\n\nthe equation equivalence to:\n\n$ (\\sqrt [3]{x^2 \\plus{} 7} \\minus{} 2) \\plus{} 3(\\sqrt {2x \\minus{} 1} \\minus{} 1) \\plus{} (\\sqrt {x} \\minus{} 1) \\equal{} 0$\n\n<=> $ \\frac {x^2 \\minus{} 1}{(x^2 \\plus{} 7)^{2/3} \\plus{} 2(x^2 \\plus{} 7)^{1/2} \\plus{} 4} \\plus{} \\frac {6(x \\minus{} 1)}{(2x \\minus{} 1)^{1/2} \\plus{} 1} \\plus{} \\frac {x \\minus{} 1}{x^{1/2} \\plus{} 1} \\equal{} 0$\n\n<=> $ (x \\minus{} 1)[\\frac {x \\plus{} 1}{(x^2 \\plus{} 7)^{2/3} \\plus{} 2(x^2 \\plus{} 7)^{1/2} \\plus{} 4} \\plus{} \\frac {6}{(2x \\minus{} 1)^{1/2} \\plus{} 1} \\plus{} \\frac {1}{x^{1/2} \\plus{} 1}] \\equal{} 0$\n\n$ \\frac {x \\plus{} 1}{(x^2 \\plus{} 7)^{2/3} \\plus{} 2(x^2 \\plus{} 7)^{1/2} \\plus{} 4} \\plus{} \\frac {6}{(2x \\minus{} 1)^{1/2} \\plus{} 1} \\plus{} \\frac {1}{x^{1/2} \\plus{} 1} > 0$ with $ x\\ge \\frac {1}{2}$\n\nthus, $ x \\minus{} 1 \\equal{} 0$ equivalence to $ x \\equal{} 1$[/quote]\r\n[color=darkblue]It's nice. Thank you very much.[/color]", "Solution_3": "$ x\\ge 1/2$. The L.H.S. is clearly strictly increasing. Therefore, only one solution exists. Guess x=1. Q.E.D." } { "Tag": [ "geometry", "3D geometry", "algebra", "polynomial", "group theory", "abstract algebra" ], "Problem": "Find all perfect squares that can be written as the sum of three consecutive perfect cubes.", "Solution_1": "[hide=\"Partially somewhere?\"]\n$ b^2 \\equal{} (c\\minus{}1)^3\\plus{}c^3\\plus{}(c\\plus{}1)^3 \\equal{} 3c^3 \\plus{} 6c \\equal{} 3c(c^2\\plus{}2)$, so $ b$ must be divisible by three. Let $ b\\equal{}3b'$, where $ b'$ is an integer.\n$ 9b'^2 \\equal{} 3c^3\\plus{}6c$\n$ 3b'^2 \\equal{} c^3\\plus{}2c$.\n[/hide]", "Solution_2": "[hide]\nLet the perfect cubes be $ (x \\minus{} 1)^3$, $ (x)^3$, and $ (x \\plus{} 1)^3$.\n\nThen $ (x \\minus{} 1)^3 \\plus{} (x)^3 \\plus{} (x \\plus{} 1)^3 \\equal{} k^2$ for some integer k.\n\nThis simplifies to $ 3x(x^2 \\plus{} 2) \\equal{} k^2$\n\nI had a mistake in there... it's fixed now but I don't know where to go from here.\n[/hide]", "Solution_3": "My approach:\r\n\r\n[hide]In general, if the first of the consecutive cubes is k^3, we have k^3 + (k+1)^3 + (k+2)^3. Since these are consecutive, this can be written with the summation powers of cubes: ((k+2)^2*(k+3)^2)/4 - ((k-1)^2*k^2)/4. Simplifying yields a polynomial function.\n\nNow we cheat and use a calculator to find values for which this holds true.\n\nIf we consider 0 and 1 as perfect cubes, 9 and 36 can be written as this sum.\n\n23, 24, and 25 yield 204\n\nI went farther on my calculator but the values all seemed to be rounded.[/hide]", "Solution_4": "[quote=\"mysmartmouth\"][hide]\nLet the perfect cubes be $ (x \\minus{} 1)^3$, $ (x)^3$, and $ (x \\plus{} 1)^3$.\n\nThen $ (x \\minus{} 1)^3 \\plus{} (x)^3 \\plus{} (x \\plus{} 1)^3 \\equal{} k^2$ for some integer k.\n\nThis simplifies to $ 3x(x^2 \\plus{} 2) \\equal{} k^2$\n\nI had a mistake in there... it's fixed now but I don't know where to go from here.\n[/hide][/quote]\r\n\r\nIt might help to analyze $ 3x$ and $ x^2 \\plus{} 2$ in terms of their factors...\r\n\r\nBut I did that for a while and only got two values for x. The third I found using a calculator.\r\n\r\nMoral: Diophantine equations= calculator abuse.", "Solution_5": "Once you expand and combine like terms you'll have $ n^2 \\equal{} 3k(k^2 \\plus{} 2)$. After a little musing I subtracted $ 9k^2$ from both sides so I could factor the thing. That gave me\r\n\\[ (n \\minus{} 3k)(n \\plus{} 3k) \\equal{} 3k(k \\minus{} 1)(k \\minus{} 2)\r\n\\]\r\nObviously the RHS is zero for $ k \\equal{} \\{0,1,2\\}$. Using that in conjunction with $ n^2 \\equal{} 9k^2$, the three ordered pairs $ (k,n)$ : $ (0,0)$, $ (1,9)$, $ (2,36)$ are easily found. I also performed some subsitution letting $ \\pm k \\equal{} \\pm \\frac {n}{3}$. I completed a square and got\r\n\\[ n((n \\minus{} 9)^2 \\minus{} 63) \\equal{} \\minus{} 81k^2\r\n\\]\r\nLooking at that thing we can say exactly where the LHS is negative and a not so bad approximation is $ n\\in (2,16)$. It's equivalent to finding integer solutions to\r\n\\[ (u \\plus{} 9)(u^2 \\minus{} 63) \\equal{} \\minus{} 81k^2\r\n\\]\r\nwith $ u$ ranging the integers from 0-7. Doesn't look like there are any. \r\n\r\nThat's all I've got I don't know how to extract n=204 from that.\r\n[b]edit:[/b] Actually I made a mistake. If someone wants, plug and chug the u-k equation again. I forgot that the u on the outside isn't squared so that butchered all of my computations. :D", "Solution_6": "If $ k$ is odd, then $ (k,k^2 \\plus{} 2) \\equal{} (k,2) \\equal{} 1$.\r\n\r\nThus $ 3k \\equal{} m^2$, $ k^2 \\plus{} 2 \\equal{} p^2$\r\n\r\n(easy to factor and solve)\r\n\r\nOR\r\n\r\n$ k \\equal{} m^2$, $ 3(k^2 \\plus{} 2) \\equal{} p^2$\r\n\r\n$ m^4\\plus{}2\\equal{}3p'^2$... $ (m^2\\plus{}1)(m\\plus{}1)(m\\minus{}1)\\equal{}3(p'\\minus{}1)(p'\\plus{}1)$\r\n\r\nnot much help...\r\n\r\nif $ k$ is even there are even more cases...", "Solution_7": "The reason elementary techniques aren't working is that this is a nonsingular [url=http://planetmath.org/encyclopedia/ArithmeticOfEllipticCurves.html]elliptic curve[/url]. In particular, it has Weierstrass form $ y^2 \\equal{} k^3 \\plus{} 18k$. If you'd like, I can give a rather computational solution using those tools.\r\n\r\nEdit: Well, not really a solution. The techniques to exhaustively find all the integer points on an elliptic curve aren't very elementary. What I can tell you is that Siegel's Theorem guarantees that the number of integer points on a cubic curve is finite. A short calculation demonstrates that the above curve has [url=http://en.wikipedia.org/wiki/Torsion_subgroup]torsion subgroup[/url] $ \\{\\mathcal{O}, (0, 0) \\}$ and [url=http://planetmath.org/encyclopedia/RankOfAnEllipticCurve.html]rank[/url] at least $ 1$ (since it has the non-torsion point $ (3, 9)$). I believe the point $ (72, 612)$ is independent, which would give this curve rank at least $ 2$. (The point $ (6, 18)$ is not independent; $ (3, 9) \\plus{} (0, 0) \\equal{} (6, \\minus{}18)$.)", "Solution_8": "Thank you all for your replies, and t0rajir0u for your explanation :) And my applogies for inaccurately guaging the difficulty of the problem." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c,d$ are positive prove that \r\n$\\frac{a^2+b^2+c^2}{a+b+c}+\\frac{b^2+c^2+d^2}{b+c+d}+\\frac{c^2+d^2+a^2}{c+d+a}+\\frac{d^2+a^2+b^2}{d+a+b}\\ge a+b+c+d$", "Solution_1": "[quote=\"payman_pm\"]$a,b,c,d$ are positive prove that \n$\\frac{a^2+b^2+c^2}{a+b+c}+\\frac{b^2+c^2+d^2}{b+c+d}+\\frac{c^2+d^2+a^2}{c+d+a}+\\frac{d^2+a^2+b^2}{d+a+b}\\ge a+b+c+d$[/quote]\r\n$\\sqrt{\\frac{a^2+b^2+c^2}{3}}\\ge \\frac{a+b+c}{3}$\r\n$\\Rightarrow \\frac{a^2+b^2+c^2}{3}\\geq \\frac{(a+b+c)^2}{3^2}$\r\n$\\Rightarrow \\frac{a^2+b^2+c^2}{a+b+c}\\ge \\frac{a+b+c}{3}$\r\n$\\Rightarrow \\frac{a^2+b^2+c^2}{a+b+c}+\\frac{b^2+c^2+d^2}{b+c+d}+\\frac{c^2+d^2+a^2}{c+d+a}+\\frac{d^2+a^2+b^2}{d+a+b}\\ge \\\\\\frac{a+b+c}{3}+\\frac{b+c+d}{3}+\\frac{c+d+a}{3}+\\frac{d+a+b}{3}=a+b+c+d$" } { "Tag": [], "Problem": "Has any pianist here played a piece from Prokofieff's Romeo and Juliet, the score he transcribed from his ballet? All of the pieces are just phenomenal music. My favorite is the Montagues and Capulets.", "Solution_1": "Cool, I played Capulet and Montague...It's a great piece...", "Solution_2": "Difficult though. I love it, been playing it for about six months and it's almost perfected. I wish I had more time to spend at the piano bench.", "Solution_3": "I like #10!\n\nI've played 2, 3, 4, and 10." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Let $MNP$ be an acute-angled triangle. We denote by $O$ the circumcenter of the triangle given.\r\n We suppose that $MO$ and $NP$ intersect in $K$ and we also suppose that on the edges $[MN]$ and $[MP]$ of the \r\n triangle, we draw the points $A$ and $B$ such that $KA=KN$ and $KB=KP$. Show that $ABNP$ is a trapezoid. ;)", "Solution_1": "Caesar, the your proposed problem is nice, but easily and wellknown !\r\n\r\n[hide=\"The first (metrical) method.\"]\n$\\frac{KN}{KP}=\\frac{c\\sin \\widehat {NMK}}{b\\sin \\widehat {PMK}}=\\frac{c\\cos P}{b\\cos N}$,\n\n$AN=2KN\\cos N$, $BP=2KP\\cos P$ $\\Longrightarrow$ \n\n$\\frac{AN}{BP}=\\frac{KN}{KP}\\cdot \\frac{\\cos N}{\\cos P}=\\frac{c\\cos P}{b\\cos N}\\cdot \\frac{\\cos N}{\\cos P}=\\frac cb=\\frac{MN}{MP}$ $\\Longrightarrow$\n\n$\\frac{MN}{MP}=\\frac{AN}{BP}\\Longrightarrow AB\\parallel NP.$ [/hide]\n[hide=\"The second (syntetical) method.\"]\nI note the reflection $M'$ of the point $M$ w.r.t. the centre $O$\n\nand the middlepoints $U$, $V$ of the segments $AN$, $BP$ respectively.\n\n$KA=KN$, $KB=KP$ $\\Longrightarrow$ $KU\\perp AN$, $KV\\perp BP$ $\\Longrightarrow$\n\n$KU\\parallel M'N$, $KV\\parallel M'P$ $\\Longrightarrow$\n\n$\\frac{MU}{MN}=\\frac{MK}{MM'}=\\frac{MV}{MP}$ $\\Longrightarrow$ $\\frac{MU}{MN}=\\frac{MV}{MP}\\Longrightarrow$\n\n$\\frac{MA+MN}{2MN}=\\frac{MB+MP}{2MP}\\Longrightarrow\\frac{MA}{MN}=\\frac{MB}{MP}$ $\\Longrightarrow$ $AB\\parallel NP.$\n[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "given a positive integer, consider the set \r\n$ A_n$={$ 1 \\le a \\le n: gcd(a,n)\\equal{}gcd(a\\plus{}1,n)\\equal{}1$}\r\nprove that the product of all the element of $ A_n$ congruent 1 modulo n", "Solution_1": "il prove that if $ a$ satisfies$ gcd(x(x\\plus{}1),n)\\equal{}1$, then so does the multiplicative inverse of $ a$ mod n \r\nLet b be the inverse of a, $ ab\\equal{}1(modn)$. Then, obviously $ gcd(b,n)\\equal{}1$. Now observe tht $ a(b\\plus{}1)\\equal{}1\\plus{}a (modn)$. Since $ gcd(a\\plus{}1,n)\\equal{}1$ it follows that $ gcd(b\\plus{}1,n)\\equal{}1$. Hence $ gcd(b(b\\plus{}1),n)\\equal{}1$\r\n\r\nNow, if a number which is its own inverse satisfies the condition, say $ t$, then $ t^2\\equal{}1(mod n)$. Since $ gcd(t\\plus{}1,n)\\equal{}1$ which forces $ n$ to divide $ t\\minus{}1$. Since $ t 1$ , $ \\forall k\\geq 1$\r\n\r\nDar ce zici de $ \\lim_{n\\to\\infty} \\sum_{k\\equal{}1}^n \\frac{k}{2^k}$ ? :lol:", "Solution_2": "[quote=\"vador\"]\ncalculati limita sirului $ \\displaystyle\\lim_{n\\to\\infty}\\sum_{k \\equal{} 1}^n \\frac {1}{k2^k}$.\n.[/quote]\r\n\r\nPentru $ x \\in (0,1)$ avem:\r\n\r\n$ \\frac { \\minus{} 1}{x \\minus{} 1} \\equal{} \\sum_{k \\equal{} 0}^{\\infty} x^k$ $ \\Rightarrow$ $ \\int \\frac { \\minus{} 1}{x \\minus{} 1}dx \\equal{} \\int \\sum_{k \\equal{} 0}^{\\infty}x^kdx$ , deci\r\n\r\n$ \\int \\frac { \\minus{} 1}{x \\minus{} 1}dx \\equal{} \\sum_{k \\equal{} 0}^{\\infty} \\frac {x^{k \\plus{} 1}}{k \\plus{} 1}$, sau scrisa sub alta forma \r\n\r\n$ \\lim_{n\\to\\infty} \\sum_{k \\equal{} 1}^n \\frac {x^k}{k} \\equal{} \\minus{} \\ln|x \\minus{} 1|$, si pentru $ x \\equal{} \\frac {1}{2}$ rezulta ca limita cautat e $ \\ln 2$. \r\n\r\nSau poti sa folosesti formula \"de-a gata\", ca sa zic asa dezvoltand dupa formula lui Taylor. vezi [url=http://en.wikipedia.org/wiki/Taylor_series]aici[/url].(parca se cere formula lui Taylor la cls 11 pt olimpiada, eu am incercat sa fac o justificare destul de naturala pentru formula.)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Show that if $a,b,c$ are any three positive numbers\r\n$(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\\geq (\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca})^{3}$", "Solution_1": "[quote=\"Albanian Eagle\"]Show that if $a,b,c$ are any three positive numbers\n$(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\\geq (\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca})^{3}$[/quote]\r\n$(a^{2}+a+1)(b^{2}+b+1)\\geq\\left(a+b+\\sqrt{ab}\\right)^{2}.$\r\nThus, $(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\\geq(a+b+\\sqrt{ab})(a+c+\\sqrt{ac})(b+c+\\sqrt{bc})\\geq$\r\n$\\geq\\left(\\sum_{cyc}\\sqrt[6]{ab}\\cdot\\sqrt[3]{a}\\cdot\\sqrt[3]{b}\\right)^{3}=\\left(\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}\\right)^{3}.$" } { "Tag": [ "trigonometry", "function", "quadratics", "algebra", "quadratic formula", "trig identities", "Law of Cosines" ], "Problem": "A teacher asks his pupil to find the greates angle of a triangle with sides 21, 41 and 50. The pupil \"solves\" the problem as follows :\r\n\r\n[i]Let C be the vertex that corresponds with the greatest angle.\nThen sin(1, let p is minimal prime divisor p|n, and T(p) minimal natural period ($T(p)=min(m| \\ p|(2^{m}-1))$. Then $T(p)\\not =1$ (p>2) and $T(p)|n$ give contradition.", "Solution_2": "[quote=\"Rust\"]Only n=1.\nIf n>1, let p is minimal prime divisor p|n, and T(p) minimal natural period ($T(p)=min(m| \\ p|(2^{m}-1))$. Then $T(p)\\not =1$ (p>2) and $T(p)|n$ give contradition.[/quote]\r\n\r\ncan you explain clearly what is $T(p)$ ? I cant get what you mean . :maybe:", "Solution_3": "T(p) is minimal natural m, suth that $p|(2^{m}-1).$ For example $T(7)=3$. T(p) is divisor p-1, therefore $T(p)\\le p-1$.", "Solution_4": "Thank's Rust.Nice solution. :)" } { "Tag": [ "geometry", "perimeter", "algebra", "polynomial", "trapezoid", "modular arithmetic", "quadratics" ], "Problem": "No calculators.\r\n\r\n1. [4] Determine $x$ if \\[x-\\frac{x}{x-\\frac{x}{x-\\frac{x}{x-\\cdots}}}= \\frac{x}{x-\\frac{x}{x-\\frac{x}{x-\\cdots}}}.\\]\r\n\r\n2. [6] If the perimeter and hypotenuse of a right triangle are 14 and 6, respectively, determine its area.\r\n\r\n3. [8] The roots of the polynomial $ax^{3}+bx^{2}+cx+a$ form a geometric series, with $a\\ne 0$. Find $b/c$.\r\n\r\n4. [9] $O$ is the center of the inscribed circle in $30^\\circ-60^\\circ-90^\\circ$ $\\triangle ABC$ with the right angle at $C$. If the circle is tangent to $AB$ at $D$, find $\\angle COD$.\r\n\r\n5. [10] Define a sequence of numbers $a_{1},a_{2},\\ldots,a_{n}$ by $a_{1}=9$ and $a_{n}= a_{n-1}^{2}-1$, $n\\ge 2$. Find the smallest $n$ such that $N$ is the multiple of 10, where \\[N = \\sum_{k=1}^{n}a_{k}.\\]\r\n\r\n6. [11] In isosceles trapezoid $ABCD$, the sum of the bases $AB$ and $CD$ is 3, and $\\angle ACD=60^\\circ$. Find the length of diagonal $AC$.\r\n\r\n7. [12] Find all integer pairs $(m,n)$ such that \\[(m+n)^{4}-(m-n)^{4}= 256.\\]\r\n\r\n8. [13] Twelve points are equally distributed on the circumference of a circle. How many right triangles have three of these points as their vertices?\r\n\r\n9. [13] Alex writes the number 16 on a blackboard. He then performs several \"replacements\" in which he erases the current number and replaces it with one of its factors. In how many ways can Alex end up with the number 1 after five replacements?\r\n\r\n10. [14] $a,b,c,d,x,y,z$ are real numbers such that\r\n\r\n\\begin{eqnarray*}ax+by+cz &=& d,\\\\ bx+cy+dz &=& a,\\\\ cx+dy+az &=& b,\\\\ dx+ay+bz &=& c. \\end{eqnarray*}\r\n\r\nIf $a\\ne c$, find all possible values of $y$.", "Solution_1": "[hide=\"#8\"]\n\nAn angle inscribed in a circle is a right angle if and only if it subtends a semicircle. Thus the opposite side is a diameter. There are 6 diameters among the 12 points. For each there are 10 other points to be the right angle. So $6*10 = 60$ right triangles.\n\n[/hide]", "Solution_2": "[hide=\"#1\"]\n\n$x=4$ lol after some explanation from chess.....thought it was 0 the first time :oops: \n[/hide]", "Solution_3": "[quote=\"7h3.D3m0n.117\"][hide=\"#1\"]\n\n$x=4$ lol after some explanation from chess.....thought it was 0 the first time :oops: \n[/hide][/quote]\n\nPlease post a solution.\n\n[hide=\"#5\"]\nNote that $9 \\equiv-1 \\pmod{10}$. $(-1)^{2}-1 = 0$, so $a_{2}\\equiv 0 \\pmod{10}$. Similarly, $0^{2}-1 =-1$, so $a_{3}\\equiv-1 \\pmod{10}$. Thus the sequence is $-1, 0,-1, 0,-1, 0, \\cdots \\pmod{10}$. We need $10$ $-1$'s to get down to $0 \\pmod{10}$. Since the sequence starts with $-1$, we need 19 terms and $N = 19$.\n[/hide]", "Solution_4": "[quote=\"E^(pi*i)=-1\"][quote=\"7h3.D3m0n.117\"][hide=\"#1\"]\n\n$x=4$ lol after some explanation from chess.....thought it was 0 the first time :oops: \n[/hide][/quote]\n\nPlease post a solution.\n\n[hide=\"#5\"]\nNote that $9 \\equiv-1 \\pmod{10}$. $(-1)^{2}-1 = 0$, so $a_{2}\\equiv 0 \\pmod{10}$. Similarly, $0^{2}-1 =-1$, so $a_{3}\\equiv-1 \\pmod{10}$. Thus the sequence is $-1, 0,-1, 0,-1, 0, \\cdots \\pmod{10}$. We need $10$ $-1$'s to get down to $0 \\pmod{10}$. Since the sequence starts with $-1$, we need 19 terms and $N = 19$.\n[/hide][/quote]\n\n[hide]$x=4$ isn't a solution?[/hide]", "Solution_5": ":rotfl: Sorry. \r\n[hide]$x=4$ [/hide]\r\nis the answer to the problem. I meant that you should post how you solved the problem instead of just the answer.", "Solution_6": "Apologies for not using Latex. my computer has some problems, and i cannot install latex. I tried few times and i kind of gave up installing it.\r\nAnyway,\r\n[hide=\"#2\"]\nHypotenuse=6, so the other sides are x and (8-x).\n(1) Using Pythagoras,\n x^2+(8-x)^2=36\n x^2-8x+14=0\nUsing quadratic formula, x=4+Square root(2) or 4-Square root(2)\n two x-values will give the same answer.\n\n(2) the area of the triangle=1/2*(x)(8-x)\n =1/2*(4-square root(2))(4+square root(2))\n =1/2*(16-2)\n =7 square units.\n[/hide]", "Solution_7": "The answer for 2 is correct but it isn't necessary to invoke the quadratic equation. Take sides $a,b,6$ and let $A$ denote its area. Then\r\n\r\n\\begin{eqnarray*}a+b+6&=&14,\\\\ a+b&=&8,\\\\ a^{2}+b^{2}+2ab&=&8^{2},\\\\ 6^{2}+4A&=&64,\\\\ A&=&\\fbox{7}. \\end{eqnarray*}", "Solution_8": "[quote=\"wonki_noh\"]Apologies for not using Latex. my computer has some problems, and i cannot install latex. I tried few times and i kind of gave up installing it.\nAnyway,\n[hide=\"#2\"]\nHypotenuse=6, so the other sides are x and (8-x).\n(1) Using Pythagoras,\n x^2+(8-x)^2=36\n x^2-8x+14=0\nUsing quadratic formula, x=4+Square root(2) or 4-Square root(2)\n two x-values will give the same answer.\n\n(2) the area of the triangle=1/2*(x)(8-x)\n =1/2*(4-square root(2))(4+square root(2))\n =1/2*(16-2)\n =7 square units.\n[/hide][/quote]\r\n\r\nyou do not need to install LaTeX in order to program it on forums like AoPS. just use the $\\$$ signs for programming", "Solution_9": "serious??\r\n\r\ncan you do the latex writing,, without actually installing it?\r\n\r\ncould you please tell me how??", "Solution_10": "[quote=\"wonki_noh\"]serious??\n\ncan you do the latex writing,, without actually installing it?\n\ncould you please tell me how??[/quote]\r\n\r\nThe forum is LaTeX enabled. Just type as you normally would in LaTeX and it will work.", "Solution_11": "[hide=\"Here is my solution to #1\"]Define\n\n\\[y = \\frac{x}{x-\\frac{x}{x-\\frac{x}{x-\\cdots}}}.\\]\nThen\n\\[y ={x\\over y}\\ \\ \\Rightarrow\\ \\ y^{2}= x.\\]\nFrom the original equation:\n\\[x-y = y\\ \\ \\Rightarrow\\ \\ x=2y.\\]\nSo we have $2y = y^{2}\\ \\ \\Rightarrow\\ \\ y=2.$ This gives\n\\[x=4.\\]\n[/hide]", "Solution_12": "[hide=\"1\"]\n$\\frac{x}{2}=\\frac{x}{x-\\frac{x}{2}}$\n\n$x = \\frac{2x}{\\frac{x}{2}}\\cdot \\frac{2}{2}$\n\n$x = \\frac{4x}{x}$\n\n$x=4$\n[/hide]", "Solution_13": "[hide=\"#4\"]\nWe know that a tangent to a circle is tangent to the radius of the circle, so $m\\angle ODB=90^{\\circ}$. Say that the circle is tangent to $BC$ at $E$ and to $AC$ at $F$. Clearly $\\angle COD=\\angle COE+\\angle EOD$. \n\nIn quadrilateral $EODB$, $m\\angle OEB=m\\angle ODB=90^{\\circ}$ and $m\\angle DBE=60^{\\circ}$. Thus $m\\angle EOD=120^{\\circ}$.\n\nIn quadrilateral $CFOE$, $m\\angle CFO=m\\angle ECF=m\\angle OEC=90^{\\circ}$, thus the quadrilateral is a rectangle. Since $EO=FO$, both being radii of the circle, the quadrilateral is in fact a square. Since $CO$ is a diagonal, $m\\angle COE=45^{\\circ}$.\n\nThus $m\\angle COD=m\\angle COE+m\\angle EOD=120^{\\circ}+45^{\\circ}=165^{\\circ}$.\n[/hide]", "Solution_14": "[hide=\"#7\"]\nSo expanding we see that this is $(m^4 + 4m^3n + 6m^2n^2 + 4mn^3 + n^4) - (m^4 - 4m^3n + 6m^2n^2 - 4mn^3 + n^4) = 256$, so $8m^3n + 8mn^3 = 256$, and $mn(m^2 + n^2) = 32$. Now we test factors and casework. We find that since $m^2$ and $n^2$ must be positive, $mn$ must be positive and all we need to look for are positive factors. \n\nIf $mn = 1$, then $m^2 + n^2 = 32$, so adding $2mn + m^2 + n^2$ yields $34$ which is not a perfect square. So we can throw out this case. \n\nIf $mn = 2$ then $m^2 + n^2 = 16$, so adding yields $m^2 + 2mn + n^2 = 20$ which is not a perfect square. So we can throw out this case.\n\nIf $mn = 4$ then $m^2 + n^2 = 8$, so adding yields $m^2 + 2mn + n^2 = 16$ which is a perfect square! Hence $m + n = \\pm 4$. If $m + n = 4$ then $m = 4 - n$, so $4n - n^2 = 4$ and $(n - 2)^2 = 0$, hence $n = 2$ and $m = 2$. $\\boxed{(2,2)}$. If $m + n = -4$ then $m = -4 - n$, so $-4n - n^2 = 4$, hence $n^2 + 4n + 4 = 0$ and $n = -2$, so $m = -2$. $\\boxed{(-2,-2)}$.\n\nIf $mn = 8$ then $m^2 + n^2 = 4$, so adding yields $m^2 + 2mn + n^2 = 20$ which is not a perfect square.\n\nIf $mn = 16$ then $m^2 + n^2 = 2$, so adding yields $m^2 + 2mn + n^2 = 34$ which is not a perfect square.\n\nIf $mn = 32$ then $m^2 + n^2 = 1$, so adding yields $m^2 + 2mn + n^2 = 65$ which is not a perfect square.\n\nSo we are done :)\n[/hide]" } { "Tag": [ "conics", "parabola", "analytic geometry", "linear algebra", "matrix", "algebra unsolved", "algebra" ], "Problem": "Consider all parabolas of the form $ y\\equal{}x^2\\plus{}2px\\plus{}q$ for $ p,q \\in \\mathbb{R}$ which intersect the coordinate axes in three distinct points. For such $ p,q$, denote by $ C_{p,q}$ the circle through these three intersection points. Prove that all circles $ C_{p,q}$ have a point in common.", "Solution_1": "The points of intersection are $ A(0,q), \\ B(x_1,0), \\ C(x_2,0)$, where $ x_1, x_2$ are the roots of the equation $ x^2\\plus{}2px\\plus{}q\\equal{}0$\r\n\r\nThe equation of the circle passing through A, B, C is\r\n\r\n$ \\begin{vmatrix}x^2\\plus{}y^2 & x & y & 1\\\\\r\nq^2 & 0 & q & 1\\\\\r\nx_1^2 & x_1 & 0 & 1\\\\\r\nx_2^2 & x_2 & 0 & 1\\end{vmatrix}\\equal{}0\\Rightarrow (x_1\\minus{}x_2)(q^2y\\plus{}qy\\minus{}q(x^2\\plus{}y^2)\\minus{}q^2\\minus{}2px)\\equal{}0$\r\n\r\nFor $ x\\equal{}0, \\ y\\equal{}1$ the equality is true $ \\forall p,q$.", "Solution_2": " I claim that the point $(0, 1)$ is on all circles $C_{p, q}$.\n\n The circle $C_{p, q}$ must pass through the three points\n $$\n (-p-\\sqrt{p^{2}-q}, 0), \\quad (-p+\\sqrt{p^{2}-q}, 0), \\quad \\text{and} \\quad (0, q).\n $$\n\n Some computation yields that the circle has the center $O$ and radius $r$ with\n $$\n O = \\left(-p, \\frac{1 + q}{2}\\right), \\quad \\text{and} \\quad r = \\frac{1}{2} \\sqrt{4 p^{2}+(q-1)^{2}}.\n $$\n giving the equation of the circle \n $$\n (x + p)^2 + \\left(y - \\frac{1 + q}{2}\\right)^2 = \\frac{4p^2 + (q - 1)^2}{4}.\n $$\n\n Letting $x = 0$ and $y = 1$ on the LHS gives\n $$\n (p)^2 + \\left(1 - \\frac{1 + q}{2}\\right)^2 = \\frac{4p^2 + (q - 1)^2}{4},\n $$\n so the point $(0,1)$ lies on all circles $C_{p, q}$.", "Solution_3": "[url=https://artofproblemsolving.com/community/c6h1276694p6699291]same?\u200d\u200d\u200d[/url]\n" } { "Tag": [], "Problem": "Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...\r\n\r\ni hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think... :oops: \r\n\r\n\r\nIf ${a}\\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\\oplus{b}=-3$ and $a=1$\r\n\r\ni know the answer...have no solution....and no clue how to do this (other than guess and check) :P", "Solution_1": "[quote=\"DaReaper\"]Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...\n\ni hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think... :oops: \n\n\nIf ${a}\\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\\oplus{b}=-3$ and $a=1$\n\ni know the answer...have no solution....and no clue how to do this (other than guess and check) :P[/quote]\r\n[hide]Just make b a variable and solve for it. So you have $(1+b)^{2}-6b=-3$. Foil and simplify until you get $b^{2}-4b+4=0$. Factoring gets you $(b-2)^{2}$. So b=2.[/hide]", "Solution_2": "wow..i guess that was simple thanks... :blush: :P", "Solution_3": "Ckck, you are completely wrong. You cannot simplify until b^2-4b +4, you can simplify to b^2-6b+4=0, in which there is no solution.\r\nThis is how it SHOULD be done:\r\n(a+b)^2-6b=-3\r\n(1+b)^2-6b=-3\r\n 1+b^2-6b=-3\r\n +3 +3 (To get rid of the stuff on the right)\r\nb^2-6b+4=0\r\n\r\nThere is no way for x to be true so it is no solution.", "Solution_4": "[quote=\"DaReaper\"]Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...\n\ni hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think... :oops: \n\n\nIf ${a}\\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\\oplus{b}=-3$ and $a=1$\n\ni know the answer...have no solution....and no clue how to do this (other than guess and check) :P[/quote]\r\n$(a+b)^{2}-6b=a^{2}+2ab+b^{2}-6b$\r\nPlug in $a=1$ to get $1+2b+b^{2}-6b=-3$ so $b^{2}-4b+4=0$ and $(b-2)^{2}=0$ so $b=2$.", "Solution_5": "[quote=\"myspacesucks\"]Ckck, you are completely wrong. You cannot simplify until b^2-4b +4, you can simplify to b^2-6b+4=0, in which there is no solution.\nThis is how it SHOULD be done:\n(a+b)^2-6b=-3\n(1+b)^2-6b=-3\n 1+b^2-6b=-3\n +3 +3 (To get rid of the stuff on the right)\nb^2-6b+4=0\n\nThere is no way for x to be true so it is no solution.[/quote]\r\n$(1+b)^{2}\\ne1^{2}+b^{2}$, in fact, in general $(a+b)^{2}\\ne(a^{2}+b^{2})$. Instead, $(a+b)^{2}=a^{2}+2ab+b^{2}.$", "Solution_6": "[quote=\"myspacesucks\"]Ckck, you are completely wrong. You cannot simplify until b^2-4b +4, you can simplify to b^2-6b+4=0, in which there is no solution.\nThis is how it SHOULD be done:\n(a+b)^2-6b=-3\n(1+b)^2-6b=-3\n 1+b^2-6b=-3\n +3 +3 (To get rid of the stuff on the right)\nb^2-6b+4=0\n\nThere is no way for x to be true so it is no solution.[/quote]\r\n\r\nI've seen you post things like that a few times now. Note that \r\n\r\n$(5+5)^{2}\\ne 5^{2}+5^{2}$\r\n$(5+5)^{2}= 5^{2}+2(5)(5)+5^{2}$\r\n\r\nIn general, $(a+b)^{2}=a^{2}+2ab+b^{2}$", "Solution_7": "[quote=\"DaReaper\"]Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...\n\ni hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think... :oops: \n\n\nIf ${a}\\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\\oplus{b}=-3$ and $a=1$\n\ni know the answer...have no solution....and no clue how to do this (other than guess and check) :P[/quote]\r\n\r\n[hide]$(a+b)^{2}-6b=-3$\n\nplug in the given $a$ value...\n\n$(1+b)^{2}-6b=-3$\n\n$1+2b+b^{2}-6b=-3$\n\n$b^{2}-4b+4=0$\n\n$(b-2)(b-2)=0$\n\n$\\boxed{b=2}$[/hide]", "Solution_8": "hee hee, stupid me!!!!!!!!!!!!!!!!\r\nsorry, i'm wacko\r\n :blush:" } { "Tag": [ "calculus", "trigonometry", "complex numbers", "\\/closed" ], "Problem": "Will you have these classes available over the summer? pre-calc+calculous?", "Solution_1": "We'll have Intermediate Algebra and Intermediate Trig/Complex numbers in the spring. These are essentially pre-calculus courses. The Algebra course will run through the summer.", "Solution_2": "If I wanted to take pre-calc over the summer, would I need to take the trig one? (I already learned algebra II in school)", "Solution_3": "[quote=\"dbkarp\"]If I wanted to take pre-calc over the summer, would I need to take the trig one? (I already learned algebra II in school)[/quote]\r\n\r\nI'm not sure what you're asking. Our trig class is a lot harder than your typical pre-calc class. (And our algebra is a lot more challenging than Algebra II in school.) If your target is preparing for AMC12/AIME, you'll want to take both of these Intermediate classes (assuming you aren't already easily passing both).", "Solution_4": "I admit my questions were a bit vague, here is what I need to know:\r\n\r\nI'm in eight grade, but taking what they call Algebra III Honors. It's a course with me and a bunch of 11th graders to learn all the hard parts of Algebra II+Some Trig. I'm going into pre-calc next year, but I want to know if I can make that calculous instead. Are there any courses on AoPS I can take to skip pre-calc over the summer.", "Solution_5": "If you want school credit, you will need ask the people at your school if they will accept the course. My guess is that they will not because the AoPS courses do not cover exactly the same material that your state likely wants you to have covered. However, the courses are still good to learn material for contests and for more knowledge in general.\r\n\r\nMy precalc course covered only the very basics of what was covered in the Intermediate Trig and Complex Numbers course...", "Solution_6": "[quote=\"dbkarp\"]I admit my questions were a bit vague, here is what I need to know:\n\nI'm in eight grade, but taking what they call Algebra III Honors. It's a course with me and a bunch of 11th graders to learn all the hard parts of Algebra II+Some Trig. I'm going into pre-calc next year, but I want to know if I can make that calculous instead. Are there any courses on AoPS I can take to skip pre-calc over the summer.[/quote]\r\n\r\nIf you want to replace your precalc with AoPS work, then I would recommend the Interm Algebra and Interm Complex/Trig classes.", "Solution_7": "ok, maybe i should just learn pre-calc at school... Thanks anyway!" } { "Tag": [ "geometry", "3D geometry", "calculus", "calculus computations" ], "Problem": "this is not the exact question because it was from a test, and i didnt have to time to copy it down, but this is basically what we dealing wit here:\r\n\r\ntheres a mountain that resembles a cone (treat it like a cone), and theres a cloud surrounding top of the mountain, (if u cut the cone side ways parallell to the base, the cloud would cover the top small cone) and the bottom of the cloud is moving up at 0.5 m/s (we r assuming the bottom of cloud is a straight line, lol). the height of the mountain, or cone, is 1350 m, diameter of base is 1200, and the question is, how fast is the area not covered by the cloud increasing when the bottom of the cloud is 800 m above the ground (meaning when the bottom part has height 800 m, and small cone on top has 550 m)? in the question it also gave us the equation to find the surface area of the cone, which i think y'all know, but i'll jsut put it here anyways...A= (pi) (r) (s), where r is radius, and s is the slanted height\r\n\r\nif anyone doesnt understand it, can some1 else make a diagram somehow to explain it? just make it simple, assume we looking at it horizontally, so just draw a triangle, the cone in 2d, representing the mountain, and a line in the mid paralell to the base representing the bottom of the cloud...\r\n\r\ni dunno about u guys, but i think that question was quite hard...so...if any1 can solve it for me, that'd be great...", "Solution_1": "Isn't this just a straightforward shadow on a cone problem that is to be solved using \"related rates\"? By the way, very nice avatar. :blush:", "Solution_2": "thx...\r\n\r\nya i think it is, i think ur supposed to find the rate of area w.r.t rate of height, by using chain rule, rate of area / rate of radius x rate of radio / rate of height, but then somehow the slanted height have to go in somewher, i dunno...i was rushed during the tset, so i wasnt thinking straight, i have no idea how to solve it now, and i just dun feel like thinking about it now...so i was hoping some ppl would solve it and post the solutions", "Solution_3": "i just tried it, and i got 840. something, i dunno if its right...can some1 verify the answer plz???" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Define a sequence by $a_1=a_2=1$ and $a_n a_{n-2} = a_{n-1}^2+2$ for $n>2$. Show that all terms of this sequence are odd integers.", "Solution_1": "It is easy to verify that the sequence is also defined by $a_{n+2} = 4a_{n+1} - a_n$.\r\nThe result is now trivial by induction.\r\n\r\nPierre.", "Solution_2": "[quote=\"pbornsztein\"]It is easy to verify that the sequence is also defined by $a_{n+2} = 4a_{n+1} - a_n$.\nThe result is now trivial by induction.[/quote]\r\n\r\nThat's the solution I saw too. But how exactly did you manage to reduce the non-linear recurrsive relation to $a_{n+2} = 4a_{n+1} - a_n$? Just by guess-and-check?", "Solution_3": "The first thing which came to my mind is that it smelt like a Fibonacci property ($F_{2n+3}F_{2n-1} = F_{2n+1}^2 + 1$).\r\nThus I looked over a recurrence of order 2, and that one is easy to find.\r\n\r\nPierre." } { "Tag": [ "AMC" ], "Problem": "are we supposed to receive an email/mail saying AMC has already received our faxed/mailed permission slip?", "Solution_1": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83445]here[/url].", "Solution_2": "http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamo-cklist.html\r\n\r\nSteve Dunbar\r\nAMC Director" } { "Tag": [ "conics", "ellipse", "Euler", "geometry proposed", "geometry" ], "Problem": "Hi all!\r\nI'd like to presant my new problem :) .\r\nI think it is rather unusual problem and also a beautiful result. :P \r\n\r\n[color=blue]Problem.Let $ABC$ be an arbitrary triangle in the plane.${\\varepsilon}_{A}$ is an elipse with focuses $B$ and $C$ and passing trough $A$.The elipses ${\\varepsilon}_{B}$ and ${\\varepsilon}_{C}$ are defining similary.Let $A'$ be the intersection point of ${\\varepsilon}_{B}$ and ${\\varepsilon}_{C}$ that lies in the angle $A$.The points $B'$ and $C'$ are defining similary.Prove that the segments $AA'$,$BB'$ and $CC'$ are intersecting at the same point $E$,which lies on the $Euler$'s line of the triangle.[/color]\r\n\r\nEnjoy it! :wink:", "Solution_1": "Dear users don't forget that you have an unsolved problem here :wink: .", "Solution_2": "I was playing with some test cases, and I found that $AA'$, $BB'$, and $CC'$ do not concur.\r\n\r\nBut the same test cases strongly suggest that the common chords of the ellipses, taken in pairs, concur at the [url=http://mathworld.wolfram.com/deLongchampsPoint.html]de Longchamps point[/url] of triangle $ABC$, which does lie on the Euler line. No proof yet, though.\r\n\r\nBut I agree, it is a very nice problem.", "Solution_3": "[quote=\"nsato\"]I was playing with some test cases, and I found that $AA'$, $BB'$, and $CC'$ do not concur.\n\nBut the same test cases strongly suggest that the common chords of the ellipses, taken in pairs, concur at the [url=http://mathworld.wolfram.com/deLongchampsPoint.html]de Longchamps point[/url] of triangle $ABC$, which does lie on the Euler line. No proof yet, though.\n\n[/quote]\r\nIt sounds like a contradiction,isn't it?\r\nHow can two of them intersect in a fixed point,but all three do not? :maybe: \r\nMaybe there is something wrong with the test cases? :blush:", "Solution_4": "I think you misunderstand me. The ellipses $\\epsilon_{A}$ and $\\epsilon_{B}$ have a common chord, as do $\\epsilon_{A}$ and $\\epsilon_{C}$, and $\\epsilon_{B}$ and $\\epsilon_{C}$. My claim is that all three common chords concur.\r\n\r\nA related problem was proposed for the 1985 IMO:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29201[/url]\r\n[url]http://www.kalva.demon.co.uk/short/soln/sh859.html[/url]", "Solution_5": "Ooops :oops: ,I see what you meant.\r\nThen can you show me a drowing that will show that the main problem is incorrect?In my drowings it is always true :huh: .", "Solution_6": "I found a reference for my conjecture:\r\n[url]http://tech.groups.yahoo.com/group/Hyacinthos/message/458[/url]" } { "Tag": [ "symmetry", "function", "geometry", "3D geometry", "algebra proposed", "algebra" ], "Problem": "$ x^3 \\plus{} 1 \\equal{} 2\\sqrt[3]{{2x \\minus{} 1}}$", "Solution_1": "[hide=\"Step 1\"]$ \\frac{x^3\\plus{}1}{2}\\equal{}\\sqrt[3]{2x\\minus{}1}$\nDo you notice something?[/hide]\n[hide=\"Step 2\"]$ f(x)\\equal{}\\frac{x^3\\plus{}1}{2}$ $ \\to$ $ f^{\\minus{}1}(x)\\equal{}\\sqrt[3]{2x\\minus{}1}$\nThe equation translates to\n$ f(x)\\equal{}f^{\\minus{}1}(x)$[/hide]\n[hide=\"Step 3\"]Using symmetry, $ f(x)\\equal{}f^{\\minus{}1}(x)$ $ \\to$ $ f(x)\\equal{}x$[/hide]\n[hide=\"Step 4\"]$ f(x)\\equal{}x$ $ \\to$ $ \\frac{x^3\\plus{}1}{2}\\equal{}x$\n$ x^3\\minus{}2x\\plus{}1\\equal{}0$ $ \\to$ $ (x\\minus{}1)(x^2\\plus{}x\\minus{}1)\\equal{}0$\nThe solutions are:\n$ x\\equal{}1$ , $ x\\equal{}\\frac{\\minus{}1 \\pm \\sqrt{5}}{2}$[/hide]", "Solution_2": "Hmm, does $ (f(x)\\equal{}f^{\\minus{}1}(x))\\Rightarrow (f(x)\\equal{}x)$ hold for all continuous functions? If yes, could you post a short explanation about that?", "Solution_3": "The solution\r\n\r\n\r\n$ x_0\\equal{}\\minus{} \\frac {1 \\plus{} \\sqrt {5}}{2}$\r\n\r\ndoesn't hold for the equation.\r\n\r\nThe equation has 2 real and 2 complex solutions. The two reals are already written, and the other two are the complex solutions with negative real parts to the following equation:\r\n\r\n$ x^6\\plus{}2 x^4\\plus{}2 x^3\\plus{} 4 x^2 \\plus{}2 x\\plus{}9\\equal{}0$", "Solution_4": "[quote=\"polskimisiek\"]Hmm, does $ (f(x) \\equal{} f^{ \\minus{} 1}(x))\\Rightarrow (f(x) \\equal{} x)$ hold for all continuous functions? If yes, could you post a short explanation about that?[/quote]\r\n\r\nI did not read the other posts, but I can said that $ (f(x) \\equal{} f^{ \\minus{} 1}(x))\\Rightarrow (f(x) \\equal{} x)$ does not hold for all continuous functions.\r\n\r\nExample of continuous functions different from $ f(x)\\equal{}x$ and such that $ f(x)\\equal{}f^{\\minus{}1}(x)$ :\r\n\r\n$ f(x)\\equal{}a\\minus{}x$\r\n\r\n$ f(x)\\equal{}\\sqrt[3]{a\\minus{}x^3}$\r\n\r\n$ f(x)\\equal{}g^{\\minus{}1}(\\minus{}g(x))$ for any bijective function $ g(x)$, $ \\mathbb{R}\\rightarrow\\mathbb{R}$", "Solution_5": "Or, a simple one:\r\n\r\n$ f(x)\\equal{}\\frac{1}{x}$", "Solution_6": "[quote=\"milin\"]Or, a simple one:\n\n$ f(x) \\equal{} \\frac {1}{x}$[/quote]\r\n\r\nwhich, unfortunately, is not continuous at $ x\\equal{}0$", "Solution_7": "True.. But adding $ f(x)\\equal{}1, x\\equal{}0$ should fix the problem.", "Solution_8": "The complex roots aren't even particularly well-defined since it's not clear which cube root to take. Anyway, I'd suggest looking at the graph of $ f(x) \\equal{} \\frac {x^3 \\plus{} 1}{2}$ and then reflecting it across the line $ y \\equal{} x$. If there are intersection points off of $ y \\equal{} x$, we can find them graphically and then show them algebraically.", "Solution_9": "[quote=\"milin\"]True.. But adding $ f(x) \\equal{} 1, x \\equal{} 0$ should fix the problem.[/quote]\r\n\r\nNot really.\r\n\r\nIt allows $ f(x)\\equal{}\\frac{1}{x}$ to be defined at $ x\\equal{}0$ but not to be continuous.", "Solution_10": "OK, so if that does not hold for every continuous function, than what are the required assumptions? And why [b]lordWings[/b] could use that here?", "Solution_11": "Although many functions can have $ a$ such that $ f(f(a))\\equal{}a$ without $ f(a)\\equal{}a$, only solution to $ f(x)\\equal{}x$ will solve $ f(f(x))\\equal{}x$, so you can know what kind of factor to pull out.\r\n\r\nIn this case, $ x^3\\plus{}1\\equal{}2\\sqrt[3]{2x\\minus{}1}$ can be written as\r\n\\[ x^3\\minus{}2x\\plus{}1\\equal{}2\\left( \\sqrt[3]{2x\\minus{}1}\\minus{}x\\right) \\equal{}\\minus{}2\\cdot \\frac{x^3\\minus{}2x\\plus{}1}{x^2\\plus{}x\\sqrt[3]{2x\\minus{}1}\\plus{}(2x\\minus{}1)^{2/3}}\\]\r\nSo any solution other than a root of $ x^3\\minus{}2x\\plus{}1$ must have\r\n\\[ x^2\\plus{}x\\sqrt[3]{2x\\minus{}1}\\plus{}(2x\\minus{}1)^{2/3}\\equal{}\\minus{}2\\]\r\nBut this has no real solutions, since $ y\\equal{}\\sqrt[3]{2x\\minus{}1}$ implies $ x^2\\plus{}xy\\plus{}y^2\\equal{}\\left(x\\plus{}\\frac{y}{2}\\right)^2\\plus{}\\frac{3}{4}y^2\\geq 0$.", "Solution_12": "Too easy\n\n Setting $ y = \\sqrt[3]{2x-1}$\n\n $ \\implies y^3 - 2x +1 \\ = \\ 0$ and $ x^3 - 2y +1 \\ = 0 \\ (*)$\n\n $ \\implies (y^3 - 2x +1)-(x^3 - 2y +1) \\ = \\ 0$\n\n $ \\iff (y-x)( y^2 + xy + y^2 + 2) \\ = \\ 0$\n\n $ \\iff y = x$\n\n replace $y$ with $x$ in $(*)$ ; we have : $ x^3 - 2x + 1 \\ = \\ 0$\n\n $ \\iff (x-1)( x^2 + x -1) \\ = \\ 0$\n\n $ \\iff x = 1 \\ ; x = \\frac{-1 \\pm \\sqrt{5}}{2}$" } { "Tag": [ "calculus", "integration", "inequalities", "function", "real analysis", "induction", "triangle inequality" ], "Problem": "This one is pretty tough and technical:\r\n Prove the existence of a sequence of positive numbers $c_k$ with the properties:\r\n 1) For any $\\lambda>0$, any $ a0$, any function $f$ of class $C^{k+1}$ such that $|f^{(k)}|\\geq 1$ for all $x\\in [a,b]$ we have ${|\\int_{a}^{b}{e^{i\\cdot\\lambda\\cdot f(x)}dx}|\\leq\\frac{c_k}{\\sqrt[k]{\\lambda}}}$. This is the problem with the greatest number of \"any\" that I have ever seen. :D", "Solution_1": "I'll take care of the easy part here :) , namely proving that there is such a $ c_1.$\r\n\r\n\r\nThis resembles the Riemann-Lebesgue lemma in the case when the function is $ \\mathcal{C}^1,$ so the proof will be similar.\r\n\r\n\r\nWLOG, we may assume that $ \\forall x \\in [a, b], \\ \\ f'(x) \\geq 1.$\r\n\r\nAn integration by parts yields :\r\n\\begin{eqnarray*} \\int_a^b e^{i \\lambda f(x)} dx & = & \\int_a^b \\frac 1{i \\lambda f'(x)} (e^{i \\lambda f(x)})' dx \\\\\r\n& = & \\frac 1{\\lambda} \\left ( \\left [\\frac 1{i f'(x)} e^{i \\lambda f(x)} \\right ]_a^b + \\int_a^b \\frac 1i e^{i \\lambda f(x)} \\frac {f''(x)}{f'^2(x)} dx \\right ) \\end{eqnarray*}\r\n\r\nTherefore we get, with a few applications of the triangle inequality :\r\n\\begin{eqnarray*} \\left | \\int_a^b e^{i \\lambda f(x)} dx \\right | & = & \\frac 1{\\lambda} \\left |\\frac 1{i f'(b)} e^{i \\lambda f(b)} - \\frac 1{i f'(a)} e^{i \\lambda f(a)} + \\frac 1i \\int_a^b e^{i \\lambda f(x)} \\frac {f''(x)}{f'^2(x)} dx \\right | \\\\\r\n& \\leq & \\frac 1{\\lambda} \\left (\\frac 1{f'(a)} + \\frac 1{f'(b)} + \\int_a^b \\frac {|f''(x)|}{f'^2(x)} dx \\right ) \\\\\r\n& = & \\frac 1{\\lambda} \\left (\\frac 1{f'(a)} + \\frac 1{f'(b)} + \\left | \\int_a^b \\frac {f''(x)}{f'^2(x)} dx \\right | \\right ) \\\\\r\n& = & \\frac 1{\\lambda} \\left (\\frac 1{f'(a)} + \\frac 1{f'(b)} + \\left | \\frac 1{f'(a)} - \\frac 1{f'(b)} \\right | \\right ) \\\\\r\n& \\leq & \\frac 2{\\lambda} \\left (\\frac 1{f'(a)} + \\frac 1{f'(b)} \\right) \\\\\r\n& \\leq & \\frac 4{\\lambda} \\end{eqnarray*}\r\n\r\nHence, we may take $ \\boxed{c_1 = 4}.$", "Solution_2": "Well, it's actually almost all the work. A proof by induction using the same arguments will prove the statement. Note however that the proof will not be completely trivial, since there will be some cases to discuss. It's a good exercise." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let $MAB$ be a triangle with constant base $AB$. Assume that the point $M$ moves\r\n\r\non to its plane such that the external squares $MAKC$, $MBHZ$, $MBLI$ that figures\r\n\r\nfulfil the condition $[AMCK]+[BMZH]+[ABLI]+4[MAB]=2002$ where with $[...]$ we \r\n\r\nsymbolise the areas. Find the locus of the point $M$.", "Solution_1": "Noone has a solution to this nice problem ?Thanks.\r\n\r\nBTW, the 34 views in two months saw that my problem didn't interest you :(", "Solution_2": "it looks hard,,, :(" } { "Tag": [ "search" ], "Problem": "Supose that the refractive index of a region in the space is proportional to the inverce of the distance from the point $ (0,0)$ (working in 2 dimensions for simplicity).\r\n\r\nIf a ray of light start at the point $ (r,\\theta)$ for $ \\frac {\\pi} {2} > \\phi> 0 > r$ , $ \\frac {\\pi} {2} < \\theta < \\pi$ , The ray will describe a Arc of a circunference and will ever end in $ (0,0)$. ($ \\phi$ is the tangent angle needed to solve this problem\r\n\r\n(Theorically speaking, 'cause at the point $ (0,0)$ $ n$ would be infinity, and at very far points $ n$ would be zero.)", "Solution_1": "[url=http://www.mathlinks.ro/viewtopic.php?search_id=682405916&t=167520]here[/url] is a similar problem posted by me" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Factorise:\r\n\r\n a^3 - 2b^3 + (a+b)^3\r\n\r\nThank you.", "Solution_1": "I think it can't be factored into 2 polynomials of $ a,b$ with integer coefficients." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "1. Cho a-b+c=0.CMR: asinx+9bsin3x-25csin5x=0. CMR co it nhat 4 nghiem phan biet thuoc (0;2pi)\r\n2. CMR: a(25sin5x-sinx)+b(49sin7x-9sin3x)=0 co it nhat 7 nghiem phan biet thuoc (0;2pi)\r\n\r\nAi nguoi Viet Nam dich. ho^ nhe'", "Solution_1": "[quote=\"Mather\"]1. Cho a-b+c=0.CMR: asinx+9bsin3x-25csin5x=0. CMR co it nhat 4 nghiem phan biet thuoc (0;2pi)\n2. CMR: a(25sin5x-sinx)+b(49sin7x-9sin3x)=0 co it nhat 7 nghiem phan biet thuoc (0;2pi)\n\nAi nguoi Viet Nam dich. ho^ nhe'[/quote]\r\n\r\n[color=darkblue]He is a new member, He can't post by English. \n\nThis is his problem:\n\n$ 1$. Let $ a \\minus{} b \\plus{} c \\equal{} 0$. Prove that this equation: $ asinx \\plus{} 9bsin3x \\minus{} 25csin5x \\equal{} 0$, has at least 4 roots distributive $ \\in (0; 2\\pi)$.\n\n$ 2$. Prove that this equation: $ a(25sin5x \\minus{} sinx) \\plus{} b(49sin7x \\minus{} 9sin3x) \\equal{} 0$, has at least 7 roots distributive $ \\in (0; 2\\pi)$.[/color]" } { "Tag": [ "geometry", "3D geometry", "analytic geometry" ], "Problem": "You need to hold your breath to get out of a room. Assume that a room is in\r\nthe shape of a cube that was 5 inches on each side. You are located 1 inch\r\nfrom the top and 2 inches from the side of one wall. The opening is on\r\nthe opposite wall 3 inches from the top and one inch from the side. Also, assume\r\nyou can hold your breath for five more seconds and can crawl 2 inches per\r\nsecond. Are you able to escape? What is the shortest pathway out?", "Solution_1": "The problem is kind of obscure.\r\n[hide]\nShortest path $3$.\nYes you can escape.\nBut that does not seem valid.\n[/hide]", "Solution_2": "\"You are located 1 inch from the top and 2 inches from the side of one wall\". So if you put this in form \r\n([i]x, y, z[/i]), you gave the x coordinates and the y coordinated but you never gave the z coordinates. \r\n\r\nAlso, you can \"crawl\" 2 inches per second. Don't you mean fly 2 inches per second? Because that looks like you have to travel in air." } { "Tag": [ "geometry", "rectangle", "AMC 10" ], "Problem": "An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle and not more than 3 units from a point of the triangle?", "Solution_1": "3 120 degree circle sectors with radii 3 and 3 3x6 rectangles:\r\n$ (3)\\dfrac{120}{360}3^2\\pi \\plus{} 3(3)(6) \\equal{} 9\\pi \\plus{} 54$\r\n\r\n2008 AMC 10 Problem 17", "Solution_2": "Surrounding the circle, there will be three 3*6 rectangles and at each corner there will be a 1/3 of a circle with radius 3. Therefore, area= 3(3x6) + 3(1/3pi3\u00b2) = 54 + 9pi" } { "Tag": [], "Problem": "This song makes me laugh everytime i listen to it. \r\n\r\nhttp://viknluda.com/foe/video/", "Solution_1": "hehehehe\r\n :rotfl: \r\n\"Oh my god look she's a white!\"\r\nSo racist yet so funny.", "Solution_2": "oh my gosh lol :rotfl:", "Solution_3": "[quote=\"Klebian\"]So racist yet so funny.[/quote]\r\n\r\nSo we're locking this discussion." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all primes p such that each nonzero element of Z (little p) is its own multiplicative inverse. You must prove that the primes you find work and are the only ones that work.", "Solution_1": "Is there something wrong with me, or is it trivial? For all $x \\in \\{1, 2, ..., p-1\\}$ we must have $x^2 \\equiv 1 \\mod{p}$ but that's equivalent to $(x-1)(x+1) \\equiv 0 \\mod{p}$ so, all residues $\\mod{p}$ are $1$ or $-1$ which implies that $p=2$ and $p=3$ are only primes which work.", "Solution_2": "for $p=2,3$ it obviously works\r\n\r\nnow $\\mathbb{Z}_p$ is a field and $a=a^{-1}$ implies $a^2=1$ which can not have more than two solution, implying $p-1 \\leq 2$ and we are done", "Solution_3": "Well, it doesn't get harder for nonprimes (looking at the invertible elements then):\r\nWhen any number $k$ divides it, by reducing $\\mod k$ shows that also $k$ itself works. Thus neither a prime $>3$ nor $9$ nor $16$ can divide a working number (the first by the above post, the other two by simple checking).\r\nThus the only numbers left are $2,3,4,6,8,12,24$, and all those work." } { "Tag": [ "algebra", "polynomial", "function", "functional equation", "algebra proposed" ], "Problem": "Find all solutions $ f$: $ R$ \u2192 $ R$ of the functional equation\r\n$ f(f(x) \\plus{} f(y) \\plus{} kxy) \\equal{} xf(y) \\plus{} yf(x)$, x, y \u2208 R.\r\nConsider the case when f is (a) a polynomial, (b) a continuous function, (c) an arbitrary\r\nfunction.", "Solution_1": "Dear ghostchungho,\r\n\r\nYou are not allowed to post this functional equation. \r\nIt is a part of the list of problems suggested by the Jury of the International Tournament of Young Mathematicians (visit http://www.itym.org/). These problems are not to be widespread on any forum and it is prohibited to discuss or publish any possible solutions to them before the 6th of July 2009.\r\n\r\nAll rights reserved.\r\n\r\nSincerely,\r\nThe ITYM Organizing Committee", "Solution_2": "I'm so sorry! \r\nSo it is a probleme of a competition!!! :lol:", "Solution_3": "Apologies accepted.\r\n\r\nIndeed, it is a research-type problem from our competition." } { "Tag": [ "MATHCOUNTS", "geometry", "3D geometry", "probability" ], "Problem": "Find a two-digit number such that the square of the sum of its digits is equal to the number.\r\nFind a three-digit number such that the cube of the sum of its digits is equal to the number.\r\nGeneralize. (Find a x-digit number such that the number^x of the sum of its digits is equal to the number.\r\n\r\n...*yawn* This'll be the last problem of the day, good night.\r\nAnnouncer: And the record holds at 15! 15 folks! :lol:", "Solution_1": "[hide]1) 81. You know the number has to be a square number due to the fact that the sum of the digits is squared. \n\n\n\n2) 512. Same theory as above. The number has to be a perfect cube.\n\n\n\n3) No clue. Haven't learned how to generalize yet. Would someone please explain it to me?[/hide]", "Solution_2": "You're correct, maybe even for the third one. I was so sleepy when I wrote this problem that I didn't quite check if there existed a generalization :)", "Solution_3": "But how [b]do[/b] you generalize? You ask for generalizatoions on a lot of the other problems you post, but I've never been taught how to do them.", "Solution_4": "meh. basically just apply what you did but with variables!", "Solution_5": "you know how in this problem, it says (2-digit) and (3-digit)? Well, that's pretty specific, choosing one particular integer from the many many integers out there, so we generalize to an n-digit integer. I don't think a nice compact formula exists for this problem, nor am I sure if one such number always exists, but that would be a possible generalization of this problem.", "Solution_6": "[quote=\"MathFiend\"]But how [b]do[/b] you generalize? You ask for generalizatoions on a lot of the other problems you post, but I've never been taught how to do them.[/quote]\r\nNotice that I usually put two problems when I say \"Generalize\"\r\nJust find the number that changed and replace them with x or y or whatever and solve the problem in term of x and y and such.\r\nExample: I flip a coin 2 times. What is the probability of getting all heads? Answer: 1/2:^2:=1/4\r\nI flip a coin 5 times, What is the probability of getting all heads? Answer: 1/25=1/32\r\nGeneralize. :arrow: I flip a coin x times, What is the probability of getting all heads? Answer: 1/2x\r\n\r\nI'll go and change all the problem that just says \"Generalize\" into a question with variables just in case.", "Solution_7": "Could someone post a solution to this? It's a very interesting problem. I doubt it's possible, though.", "Solution_8": "Actually, it ONLY works for one more: 7^4's digits add up to 7.\r\nAfter that, it never works. When you get to 5, only 7^5, 8^5, and 9^5 have 5 digits, so it's easy to check all those.\r\n\r\nCuriously, for n>21, NO integer has n digits when you take it to the nth power..." } { "Tag": [], "Problem": "How do you convert a base n number to base n^2?\r\n\r\nLike how would you quikly convert (for example) 110110 in base 3 to a base 9 number?", "Solution_1": "Well, let $ N$ be the number in base $ b$, and let the digits be $ d_n,d_{n\\minus{}1},\\cdots,d_1,d_0$, so \\[ N\\equal{}d_0\\plus{}b\\cdot d_1\\plus{}b^2\\cdot d_2\\plus{}b^3\\cdot d_3\\plus{}\\cdots \\plus{} b^n d_n\\]\r\nTo convert to base $ b^2$, observe that \\[ N\\equal{}(d_0\\plus{}b\\cdot d_1)\\plus{}b^2(d_2\\plus{}b\\cdot d_3)\\cdots\\]\r\n\r\nSo each digit in the base $ b^2$ number is the combination of two digits in the base $ b$ number, starting at the right side. \r\n\r\nIn your example, we group 11 01 10 and since it's originally in base 3 the digits are $ (3\\cdot 1\\plus{}1),(3\\cdot 0\\plus{}1),(3\\cdot 1\\plus{}0)$, so in base 9 the number is 413." } { "Tag": [ "modular arithmetic", "trigonometry" ], "Problem": "If $R_n = \\frac{1}{2}(a^n + b^n)$ where $a = 3 + 2\\sqrt{2}$, $b = 3-2\\sqrt{2}$, and $n = 0,1,2,\\ldots,$ then $R_{12345}$ is an integer. Its units digit is:\r\n(A) 1\r\n(B) 3\r\n(C) 5\r\n(D) 7\r\n(E) 9", "Solution_1": "[hide]$R_n=\\frac{1}{2}(a^n+b^n)=\\frac{1}{2}((a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2}))=6R_{n-1}-R_{n-2}$.\n\n$R_0=1,\\ R_1=3$. Then by recursion, look at the units digit of $R_n$: $R_2\\equiv 7,\\ R_3\\equiv 9,\\ R_4\\equiv 7,\\ R_5\\equiv 3,\\ R_6\\equiv 1,\\ R_7\\equiv 3\\pmod{10}$.\n\nSince the units digits of $R_6$ and $R_7$ are the same as $R_0$ and $R_1$ and the recursion only looks at the last two numbers in the sequence, the units digit cycles every 6.\n\n$12345\\equiv 3\\pmod{6}\\Rightarrow$ units digit is 9 $\\Rightarrow \\boxed{E}$.[/hide]", "Solution_2": "[hide]\nLet $w=3 + 2\\sqrt2 i=r \\text{ cis } \\theta$ and $v = 3 - 2\\sqrt2 i= r \\text{ cis} (- \\theta)$. So we want $\\frac{\\Re(s) + \\Im(s)}{2}$ where $s = w^n + v^n$. By deMoivre, $s = 2r^n \\cos n\\theta$, so the thing we want just turns into $r^n \\cos n\\theta$. We can find $\\cos \\theta$ by constructing a right triangle with sides $3$ and $2\\sqrt2$ and hypotenuse $\\sqrt{17}$, getting $\\cos \\theta = \\frac{3}{\\sqrt{17}}$ and $\\sin \\theta = \\frac{2 \\sqrt2}{\\sqrt{17}}$. Using our definition of $w$ from above, we see $3 + 2\\sqrt2i = r \\left( \\frac{3}{\\sqrt{17}} + \\frac{2 \\sqrt2}{\\sqrt{17}}i \\right)$, so $r=\\sqrt{17}$. Now we want the units digit of $17^{\\frac{12345}{2}} \\cdot \\cos(12345 \\theta)$... and I can't find the latter part. I edited since I forgot about the 12345 coefficient before, which simplified very nicely and came out to one of the choices (B), but is probably not right...\n[/hide]" } { "Tag": [ "induction", "modular arithmetic", "limit", "ratio", "logarithms", "geometry", "trigonometry" ], "Problem": "It seems that Slizzel already took the idea of Intermeidate Problem Solving Marathon so I'm trying to do something even \"harder.\"\r\n\r\nThis is proof marathon so there will be a proof question and you have to give your answer with full proof. I think it would be really good idea to see a lot of good proof. I'm not sure how hard they have to be as long as someone can answer it. But let's not try to go too hard.\r\n\r\nWhen posting problem, please post problem and hint in case it's too hard. :D And put source as well.\r\n\r\n[size=150]Proof 1[/size]\r\n\r\nFMO 2001\r\n\r\nProve that for every natural number $n$ greater than 1 satisfies:\r\n\r\n$(n-1)^2$ is a divisor of $n^{n-1} - 1$.\r\n\r\n[hide=\"hint\"]\nI think induction will be nice in this one.[/hide]", "Solution_1": "I like this proof, although it's probably not one that most people would think of.\r\n\r\nWe have\r\n\r\n${n^{n-1}-1=(n-1)(n^{n-2}+n^{n-3}+\\ldots+1})$\r\n\r\nSo we need to show that the last factor is divisible by $n-1$. Towards that end, subtracting $2(n-1)$ from the last factor doesn't change anything. After we do this, the last factor becomes\r\n\r\n$n^{n-2}+n^{n-3}+\\ldots +n^2-(n-3)$. This can be rewritten as\r\n\r\n$(n^{n-2}-1)+(n^{n-3}-1)+(n^{n-4}-1)+\\ldots +(n^2-1)$\r\n\r\nand it's obvious that all of those terms are divisible by $n-1$\r\n\r\nSo we're done!", "Solution_2": "wouldnt this be more appropriate for Pre-Olympiad??", "Solution_3": "Hope you don't mind ;)", "Solution_4": "I got lost after the word \"So\" and where you subtracted $2(n-1)$.\r\n\r\nI was thinking of induction.. Except I stuck on $k+1$ because:\r\n\r\n$(k+1-1)^2 = k^2$ and $k^{k+1-1} - 1 = k^k - 1$...\r\n\r\nThis was the final round of FMO... It should be hard.. :D", "Solution_5": "[quote=\"Silverfalcon\"]I got lost after the word \"So\" and where you subtracted $2(n-1)$.\n[/quote]\r\nblahblahblah's proof is perfectly valid.\r\n\r\nTo help you to understand his proof, I am trying to explain in a different way:\r\n\r\n$n^k\\equiv 1\\pmod{n-1}$ since $n\\equiv 1\\pmod{n-1}$.\r\n\r\nSo ${n^{n-2}+n^{n-3}+\\cdots+1}\\equiv 1+1+\\cdots+1\\equiv n-1 \\equiv 0 \\pmod{n-1}$", "Solution_6": "Yes, the key is to break up $(n-3)$ into $(n-3)$ ones. We can do this because the numbers of the form $n^j$ are indexed by $j=2$, $j=3$, $j=n-4$, $j=n-3$, $j=n-2$, etc. As I said, it's not obvious, and its certainly not the way I would have done it back in high school when I was starting on proofs, but it's both direct and cute, in my opinion.", "Solution_7": "[quote=\"Silverfalcon\"]I got lost after the word \"So\" and where you subtracted $2(n-1)$.\n\nI was thinking of induction.. Except I stuck on $k+1$ because:\n\n$(k+1-1)^2 = k^2$ and $k^{k+1-1} - 1 = k^k - 1$...\n\nThis was the final round of FMO... It should be hard.. :D[/quote]\r\n\r\n[hide]I think you forgot that you are trying to prove $k^2$ is a divisor of $(k+1)^k-1$.\n\nAfter this, you can do as blahblahblah did and make $(k+1)^k-1=(k+1-1)((k+1)^{k-1}+(k+1)^{k-2}+...+1)$.\n\nAgain, since the $k$ on top will cancel we are proving that $((k+1)^{k-1}+(k+1)^{k-2}+...+1)$ is divisble by $k$. This obvious since every term will have a $k$ in it except for the $1$ (use binomial expansion). But then since the first term will make a $1^{k-1}$ and the second will make $1^{k-2}$ all the way to $1^0$, this adds up to $k$. So thus, it's divisible.[/hide]", "Solution_8": "Here's a classical one that everyone should like. :)\r\n\r\nUhh...I guess Proof 2:\r\n\r\nProve that there exists an infinite number of primes.", "Solution_9": "Suppose there are a finite amount of primes\r\n$p_1, p_2, p_3, ... , p_k$\r\nThen $p_1*p_2*p_3*....*p_k+1$ is clearly not divisible by any of the primes.\r\nThus $p_1*p_2*p_3*....*p_k+1$ is a new prime.\r\nProved by contradiction\r\n\r\n[b]Proof 3:[/b]\r\nProve that :sqrt: 2 is irrational", "Solution_10": "Assume $\\sqrt{2}$ is rational, so that $\\sqrt{2}=\\frac{a}{b}$ for relatively prime integers $a,b$.\r\nThen $2=\\frac{a^2}{b^2}$, and $a^2=2b^2$. But then $2|a$, so $a=2a_0$ for some integer $a_0$ and $4a_0^2|b^2$, so $2|b$, and $a,b$ are not relatively prime. Hence, $\\sqrt{2}$ is not rational.\r\n\r\n\r\nProof 4:\r\nProve that the $n$th triangular number is composite for $n>2$.", "Solution_11": "The nth triangular number is equal to (1+2+...+n), which is equal to n(n+1)/2. This is obviously a whole number, and it has two factors: either has n/2 and n+1, or n and (n+1)/2, depending on which is even.\r\nProve: Lim([n+1]/n)^nr=Lim([n+r]/n)^n as n goes to infinity.", "Solution_12": "[quote=\"K81o7\"]The nth triangular number is equal to (1+2+...+n), which is equal to n(n+1)/2. This is obviously a whole number, and it has two factors: either has n/2 and n+1, or n and (n+1)/2, depending on which is even.\nProve: Lim([n+1]/n)^nr=Lim([n+r]/n)^n as n goes to infinity.[/quote]\r\nfirst, make it clear what exactly does \"r\" stands for, also, assuming \"r\" is a constant, then you need a parranthesis around \"nr\"--that's what i think... ;)", "Solution_13": "Okay..\r\n\r\nProve that Lim([n+1]/n)^[nr]=Lim([n+r]/n)^n as n goes to infinity given that r is a natural number", "Solution_14": "there we go :lol: \r\nthe $\\lim$ on both sides is equal to $1$. because on the left side you will have something like this: $\\frac{n^{nr}+....}{n^{nr}}$ where the $\\lim$ is one (the ratio of the highest degree of coefficents)\r\nsame thing for the right side, the $\\lim$ would be the same$=1$", "Solution_15": "OK i got it. writing $b=\\frac{1}{a}$, and subing this in, multiply across by $a^3$ and we get something like this:\r\n\r\n$a^6-2a^4+a^3-a+1 \\geq 0$ By inspection and dividing out factors, we find three obvious roots: $-1$, and a double root $1$, taking these out we get:\r\n\r\n$ \\leftrightarrow (a^3+a^2+1)(a+1)(a-1)^2 \\geq 0$\r\nWhich is obvious", "Solution_16": "that's it i did it the same way. there still alot more ways, i know of 4 elementary solutions.\r\nnevertheless, next proof plz:\r\n\r\npeeta", "Solution_17": "[quote=\"seamusoboyle\"]Firstly note that one of $a$ or $b$ is $ \\geq 1$\nWe can transform this inequality into:\n\n$(a+b)(a-b)^2+1-a \\geq 0$ if it is b that is greater than or equal to 1, this holds because a would then be less than or equal to 1.\n\nNow the other way round: if b is less than equal to 1, set the $1$ in that expression to $ab$ and we get:\n\n$(a+b)(a-b)^2+a(b-1) \\geq 0$ which holds in this case\n\n[size=150]Problem the next:[/size]\n\nThe series\n\n$f(n)=\\lfloor n + \\sqrt{2n} + \\frac{1}{2} \\rfloor$ misses exactly the triangular numbers[/quote]\r\n\r\nThis one?", "Solution_18": "ok i got it now, finally.\r\n\r\nthe only problem is that i am using a lemma, which is kind of obvious, when u look at it, but it's just not fun to proof. \r\n\r\nLemma 1:\r\n$g(n):=$ is the function that misses all triangle numbers\r\nso i let $n=\\frac{k^2+k}{2}-m$ with $0\\le m 8m\\\\\r\nk>m$\r\nwhich is stated before\r\n\r\n2.\r\n$\\frac{1}{2}\\cdot\\left( \\sqrt{2n}+1\\right)^2< \\frac{1}{2}\\cdot\\left(k^2+3k-2m\\right) +1\\\\\r\nk^2+k-2m+1+2\\sqrt{k^2k-2m} angle(ANM) \\geq 60 =angle(MAN) $\r\n\r\nSo $|AM| >= |MN|$ ... and so on ...\r\n\r\n :cool: I like the innocent questions :cool:", "Solution_2": "All right. Thanks, Diogene, but it was the straightforward solution which is the ugly solution I was talking about. \r\n\r\nIs there not an other one? I just can't believe it!!!", "Solution_3": ":cool: Really ? What is it so ugly ?!! :cool:", "Solution_4": "It is just a rather long solution. If you consider the same problem for an 2n-gon: you have a regular 2n-gon and two points inside and you try to find the largest possible distance between them.\r\n\r\nThen you just draw the circumcircle and the larges possible distance must be 2r(with r as radius).\r\n\r\nI awaited a similiar prove for a (2n+1)-gon or at least for a triangle. Or perhaps an unawaited proof using projections, reflections or something like that.\r\n\r\nThe other disadvantage of our solution is that it can't be generalized for a (2n+1)-gon :( .\r\n\r\nMisha", "Solution_5": "perhaps I dont understand the question: can't you draw a circle of radius 1 centered on p, it will contain the triangle...", "Solution_6": "Right, but you can't just assert that the circle contains the triangle; you have to prove it.", "Solution_7": "umm :P isn't this a bit very trivial? :P\r\n\r\nI mean, the triangle ABC has a distance 1 between A and C, and it only decreases when moving the points away from the sides...\r\n\r\nPlus you don't need to prove anything about the circle with radius 1, not on any good math contest. That's just too trivial :P", "Solution_8": "I agree that the problem is easy, but that doesn't eliminate the need to prove assertions, especially when that is what the original proposer is asking for.", "Solution_9": "Misha, there is absolutely no wonder that the problem you posted is not trivial. Concerning such problems from convexity geometry, I advise you to take a look at [url=http://groups.google.com/groups?hl=de&lr=&ie=UTF-8&threadm=2mc5ifFl2lufU1%40uni-berlin.de&rnum=3&prev=/groups%3Fq%3Dpolygons%2Bhave%2Bears%26hl%3Dde%26lr%3D%26ie%3DUTF-8%26selm%3D2mc5ifFl2lufU1%2540uni-berlin.de%26rnum%3D3]this de.sci.math thread[/url]. One of the problems discussed there was to prove that every polygon (not necessarily convex) has a diagonal which completely lies inside the polygon. Trivial? By no means; it could rather be a pretty good olympiad problem if one wouldn't have to expect so many students bombarding the jury with justified question on how the \"interior\" of a non-convex polygon is defined.\r\n\r\nYour problem has a much simpler outset, an equilateral triangle, but it is clear that it is actually a special case of the following problem: Given an arbitrary polygon $A_1A_2...A_n$ (not necessarily convex, but not self-intersecting), and two points P and Q inside it. (Here, \"inside\" can also mean \"on the boundary\".) Prove that the distance PQ is not greater than the greatest of the pairwise distances $A_iA_j$ between two vertices of the polygon.\r\n\r\nIn order to prove it, we note that since the points P and Q lie inside the polygon $A_1A_2...A_n$, the line PQ intersects the boundary of the polygon at two points, which we may call U and V. (If the polygon is concave, the line PQ may intersect its boundary at more than two points; in this case, take the leftmost one and the rightmost one.) Clearly, PQ $\\leq$ UV. So it remains to show that the distance UV is not greater than the greatest of the distances $A_iA_j$. In fact, there is a pretty easy lemma saying that if C is a point on a segment AB, and O is another point in the plane, then the distance OC is less or equal to the greater of the two distances OA and OB. Now, the points U and V both lie on the boundary of the polygon $A_1A_2...A_n$, i. e. each of these two points lies on one side of this polygon. Call $A_uA_{u+1}$ the side on which the point U lies, and call $A_vA_{v+1}$ the side on which the point V lies. Then, the lemma above, we see that the distance UV is less or equal to the greater of the distances $UA_v$ and $UA_{v+1}$. We assume, for instance, that the greater of the distances $UA_v$ and $UA_{v+1}$ is $UA_{v+1}$. Then, UV $\\leq$ $UA_{v+1}$. Now, applying the lemma again (this time we use that the point U lies on the segment $A_uA_{u+1}$), we see that the distance $UA_{v+1}$ is less or equal to the greater of the distances $A_uA_{v+1}$ and $A_{u+1}A_{v+1}$. This time, let's assume that the greater of the distances $A_uA_{v+1}$ and $A_{u+1}A_{v+1}$ is $A_uA_{v+1}$. Then, $UA_{v+1} \\leq A_uA_{v+1}$, and we get the chain of inequalities\r\n\r\n$PQ \\leq UV \\leq UA_{v+1} \\leq A_uA_{v+1}$.\r\n\r\nAnd $A_uA_{v+1}$ is, of course, less or equal to the greatest of the pairwise distances $A_iA_j$. This completes the proof.\r\n\r\nI hope I didn't make stupid errors above; I am really unexperienced in this topic of geometry...\r\n\r\n Darij", "Solution_10": "Here's another solution to darij's more general problem, using vectors. Because $P$ is in the convex hull of the $A_i$, we have $\\textstyle P = \\sum_i p_i A_i$ for some probability distribution $p$. Similarly $\\textstyle Q = \\sum_j q_j A_j$ for some probability distribution $q$. Thus\r\n\\begin{eqnarray*}\r\n|P - Q|\r\n & = & \\bigl| \\sum_i p_i A_i - \\sum_j q_j A_j \\bigr| \\\\\r\n & = & \\bigl| \\sum_{i, j} p_i q_j (A_i - A_j) \\bigr| \\\\\r\n & \\leq & \\sum_{i, j} p_i q_j | A_i - A_j | \\\\\r\n & \\leq & \\max_{i, j} | A_i - A_j |.\r\n\\end{eqnarray*}", "Solution_11": "Thanks a lot for the solutions the more general problem: I heard of it before.\r\n\r\nThe story behind my first question here is: I considered how to explain the solution of my problem to a beginner and then I understood that I've never thought about it properly (I just mentioned this fact as obvious in my solutions to some more complicated olympiad problems). Then I found the solution which was posted at the beginning of the topic and I couldn't believe that there is no beatiful argumentation as there often are in geo-combinatorial problems. But it seems that it was hopeless. \r\n\r\nBut it is still an interesting problem and I'll look up the link Darij posted.\r\n\r\nMisha", "Solution_12": "[quote=\"Misha123\"]Hello! \nHere is a problem which seems to be trivial. The straightforward solution I have seen is very ugly so I search for another(nice!!!) solutions.\n\nYou have an equilateral triangle of side length 1 and two points P, Q inside. Show that the distance PQ is less or equal to 1.\n\nSorry if there is a one-sentence solution. I just could not find it.\n\nMisha[/quote]\r\n\r\nActually, we can go for a less rigorous geomerical proof .\r\n\r\nLet's consider the following cases.\r\n\r\n1> The points P, Q and any one of the three vertices are collinear.\r\nLet's say A,P,Q in triangle ABC are collinear. Let APQ meet BC at D\r\n\r\nThen PQ < AD P, Q are non-collinear with all the three vertices.\r\n\r\nProduce PQ both ways . It will intersect exactly two sides . With no loss of generality, let's say PQ meets AB at D and BC at E. Then constructing an equilateral triangle with BD as one of the sides,we observe that PQ The points P, Q and any one of the three vertices are collinear.\nLet's say A,P,Q in triangle ABC are collinear. Let APQ meet BC at D\n\nThen PQ < AD P, Q are non-collinear with all the three vertices.\n\nProduce PQ both ways . It will intersect exactly two sides . With no loss of generality, let's say PQ meets AB at D and BC at E. Then constructing an equilateral triangle with BD as one of the sides,we observe that PQ x, and x in E. Is the converse of this true?", "Solution_1": "n.t.tuan also posted the same problem. its just few problems earlier than this. please check that out.", "Solution_2": "Check out this topic:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=135394", "Solution_3": "What does the converse of that statement look like and how do we show that it is true?" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Show that there exists an integer $n$ with the following properties:\r\n\r\n(i) The binary expansion on $n$ has precisely 2004 $0$s and 2004 $1$s.\r\n(ii) 2004 divides $n$.", "Solution_1": "Just note that the number of $1$s in $2004$ binary representation is $7$ and for $3*2004$ it's $9$.\r\nNow write $2004=7a+9b$ for some $a,b\\in \\mathbb N$.\r\nNow simply put $a$ copies of $2004$ binary representation in a row and put next to them $b$ copies of $3*2004$ binary representation.\r\nAlso note that number of $0$s in $2004$ binary rep. is less than number of $1$s, and this holds for $3*2004$ too.\r\nNow just put enough $0$s right to the number you constructed already.", "Solution_2": "does this solution work?\r\nLooking at the permutations of the number in binary, there are exactly $\\binom{4007}{2003}$ numbers that satisfy the first requirement. We can assume that these numbers are spread out equally mod 2004, and are certain that at least one is divisible by 2005." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Is graph paper allowed in Mathcounts Competitions?", "Solution_1": "unless they suddenly change the rules for the 09 competitions, you can't use graph paper in mathcounts", "Solution_2": "You can always just take scrap paper and draw lines on it, though.", "Solution_3": "You're allowed as much scratch paper, so make use of it. There's nothing wrong about writing stuff down on your paper--make boxes for each of the problems, etc." } { "Tag": [ "induction", "inequalities", "number theory proposed", "number theory" ], "Problem": "Find all positive integers $ n$ such that $ 8^n \\plus{} n$ is divisible by $ 2^n \\plus{} n$.", "Solution_1": "$ x^3 \\plus{} y^3 \\equal{} (x \\plus{} y)(x^2 \\minus{} xy \\plus{} y^2) \\implies 2^n \\plus{} n|8^n \\plus{} n^3 \\implies 2^n \\plus{} n | 8^n \\plus{} n^3 \\minus{} (8^n \\plus{} n) \\equal{} n^3 \\minus{} n$ \r\nAnd it implies that $ n^3 \\minus{} n \\equal{} 0$ or $ |n^3 \\minus{} n| \\ge 2^n$.\r\nThe first case gives $ n \\equal{} 1$, which is a solution.\r\nThe second case: for $ n \\ge 10$, $ 2^n > n^3 > n^3 \\minus{} n$, by induction: $ 1024 > 1000$, and $ 2^n > n^3$ implies that $ 2^{n \\plus{} 1} > 2n^3 \\ge (n \\plus{} 1)^3$ because $ 3n^2 \\plus{} 3n \\plus{} 1 \\le 3n^2 \\plus{} 3n^2 \\plus{} 3n^2 \\equal{} 9n^2 < n^3$ for $ n \\ge 10$.\r\nSo we only need to consider $ 2 \\le n \\le 9$. The valid solutions are $ n \\equal{} 2,4,6$. When we include the 1st case, the set of solutions is $ n \\equal{} 1,2,4,6$.", "Solution_2": "[quote=\"kunny\"]Find all positive integers $ n$ such that $ 8^n \\plus{} n$ is devisible by $ 2^n \\plus{} n$.[/quote]\r\nWe have $ 8^n \\plus{} n \\equiv n^3 \\minus{} n \\bmod 2^n \\plus{} n$ so $ 2^n \\plus{} n \\mid n^3 \\minus{} n$. From here it is easy to use inequalities to obtain all solutions.", "Solution_3": "[hide=Solution]Note that $$8^n + n\\equiv (2^n)^3 + n\\equiv (-n)^3 + n \\equiv n - n^3 \\pmod{2^n+n}.$$ Hence, we wish to find $n$ such that $2^n+n|n^3-n$. Note that for $n\\ge 10$, we have that $2^n > n^3$. Thus, we only need to try values below $10$. Doing so, the only solutions that work are $\\boxed{n=1,2,4,6}$.[/hide]", "Solution_4": "[quote=Japan MO Finals 2009 P1]Find all positive integers $ n$ such that $ 8^n \\plus{} n$ is divisible by $ 2^n \\plus{} n$.[/quote]\n[b][color=#000]Solution:[/color][/b] Note that, $x+y \\mid x^3+y^3$ Hence,\n$$\\frac{8^n+n}{2^n+n}=\\frac{2^{3n}+n^3+n-n^3}{2^n+n} \\implies \\frac{n^3-n}{2^n+n} \\in \\mathbb{Z} \\implies n^3 \\geq 2^n+~2n$$\nWhich is [url=https://math.stackexchange.com/questions/438260/proving-by-induction-2n-n3-for-any-natural-number-n-9]true for 1 < n < 10[/url]. Also, $n^3-n$ $ =$ $0$ when $n=1$. Quickly checking the permissible values for $n$, we get $n=1,2,4,6$ $\\qquad \\blacksquare$" } { "Tag": [ "linear algebra", "linear algebra unsolved" ], "Problem": "hello everyone.\r\n\r\nLooking at a engine tuning software for a specific car.\r\n\r\nIn the software the engine computer is looking to do all mathematical calculations based on what injectors are in there currently. well to run larger injectors we have to do some mathematics, problem is that the injectors in there now flow 682cc/min at a fuel pressure of static 58psi. the company uses N-heptane which has a specific gravity of 685 kg/m^3 for their injector ratings. \r\n\r\nNow the aftermarket larger injectors from RC engineering are 750cc/min at static 43.5psi. problem is that RC engineering uses a denser fluid which has a SG of 720 kg/m^3.\r\n\r\nI need to convert the 750cc/min @ 43.5psi injector to flow using n-heptane which will be a lower flow in the end. I have turned the fuel pressure down to 43.5psi. \r\n\r\n[b]So the test fluids:[/b]\r\n\r\nN-heptane has a SG of 685 kg/m^3\r\nGasoline has a SG of 720 kg/m^3\r\n\r\n[b]Injectors:[/b]\r\n\r\nFactory injectors - 682cc/min @ static pressure of 58 psi. (test fluid, N-heptane)\r\nAfter market injectors - 750cc/min @ static 43.5 psi. (test fluid, Gasoline)\r\n\r\nNow here is a formula to get lb/hr using n-heptane:\r\n682cc/min * 60min/hr * (1m/100cm)^3 * 685kg/m^3 *2.2lb/kg = 61.7lb/hr\r\n\r\nNow here is a formula to get lb/hr using gasoline:\r\n682cc/min * 60min/hr * (1m/100cm)^3 * 720kg/m^3 *2.2lb/kg = 64.8lb/hr\r\n\r\n1 lb/hr gasoline = 10.5 cc/min (@720 kg/m3) \r\n(1 (pound / hr)) / (720 (kg / (m^3)))\r\n\r\n1 lb/hr N-heptane = 11.03 cc/min (@685 kg/m3)\r\n (1 (pound / hr)) / (685 (kg / (m^3)))\r\n\r\nNow am I to assume:\r\n750cc/min * 60min/hr * (1m/100cm)^3 * 685kg/m^3 *2.2lb/kg = 67.8lb/hr\r\n\r\nto convert using n-heptane for lb/hr to cc/min:\r\n\r\n67.8 lb/hr * 11.03 cc/min = 747.9 cc/min (doesn't make sense to me)\r\n\r\nnow most online calculators to convert lb/hr to cc/min will use a factor of 10.5. this translates to the aftermarket injector company's test fluid having a specific gravity of 720 kg/m^3 as seen above...\r\n\r\nanyone know where im going wrong here.", "Solution_1": "anyone? a little help?", "Solution_2": "my guess is that this is the wrong forum .." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Thomas J. Mildorf in his paper over the olympiad inequalities\r\n\r\nhttp://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf\r\n\r\nwrite to prove $ 9a^2b^2c^2 \\leq (a^2b \\plus{} b^2c \\plus{} c^2a)(ab^2 \\plus{} bc^2 \\plus{} ca^2)$ :\r\n\r\nRearrange the terms of each factor and apply Cauchy,\r\n\r\n$ (a^2b \\plus{} b^2c \\plus{} c^2a)(ab^2 \\plus{} bc^2 \\plus{} ca^2) \\geq (\\sqrt [3]{a^2b^2c^2} \\plus{} \\sqrt [3]{a^2b^2c^2} \\plus{} \\sqrt [3]{a^2b^2c^2})^2 \\equal{}$\r\n\r\n$ 9a^2b^2c^2$.\r\n\r\nI don't understand how rearrange the terms and apply Cauchy to obtain \r\n\r\nthis. Thanks in advance for some hint. \r\n\r\nInstead of $ \\sqrt [3]{a^2b^2c^2}$ I think is $ \\sqrt {a^2b^2c^2}$?", "Solution_1": "[quote=\"Bob75\"]\n\nInstead of $ \\sqrt [3]{a^2b^2c^2}$ I think is $ \\sqrt {a^2b^2c^2}$?[/quote]\r\n\r\nyes.", "Solution_2": "$ (a^2b\\plus{}b^2c\\plus{}c^2a)(ab^2\\plus{}bc^2\\plus{}ca^2)\\geq 3\\cdot 3\\sqrt[3]{a^6b^6c^6}\\equal{}9a^2b^2c^2$.\r\nFrom AM-GM. I think the document had a small error in it.", "Solution_3": "From AM-GM is obvious, but my question is about using rearrangement and Cauchy? Thanks in advance.", "Solution_4": "We have to prove that $ (a^2b \\plus{} b^2c \\plus{} c^2a)(ab^2 \\plus{} bc^2 \\plus{} ca^2)\\geq 9a^2b^2c^2$\r\n$ (a^2b \\plus{} b^2c \\plus{} c^2a)(ab^2 \\plus{} bc^2 \\plus{} ca^2) \\equal{} (a^2b \\plus{} b^2c \\plus{} c^2a)(bc^2 \\plus{} ca^2 \\plus{} ab^2)$\r\n$ \\geq (abc \\plus{} abc \\plus{} abc)^2\\equal{}9a^2b^2c^2$\r\n :)", "Solution_5": "Thanks you very much for your answer, and Cauchy is useless because only with the rearrangement we solve it easy!" } { "Tag": [ "integration", "function", "Functional Analysis", "real analysis", "real analysis unsolved" ], "Problem": "I have been completely lost as I studied orthogonal projections. How to do this one:\r\n\r\nLet $L^{2}([-3,3])$ be a Hilbert space with norm given by $\\parallel f\\parallel _{2}: =\\left (\\int_{-3}^{3}|f(t)|^{2}\\,dt\\right )^{1/2}$. Let $M$ be a subspace\r\n\\[M: =\\{f\\in L^{2}([-3,3])|f(t)=0 \\text{ for almost every }t\\in [0,3]\\}\\]\r\nand $N$ 1-dimensional subspace spanned by the constant function 1. \r\nFind orthogonal projections from $L^{2}([-3,3])$ to $M$ and to $N$.", "Solution_1": "Orthogonal projection to $M$ zeros the function on $(0,3]$ and leaves it alone on the other half; it's pretty easy to see that this gets you the minimum distance, and leaves a difference orthogonal to $M$. Orthogonal projection to $N$ maps a function $f$ to its average value. That requires a bit of calculation, but it's not hard to show it works.", "Solution_2": "[quote=\"jmerry\"]Orthogonal projection to $N$ maps a function $f$ to its average value. That requires a bit of calculation, but it's not hard to show it works.[/quote]\r\nCan someone elaborate this more. I'm relative new in functional analysis and I don't see how to prove this.", "Solution_3": "Let $u$ be an element of a Hilbert space $H$ - or inner product space, since completeness will play no role. We want to find the orthogonal projection from $H$ to the one dimensional subspace spanned by $u.$\r\n\r\nI'll do this algebraically: Given $f\\in H,$ we want to find $x$ that minimizes\r\n\r\n$\\|f-xu\\|^{2}=\\langle f-xu,f-xu\\rangle=\\|f\\|^{2}-\\langle xu,f\\rangle-\\langle f,xu\\rangle-\\langle xu,xu\\rangle$\r\n\r\n$=\\|u\\|^{2}x\\overline{x}-x\\overline{\\langle f,u\\rangle}-\\overline{x}\\langle f,u\\rangle+\\|f\\|^{2}$\r\n\r\n$=\\|u\\|^{2}\\left(x\\overline{x}-\\frac{\\overline{\\langle f,u\\rangle}}{\\|u\\|^{2}}x-\\frac{\\langle f,u\\rangle}{\\|u\\|^{2}}\\overline{x}\\right)+\\|f\\|^{2}$\r\n\r\n$=\\|u\\|^{2}\\left\\|x-\\frac{\\langle f,u\\rangle}{\\|u\\|^{2}}\\right\\|^{2}-\\frac{|\\langle f,u\\rangle|^{2}}{\\|u\\|^{2}}+\\|f\\|^{2}$\r\n\r\nThis will be minimized by setting $x=\\frac{\\langle f,u\\rangle}{\\|u\\|^{2}}.$ The projection then sends $f$ to $\\frac{\\langle f,u\\rangle}{\\|u\\|^{2}}\\, u.$\r\n\r\nFor your particular case, you have $H=L^{2}(-3,3)$ and $u=1.$\r\n\r\nThen $\\|u\\|^{2}=\\int_{-3}^{3}1=6$ and $\\langle f,u\\rangle=\\int_{-3}^{3}f.$ The orthogonal projection sends $f$ to the constant function whose value is $\\frac16\\int_{-3}^{3}f.$\r\n\r\n(Which is exactly what jmerry said.)\r\n\r\nThe argument would be very similar for the orthogonal projection onto some finite dimensional space, provided that we have already gone to the trouble of finding an orthogonal basis for that finite dimensional space." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "A sequence is defined by $ a_1\\equal{}a_2\\equal{}a_3\\equal{}1$ and $ a_{n\\plus{}3}\\equal{}\\minus{}a_n\\minus{}a_{n\\plus{}1}$ for all $ n \\in \\mathbb{N}$. Prove that this sequence is not bounded , i.e. that for every $ M \\in \\mathbb{R}$ there is an $ n$ for which $ |a_n|>M.$", "Solution_1": "$ a_n\\equal{}Ax_0^n\\plus{}Bx_1^n\\plus{}Cx_2^n$, were $ x_0,x_1,x_2$ are roots of $ x^3\\plus{}x\\plus{}1\\equal{}0$ and $ x_0<\\minus{}1$ real root. Therefore complex roots $ |x_1|\\equal{}|x_2|<1$.\r\nIf $ A\\equal{}0$ $ a_n\\to 0$, when $ n\\to \\infty$ give contradition with $ a_n$ is integer. Therefore for big n $ |a_n|to \\infty$ as $ |Ax_0^n|$." } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let $ABC$, a triangle such that $P$ is an interior point, and $PA+PB+PC\\sqrt2$ is minimum. Find $\\angle APB$.", "Solution_1": "Let $Q$ the exterior point to the triangle $ABC$ such that $AQB$ is isosceles right angle at $Q$. We have $AQ\\sqrt{2}=BQ\\sqrt{2}=AB$.\r\nBecause Ptolomeo's inequality at cuadrilateral $BPAQ$,\r\n$PA+PB+PC\\sqrt{2}\\geq (PQ+PC)\\sqrt{2}\\geq CQ\\sqrt{2}$\r\nThen, $C$, $P$ and $Q$ are collinear; and $BPAQ$ is cyclic.\r\nIn particular, $2 max(f)=1\r\n\r\nLet $T(x_{1},..x_{n})= (x_1,S_1)+...+(x_n,S_n)$\r\nThen $T \\leq x_1+...+x_n \\Rightarrow f\\leq 1$ \r\nAlso we can consider $(x_{1},..x_{n})=1$ (in another case we can devide each number \r\nby the greatest common divisor and we'll get this condition)\r\n\r\nLet n=2 then we have: $T=(x_1,x_2)+(x_2,x_1)=2 \\Rightarrow f= \\frac{2}{x_1+x_2} \\leq \\frac{2}{3} (\\mbox {equality for } x_1=1 ,\\ x_2=2)$\r\n\r\nIf n>2 then let's construct such sequence: $x_1=1 ,\\ x_2=2 ;\\ \\mbox{ for } i \\geq 3 : \\ x_i=3*2^{i-3} \\Rightarrow$\r\nfor $i \\geq 3 : \\ (x_i,S_i)=(x_i\\mbox{ } , \\mbox{ }x_i(2+2^2+...+2^{n-i})+3(2^{i-4}+...+1)+3)=$\r\n$(x_i,3(2^{i-4}+...+1+1))=(x_i,3*2^{i-3}) =x_i$\r\n\r\nfor $i=2 : \\ (x_2,S_2)=(2,2*(...)+3+1)=2$\r\nfor $i=1 : \\ (x_1,S_1)=(1,S_1)=1 \\Rightarrow$\r\n$T=x_1+...+x_n \\Rightarrow f=1$\r\nV :)" } { "Tag": [ "inequalities", "geometry unsolved", "geometry" ], "Problem": "Let P be an internal point of triangle ABC. The line through P parallel to AB meets BC at L, the line through P parallel to BC\r\nmeets CA at M, and the line through P parallel to CA meets AB at N. Prove that\r\n$ \\frac{BL*CM*AN}{LC*MA*NB} \\le \\frac{1}{8}$.\r\nand locate the position of P in triangle ABC when equality holds.", "Solution_1": "[quote=\"Machu Picchu\"]Let P be an internal point of triangle ABC. The line through P parallel to AB meets BC at L, the line through P parallel to BC\nmeets CA at M, and the line through P parallel to CA meets AB at N. Prove that\n$ \\frac {BL*CM*AN}{LC*MA*NB} \\le \\frac {1}{8}$.\nand locate the position of P in triangle ABC when equality holds.[/quote]\r\n\r\nI think this problem is not difficult\r\nThe line through P parallel to CA meets AB at N and meets BC at E\r\nLet: $ x \\equal{} BL, y \\equal{} LE, z \\equal{} EC$ \r\n$ \\frac {BL.CM.AN}{LC.MA.NB} \\le \\frac {1}{8}$\r\n $ \\Leftrightarrow \\frac{xyz}{(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)} \\le \\frac {1}{8}$ :P" } { "Tag": [ "vector", "algebra", "polynomial", "linear algebra", "matrix", "induction", "invariant" ], "Problem": "Let V be a real vector space and T an operator on V. Prove that for any real a,b so that $ a^2<4b$, $ dimnull(T^2\\plus{}aT\\plus{}b)^k$ is even for any positive integer k.", "Solution_1": "Let $ P$ be a real polynomial with no real roots. We prove $ dim(ker(P(T)))$ is even.\r\nConsidering the Jordan's theory, it is sufficient to show this result in the 2 following cases:\r\n1) $ T\\equal{}\\lambda{I_n}\\plus{}N$ where $ N$ is the nilpotent matrix of order $ n$ ($ N^{n\\minus{}1}\\not\\equal{}0$) and $ \\lambda$ is a real number:\r\n$ P(T)\\equal{}P(\\lambda)I_n\\plus{}P'(\\lambda)N\\plus{}\\cdots$. $ P(\\lambda)\\not\\equal{}0$ then $ P(T)$ is invertible.\r\n2) $ T\\equal{}diag(\\lambda{I_n}\\plus{}N,\\bar{\\lambda}I_n\\plus{}\\bar{N})$ where $ \\lambda$ is a non real number. $ P(T)\\equal{}diag(P(\\lambda)I_n\\plus{}P'(\\lambda)N\\plus{}\\cdots,P(\\bar{\\lambda})I_n\\plus{}P'(\\bar{\\lambda})\\bar{N}\\plus{}\\cdots)$.\r\nIf $ P(\\lambda)\\not\\equal{}0$ then $ P(\\bar{\\lambda})\\not\\equal{}0$ and $ P(T)$ is invertible.\r\nIf $ P(\\lambda)\\equal{}0$ then $ P(\\bar{\\lambda})\\equal{}0$ and we have the same result for the derivatives.\r\nThus or $ P(T)\\equal{}0$ and $ dim(ker(P(T)))\\equal{}2n$ is even or $ P(T)\\equal{}diag(P^{(k)} (\\lambda)N^k/k!\\plus{}\\cdots,P^{(k)} (\\bar{\\lambda})\\bar{N}^k/k!\\plus{}\\cdots)$ where $ P^{(k)} (\\lambda)\\not\\equal{}0,P^{(k)} (\\bar{\\lambda})\\not\\equal{}0$. Thus $ dim(ker(P(T)))\\equal{}2k$ is even. QED.", "Solution_2": "What is Jorden Theory?\r\n\r\nAlso, this is an exercise appear on a very basic linear algebra book, so is there any more elementary solution?", "Solution_3": "see the Jordan normal form in:\r\nhttp://en.wikipedia.org/wiki/Jordan_normal_form\r\nHere I use also this fact:\r\nLet $ A$ be a real matrix; if in the Jordan normal form of $ A$ we find a non real Jordan block then its conjugate is also in the Jordan form with the same order of multiplicity.\r\n\r\n\r\nAnother method:\r\n$ P(t)\\equal{}u.\\Pi(t\\minus{}\\lambda_k)^{a_k}$ and $ ker(P(T))$ is the direct sum of the $ ker((T\\minus{}\\lambda{I})^{a_k})$. It remains to see that $ ker((T\\minus{}\\lambda{I})^{a_k})$ and $ ker((T\\minus{}\\bar{\\lambda}{I})^{a_k})$ are of same degree.", "Solution_4": "I rewrite the 2 last lines of my last post\r\n\r\n$ ker((T\\minus{}\\lambda_k{I})^{a_k})$. It remains to see that $ ker((T\\minus{}\\lambda_k{I})^{a_k})$ and $ ker((T\\minus{}\\bar{\\lambda_k}{I})^{a_k})$ are of same dimension.", "Solution_5": "A bit cleaner:\r\n\r\nIt suffices to show that, for $ c$ a nonzero real number, that $ \\ker (T^2 \\plus{} c^2)$ has even dimension (by induction and completing the square). Letting $ \\phi \\equal{} T^2 \\plus{} c^2$, we do this by showing that $ \\ker \\phi$ is actually a $ \\mathbb{C}$ vector space. Endow $ \\ker \\phi$ with a $ \\mathbb{C}$ action by the relation $ i v \\equal{} (T/c) v$. Then since $ T^2 v \\equal{} \\minus{}c^2 v$, it follows that $ i^2 v \\equal{} \\minus{} v$; therefore we obtain a faithful $ \\mathbb{C}$-action, so $ \\ker \\phi$ is even dimensional as a real vector space.", "Solution_6": "Nukular is right.\r\nI assumed these facts: \r\n1) Let $ v,\\bar{v}$ be a complex basis of the complex plane $ \\Pi$; then $ Re(v),Im(v)$ is also a basis of $ \\Pi$. But $ Re(v),Im(v)$ is a real basis of a real plane $ P$ which is included in $ \\Pi$.\r\n2) If $ T$ is a $ \\mathbb{R}$-linear application:$ \\mathbb{R}^n\\rightarrow\\mathbb{R}^n$ then we can associate a $ \\mathbb{C}$-linear application:$ \\mathbb{C}^n\\rightarrow\\mathbb{C}^n$ as follows:\r\n$ T(v)\\equal{}T(Re(v))\\plus{}iT(Im(v))$.\r\nMoreover $ dim(ker(T))$ (as a $ \\mathbb{R}$-linear application)=$ dim(ker(T))$ (as a $ \\mathbb{C}$-linear application)", "Solution_7": "Your second induced map can also be written, for real vector spaces $ V$, as \r\n\r\n$ T \\otimes_{\\mathbb{R}} id : V \\otimes_{\\mathbb{R}} \\mathbb{C} \\to V \\otimes_{\\mathbb{R}} \\mathbb{C}$. \r\n\r\nSecondary question: Say we are not working over $ \\mathbb{R}$, but rather $ \\mathbb{Q}$. If we have an irreducible polynomial $ P(T)$ of degree $ n$, does it follow that $ \\dim \\ker(P(T))$ is divisible by $ n$?", "Solution_8": "Hi Nukular,\r\nLet $ T\\in\\mathcal{M}_k(\\mathbb{Q})$ and $ P$ be an irreducible polynomial over $ \\mathbb{Q}$ of degree $ n$. \r\nLet $ E\\equal{}ker(P(T))$; we assume $ E\\not\\equal{}\\{0\\}$; $ E$ is a $ T$-invariant $ \\mathbb{Q}$-subspace. Let $ U\\equal{}T_{|E}: E\\rightarrow{E}$; then $ P(U)\\equal{}0$. If $ \\lambda$ is an $ U$-eigenvalue then $ P(\\lambda)\\equal{}0$ and $ P$ is the $ \\lambda$-minimal polynomial. Thus the $ U$-characteristic polynomial (over $ \\mathbb{Q}$) is in the form $ P^l$ where $ n.l\\equal{}dim(E)$ and $ n$ is a divisor of $ dim(E)$." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\n\r\nI have yet another question about something in my algebra book.\r\nThe context is : projective plane curves.\r\n\r\nLet $k$ be an algebraically closed field. Let $f$ be a polynomial in $k[x,y,z]$, homogeneous, and of degree $d$.\r\nWe know we can write $f$ as $f_{0}z^{d}+\\cdots f_{d}z^{0}$, where the $f_{i}$ are homogeneous polynomials in $x,y$ of degree $i$.\r\n\r\nLet $R$ be the ring of fractions of homogeneous polynomials of the same degree, such that the denominator does not become zero for $x=0,y=0,z=1$\r\nLet $m_{P}$ be the ideal in this ring generated by $\\frac{x}{z}, \\frac{y}{z}$\r\n\r\nProve that the smallest $i$ such that $\\frac{f}{z^{d}}\\in m_{P}^{i}$ is the smallest $i$ such that $f_{i}\\neq 0$.\r\n\r\nSeems trivial, I mean : if $f_{0},\\cdots, f_{r-1}$ are all zero, then $\\frac{f}{z^{d}}$ is in $m_{P}^{r}$, but what about the converse...\r\n\r\nI feel silly for asking something like this, but I've been playing with these fractions for quite some time now and I still haven't found a satisfactory answer...\r\nAll hints welcome... :huh:", "Solution_1": "Hmm, I'm still in doubt about this. Is this trivial or very hard? :maybe:" } { "Tag": [ "email", "geometry" ], "Problem": "[i][b]WIN ETERNAL AoPS FAME AND GLORY![/b][/i]\r\n\r\n\r\n\r\nThis is my rather difficult national competition that I wrote. Because it's a little harder than mathcounts(but covers the same topics) I'm not sure if I should post it here or not. So if the mods want to give me a little nudge, that's ok.\r\n\r\nAnyways, you're on your own honor for this, and the same rules apply for the sprint and target as they do for any other competition. PM your answers to me using the answer sheet for sprint, [b]WHICH DOESN'T SAY THE ANSWERS, BUT HELPS YOU EXPRESS THEM IN THE WAY I WANT! [/b]but I never made an answer sheet for target, and have fun. Those of you who send me your answers will recieve the team round as a little bonus. [b]And your results will come to you as soon as possible after you take it.[/b] Requested solutions will also be PMed to you.\r\nAs well, for those of you who don't post here, you can email me at nma4@comcast.net\r\n\r\nEratta:\r\nSprint #7: Please eliminate the words \"Can You\".(Thanks Nathan :roll: )\r\n\r\nSprint # 10, assume what matters is who goes against who in the first round, who wins, who gets second, and who gets third.\r\n\r\nSprint#22: Someone who has taken it has pointed out that there is no answer, so credit for everyone!\r\n\r\n\r\nSprint #25:Assume that there are an infinite number of cereal boxes.\r\n\r\nTarget #3: A set of dominoes has every distinict combination of two integers from 0-6 once. \r\n\r\nTarget#6:A doolar is a dollar!\r\n\r\nTarget#8: There can only be one next closest city, if there isn't then no lines should be drawn.", "Solution_1": "Anyone gonna take it? Anyone???", "Solution_2": "It looks like not a lot of people want to do math after Nats, but if anyone else participates I'll join in.", "Solution_3": "I'm doing it.", "Solution_4": "Ill do it", "Solution_5": "awesome, it's good practice for next year as well if you're eligible, because it'll probably be closer to the level of difficulty than most old nats are", "Solution_6": "Also, if you only do one or two problems, you can still PM or email me the answers, or ask for solutions", "Solution_7": "So far, nobody has sent me their answers, so if you have them, send them, and you're results and solutions will be near instant!!! Also, it's really hard, so don't be embarassed if you only have a few answers. Sergei only got a three on the target, and we never officially did the sprint.", "Solution_8": "So far, still no answers!", "Solution_9": "EDIT: Sorry cant do it now", "Solution_10": "Do it whenever, it would just be good to have a sense of how hard it was.", "Solution_11": "I'll do it, but I haven't looked at any problems yet...", "Solution_12": "Cool", "Solution_13": "I have now made the tests (not the answer key) into .pdf format. They also now have lines to write your answers on. I hope that this will be a help to some people.", "Solution_14": "Thank you. [u][b]Just so everyone knows, the Answer page, doesn't have the answers to the questions, just it tells you how I want the answer expressed.[/b][/u]", "Solution_15": "So this is a competition (you'll post the results)?", "Solution_16": "Yeah", "Solution_17": "Okay, maybe I'll actually try now.", "Solution_18": "Yeah, you could win fame and glory!", "Solution_19": "Just because you edited your first post......", "Solution_20": "Shh. :roll:", "Solution_21": "Okay, I'll be [size=200][color=darkred][b]quiet![/b][/color][/size]:rotfl:", "Solution_22": "Just take the test!", "Solution_23": "Are there any geometry questions on the test? :P JK I'll take it :D (but probably tomorrow or something, cause I'm kinda tired now ;) )", "Solution_24": "I'm exausted from all that \"math\" :sleeping: :stretcher: :starwars: .....I'm gonna retire for the night.", "Solution_25": "Nathan, we know we share our hatred of geometry, so there's one question that's geometry, and it's fairly easy, and there's one question that one solution is geometry, so it's not bad at all.", "Solution_26": "Just some Errata for those who don't look at the first post:\r\n\r\nSprint #7: Please eliminate the words Can You.(Thanks Nathan :roll: )\r\n\r\nSprint # 10, assume what matters is who goes against who in the first round, who wins, who gets second, and who gets third.\r\n\r\nSprint#22: Someone who has taken it has pointed out that there is no answer, so credit for everyone!\r\n\r\nSprint #25:Assume that there are an infinite number of cereal boxes.\r\n\r\nTarget#6:A doolar is a dollar!", "Solution_27": "I just took this test and it is really good. So if you don't have anything better to do for 1 hour and 4 minutes, take this test!!!\r\n\r\n:D\r\n\r\n(It's really hard too ;) )", "Solution_28": "Aww, thanks Nathan :D", "Solution_29": "[quote=\"mathgeek2006\"]I'm exausted from all that \"math\" :sleeping: :stretcher: :starwars: .....I'm gonna retire for the night.[/quote]\r\n\r\nThat starwars emoticon is sick.", "Solution_30": "Also, those of you who send me your answers get the team round as a bonus. I might add bonus points to your score accordingly. Take half an hour for the team round if you do it by yourself, you won't finish it but still try!" } { "Tag": [ "probability" ], "Problem": "Here's how it works: one person gives a problem, another solves it and gives another problem. But all the problems must be numbered, and each problem must be associated with that number. Some of the annual competitions would do something like this with the year. For example, NYSML had this as an individual problem this year: \"Find the sum of all the positive factors of 2006.\" See what I mean? Problem [i]n[/i] is centered around the number [i]n[/i]. Well, let's do it!\r\n\r\nI'll start with problem 1, which will be centered around the number 1.\r\n\r\n[b]PROBLEM 1[/b]\r\nIf $G=1+\\sqrt{1+\\sqrt{1+...}}$ and $R=1+\\frac1{1+\\frac1{1+...}}$, what is $GR$?", "Solution_1": "[hide=\"Hint\"]$G=1+\\sqrt{1+\\sqrt{1+...}}=1+\\sqrt{1+G}$\n$R=1+\\frac1{1+\\frac1{1+...}}=1+\\frac1R$[/hide]\n\n[hide=\"Bigger Hint\"]Solve for $G-1$, and then use that to find $G$.[/hide]\n\n[hide=\"Gives It Away\"]$x^2-x-1=0$[/hide]", "Solution_2": "[hide] We see that (g-1)^2=g, so g^2-3g+1=0, so g=$\\frac{3+\\sqrt{5}}{2}$. $\\frac{1}{r-1}=r$, so $r=\\frac{1+\\sqrt{5}}{2}$. Thus $gr=2+\\sqrt{5}$[/hide]\r\n\r\n[b]Problem 2[/b]\r\n\r\nI draw 2 cards from a set of 2222 cards numbered 1 through 2222 and multiply the numbers on the cards. What is the probability that the product obtained is divisible by 22 or 222 or both?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "If $ x_{1},x_{2},...,x_{n}\\in [a,b]$ with $ 00$ and are integers and let this be $Y$\nand then let $X$ equal the number of solutions to $b_{1}+b_{2}+\\dots+b_{7}=12$ where all $b_{i}\\geq{0}$ and are integers.\nso the probability would just be $\\frac{Y}{X}$ but the book says that this is incorrect. what did i do wrong?\n[/hide]", "Solution_1": "[quote=\"maokid7\"]If you receive 12 phone calls in one week what is the probability that you received atleast 1 call each day.\n[hide]\ni was thinking find the number of solutions to $a_{1}+a_{2}+\\dots+a_{7}=12$ where $a_{i}>0$ and are integers and let this be $Y$\nand then let $X$ equal the number of solutions to $b_{1}+b_{2}+\\dots+b_{7}=12$ where all $b_{i}\\geq{0}$ and are integers.\nso the probability would just be $\\frac{Y}{X}$ but the book says that this is incorrect. what did i do wrong?\n[/hide][/quote]\n\nHere's what I did... the probability seems a little low... well actually very low \n\n[hide]Count the number of ways to not receive a phone call every day. Exclude one day so you have 6 days and 12 calls. The number of ways is the number of non-negative solutions to $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=12$. Then you include the seventh day but you put a call in each day so that you know that every day you got a phone call. So you have 5 calls left and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}=5$. The total number of ways to distribute the calls is $\\binom{11}{6}+\\binom{17}{5}$. The number of ways to not get a call one day is $\\binom{17}{5}$. So then the answer is $1-\\frac{\\binom{17}{5}}{\\binom{11}{6}+\\binom{17}{5}}=\\boxed{\\frac{33}{475}}$?[/hide]", "Solution_2": "[quote=\"ch1n353ch3s54a1l\"][quote=\"maokid7\"]If you receive 12 phone calls in one week what is the probability that you received atleast 1 call each day.\n[hide]\ni was thinking find the number of solutions to $a_{1}+a_{2}+\\dots+a_{7}=12$ where $a_{i}>0$ and are integers and let this be $Y$\nand then let $X$ equal the number of solutions to $b_{1}+b_{2}+\\dots+b_{7}=12$ where all $b_{i}\\geq{0}$ and are integers.\nso the probability would just be $\\frac{Y}{X}$ but the book says that this is incorrect. what did i do wrong?\n[/hide][/quote]\n\nHere's what I did... the probability seems a little low... well actually very low \n\n[hide]Count the number of ways to not receive a phone call every day. Exclude one day so you have 6 days and 12 calls. The number of ways is the number of non-negative solutions to $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=12$. Then you include the seventh day but you put a call in each day so that you know that every day you got a phone call. So you have 5 calls left and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=5$. The total number of ways to distribute the calls is $\\binom{10}{5}+\\binom{17}{5}$. The number of ways to not get a call one day is $\\binom{17}{5}$. So then the answer is $1-\\frac{\\binom{17}{5}}{\\binom{10}{5}+\\binom{17}{5}}=\\boxed{\\frac{9}{230}}$?[/hide][/quote]\r\n\r\nI don't think that's quite right. For one thing, you're only distributing the last $5$ calls among $6$ days, which I don't get. Shouldn't it just be all $7$ days?\r\n\r\nHere's how I did it (maybe I'm wrong?):\r\n\r\n# of ways to get at least $1$ call each day: Assign $1$ call to each day. Then distribute $5$ among the rest of the $7$ days. This is $\\binom{11}{5}$.\r\n\r\n# of total ways to get $12$ calls in a week: Distribute $12$ among $7$ days. This is $\\binom{18}{6}$.\r\n\r\nThus the probability is $\\frac{\\binom{11}{5}}{\\binom{18}{6}}= \\frac{11}{442}$.", "Solution_3": "Agh I screwed the second one up.....edited :ninja:", "Solution_4": "why do u do 11 choose 5 and 18 choose 5 and not 18 choose 6?", "Solution_5": "[quote=\"maokid7\"]why do u do 11 choose 5 and 18 choose 5 and not 18 choose 6?[/quote]\r\n\r\nOops. You're right; I went through a little too quickly.", "Solution_6": "try \\binom{}{}" } { "Tag": [ "LaTeX", "real analysis", "real analysis unsolved" ], "Problem": ":) but I can't slove it. ;) \r\n$f\\in C^3[a,b]$\r\nprove there exist a $k$ such that:\r\n$f(b)=f(a)+\\frac{(b-a)}{2} (f'(a)+f'(b) )-\\frac{1}{12}(b-a)^3f'''(k)$\r\nhelp me.thank you", "Solution_1": "Let $F(x)=f(x)-f(a)-\\frac{1}{2}(x-a)(f'(a)+f'(x))$ and $G(x)=(x-a)^3$.\r\nTry to apply Cauchy's mean-value theorem for two times..", "Solution_2": "[quote=\"zhaobin\"]$f\u2208C^3[a,b]$[/quote]Use \\in for that symbol in $\\text{\\LaTeX}$: $f\\in C^3[a,b]$", "Solution_3": "thank you,liyi and jmerry.I used with TAYLOR formal to Expansive but I failed :) ;)", "Solution_4": "I found this problem in one of my college exam & I am pretty sure that it can be done using finite Taylor series expansion.", "Solution_5": "[quote=\"the game\"]I found this problem in one of my college exam & I am pretty sure that it can be done using finite Taylor series expansion.[/quote]\r\ncan you show the Taylor series expansion.\r\nthank you in advance :)", "Solution_6": "Zhaobin!\r\n\r\nPlease, read http://www.mathlinks.ro/viewtopic.php?p=140098 and name problems properly next time!\r\n\r\nMyth", "Solution_7": "[quote=\"Myth\"]Zhaobin!\n\nPlease, read http://www.mathlinks.ro/viewtopic.php?p=140098 and name problems properly next time!\n\nMyth[/quote]\r\nthank you for reminding me.Myth!\r\nbut which name can be proper.\r\ndo you think \"$a median problem$\" will be better :)", "Solution_8": "No, I don't think so. Try better!", "Solution_9": "can you give an explame? :)", "Solution_10": "Asked by Leitingok in 2013 \n\n[url= https://math.stackexchange.com/questions/295246/application-for-mean-value-theorem] solution by saz [/url] \nhttps://math.stackexchange.com/questions/295246/application-for-mean-value-theorem" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let $(x_{n})_{n \\geq 1}$ be a squece of real numbers such that $\\forall n\\in \\mathbb N$ ,$n\\geq 2$ we have: \\[x_{1}-C_{n}^{1}\\cdot x_{2}+C_{n}^{2}\\cdot x3-...+(-1)^{n}\\cdot C_{n}^{n}\\cdot x_{n+1}=0\\]\r\nProve that:\r\n\\[x_{1}^{k}-C_{n}^{1}\\cdot x_{2}^{k}-...+(-1)^{n}\\cdot C_{n}^{n}\\cdot x_{n+1}^{k}=0 , \\forall k\\in N , n\\geq k+1\\]", "Solution_1": "[quote=\"Svejk\"]Let $(x_{n})_{n \\geq 1}$ be a squece of real numbers such that $\\forall n\\in \\mathbb N$ ,$n\\geq 2$ we have:\n(*) \\[x_{1}-C_{n}^{1}\\cdot x_{2}+C_{n}^{2}\\cdot x3-...+(-1)^{n}\\cdot C_{n}^{n}\\cdot x_{n+1}=0 \\]\nProve that:\n\\[x_{1}^{k}-C_{n}^{1}\\cdot x_{2}^{k}-...+(-1)^{n}\\cdot C_{n}^{n}\\cdot x_{n+1}^{k}=0 , \\forall k\\in N , n\\geq k+1 \\]\n[/quote]\r\nMust be (*) for all $n\\ge 1$." } { "Tag": [ "Product", "number theory", "IMO", "imo 1976", "maximization" ], "Problem": "Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is $1976.$", "Solution_1": "Did I understand this question well:\r\n You have to find largest number which can be represented as $xy$ s.t. $x+y=1976$?\r\nBecause if that is a question then it is trivial:\r\n $x+y\\geq 2\\sqrt{xy}$ so $(x+y)^2\\geq 4xy$ so that number is $\\frac{1976^2}4=976144$\r\nm@re", "Solution_2": "You can get a number which is MUCH bigger. For example, you could get $2^{988}$ :)", "Solution_3": "The biggest number you can find is $2.3^{658}$.\r\n\r\nProof: Assume there was any factor 5 or more. we can split 5=2+3 which restults in a factor 6>5. Same for numbers larger than 5. So there are none.\r\n\r\nThus we have the optimal number is of the form $2^m.3^n$. And since $3^2>2^3$, we need as much as possible into 3's, thus we end up with $2.3^{658}$.", "Solution_4": "Well, that's why I asked did I understand this question :D", "Solution_5": "Maybe we can find more: If $x_1,x_2,...,x_n$ are reals with $\\sum x_i = 1976$, what is the maximum value $\\prod x_i$ can obtain? Or maybe easier to start with, what is the value of $n$?", "Solution_6": "Simple computations gives that when $n$ is given, it's optimal to take $x_1=x_2=...=x_n$.\r\nSo we want $(\\frac{a}{n})^n$ ($a=1976$ here) to be as big as possible.\r\nDeriving $f(x)=(\\frac{a}{x})^x=e^{x \\ln \\frac{a}{x}}$ gives $f'(x)=e^{x \\ln \\frac{a}{x}} (ln(\\frac{a}{x})-1)$, so it attains it's only maximum at $x=\\frac{a}{e}$. Now $n$ has to be the next greater or smaller integer (it's monotonous on both sides).", "Solution_7": "I just base cashed using AM-GM;\r\n\r\n$ \\left(\\frac{1976}{n}\\right)^n \\ge a_1 a_2 ... a_n$\r\n\r\nThen, as the product was an integer n must be a factor of 1976 and the prime factors of 1976 are;\r\n\r\n2,2,2,13,19 then consider all combinations of these.\r\n\r\nHowever, I don't know if I am missing something - it just seems a bit too simple", "Solution_8": "Exactly the same except that i explained a bit more on why we chose (2 and 3 and 3)not (4 and 4) or (5 and 3) or(1 and 4 and 3)", "Solution_9": "how do you know you must do an additional 2? how do you know that we should not separate the 2's and distribute them onto 3's?", "Solution_10": "what we know that is that $3^4>4^3$, but this does not mean $3\\times3\\times2>4\\times4$ (without calculation)", "Solution_11": "So what is the answer here?", "Solution_12": "[quote=Peter]Proof: Assume there was any factor 5 or more. we can split 5=2+3 which restults in a factor 6>5. Same for numbers larger than 5. So there are none.[/quote]\n\nFor this we first need to prove this claim. Anyone here please prove this.\n", "Solution_13": "The critical part is it says about positive integers. If positive reals, then the answer is $e^{\\frac{1976}{e}}$. As it says for positive integers then have to work a little.\nWe get from AM-GM that $\\left(\\frac{1976}{n}\\right)^n\\geq t$ where $t$ is the product of $n$ integers. To maximize the value of $f(x)=\\left(\\frac{1976}{x}\\right)^x$ we have to find the first derivative of this then have to make it zero.\n\nApplying chain rule we have $f'(x)=e^{xln\\frac{1976}{x}}\\cdot f'(xln\\frac{1976}{x})=e^{xln\\frac{1976}{x}}\\cdot (ln\\frac{1976}{x}-1)$.\nHence we get $ln\\frac{1976}{x}-1=0\\Longrightarrow x=\\frac{1976}{e}$\n\nWe have to do some case work. We have find the integer solutions of the plot of $f(x)$ and the highest $y$ value is finally the answer. The value $e^{\\frac{1976}{e}}$ is useful for the reason that the nearest values of it will be actually the answer (I think). The rest is little boring therefore I didn't find.", "Solution_14": "If @above's solution is correct, then we have a nice way of giving explicit computations for $\\lfloor e^{\\frac{n}{e}}\\rfloor$ for all positive integers.", "Solution_15": "[quote=starchan]If @above's solution is correct, then we have a nice way of giving explicit computations for $\\lfloor e^{\\frac{n}{e}}\\rfloor$ for all positive integers.[/quote]\nI have found error in my solution. Whatever. you can show the way of computing the value of $\\lfloor e^{\\frac{n}{e}}\\rfloor$ what you have found.\n", "Solution_16": "[quote=Abdullahil_Kafi]The critical part is it says about positive integers. If positive reals, then the answer is $e^{\\frac{1976}{e}}$. As it says for positive integers then have to work a little.\nWe get from AM-GM that $\\left(\\frac{1976}{n}\\right)^n\\geq t$ where $t$ is the product of $n$ integers. To maximize the value of $f(x)=\\left(\\frac{1976}{x}\\right)^x$ we have to find the first derivative of this then have to make it zero.\n\nApplying chain rule we have $f'(x)=e^{xln\\frac{1976}{x}}\\cdot f'(xln\\frac{1976}{x})=e^{xln\\frac{1976}{x}}\\cdot (ln\\frac{1976}{x}-1)$.\nHence we get $ln\\frac{1976}{x}-1=0\\Longrightarrow x=\\frac{1976}{e}$\n\nWe have to do some case work. We have find the integer solutions of the plot of $f(x)$ and the highest $y$ value is finally the answer. The value $e^{\\frac{1976}{e}}$ is useful for the reason that the nearest values of it will be actually the answer (I think). The rest is little boring therefore I didn't find.[/quote]\n\nUmm you cannot have a nonintegral number of reals. $x$ must remain integer no matter what. You cannot have $x = \\frac{1976}{e}$ because you cannot have those many positive reals. (this number should be an integer no matter what)", "Solution_17": "Suppose some positive integer $k\\ge 5$ is there. Then replace $k$ with $2$ and $k-2$. We have $2(k-2)=2k-4>k$, so it is more optimal. If some $k=4$ is there, then replace it with two $2$'s. Thus, optimally, all these positive integers are either $2$ or $3$. Now we can replace three twos with two threes, because $3^2>2^3$. So the greatest is $\\boxed{2\\cdot 3^{658}}$. ", "Solution_18": "Let $x_1, x_2, \\dots, x_n$ be positives summing to $1976$ and $M$ be the max product of $x_1x_2 \\dots x_n$. By observation we find that $M$ is of the form $2^r3^s$ implying that $2r+3s=1976$. It's obvious that to get max we must have as many 3's more than 2's possible (since $3^2> 2^3$). So, $2r \\equiv 1976 ( \\text{mod } 3) \\implies (k,l)= (1,658) \\implies M=2 \\cdot 3^{658}.$ $\\square$" } { "Tag": [ "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "$E_n$ is the vectorial space of homogenous polynomial $P(X,Y) \\in R[X,Y]$ \r\nof degree $n$.\r\n$A_n$ the subset of $E_n$ of those polynomial that are multiple of $(X^2+Y^2)$\r\n$H_n$ the subset of $E_n$ of polynomial with nul laplacian.\r\n\r\nShow that $E_n$ is the direct product of $A_n$ and $H_n$", "Solution_1": "Alekk Does the problem is true with three variables\r\n\r\n\r\n$E_n$ is the vectorial space of homogenous polynomial $P(X,Y,Z) \\in R[X,Y,Z]$ \r\nof degree $n$.\r\n$A_n$ the subset of $E_n$ of those polynomial that are multiple of $(X^2+Y^2+Z^2)$\r\n$H_n$ the subset of $E_n$ of polynomial with nul laplacian.\r\n\r\nthen $E_n$ is the direct product of $A_n$ and $H_n$" } { "Tag": [ "integration", "real analysis", "real analysis unsolved" ], "Problem": "Show that:\r\n\r\n$ L(1/2, \\chi) \\equal{} \\sum_{n\\equal{}1}^A \\frac{\\chi(n)}{\\sqrt{n}} \\plus{} O(A^{\\minus{}1/2})$", "Solution_1": "I will assume that $ \\chi$ is a nontrivial Dirichlet character, so that $ s_n = \\sum_{k}^{n} \\chi(k)$ is bounded. That is, there is $ M > 0$ such that $ |s_n| \\leq M$. Then for $ s > 0$, summation by parts shows that\r\n\r\n\\begin{eqnarray*}\r\n\\left| \\sum_{k=m}^{n} \\frac{\\chi(k)}{k^s} \\right|\r\n& = & \\left| \\sum_{k=m}^{n} \\frac{s_k - s_{k-1}}{k^s} \\right| \\\\\r\n& = & \\left| \\frac{s_n}{n^s} - \\frac{s_{m-1}}{m^s} + \\sum_{k=m}^{n-1} s_k \\left( \\frac{1}{k^s} - \\frac{1}{(k+1)^s} \\right) \\right| \\\\\r\n& \\leq & \\frac{|s_n|}{n^s} + \\frac{|s_{m-1}|}{m^s} + \\sum_{k=m}^{n-1} |s_k| \\left| \\frac{1}{k^s} - \\frac{1}{(k+1)^s} \\right| \\\\\r\n& \\leq & \\frac{|s_n|}{n^s} + \\frac{M}{m^s} + M \\sum_{k=m}^{n-1} \\frac{s}{k^{s+1}} \\\\\r\n& \\leq & \\frac{|s_n|}{n^s} + \\frac{M}{m^s} + M \\int_{m-1}^{\\infty} \\frac{s}{x^{s+1}} \\, dx \\\\\r\n& = & \\frac{|s_n|}{n^s} + \\frac{M}{m^s} + \\frac{M}{(m-1)^s}\r\n\\end{eqnarray*}\r\n\r\nThus taking $ \\limsup_{n\\to\\infty}$ to both side, we have $ \\sum_{k=m}^{\\infty} \\frac{\\chi(k)}{k^s} = O\\left( \\frac{1}{m^s} \\right)$." } { "Tag": [ "MATHCOUNTS", "probability", "percent", "ratio", "conditional probability" ], "Problem": "(this is not a mathcounts problem, but a good probability problem from a math/ed class I had in college)\r\n\r\nAn auto accident occured outside the window on an elderly lady's apartment. The car in question sped away but not before the woman looked out her window and saw the car. When the police questioned her she told the police the car was definitely a Taxi. Since in her city all cabs are either Green Cabs or Blue Cabs, the officer asked her which company the cab came from and she said the cab was a Blue Cab.\r\n\r\nWhen the case eventually came to a jury trial against the Blue Cab company the lawyers in an effort to question the witness had her eyesight tested by having her identify the color of several cabs going past her window in a similar fashion. She correctly identified the cab color 80% of the time.\r\n\r\nIn this city 85% of all cabs are Green and the other 15% are Blue. It is your job as a member of the jury to determine the probability of the woman having correctly identified the cab in the accident.", "Solution_1": "I've seen the answer to this one in a book I read. Jump in here; try it out. Just remember to think carefully.", "Solution_2": "I would post, but if I did, frost (my coach) would kill me.\r\nAnyways, hint:\r\n[hide=\"hint-dont read unless REALLY stymed!!!\"]\nIt says probability in percents right? But who says its x over 100?[/hide]", "Solution_3": "Well, maybe the old lady was color blind to green or something :D", "Solution_4": "This is probably wrong:\r\n\r\n72% :?::?::?::?::?::?::?::?::?::?::?::?::?::?::?:", "Solution_5": "not 72%, your best off as a common fraction\r\n\r\nTry making a probability tree and thinking of it in terms of favorable/possible\r\n\r\nKeep trying", "Solution_6": "umm...wouldn't this just be 80%?", "Solution_7": "I think this is a classic example of [b]conditional probability[/b].", "Solution_8": "Not 80% this is a problem that is not as easy as it looks, yet in the end not all that hard either :) \r\n\r\nIt is conditional probability. The problem doesn't just ask for the probability that she is right, but the probability that she says \"Blue\" and is right.", "Solution_9": "[quote=\"frost13\"]It is conditional probability. The problem doesn't just ask for the probability that she is right, but the probability that she says \"Blue\" and is right.[/quote]\r\n\r\nTo nitpick, you're not really looking for a simple \"and\" here, what you're looking for, in the traditional language of conditional probability problems is the probability that the cab was blue, [i]given that [/i]she said \"blue\".\r\n\r\nI hope frost won't mind if I add this hint:\r\n[hide]So, you need to consider, what are all the cases in which she might say blue, and in what proportion of those is the cab actually blue.[/hide]", "Solution_10": "if she identifies the right color 80% of the time and 15% of all cabs are blue so would it no be: 4/5*3/20 = 3/25 = 12%? probably wrong...", "Solution_11": "Shake, you've got half the answer :) Closest one yet.\r\n\r\nMathmom was right too!! What you have is the probability she says blue and is right.\r\n\r\nIf you define probability as favorable / possible, or success / attempts, then Shake has the favorable or success part. Keep going.", "Solution_12": "It IS a conditional probability problem!\r\n\r\n[hide]\nLet $G$ = the cab is Green, $B$ = the cab is Blue.\n\nWe are given the following probabilities:\n$Pr(G) = 0.85,\\;Pr(B) = 0.15$\n$Pr(\\text{Right}) = 0.80,\\;Pr(\\text{Wrong}) = 0.20$\n\nThere are four cases:\n[color=yellow]. . . [/color]$(1)\\;Pr(G \\land \\text{said Green}) = (0.85)(0.80) = 0.68$\n[color=yellow]. . . [/color]$(2)\\;Pr(G \\land \\text{said Blue})\\;\\; = (0.85)(0.20) = 0.17$\n[color=yellow]. . . [/color]$(3)\\;Pr(B \\land \\text{said Blue})\\;\\; = (0.15)(0.80) = 0.12$\n[color=yellow]. . . [/color]$(4)\\;Pr(B \\land \\text{said Green}) = (0.15)(0.20) = 0.03$\n\nIn cases (2) and (3), she said the cab was Blue.\n[color=yellow]. . . [/color]Hence, $P(\\text{said Blue}) = 0.17 + 0.12 = 0.29$\n\n\nThe problem asks \"What is the probability that the cab $is$ Blue, given that she said Blue?\"\n[color=yellow]. . . [/color]That is: $Pr(B\\;|\\;\\text{said Blue})$\n\nBayes' Theorem says: $Pr(B\\;|\\;\\text{said Blue}) \\;= \\;\\frac{Pr(B \\land \\text{said Blue})}{Pr(\\text{said Blue})} $\n\n[color=yellow]. . . [/color]The numerator is Case (3): 0.12\n[color=yellow]. . . [/color]The denominator was determined above: 0.29\n\n\nTherefore: $Pr(B\\;|\\;\\text{said Blue}) \\;= \\;\\frac{0.12}{0.29} = 0.413793 \\;\\approx \\;41.4\\%$\n\n[size=75]Solution restored by moderator rcv on Sun Feb 13, 2005 15:52 EST[/size][/hide]", "Solution_13": "since we have the probability that she correctly identifies the cab and says its blue (which is correct)\r\nnow i guess we have to find the probability that she identifies the cab and says its green: .85 * .8 = 68%\r\nso it seems that she correctly identifies the car as blue 12% of the time, correctly identifies the car as green 68% if the time, and is completely wrong 20% of the time...", "Solution_14": "getting close still shake, but if being RIGHT about BLUE is favorable, the possible wouldn't involve identifying GREEN.", "Solution_15": "she has a 12% chance of identifying the car correctly provide that she says blue.\r\nbut what about the probability that she identifies the car incorrectly but still says blue: 20% * 15% = 3%\r\nso the ratio of these probabilities is: 4:1\r\nis this any good?", "Solution_16": "shake, now you have the right idea but wrong multiplication. If she incorrectly says blue, what color is the cab? You put 15%, but those are the cabs that are blue. If she says it wrong and says blue, that means the cab was green, not blue, so not 15%. Almost there, but use a fraction for final answer.\r\n\r\nkeep going", "Solution_17": "85% of the cabs are green... so the probability that she incorrectly says that the cab was blue is: 1/5*17/20 = 17% \r\nso our ratio is: \r\n(3/20*4/5) / (1/5*17/20) = 12/17 \r\nso the probability that she correctly identified the car is 12/29....\r\nso as a member of the jury... she does not seem that reliable", "Solution_18": "Good Job Shake, your persistence paid off!!! That is correct", "Solution_19": "thanks...\r\nto rcv: how about soroban's solution now that a middleschooler has gotten it.", "Solution_20": "[quote=\"shake9991\"]thanks...\nto rcv: how about soroban's solution now that a middleschooler has gotten it.[/quote]\r\nSoroban's solution has been restored.\r\n\r\n[hide=\"Some comments on the solutions\"]I think this was an excellent example of a problem, generally introduced at a high-school or college-level course, that [b]can[/b] be solved by middle schoolers, using middle school techniques. Thanks to Coach Frost for providing the question and the necessary guidance to get to a solution.\n\nSoroban's solution is very good. However, it uses a theorem that I would not introduce to middle school students.[/hide]", "Solution_21": "[quote=\"Soroban\"]It IS a conditional probability problem!\n\n[hide]\nLet $G$ = the cab is Green, $B$ = the cab is Blue.\n\nWe are given the following probabilities:\n$Pr(G) = 0.85,\\;Pr(B) = 0.15$\n$Pr(\\text{Right}) = 0.80,\\;Pr(\\text{Wrong}) = 0.20$\n\nThere are four cases:\n[color=yellow]. . . [/color]$(1)\\;Pr(G \\land \\text{said Green}) = (0.85)(0.80) = 0.68$\n[color=yellow]. . . [/color]$(2)\\;Pr(G \\land \\text{said Blue})\\;\\; = (0.85)(0.20) = 0.17$\n[color=yellow]. . . [/color]$(3)\\;Pr(B \\land \\text{said Blue})\\;\\; = (0.15)(0.80) = 0.12$\n[color=yellow]. . . [/color]$(4)\\;Pr(B \\land \\text{said Green}) = (0.15)(0.20) = 0.03$\n\nIn cases (2) and (3), she said the cab was Blue.\n[color=yellow]. . . [/color]Hence, $P(\\text{said Blue}) = 0.17 + 0.12 = 0.29$\n\n\nThe problem asks \"What is the probability that the cab $is$ Blue, given that she said Blue?\"\n[color=yellow]. . . [/color]That is: $Pr(B\\;|\\;\\text{said Blue})$\n\nBayes' Theorem says: $Pr(B\\;|\\;\\text{said Blue}) \\;= \\;\\frac{Pr(B \\land \\text{said Blue})}{Pr(\\text{said Blue})} $\n\n[color=yellow]. . . [/color]The numerator is Case (3): 0.12\n[color=yellow]. . . [/color]The denominator was determined above: 0.29\n\n\nTherefore: $Pr(B\\;|\\;\\text{said Blue}) \\;= \\;\\frac{0.12}{0.29} = 0.413793 \\;\\approx \\;41.4\\%$\n\n[size=75]Solution restored by moderator rcv on Sun Feb 13, 2005 15:52 EST[/size][/hide][/quote]\r\nthat's elegant... Bayes' theorem probably could be helpful for National Mathcounts only", "Solution_22": "I have a similar problem at school that I will post soon and see if you can make the connection between the two. Thanks to rcv for moderating the discussion and good job shake for sticking with it", "Solution_23": "No way i coulda solved these problems without help" } { "Tag": [ "linear algebra", "matrix", "vector" ], "Problem": "Prove that if $ A\\in M_{n\\times n}$ has $ n$ distinct eigenvalues, then $ A$ is diagonalizable.", "Solution_1": "Choose $ n$ eigen-vectors $ v_1, v_2, ..., v_n$. Clearly they are linearly independent. And the matrix $ (v_1 v_2 ... v_n)$ diagonalises A.", "Solution_2": "Start by noting that if a matrix is diagonalizable that means that it is similar to a diagonal matrix. The complete theorem is:\r\n\r\nA square matrix $ A \\in M_{n \\times n}$ is similar to a diagonal matrix if and only if A has n linearly independent eigenvectors.\r\n\r\nNotice that it is not necessary for A to have n disctinct eigenvalues. To prove the above theorem, go through the following steps:\r\n\r\n1) Show that two similar matrices share the same eigenvalues.\r\n\r\n2) Show that every $ n \\times n$ diagonal matrix has $ n$ linearly independent eigenvectors.\r\n\r\n3) Show that if a matrix $ A \\in M_{n \\times n}$ is similar to a diagonal matrix, then it has $ n$ linearly independent eigenvectors. \r\n\r\n4) Show that if a matrix$ A \\in M_{n \\times n}$ has $ n$ linearly independent eigenvectors, then A is similar to a diagonal matrix. This will complete the proof." } { "Tag": [ "LaTeX", "inequalities", "triangle inequality" ], "Problem": "Find integers $ x,y,z$ such that $ xy\\plus{}yz\\plus{}zx\\equal{}2\\plus{}xyz$", "Solution_1": "Assume WLOG that x>=y>=z. If z=1, then xy+x+y=2+xy, so x+y=2, x=y=1.\r\nSo (1,1,1) is a solution. If z=2, then xy+2x+2y=2+2xy, so xy+2=2x+2y, (x-2)(y-2)=2. Thus x=4, y=3. So (4, 3, 2) is a solution.\r\nIf z>=3, then xyz+2>3xy>xy+xz+yz.\r\n\r\nTherefore, the only solutions are (1,1,1), (4,3,2) and permutations of those.", "Solution_2": "sorry, but what is WLOG? Also, what do you mean by \"permutations of these\"?\r\n\r\nAlso, the problem just says \"intergers\", not \"positive intergers", "Solution_3": "[quote=\"abcak\"]sorry, but what is WLOG? Also, what do you mean by \"permutations of these\"?\n\nAlso, the problem just says \"intergers\", not \"positive intergers[/quote]\n\nWLOG = Without loss of generality\n\nby permutations of these he means that $ (2,3,4),(4,2,3),(2,4,3)\\cdots$ are also answers because they all have the same numbers but in different order.\n\nSolution $ \\text{\\LaTeX}$'xed:\n\n[quote=\"bpgbcg\"]Assume WLOG that $ x\\ge y\\ge z$. If $ z=1$, then $ xy+x+y=2+xy$, so $ x+y=2$, $ x=y=1$.\nSo $ (1,1,1)$ is a solution. If $ z=2$, then $ xy+2x+2y=2+2xy$, so $ xy+2=2x+2y$, $ (x-2)(y-2)=2$. Thus $ x=4, y=3$. So $ (4, 3, 2)$ is a solution.\nIf $ z\\ge3$, then $ xyz+2>3xy>xy+xz+yz$.\n\nTherefore, the only solutions are $ (1,1,1), (4,3,2)$ and permutations of those.[/quote]", "Solution_4": "Actually, I think there are $ sideways8$ solutions.\r\ni.e., z=1, x+y=1 and permutations as bpgbcg pointed out.\r\n\r\nAlso, i don't get the contradiction when z=3+", "Solution_5": "If $ z \\geq 3$, then $ xyz \\plus{} 2 > 3xy$. Furthermore, $ xy \\plus{} xz \\plus{} yz < 3xy$ (since $ x \\leq y \\leq z$), so $ xy \\plus{} xz \\plus{} yz \\equal{} xyz \\plus{} 2 > 3xy > xy \\plus{} xz \\plus{} yz$, a contradiction.", "Solution_6": "[quote=\"abcak\"]$ sideways8$[/quote]\r\n\r\n\\infty = $ \\infty$.", "Solution_7": "We get all sorts of solutions if we don't assume the numbers are positive. Some examples:\r\n\r\n$ x\\equal{}0 \\implies yz\\equal{}2$\r\n$ x\\equal{}1 \\implies y\\plus{}z \\equal{} 2$\r\n\r\nI don't see any other families immediately. Maybe it will help to rewrite the equation as:\r\n\r\n$ \\frac{1}{x}\\plus{}\\frac{1}{y}\\plus{}\\frac{1}{z} \\equal{} 1\\plus{}\\frac{2}{xyz}$", "Solution_8": "By shifting the variables by $ 1$ (i.e. $ x\\equal{}a\\plus{}1$), then the condition is $ abc\\equal{}a\\plus{}b\\plus{}c$.\r\n\r\nSo $ a(bc\\minus{}1)\\equal{}b\\plus{}c$. i) $ bc\\equal{}1$, $ b\\plus{}c\\equal{}0$ no solution.\r\n\r\nii) Else, $ |b|\\plus{}|c|\\ge |b\\plus{}c|\\ge |bc\\minus{}1|\\ge |bc|\\minus{}1$ where we use the triangle inequality, the fact that $ |a|\\ge 1$, and the triangle inequality again. Then we have\r\n\r\n$ 2\\ge (|b|\\minus{}1)(|c|\\minus{}1)$\r\n\r\nSo anyways, $ b\\equal{}0$ gives $ a\\equal{}\\minus{}c$ so $ (t,0,\\minus{}t)$ is a solution.\r\n\r\nThe rest is easy casework (and there are a finite number of other solutions)." } { "Tag": [ "function", "geometry", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "given $ a,b,d>0$ consider function $ f(x)\\equal{}\\sqrt{a^2\\plus{}x^2}\\plus{}\\sqrt{b^2\\plus{}(d\\minus{}x)^2}$ on $ [0,d]$\r\n\r\n$ f$ continuous and $ [0,d]$ compact so $ f$ has minimum.\r\n\r\nfind it, but don't use the well known fact $ f'(x)\\equal{}0$ at extreme-point", "Solution_1": "[hide=\"Hint.\"]Mirror. [hide=\"More explicitly.\"]Where does the line segment from $ (0,a)$ to $ (d,\\minus{}b)$ go?[/hide][/hide]", "Solution_2": "[quote=\"primenumber\"]given $ a,b,d > 0$ consider function $ f(x) \\equal{} \\sqrt {a^2 \\plus{} x^2} \\plus{} \\sqrt {b^2 \\plus{} (d \\minus{} x)^2}$ on $ [0,d]$\n\n$ f$ continuous and $ [0,d]$ compact so $ f$ has minimum.\n\nfind it, but don't use the well known fact $ f'(x) \\equal{} 0$ at extreme-point[/quote]\r\nSimple geometry :)\r\nI would like to see a calculus explanation too though..\r\n\r\nLet $ |AO| \\equal{} a$, $ |OB| \\equal{} b$, $ |OD| \\equal{} d$, $ |OX| \\equal{} x$\r\n\r\nSo the $ |AX| \\equal{} \\sqrt {a^2 \\plus{} x^2}$ and $ |DX| \\equal{} \\sqrt {b^2 \\plus{} (d \\minus{} x)^2}$ So $ |AX| \\plus{} |XD| \\ge |AD| \\equal{} \\sqrt {(a \\plus{} b)^2 \\plus{} d^2}$, so the minimum is $ \\sqrt {(a \\plus{} b)^2 \\plus{} d^2}$\r\n\r\nEdit: Dayum... How do I insert my geogebra picture..? \"Insert picture\" doesn't work... :S", "Solution_3": "[quote=\"Mathias_DK\"]I would like to see a calculus explanation too though..[/quote]\r\nI just ground through it. Take the derivative of that $ f,$ set $ f'(x)\\equal{}0.$ Put one square root on each side of the equation. Square. Clear fractions. Cancel terms so that an apparent quartic equation turns out to be quadratic. Solve. Put results back into expression for $ x.$\r\n\r\nLots of algebraic manipulation, with not a lot of insight. If you really want to see it, do it yourself. Since you already know what the answer is, it should be possible to de-bug what you have.", "Solution_4": "thank you :)" } { "Tag": [ "number theory", "superior algebra", "superior algebra unsolved" ], "Problem": "hi\r\n\r\nI have briefly started looking at 7-adic residues and have found that the only non-trivial ones of 29 are -12,12. Why is it that 29 has so few? Is it because it is the smallest prime of the form 7k+1? Any help or pointing to relevant theorems would be greatly appreciated.\r\n\r\nThank you very much", "Solution_1": "what do you mean by \"$ 7$-adic residues of $ 29$\"?", "Solution_2": "if there exists an x s.t. x^7 is congruent to a (mod 29) then a is a 7-adic residue of 29.", "Solution_3": "I think it can be explained elementarily by Fermat's Little Theorem.\r\n\r\nLet $ x^7 \\equiv a (mod 29)$,\r\n\r\nSince by Fermat's Little Theorem, $ x^{28} \\equiv 1 (mod 29)$ if $ x$ isn't a multiple of 29,\r\ni.e. $ a^4 \\minus{} 1 \\equiv 0 (mod 29)$, so it has at most 4 roots, which are 1, - 1, 12, -12 if the calculation of amitie is correct.", "Solution_4": "thank you!!", "Solution_5": "I wouldn't call this a $ 7$-adic residue. It's an $ 7$-th power or something like that.", "Solution_6": "\"$ 7$-th power residue\" is the common nomenclature. Of course it's quite unrelated to anything $ 7$-adic, which is what you get by investigating solutions modulo $ 7$, $ 7^2$, $ 7^3$ etc. Here you are interested in solutions mod $ 29$.", "Solution_7": "ah ok. i was just using the definition from here: http://mathworld.wolfram.com/ReciprocityTheorem.html", "Solution_8": "Just an advice: MathWorld is very often a very unreliable source (for some topics, more than half of the entries have errors)." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Find the max value of \r\n\r\n$ \\sqrt[3]{4x\\minus{}3\\plus{}\\sqrt{16\\minus{}24x\\plus{}9x^{2}\\minus{}x^{3}}} \\plus{} \\sqrt[3]{4x\\minus{}3\\minus{}\\sqrt{16\\minus{}24x\\plus{}9x^{2}\\minus{}x^{3}}}$\r\n\r\nwhen x \u20ac (-1,1).", "Solution_1": "$ f(x)_{max,x\\in [\\minus{}1;1]}\\equal{}2$ when $ x\\equal{}1$ :wink:", "Solution_2": "[quote=\"zweig\"]Find the max value of \n\n$ \\sqrt [3]{4x \\minus{} 3 \\plus{} \\sqrt {16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}}} \\plus{} \\sqrt [3]{4x \\minus{} 3 \\minus{} \\sqrt {16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}}}$\n\nwhen x \u20ac (-1,1).[/quote]\r\n\r\n\r\n$ \\sqrt {16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}}\\equal{}\\sqrt{(1\\minus{}x)(4\\minus{}x)^{2}}\\equal{}[\\minus{}1 \r\n\r\n$ \\sqrt [3]{4x \\minus{} 3 \\plus{} \\sqrt {16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}}} \\plus{} \\sqrt [3]{4x \\minus{} 3 \\minus{} \\sqrt {16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}}}\\equal{}\\sqrt{1\\minus{}x}\\plus{}1\\plus{}1\\minus{}\\sqrt{1\\minus{}x}\\equal{}2$", "Solution_3": "If I understood it correctly, you proved it is [b]always[/b] 2, right?\r\n[quote=\"zaya_yc\"]\n$ \\sqrt [3]{4x \\minus{} 3 \\plus{} \\sqrt {16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}}} \\equal{} \\sqrt [3]{4x \\minus{} 3 \\plus{} (4 \\minus{} x)\\sqrt {1 \\minus{} x}} \\equal{} \\sqrt {1 \\minus{} x} \\plus{} 1$ [/quote]\r\nThis step was a nice observation :lol:", "Solution_4": "We set $ A \\equal{} 4x \\minus{} 3$ and $ B \\equal{} 16 \\minus{} 24x \\plus{} 9x^{2} \\minus{} x^{3}$\r\n\r\nObserving that $ \\sqrt [3]{A \\pm \\sqrt {B}} \\equal{} \\sqrt [3]{(A \\pm \\sqrt {B})\\cdot1\\cdot1}$ then we obtain from AM-GM :\r\n\r\n$ \\sqrt [3]{A \\plus{} \\sqrt {B}} \\plus{} \\sqrt [3]{A \\minus{} \\sqrt {B}} \\leq \\frac {(A \\plus{} \\sqrt {B} \\plus{} 2) \\plus{} (A \\minus{} \\sqrt {B} \\plus{} 2)}{3} \\equal{} \\frac {2A \\plus{} 4}{3} \\equal{} \\frac {8x \\minus{} 2}{3}$\r\n\r\nBut, $ \\frac { \\minus{} 10}{3}\\leq\\frac {8x \\minus{} 2}{3}\\leq 2$ when x \u20ac [-1;1], indeed by transitivity we obtain :\r\n\r\n$ \\sqrt [3]{A \\plus{} \\sqrt {B}} \\plus{} \\sqrt [3]{A \\minus{} \\sqrt {B}} \\leq 2$,\r\n\r\nwhich is the max, reached when $ x \\equal{} 1$." } { "Tag": [ "algebra", "polynomial", "number theory solved", "number theory" ], "Problem": "Consider a polynomial P(x) with real coefficients. Prove that if there exist infinitely many pairs of rationals (a,b) s.t. P(a)=b then all of the coefficients of P are rational.", "Solution_1": "Probably I am idiot, but this sounds too easy. Just consider a suficiently large linear system ( that is, the number of equations is the number of coefficients) and express the coefficients in terms of Vandermonde determinants and other determinants will all entries rational. Probably I misuderstood the question.", "Solution_2": "Or we can use Lagrange formula.... :lol: 0", "Solution_3": "sorry i'm the idiot here :blush:. i was too concerned trying to find a counter-example to see the obvious solution." } { "Tag": [], "Problem": "[b]The Problem[/b]\r\n\r\nDuring the holiday season, mall traffic is at its peak! Last minute shoppers, bargain hunters and tiny tots rushing to tell Santa their wish lists are filling up mall aisles everywhere! The Martins took their kids to the mall to see Santa. Mr. Martin had never seen a line so long! At 11:32 am he counted that there were 61 kids ahead of his and that Santa was spending 2 minutes 15 seconds, on average, with each child. If Santa maintained this rate, at what time would the Martin kids get to sit on Santa\u2019s lap? \r\n\r\n\r\n--------------------------------------------------------------------------------\r\n\r\nMore hours = More profit: On average, the Gadget Store profits $600 per hour it is open during regular weeks. Its regular hours are 9-5 every day of the week. Last week, due to the holidays, the store stayed open 3 hours longer per day and averaged 20% more profit per hour. How many more dollars of profit did The Gadget Store make during the holiday week than the average regular week? \r\n\r\n\r\n\r\n--------------------------------------------------------------------------------\r\n\r\nDoes The Early Bird Get The Bargain?: Last week I witnessed the classic \u201cpercent discount\u201d mistake at a local store. The store had everything on sale for 60% off, plus an additional 20% off you shopped before noon. The woman in front of me in line was shocked when the cashier gave her the final price for her purchase. She said, \u201cThat\u2019s not 80% off the price on the tag!\u201d The cashier then explained to her that she should only take 60% off of the original price, and then take 20% of that subtotal off. Because it sounds better, the store advertises \u201c60% off, plus an additional 20% off\u201d, but it\u2019s really equal to a one-time discount of x%. What is the value of x?", "Solution_1": "How can you not know these problems, JEFF?! :D \r\n\r\nsomeone else can do 1, because i think jeff should know it :roll: \r\n[hide=\"2\"]It was open for 8 hours a day on weekdays, and 11 hours during the holiday. It average 600*1.2=720 dollars an hour. \n\nIt made 33600 every week on normal weeks, during the holiday, it make 55440. \n55440-33600=21840, assuming it was open on Sunday. [/hide]\n\n[hide=\"3\"]0.8*0.4=0.32, or 68% off. (0.8=20% off, 0.4=60% off). [/hide]", "Solution_2": "do u mean 61 children[b] including or not including [/b]the child sitting on Santa's lap?\r\nalso, did a child start sitting on Santa's lap the instant Mr. Martin looked at the clock? well, to mi understanding [hide=\"answer #1\"]2 min and 15 secs times 61 is 122 min and 915 secs, which is 137 min 15 secs or 2 hours 17 min and 15 secs. 11:32 + 2:17:15 is 13:49:15 or 1:49:15 PM.[/hide]", "Solution_3": "1. 2.25 minutes per child and 61 children equals 120+15+2.25=137.25 minutes or\r\ntwo hours and 17 minutes 15 seconds" } { "Tag": [], "Problem": "Pete thinks of a number. He doubles it, adds 10, multiplies by 4,\nand ends up with 120. What was his original number?", "Solution_1": "let $ x$ be the original number:\r\nOur equation would then be $ 4(2x\\plus{}10)\\equal{}120$, dividing by 4 so our equation is now $ 2x\\plus{}10\\equal{}30 \\Longrightarrow 2x\\equal{}20 \\Longrightarrow x\\equal{}\\boxed{10}$", "Solution_2": "Or, as an alternative, we can just work backwards!\r\n\r\nLet's start with $ 120$. Pete multiplied by $ 4$, so now we divide.\r\n\r\n$ \\frac{120}{4} \\equal{} 30$\r\n\r\nHe added $ 10$, so we subtract.\r\n\r\n$ 30 \\minus{} 10 \\equal{} 20$\r\n\r\nHe doubled it so we \"un-double\" it.\r\n\r\n$ \\frac{20}{2} \\equal{} 10$\r\n\r\nOur answer is $ \\boxed{10}$" } { "Tag": [], "Problem": "a word has n characters. some of the characters are repeated. how do you find the number of permutations of r-letter words( n things taken r at a time) where r is smaller than n.", "Solution_1": "As far as I know, there is no way of working this out other than splitting it up into several cases. Maybe someone can say I'm wrong?", "Solution_2": "The only way that I know of where you can find permutations of a word with repeated letters is with ALL of the letters being used. \r\n\r\nThe formula is (number of elements!)/(number of each element!) \r\n\r\nFor example, is mississippi, there are 1 m's, 4 i's, 4 s's, and 2 p's, with 11 letters total. The number of permutations would be (11!)/(1!4!4!2!) = 34,650.\r\n\r\nHope this helps. Now, you might be able to break this down into different cases involving r letters." } { "Tag": [ "real analysis", "integration", "real analysis unsolved" ], "Problem": "Show that if $ \\parallel{} f_n \\parallel{}_1 \\leq 2^{\\minus{}n}$ for every $ n \\geq 1$ then $ (f_n)_n$ converges to zero a.e.\r\n\r\nThis should follow from\r\n\r\nIf $ (f_n)_n$ is a sequence in $ L^1[0, 1]$ is such that $ \\sum_{n\\equal{}1}^{\\infty} \\parallel{} f_n \\parallel{}_1 < \\infty$ then $ \\sum_{n\\equal{}1}^{\\infty} | f_n(s) | < \\infty$ for almost every $ s \\in [0, 1]$.\r\n\r\n\r\nHowever, I don't see how to show this. I would appreciate some hints on how to proceed.", "Solution_1": "Suppose there is a set $ A\\subset [0,1]$, with Lebesgue measure $ m(A) > 0$, such that $ \\sum_{n \\equal{} 1}^\\infty |f_n(s)| > \\sigma$ for all $ s\\in A$. Show that $ \\sum_{n \\equal{} 1}^\\infty \\Vert f_n \\Vert_1 \\equal{} \\sum_{n \\equal{} 1}^\\infty \\int_A|f_n(s)|ds \\plus{} \\sum_{n \\equal{} 1}^\\infty \\int_{A^c}|f_n(t)|dt > \\sigma m(A)$. What happens as $ \\sigma \\to \\infty$?\r\n\r\n[hide=\"remark\"]This proof seems a bit roundabout, as you're proving a rather general statement in order to then prove a simpler very specific one. Here's what I think is a slightly more natural proof of the first statement: \n\nSuppose the contrary, that there exists $ \\epsilon > 0$ such that $ |f_n| > \\epsilon$ for all $ n$ on a set $ A\\subset[0,1]$ with $ m(A) > 0$. Then $ \\Vert f_n\\Vert_1\\geq \\int_A |f_n(s)|ds\\geq \\epsilon m(A)$ for all $ n$, contradiction.[/hide]" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Prove that all norms in $ \\mathbb{R}^n$ are equivalent.", "Solution_1": "I'm not sure I understand, I mean, I always thought that there existed different norms because we want to prove that there are metric spaces with different norms.\r\n\r\nEdit: Ah ok, thanks jmerry for the clarification on the terminology.", "Solution_2": "\"Equivalent\" doesn't mean that they're isometric- it means that they induce the same topology. Two norms $ \\|\\cdot\\|_a$ and $ \\|\\cdot\\|_b$ are equivalent if $ C_1\\|x\\|_a\\le \\|x\\|_b\\le C_2\\|x\\|_a$ for some constants $ C_1,C_2$ and all $ x$.", "Solution_3": "One can compare with the $ l_1$ norm of $ R^n$. For an arbitrary norm,\r\n$ \\parallel{}a_1e_1 \\plus{} ... \\plus{} a_ne_n\\parallel{} \\leq |a_1\\parallel{}|e_1\\parallel{} \\plus{} ... \\plus{} |a_n\\parallel{}|e_n\\parallel{} \\leq \\max_{1 \\leq i \\leq n} \\parallel{}e_i\\parallel{} (|a_1| \\plus{} ... \\plus{} |a_n|)$\r\nOn the other hand, we note that with respect to $ l_1$, $ \\parallel{}.\\parallel{}$ is continuous. since $ S^{n\\minus{}1}$ is compact, $ \\parallel{}.\\parallel{}$ has a minimum on $ S^{n\\minus{}1}$ (distance between $ l_1$), say $ \\parallel{}x\\parallel{} \\ge c$ for any $ x \\in S^{n\\minus{}1}$. Then in general $ \\parallel{}x\\parallel{} \\ge c|x|_1$ (in $ l_1$ norm).\r\n\r\nThis shows the equivalence of any norm with $ l_1$ norm, since equivalence is an equivalence relation we are done." } { "Tag": [], "Problem": "So I have a pretty decent grasp of the basic principles of chemistry - reaction mechanisms, bonding patterns, solubility rules, and a little bit of electro and thermochemistry. My problems arise when I get questions which ask me about common substances not bonding because there exists some miniscule exception to standard bonding theories that the two substances exhibit. Do any of you have any good recommendations for books / texts that I can use to reinforce these basic principles and exceptions? A book at or below olympiad level would suit me best.\r\n\r\nThanks!", "Solution_1": "umm... This has been answered at least 10 times on this forum so far....\r\n\r\nIf you wanna get good at chemistry, u gotta do hard problems, and a lot of them. Whatever source can supply that is fine. Many people choose atkins' chem book. I liked zuhmdal myself. whatever floats your boat.", "Solution_2": "Thanks!!\r\n\r\nyea.. sorry, I should've browsed the forums a little bit before I posted my question.. :blush:" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": ":D \r\n\r\nIf $ \\beta\\equal{}(b_{1},b_{2},...,b_{n})'$\r\n\r\nthen $ \\beta\\beta'\\equal{}$$ \\begin{bmatrix}b_{1}b_{1}& b_{1}b_{2}&... & b_{1}b_{n}\\\\ b_{2}b_{1}& b_{2}b_{2}& ... & b_{2}b_{n}\\\\ ... & ... & ... & ...\\\\b_{n}b_{1}& b_{n}b_{2}& ... & b_{n}b_{n}\\end{bmatrix}.$\r\n\r\n[color=red]Find the EigenValues of the above matrix [/color]:lol: \r\n\r\nOK,Assume $ b_{1} \\not\\equal{} 0$\r\n\r\nwe have:$ \\begin{bmatrix}b_{1}b_{1}& b_{1}b_{2}&... & b_{1}b_{n}\\\\ b_{2}b_{1}& b_{2}b_{2}& ... & b_{2}b_{n}\\\\ ... & ... & ... & ...\\\\b_{n}b_{1}& b_{n}b_{2}& ... & b_{n}b_{n}\\end{bmatrix}$ $ \\simeq$ $ \\begin{bmatrix}b_{1}b_{1}& 0&... & 0\\\\ 0& 0& ... & 0\\\\ ... & ... & ... & ...\\\\0& 0& ... & 0\\end{bmatrix}$\r\n\r\nOK, Find the EigenValues of its similar matrix is easy. But I can't Find the eigenvalues of $ \\beta\\beta'$. Is there some relationship between $ \\beta\\beta'$ and its similar matrix? :roll:", "Solution_1": "Be careful about the words you are using. It doesn't appear to me that you really mean similar (i.e. same linear transformation, different matrix representation using another basis). It seems that you really want to say row echelon form.\r\n\r\nAnyway, to compute the characteristic polynomial for this matrix. Notice that\r\n1. rank = 1 (unless all $ b_i \\equal{} 0$, let's say this is not the case)\r\n2. $ \\beta$ is an eigenvector for this matrix. What is the corresponding eigenvalue $ \\lambda$?\r\n3. Do you know what the eigenvalues are and their multiplicities now?", "Solution_2": "$ \\beta\\beta',$ or as I'd prefer to write it, $ \\beta\\beta^T,$ is an $ n\\times n$ matrix of rank $ 1.$ That means that $ 0$ is an eigenvalue of it of geometric multiplicity $ n-1.$\r\n\r\nThe algebraic multiplicity of an eigenvalue is always greater than or equal to the geometric multiplicity. So at least $ n-1$ of the eigenvalues (considered algebraically) is zero. But the sum of the eigenvalues is the trace of the matrix, and the trace of $ \\beta\\beta^T$ is $ b_1^2+b_2^2+\\cdots+b{}^n{}^2.$\r\n\r\nThe eigenvalues are $ 0,0,\\dots,0,b_1^2+b_2^2+\\cdots+b{}^n{}^2.$\r\n\r\nThe action you called \"similar\" in your post was not, in fact, true matrix similarity.\r\n\r\n$ B$ is similar to $ A$ if and only if there exists an invertible $ P$ such that $ B=P^{-1}AP.$\r\n\r\nBy contrast, $ B$ is row equivalent to $ A$ if and only if there exists an invertible $ P$ such that $ B=PA.$ What you used above wasn't even row equivalence (or column equivalence) but a combination of both: $ B=PAQ$ for invertible $ P$ and invertible $ Q.$\r\n\r\nIf two matrices are similar, then they have the same trace, the same determinant, the same characteristic polynomial, and the same eigenvalue.\r\n\r\nRow equivalence preserves none of these things. Two row equivalent matrices will have different traces, different determinants, different eigenvalues. True, you can use row reduction to compute $ \\det A,$ but that involves keeping track of the determinant of $ P.$" } { "Tag": [ "calculus", "integration", "function", "trigonometry", "calculus computations" ], "Problem": "\\[ \\int {\\frac{{x^2 }}{{\\left( {x\\sin x \\plus{} \\cos x} \\right)^2 }}} dx\r\n\\]", "Solution_1": "[quote=\"Xaenir\"]\n\\[ \\int {\\frac {{x^2 }}{{\\left( {x\\sin x \\plus{} \\cos x} \\right)^2 }}} dx\n\\]\n[/quote]\r\n\r\n\r\nlook here please\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=168932\r\n\r\nand here :\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=106333", "Solution_2": "hello, i have got $ \\int\\frac{x^2}{(x\\sin(x)\\plus{}\\cos(x))^2}\\,dx\\equal{}\\frac{\\sin(x)\\minus{}x\\cos(x)}{\\cos(x)\\plus{}x\\sin(x)}\\plus{}C$.\r\nSonnhard." } { "Tag": [ "limit", "integration", "geometry", "rectangle", "function", "logarithms", "calculus" ], "Problem": "Evaluate\r\n$\\lim_{n\\to\\infty}1/n*[1/(1+(1/n))+1/(1+(2/n))+1/(1+(3/n))...1/(1+(n/n)]$\r\n\r\nHow would you do this?\r\nI got to $\\lim_{n\\to\\infty}\\int^{n}_{1}{1/(x+n)}dx$", "Solution_1": "It's a\r\n[hide=\"cow.\"]Riemann sum in disguise on the interval $[0,1]$.[/hide]", "Solution_2": "care to explain? thanks.", "Solution_3": "Consider using rectangles of width $\\frac{1}{n}$ on the interval $[0,1]$. What function would give you heights of\r\n\r\n$\\frac{1}{1+\\frac{1}{n}}, \\frac{1}{1+\\frac{2}{n}}, \\ldots, \\frac{1}{1+\\frac{n}{n}}$?\r\n\r\nAlternatively, write the result in summation notation and try to match it to\r\n\r\n$\\lim_{n \\rightarrow \\infty}\\frac{1}{n}\\sum_{k=1}^{n}f\\left(\\frac{k}{n}\\right)$.", "Solution_4": "$\\int^{n}_{1}\\frac{1}{x+n}\\,dx = \\ln\\left(\\frac{2n}{n+1}\\right)$,\r\n\r\nso \r\n\r\n$\\lim_{n\\to\\infty}\\int^{n}_{1}\\frac{1}{x+n}\\,dx = ...$." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Consider the finite field $ F = GF(2^h)$ with $ h$ a positive integer. A polynomial $ p(x)$ is called a normal permutation polynomial iff $ x\\mapsto p(x)$ defines a permutation on the elements of $ F$ with $ p(0) = 0$ and $ p(1) = 1$. \r\n\r\nIf $ p(x)$ and $ \\frac {p(x)}{x}$ are both normal permutation polynomials, show that $ f: F^*^2\\to F^*: (x,y)\\to \\frac {1}{p(x)}\\frac {1}{p(y)} + \\frac {x}{p(x)}\\frac {y}{p(y)}$ is not surjective.\r\n\r\n[i][size=75]Remark: if f(x) is a normal permutation polynomial, then so is 1/f(x) with 1/0=0. Here F* denotes the nonzero elements of F.[/size][/i]", "Solution_1": "No one? Is it too hard or too ugly? :maybe:" } { "Tag": [], "Problem": "A regular octagon with sides of length $ 1$ has $ 20$ diagonals. Compute the sum of the [u]squares[/u] of the lengths of these diagonals.", "Solution_1": "[hide=\"Answer\"]\n\nUsing Pythagorean and counting diagonals got me $ 24 \\plus{} 16\\sqrt {2} \\plus{} 16 \\plus{} 8\\sqrt {2} \\plus{} 16 \\plus{} 8\\sqrt {2} \\equal{} 56 \\plus{} 32\\sqrt {2}$[/hide]" } { "Tag": [], "Problem": "There is a number, which when divided by $ \\frac{2}{3}$ of $ \\frac{4}{5}$ of $ 1\\frac{1}{2}$ will produce 1. What is the square of the number? Express your answer as a common fraction.", "Solution_1": "You can tell by the wording that we are supposed to evaluate \"$ \\frac {2}3$ of $ \\frac {4}5$ of $ 1\\frac {1}{2}$\" before dividing \"a number\" by that. Note that the word \"of\" means \"$ \\times$\". Call \"a number\" $ x$. Knowing that $ 1\\frac {1}2 \\equal{} \\frac {3}{2}$, we have:\r\n\\[ \\frac {x}{\\cancel{\\frac {2}{3}}\\times\\frac {4}{5}\\times\\cancel{\\frac {3}{2}}} \\equal{} 1\\]\r\nSimplifying:\r\n\\[ \\frac {x}{\\frac {4}{5}} \\equal{} 1\\implies 5x \\equal{} 4\\implies x \\equal{} \\frac {4}{5}\\]\r\nSo our final answer is $ x^2 \\equal{} \\left(\\frac {4}{5}\\right)^2 \\equal{} \\frac {4^2}{5^2} \\equal{} \\boxed{\\frac {16}{25}}$" } { "Tag": [ "number theory", "greatest common divisor", "relatively prime" ], "Problem": "Given that $ A \\equal{} \\{\\frac{n}{24}\\}$, $ n$ is a natural number, gcd$ (n,24) \\equal{} 1$, and $ \\frac{n}{24} < 2$, what is the sum of the elements of A?", "Solution_1": "24 = 2^3 * 3\r\nThat means it must be relatively prime if it isn't divisible by 2 and/or 3\r\n\r\nNumbers less than 48 that satisfy this:\r\n5,7,11,13,17,19,23,25,29,31,35,37,41,43,47\r\n\r\nSum: 383\r\nSum of members in set: 383/24", "Solution_2": "By definition, 1 is relatively prime to all whole numbers, so the answer is $ \\frac{383\\plus{}1}{24}\\equal{}\\boxed{16}$", "Solution_3": "just feel like saying you can pair the terms that sum to 2", "Solution_4": "ok thanks!" } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "Could someone please help me with this\r\nInt[0,2]Int[x,2] y^2sin(xy)dydx", "Solution_1": "change of limits , [0,2] x [x,2] = [0,y]x[0,2],\r\n\r\n$ \\int_0^2 \\int_x^2 y^2 \\sin (xy) dy dx = \\int_0^2 \\int _0^y y^2 \\sin (xy) dx dy =$ \r\n${ \\int_0^2 y^2 \\left( 1-\\frac{\\cos (y^2) }{y}\\right)dy=\\int_0^2 y^2 - \\frac{2y \\cos (y^2)}{2} dy= \\frac{8}{3}-\\frac{ \\sin 4}{2}}$" } { "Tag": [ "algebra", "polynomial", "complex numbers" ], "Problem": "Does \r\n$x^4+x+1$ divide $x^{100}+x^{25}+1$?", "Solution_1": "Would you like to post a solution? I remain perplexed :huh: If I can show the roots of the first equation are related to the $25^{th}$ roots of unity, I have a proof...", "Solution_2": "[hide]yes...$y=x^25$, then say $r$ is a root of the first, then the roots of the second are\n\n$x^25=r \\implies (x-r)(x^24...)=0$, then since all roots of the first are also of the second, result follows[/hide]", "Solution_3": "[quote=\"Altheman\"][hide]yes...$y=x^25$, then say $r$ is a root of the first, then the roots of the second are\n\n$x^25=r \\implies (x-r)(x^24...)=0$, then since all roots of the first are also of the second, result follows[/hide][/quote]\r\nAre you saying that r has to be a root of $x^{25} = r$? That is wrong. In fact, if $|r| \\ne 1,0$ it cannot possibly be true, since the length of each root will be $|r|^{1/25}$. \r\n\r\nIt is obvious that we have no roots to $x^4 + x + 1 = 0$ for which $|r| = 0$.\r\n\r\nAssume that we have a root $r$ for which $|r| = 1$. Then we have that $r^4 + r + 1 = 0, |r^4| = |r| = |1| = 1$. It is not hard to show that that for this to be true we MUST have one of the following:\r\nCase 1: $1 = e^0, r = e^{\\frac{2\\pi}{3}i}, r^4 = e^{\\frac{4\\pi}{3}i}$.\r\nCase 2: $1 = e^0, r^4 = e^{\\frac{2\\pi}{3}i}, r = e^{\\frac{4\\pi}{3}i}$.\r\nThe geometric interpretation of this is that if we have three complex numbers on the unit circle and the sum of them is 0, they must lie in a equilateral triangle.\r\n\r\nYou can easily check (just take the respective $r$ to the power four) that neither of these two cases is possible. Therefore, there is no root $r_j$ to $x^4 + x + 1 = 0$ with length 1 or 0. \r\n\r\nSince we have concluded that every solution to $x^{100} + x^{25} + 1 = 0$ must be on the form $x^{25} = r_j$, we've also concluded that the two polynomials does not have a single root in common (using the comment in the beginning). They therefore also cannot divide each other.", "Solution_4": "you are right... i don't know what i was doing with that factorization...i was thinking of $x^{25}-r^{25}=0$...but after my mistake, we see that they do not have common roots" } { "Tag": [ "function", "trigonometry", "limit", "calculus", "calculus computations" ], "Problem": "Does there exist a real-valued function $ f(x)$ on $ \\mathbb{R }$ which is differentiable at every $ x\\in \\mathbb{R}$, but $ f '(x)$ is not continuous ?", "Solution_1": "$ f(x)\\equal{}x^2\\sin(\\frac{1}{x})$ for $ x\\neq 0$, \r\n$ f(x)\\equal{}0$ for $ x\\equal{}0$.", "Solution_2": "[quote=\"feniks\"]$ f(x) \\equal{} x^2\\sin(\\frac {1}{x})$ for $ x\\neq 0$, \n$ f(x) \\equal{} 0$ for $ x \\equal{} 0$.[/quote]\r\n$ \\lim_{x \\to 0} x \\sin \\left( \\frac{1}{x}\\right) \\equal{} 0$\r\nAs, $ x \\to 0, \\; \\frac{1}{x} \\to \\infty$ but $ f(\\theta) \\equal{} \\sin \\theta \\in [\\minus{}1, 1]$.\r\nHence, any number within $ [\\minus{}1, 1] \\times \\lim_{x \\to 0}x\\equal{}0$.", "Solution_3": "[quote=\"Rajiv\"][quote=\"feniks\"]$ f(x) \\equal{} x^2\\sin(\\frac {1}{x})$ for $ x\\neq 0$, \n$ f(x) \\equal{} 0$ for $ x \\equal{} 0$.[/quote]\n$ \\lim_{x \\to 0} x \\sin \\left( \\frac {1}{x}\\right) \\equal{} 0$\nAs, $ x \\to 0, \\; \\frac {1}{x} \\to \\infty$ but $ f(\\theta) \\equal{} \\sin \\theta \\in [ \\minus{} 1, 1]$.\nHence, any number within $ [ \\minus{} 1, 1] \\times \\lim_{x \\to 0}x \\equal{} 0$.[/quote]\r\n\r\nSorry,what means?", "Solution_4": "The value of the function is $ 0$, at $ x\\equal{}0$, hence you need not add an extra \"$ f(0)\\equal{}0$, atleast I feel so.", "Solution_5": "[quote=\"Rajiv\"]The value of the function is $ 0$, at $ x \\equal{} 0$, hence you need not add an extra \"$ f(0) \\equal{} 0$, atleast I feel so.[/quote]\r\n\r\n\r\nI think feniks' discription is conservatively careful, do you agree?\r\n\r\n :lol: :D", "Solution_6": "[quote=\"Rajiv\"]The value of the function is $ 0$, at $ x \\equal{} 0$[/quote]\r\nHow can you know what $ f(0)$ is if it involves an expression $ \\frac{1}{x}$?\r\nThe limit as x approaches 0 is 0, but $ \\frac{1}{x}$ is still not defined at $ x\\equal{}0$.", "Solution_7": "[quote=\"BlueVelvet\"][quote=\"Rajiv\"]The value of the function is $ 0$, at $ x \\equal{} 0$[/quote]\nHow can you know what $ f(0)$ is if it involves an expression $ \\frac {1}{x}$?\nThe limit as x approaches 0 is 0, but $ \\frac {1}{x}$ is still not defined at $ x \\equal{} 0$.[/quote]\r\nAlthough $ \\frac {1}{x} \\to \\infty$ as $ x \\to 0$, we have a sine function which is bounded within $ [ \\minus{} 1, 1]$. Hence $ \\sin \\left( \\frac {1}{x}\\right)$ is defined, finite. Hence you can multiply it with $ x$ to get a determined value, $ 0$.", "Solution_8": "See the problem is difference between [i][b]functional value[/b][/i] and [b][i] limiting value[/i][/b]. a function may have limiting value at a certain point but not be continuous. suppose we define $ f(0) \\equal{} 5$. then our function is not continuous, but has limiting value at $ x \\equal{} 0$.", "Solution_9": "Ok. I get it.", "Solution_10": "[quote=\"Rajiv\"]a sine function which is bounded within $ [ \\minus{} 1, 1]$. Hence $ \\sin \\left( \\frac {1}{x}\\right)$ is defined, finite. [/quote]\r\nThis just doesn't make sense. Of course, whatever you put in the sine, the outcome will be between 1 and -1. But the problem is that we don't know [i]what [/i]to put in the sine. Forget about the $ x$ in front of the sine, and consider $ f(x) \\equal{} \\sin\\left(\\frac {1}{x}\\right)$. This isn't defined at $ x \\equal{} 0$. Multiplying by $ x$ does not change that. You seem to be confusing this with what [i]is \"reasonable\" to define, considering what value the function approaches near the undefined value[/i].\r\n\r\nAs another example, consider $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{x \\minus{} 1}$. In this way, $ g(1)$ is not defined because we would be dividing by zero. You might be tempted to say $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{x \\minus{} 1} \\equal{} \\frac {(x \\minus{} 1)(x \\plus{} 1)}{x \\minus{} 1} \\equal{} x \\plus{} 1$, so that $ g(1) \\equal{} 2$, but this is not contained in the formula. Or even better, the function $ h(x) \\equal{} \\frac {x}{x}$. It equals 1 everywhere, except at zero: there it's not defined.\r\nTo define a function properly, you just have to be explicit what value the function assigns to each point in its domain. No limits involved.\r\n\r\n(sorry if it's offtopic)", "Solution_11": "[quote=\"BlueVelvet\"][quote=\"Rajiv\"]a sine function which is bounded within $ [ \\minus{} 1, 1]$. Hence $ \\sin \\left( \\frac {1}{x}\\right)$ is defined, finite. [/quote]\nThis just doesn't make sense. Of course, whatever you put in the sine, the outcome will be between 1 and -1. But the problem is that we don't know [i]what [/i]to put in the sine. Forget about the $ x$ in front of the sine, and consider $ f(x) \\equal{} \\sin\\left(\\frac {1}{x}\\right)$. This isn't defined at $ x \\equal{} 0$. Multiplying by $ x$ does not change that. You seem to be confusing this with what [i]is \"reasonable\" to define, considering what value the function approaches near the undefined value[/i].\n\nAs another example, consider $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{x \\minus{} 1}$. In this way, $ g(1)$ is not defined because we would be dividing by zero. You might be tempted to say $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{x \\minus{} 1} \\equal{} \\frac {(x \\minus{} 1)(x \\plus{} 1)}{x \\minus{} 1} \\equal{} x \\plus{} 1$, so that $ g(1) \\equal{} 2$, but this is not contained in the formula. Or even better, the function $ h(x) \\equal{} \\frac {x}{x}$. It equals 1 everywhere, except at zero: there it's not defined.\nTo define a function properly, you just have to be explicit what value the function assigns to each point in its domain. No limits involved.\n\n(sorry if it's offtopic)[/quote]\r\ncheck post 2 carefully, function has been piecewise defined, [b]so it is reasonable[/b] :wink: :)", "Solution_12": "[quote=\"BlueVelvet\"][quote=\"Rajiv\"]a sine function which is bounded within $ [ \\minus{} 1, 1]$. Hence $ \\sin \\left( \\frac {1}{x}\\right)$ is defined, finite. [/quote]\nThis just doesn't make sense. Of course, whatever you put in the sine, the outcome will be between 1 and -1. But the problem is that we don't know [i]what [/i]to put in the sine. Forget about the $ x$ in front of the sine, and consider $ f(x) \\equal{} \\sin\\left(\\frac {1}{x}\\right)$. This isn't defined at $ x \\equal{} 0$. Multiplying by $ x$ does not change that. You seem to be confusing this with what [i]is \"reasonable\" to define, considering what value the function approaches near the undefined value[/i].\n\nAs another example, consider $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{x \\minus{} 1}$. In this way, $ g(1)$ is not defined because we would be dividing by zero. You might be tempted to say $ g(x) \\equal{} \\frac {x^2 \\minus{} 1}{x \\minus{} 1} \\equal{} \\frac {(x \\minus{} 1)(x \\plus{} 1)}{x \\minus{} 1} \\equal{} x \\plus{} 1$, so that $ g(1) \\equal{} 2$, but this is not contained in the formula. Or even better, the function $ h(x) \\equal{} \\frac {x}{x}$. It equals 1 everywhere, except at zero: there it's not defined.\nTo define a function properly, you just have to be explicit what value the function assigns to each point in its domain. No limits involved.\n\n(sorry if it's offtopic)[/quote]\r\n\r\nThanks, I get it (as I already said before!!!)", "Solution_13": "[quote=\"ith_power\"]check post 2 carefully, function has been piecewise defined, [b]so it is reasonable[/b] :wink: :)[/quote]My message was still directed to the quote [quote=\"Rajiv\"]The value of the function is $ 0$, at $ x \\equal{} 0$, hence you need not add an extra \"$ f(0) \\equal{} 0$, atleast I feel so.[/quote] The original defintion in post 2 is of course correct.\n[quote] (as I already said before!!!)[/quote]\r\nSorry ;) I thought my last post added something to the discussion.\r\n\r\n(PS: I think it's a bit overdone to quote the whole message)" } { "Tag": [ "function", "LaTeX", "number theory", "prime numbers", "number theory unsolved" ], "Problem": "Let $p , q , r$ be distinct prime numbers and let $A$ be the set \r\n $A=\\{ p^aq^br^c : 0\\leq a , b,c \\leq 5 \\}$ . \r\n Find the smallest integer $n$ such that any $n$-element subset of $A$ contains two distinct elements $x,y$ such that $x$ divides $y$", "Solution_1": "Let $A=\\{p_1 ^{a_1} \\dots p_k^{a_k} ,a_i\\leq m\\}, \\ and \\ \\deg(x) \\ mean \\ function \\ A \\to Z_+ \\ \\sum a_i .$ We have k=3,m=5. Let \\[ B_n=\\{y=p_1^{b_1} \\dots p_k^{b_k}, \\ b_i \\leq m, \\deg(y)=\\sum b_i =n\\} \\] \r\nEasyly to show, that for any x exist y, that $\\deg(y)=n \\ \\ y|x \\ if \\deg(x)>n, \\ x|y \\ if \\ \\deg(x)\\left(1+\\frac{1}{n+1}\\right)\\left\\{\\left(1+\\frac{1}{n}\\right)^n-\\left(1+\\frac{1}{n}\\right)\\right\\}>0$\r\n\r\n$\\Longleftrightarrow \\left(1+\\frac{1}{n+1}\\right)^{n+2}\\geq \\left(1+\\frac{1}{n+1}\\right)\\left(1+\\frac{1}{n}\\right)^n,$ yielding $\\left(1+\\frac{1}{n+1}\\right)^{n+2}>\\left(1+\\frac{1}{n+1}\\right)^{n+1}$\r\n\r\nwhich completes the proof by induction.", "Solution_4": "[quote=\"Lovasz\"]Why must we use induction? It would be better if using AM-GM or cansider the function $f(x)=\\left(x+\\frac1{x}\\right)^x$. They're shorter.[/quote]\r\nBecause using induction is a part of our problem.", "Solution_5": "[quote=\"kunny\"]For $n=1,$ L.H.S=$\\frac{1}{2},$ R.H.S.=$\\frac{9}{4},$ yielding the claim is true.[/quote]\r\n$\\displaystyle \\left(1+\\frac{1}{(1)}\\right)^{(1)}=(1+1)^1=2$", "Solution_6": "Thanks, hpasten. :lol:", "Solution_7": "Sorry for that I have just found mistakes in my solution. I'm confusing! :?: \r\nBut I found antoher proof without using induction or A.M.-G.M.", "Solution_8": "How to Prove this using am-gm?\n", "Solution_9": "We have the following: \\[1+\\frac{1}{n+1} = \\frac{1 + n(1+\\frac{1}{n})}{n+1} \\geq \\sqrt[n+1]{1\\cdot (1+\\frac1n)^n}\\] which is what we want. ", "Solution_10": "Alternative solution: you can study the monotony of the $f(x)=\\frac{\\ln (1+x)}{x}$ using the derivative.", "Solution_11": "Is there an easy way to proof that it converges to $e$ when we deine $e$ as the factorial sum", "Solution_12": "[quote=hpasten]Show by induction that \n$\\displaystyle \\left(1+\\frac{1}{n}\\right)^n\\le \\left(1+\\frac{1}{n+1}\\right)^{n+1}$\nfor every $n\\in\\mathbb{N}$\n\nI know it`s too easy, for example using $AM\\ge GM$, but here you must use induction.[/quote]\n\nAn alternate solution is as follows:\nLet's define the sequence $\\left(x_{n}:=\\left(1+\\frac{1}{n} \\right)^{n}: n \\in \\mathbb{N} \\right)$.Clearly it suffices to prove that $(x_{n})$ is nondecreasing.. But we already know that $(x_{n})$ is bounded , nondecreasing sequence converging to $e$.", "Solution_13": "The monotony it is enought for prove the inequality.", "Solution_14": "[quote = Lukas8r20]\nIs there an easy way to proof that it converges to $e$ when we deine $e$ as the factorial sum\n[/quote]\nSee e.g. [url = https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function]here[/url] for the equivalence of the different definitions/characterizations for $e$." } { "Tag": [ "trigonometry", "AMC" ], "Problem": "Source: Furman University Wylie Math Tournament Senior Exam 2001 #21\r\n\r\nWithout using a calculator/tables, evaluate:\r\ncos 87:^o:/sin 1:^o: - sin 87:^o:/cos 1:^o:.", "Solution_1": "(I omitted all the degree simbols since there are way to many of them.)\n\n[hide]cos87/sin1-sin87/cos1\n\n(cos1cos87-sin1sin87)/sin1cos1\n\n(cos1sin3-sin1cos3)/sin1cos1\n\n\n\ncos1sin3=1/2*(sin4-sin(-2))\n\nsin1cos3=1/2*(sin4+sin(-2))\n\nsin1cos1=1/2*(sin2+sin0)=1/2*sin2\n\n\n\n(1/2*(sin4-sin(-2))-1/2*(sin4+sin(-2)))/(1/2*sin2)\n\n(1/2*sin4-1/2*sin(-2)-1/2*sin4-1/2*sin(-2))/(1/2*sin2)\n\n-sin(-2)/(1/2*sin2)\n\nsin2/(1/2*sin2)\n\n2[/hide]\n\nAm I right?", "Solution_2": "Solution (in spoiler):\n\n\n\n[hide]cos 87/sin 1 - sin 87/cos 1.\n\n(cos 1 cos 87 - sin 87 sin 1) / sin 1 cos 1\n\n\n\nSince sin x = cos (90-x) and cos x = sin (90-x),\n\n\n\n(sin 89 cos 87 - sin 87 cos 89) / sin 1 cos 1\n\nsin (89 - 87) / sin 1 cos 1\n\nsin 2 / sin 1 cos 1\n\n\n\nSince sin 2 = sin (2 * 1) = 2 sin 1 cos 1,\n\n\n\n2 sin 1 cos 1 / sin 1 cos 1\n\n2[/hide]", "Solution_3": "How did you get from\n\nyour solution wrote:[hide](sin 89 cos 87 - sin 87 cos 89) / sin 1 cos 1[/hide]\nto\nyour solution wrote:[hide]sin (89 - 87) / sin 1 cos 1[/hide]\n\n?", "Solution_4": "He got there using the identity sin (x-y) = sin x cos y - sin y cos x", "Solution_5": "I see...*sigh* I need more practice wth trigs...", "Solution_6": "don't we all?", "Solution_7": "Well there are people in this forum who are in colledge/university/etc. now...\r\nFor us, yes. [b]very[/b]", "Solution_8": "Yes. You are both right.", "Solution_9": "Nah. Trig is fun. More straightforward than algebra, if you ask me." } { "Tag": [], "Problem": "\u0397 \u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03af\u03b1 $ a_{0},a_{1},...,a_{n}$ \u03bf\u03c1\u03af\u03b6\u03b5\u03c4\u03b1\u03b9 \u03c9\u03c2 \u03b5\u03be\u03ae\u03c2: $ a_{0}\\equal{}\\frac{1}{2}$ \u03ba\u03b1\u03b9 $ a_{k\\plus{}1}\\equal{}a_{k}\\plus{}\\frac{1}{n}a^{2}_{k}$ \u03b3\u03b9\u03b1 $ k\\equal{}0,1,...,n\\minus{}1$. \u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 $ 1\\minus{}\\frac{1}{n} with(plots):\n> E:=implicitplot(x=-y^2+6*y,x=-2..10,y=-7..10):\n> F:=implicitplot(x=2*y,x=-2..10,y=-2..10):\n> display({E, F});[/code]\r\n\r\n :D" } { "Tag": [ "calculus", "derivative", "integration" ], "Problem": "Do you any good physics text book that could be comparable to the Art of Problem Solving series?\r\n\r\nI'm looking for something that is beyond high school level, and focused more on contests and such.", "Solution_1": "Most physiscs books aren't aimed at contests. To do well on a physics competition, you really have to know the physics. The more problems you do, the better you'll do on competitions.\r\n\r\nThe Halliday and Resnick book uses calculus, but the problems are very good.", "Solution_2": "does UsaPho/ipho use calculus? :D \r\nand my brother said that halliday and resnick is too advanced for beginners is this true?", "Solution_3": "IPhO does not use calculus as far as I know.\r\n\r\nHalliday and Resnick is the standard book in most freshman year physics classes so I as soon as you know calculus (basic differentiation and integrals), you can use that book. The Halliday, Resnick, and Krane book is a bit tougher, but still good." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove or Disprove \r\na- A linear Transformation T :R^3 --> R^3 WITH EIGENVALUES -7,2, 4 SUCH THAT R^3 IS NOT A THE Direct Sum of the eigenspaces of T.\r\n\r\nb- A linear Transformation T :R^3 --> R^3 WITH EIGENVALUES -8, 4 \r\nWhere 4 has multiplicity two \r\n SUCH THAT R^3 IS NOT have a basis of the eigenspaces of T.\r\n\r\nThank u in Advance", "Solution_1": "For (b). we might as well give the matrix of the transformation in Jordan form:\r\n\r\n$\\begin{pmatrix}4&1&0\\\\0&4&0\\\\0&0&-8\\end{pmatrix}$\r\n\r\nThe largest set of independent eigenvectors we can find has two elements and is thus not a basis for $\\mathbb{R}^3.$\r\n\r\nBut for (a):False. A linear tranformation from $\\mathbb{R}^3$ to $\\mathbb{R}^3$ might as well be a $3\\times 3$ matrix. As long as such a matrix has three distinct (real) eigenvalues, the three eigenvectors form an independent set, and thus span $\\mathbb{R}^3.$ So the direct sum of the eigenspaces will be $\\mathbb{R}^3$ and there will be no counterexample." } { "Tag": [ "LaTeX" ], "Problem": "cac ban cho hoi LaTex la gi. Toi nghe noi no dung de go ki tu toan hoc. Cac ban co the bay toi cach su dung duoc khong", "Solution_1": "Ban co the vao Latex help o goc tren man hinh , hoac co the vao trang web toan TV http://www.diendantoanhoc.net", "Solution_2": "nh\u01b0ng t\u00f4i n\u00ean download ca\u0301i na\u0300o.Ca\u0309m \u01a1n.(T\u00f4i \u0111a\u0303 th\u01b0\u0309 nh\u01b0ng chua ti\u0300m ra )", "Solution_3": "Latex la 1 chuong trinh de go ky hieu Toan . Ban khong can phai dowload cai gi het , chi can hoc cach viet cau lenh de xuat hien cac ky hieu do . Ban co the click vao cac cong thuc toan (in dam) trong cac bai viet cua members de xem cach viet ." } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "A pentagone ABCDE has AB=BC, CD=DE, $ \\widehat{ABC}\\equal{}120$\r\ny $ \\widehat{CDE}\\equal{}60$ y BD=2. Calculate the area of the pentagone.", "Solution_1": "See here a very similar problem...\r\nit is solved in the same way as yours\r\n\r\nDaniel", "Solution_2": "Thank you but I can not see the link.", "Solution_3": "There's no link and this should be moved into geometry proposed&own problems.", "Solution_4": ":blush: my mistake\r\nHere is it\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=863241#863241\r\n\r\n\r\nDaniel", "Solution_5": "Thank you a lot." } { "Tag": [ "MATHCOUNTS", "quadratics" ], "Problem": "Find the smallest real value of x that satisfies the equation:\r\n\r\n(x+5)(x:^2:-x-11)=x+5", "Solution_1": "[hide]\n\n-5?[/hide]", "Solution_2": "that's so correct that it's like...correct. ya", "Solution_3": "Syntax Error wrote:Find the smallest real value of x that satisfies the equation:\n\n(x+5)(x:^2:-x-11)=x+5\n\n\n\nsome people like to put their reasoning you know, but since im an impatient moderator\n\n\n\nReasoning:[hide]sigh, so you want it to equal 0 preferably, so then (-5+5)(doesnt matter about this part)=-5+5\n\n\n\nso 0=0\n\nyah![/hide]", "Solution_4": "You really [b]are[/b] impatient. Not only was your last post doubled, the whole thread was doubled too.\r\n\r\nAlso I do agree AoPS is [b]unusually[/b] groggy (slow) currently...\r\nand it stopped working for, like, 3 times already.", "Solution_5": "what the heck...some weird reasoning out there\r\nSyntax, try this problem...\r\n\r\n(x+5)(x:^2+5x-5)=(x+5). What's the smallest possible value of x?", "Solution_6": "-6. In this case the right side is x-5, not 0, so the answer is -5. (It won't work with the roots of the quadratic)", "Solution_7": "[hide]-6[/hide]", "Solution_8": "yes. that's right. however you would get the wrong answer if you used the same answer you did before.", "Solution_9": "just take the roots of the second one:x:^2:+5x-6=0\r\n\r\nto get: -5 :pm: :sqrt: (25+24) all divided by 2\r\n\r\nto get 1 or -6\r\n\r\n\r\npfft, i solved it before tare, but my internet connection is taking so long to submit stuff and stuff. pfft", "Solution_10": "Factoring is easier though\r\n\r\nx :^2: +5x -6 = (x+6)(x-1) so the smallest is x = -6\r\n\r\nand didn't umm... Mystic's problem have x :^2: +5x - 5???", "Solution_11": "[quote=\"Syntax Error\"]just take the roots of the second one:x:^2:+5x-6=0\n\nto get: -5 :pm: :sqrt: (25+24) all divided by 2\n\nto get 1 or -6\n\n\npfft, i solved it before tare, but my internet connection is taking so long to submit stuff and stuff. pfft[/quote]\r\n\r\nI know. It (AoPS server) can't handle all the posts. Mr. Rusczyk must be moving the data to the new server or something.", "Solution_12": "yeah, now all of a sudden, my thingie is fast again", "Solution_13": "OK, now that the problem is solved I'm going to change this to Normal, is that OK with you?", "Solution_14": "yeah, its just for people to see it", "Solution_15": "[hide]-5 satisfies the equality. lets see if theres anything else. set x doesn't equal -5 and divide that term out.\n\nthen x^2-x-12=0=(x-4)(x+3) --> x=-3 is another small one, but not quite as small.\n\nanswer: -5.[/hide]" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "Cramming for an exam tomorrow and looking through the past papers.\r\n\r\nI'm not too sure how to approach the following problem. (There's something similar in every past paper)\r\n\r\n\r\n\r\n\r\nShow that $ 1\\plus{}i$ is a solution of the equation\r\n\r\n$ x^6 \\plus{} 2x^5 \\plus{} 3x^4 \\plus{} 2x^3 \\plus{} 3x^2 \\plus{} 2x \\plus{} 2 \\equal{} 0$\r\n\r\nand find all other solutions (real or complex) of this equation.\r\n\r\n\r\n\r\n\r\nAny help as to how to do this type of question would be appreciated. (Also, am I in the right forum?)", "Solution_1": "hello, if $ 1\\plus{}i$ is a solution of your equation then $ 1\\minus{}i$ is also a solution.\r\nSonnhard.", "Solution_2": "So if a complex number is a solution, it's conjugate is also a solution?", "Solution_3": "Only if the coefficients of [i]your[/i] polynomial happen to be real... ;)", "Solution_4": "Cool\r\n\r\nSo how do I go about finding the other solutions? Would I be right in thinking there be 4 more to find?\r\n\r\nAm I dividing the expression by the factor $ x\\minus{} (1 \\plus{} i)$ perhaps?\r\n\r\nThat looks painful.", "Solution_5": "[quote=\"rudyard\"]Cool\n\nSo how do I go about finding the other solutions? Would I be right in thinking there be 4 more to find?\n\nAm I dividing the expression by the factor $ x \\minus{} (1 \\plus{} i)$ perhaps?\n\nThat looks painful.[/quote]\r\n\r\n$ (x\\minus{}(1\\plus{}i))(x\\minus{}(1\\minus{}i))\\equal{}(x^2\\minus{}2x\\plus{}2)$\r\nSo we only need to divide the expression by the factor $ (x^2\\minus{}2x\\plus{}2)$", "Solution_6": "[quote=\"stephencheng\"]\n\n$ (x \\minus{} (1 \\plus{} i))(x \\minus{} (1 \\minus{} i)) \\equal{} (x^2 \\minus{} 2x \\plus{} 2)$\nSo we only need to divide the expression by the factor $ (x^2 \\minus{} 2x \\plus{} 2)$\n\n[/quote]\r\n\r\nThankyou very much! That rings a bell from a 9am lecture I dozily attended. \r\n\r\nIs enough for me to crack on and get a few more marks anyway, I can do polynomial division without that pesky $ i$ happily enough." } { "Tag": [ "arithmetic sequence" ], "Problem": "In each blank below a single digit is inserted such that the following three-digit numbers, in this order, form and arithmetic sequence:\r\n\r\n1 _ _, _ _ 9, 2 _ 2, _ 6 _, 2 _ _, _ 3 _\r\n\r\nWhat is the value of the next number in the sequence?", "Solution_1": "[hide]\nI think the answer is $264$.\n\nBetween_ _ 9 and 2_2 there is an increase of a number whose ones digit is 3. With this you can fill in the numbers.\n\n1_6, _ _ 9, 2_2, _65, 2_8, _31 \nFrom here it is easy to finish the problem\n[/hide]", "Solution_2": "Oh! I know!\r\n\r\n[hide]You know it increases by the number _$3$ because there is a difference of $3$ between $6$ and $9$...The first two ones digits...\n\nYou also know that _$65$ is $265$ because both of the ones that surround it are $2$__... Given this, the last term has to be $331$, since it has to be greater than $265$. Doing $331-265$ gets you $66$, and you can divide by $2$ to get a difference of $33$. Therefore, the numbers must be $166, 199, 232, 265, 298, and 331$. You can get this by adding $33$ to $265$ to get $298$, subtracting it over and over to get $232, 199, and 166$, and all of these numbers work with what they give you. And then you add $33$ to $331$, to get $331+33=364$, the answer.[/hide]", "Solution_3": "[hide] i am pretty sure that 364 is infact the correct answer.[/hide]", "Solution_4": "[quote=\"footballrocks41237\"]Oh! I know!\n\n[hide]You know it increases by the number _$3$ because there is a difference of $3$ between $6$ and $9$...The first two ones digits...\n\nYou also know that _$65$ is $265$ because both of the ones that surround it are $2$__... Given this, the last term has to be $331$, since it has to be greater than $265$. Doing $331-265$ gets you $66$, and you can divide by $2$ to get a difference of $33$. Therefore, the numbers must be $166, 199, 232, 265, 298, and 331$. You can get this by adding $33$ to $265$ to get $298$, subtracting it over and over to get $232, 199, and 166$, and all of these numbers work with what they give you. And then you add $33$ to $331$, to get $331+33=364$, the answer.[/hide][/quote]\r\n\r\nI don't get it. Can you explain in more detail?", "Solution_5": "I can't explain very well...", "Solution_6": "[hide=\"oops\"]\nI forgot I was carrying over into 300 :blush: .\n[/hide]", "Solution_7": "[hide=\"solution\"]We look at the units digits, and see that we can fnd the other units digits.\n\n1-6,--9,2-2,-65,2-8,-31\n\nAnd, looking at the tens and hundreds digits, we can determine some hundreds digits.\n\n1-6,--9,2-2,265,2-8,331\n\nNow we can find the common difference.\n\n331-265=66\n\n33=d\n\n331+33=364[/hide]" } { "Tag": [ "limit", "real analysis", "real analysis theorems" ], "Problem": "if $\\lim_{x\\rightarrow a}f(x)=L_{1},\\lim_{x\\rightarrow a}g(x)=L_{2}$\r\n$\\lim_{x\\rightarrow a}(f(x).g(x))=\\lim_{x\\rightarrow a}f(x).\\lim_{x\\rightarrow a}g(x)=L_{1}.L_{2}$\r\nplease give some proof for this", "Solution_1": "Those are all questions you can prove easily by using the $\\epsilon$-definition of a limit. Have you tried ?", "Solution_2": "i have tried but i can't prove some parts!!can you send our proof?", "Solution_3": "no one show me the proof??!!", "Solution_4": "This is standard textbook material - do you have a textbook that you are looking at?", "Solution_5": "i have 'silverman' but it doesn't have the proof of this theory!!", "Solution_6": "I'll not wright all proof, but I put an idea. First you have to pay attention that $f(x)$ and $g(x)$ are bounded, let say $|f(x)|zero... \r\n\r\nCan, is my proof correct?", "Solution_3": "[quote=\"shaaam\"]\n$ \\sum_{cyc}{4a^3(b \\plus{} c \\minus{} a)} \\equal{} \\sum_{cyc}{\\frac {(x \\plus{} z \\minus{} y)^3(y \\minus{} x \\minus{} z)}{4}} \\equal{} \\sum_{cyc}{\\frac { \\minus{} (x \\plus{} y \\minus{} z)^4}{4}}$[/quote]\r\nI read your proof, and I think you got mistake here. It must be\r\n\\[ \\sum \\frac{4a^3(b\\plus{}c\\minus{}a)}{(b\\plus{}c)^2}\\]\r\n;) :)", "Solution_4": "[quote=\"can_hang2007\"][quote=\"shaaam\"]\n$ \\sum_{cyc}{4a^3(b \\plus{} c \\minus{} a)} \\equal{} \\sum_{cyc}{\\frac {(x \\plus{} z \\minus{} y)^3(y \\minus{} x \\minus{} z)}{4}} \\equal{} \\sum_{cyc}{\\frac { \\minus{} (x \\plus{} y \\minus{} z)^4}{4}}$[/quote]\nI read your proof, and I think you got mistake here. It must be\n\\[ \\sum \\frac {4a^3(b \\plus{} c \\minus{} a)}{(b \\plus{} c)^2}\n\\]\n;) :)[/quote]\r\n\r\nwell, i think it doesn't matter. because the denominator will be $ y^2$/ But the result will not change. Am i wrong?\r\n\r\n[b]EDIT: [/b] I am really sorry for my mistake.. \r\n\r\nP.S: can you the write expression i showed in second post as sum of the squares?", "Solution_5": "[quote=\"shaaam\"][quote=\"can_hang2007\"][quote=\"shaaam\"]\n$ \\sum_{cyc}{4a^3(b \\plus{} c \\minus{} a)} \\equal{} \\sum_{cyc}{\\frac {(x \\plus{} z \\minus{} y)^3(y \\minus{} x \\minus{} z)}{4}} \\equal{} \\sum_{cyc}{\\frac { \\minus{} (x \\plus{} y \\minus{} z)^4}{4}}$[/quote]\nI read your proof, and I think you got mistake here. It must be\n\\[ \\sum \\frac {4a^3(b \\plus{} c \\minus{} a)}{(b \\plus{} c)^2}\n\\]\n;) :)[/quote]\n\nwell, i think it doesn't matter. because the denominator will be $ y^2$/ But the result will not change. Am i wrong?[/quote]\r\nI dont think so, since I think you got mistake in calculation, it must be\r\n\\[ \\sum \\frac{4a^3(b\\plus{}c\\minus{}a)}{(b\\plus{}c)^2} \\equal{}\\sum \\frac{(x\\plus{}z\\minus{}y)^3(3y\\minus{}x\\minus{}z)}{4y^2}\\]\r\n;):)" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Prove that $ |\\left|\\leq 24$.", "Solution_1": "I believe this is false.\r\n\r\nLet $ x \\equal{} (1,2)(3,4,5,6)$ and $ y \\equal{} (1,2,3)(4,6,5)$ be elements of $ S_6$ of orders $ 4$ and $ 3$ respectively. We have $ xy \\equal{} (1,3,6)$ of order $ 3$. Thus $ x^4 \\equal{} y^3 \\equal{} (xy)^3 \\equal{} 1$. However, the group $ \\langle x,y \\rangle$ is $ A_6$, a group of size $ 360$.\r\n\r\nEDIT: I found those particular elements using GAP, but I suspected the claim was unlikely since triangle groups $ \\langle x,y | x^l \\equal{} y^m \\equal{} (xy)^n \\rangle$ with $ \\frac {1}{l} \\plus{} \\frac {1}{m} \\plus{} \\frac {1}{n} < 1$ are hyperbolic and \"large\". You should expect finite groups when $ \\frac {1}{l} \\plus{} \\frac {1}{m} \\plus{} \\frac {1}{n} > 1$ (the \"spherical\" case).\r\n\r\nNow that I think about it, the group $ \\langle x,y | x^4 \\equal{} y^3 \\equal{} (xy)^2 \\equal{} 1\\rangle$ is spherical, and according to GAP indeed has size $ 24$. Is this what you meant?" } { "Tag": [ "geometry", "similar triangles" ], "Problem": "EF is parallel to BC\r\nBD=DC\r\nProve that EK=KF\r\n\r\nRefer to picture below.", "Solution_1": "Bye bye, save proofs for getting started, not the basics. :D", "Solution_2": "Similar triangles, as before: \r\n\r\nAKE and ABD are similar so BD=EK*k\r\nAKF and ACD are similar so CD=KF*k \r\n\r\nGiving \r\n\r\nBD=CD\r\nEK*k=KF*k \r\nEK=KF" } { "Tag": [ "Support", "induction", "symmetry", "strong induction" ], "Problem": "I think I know the answer, but I'm looking for an easy proof to this problem:\r\n\r\nTwenty points are marked on the circumference of a circle. Two players play the following game. On each turn, one connects two of the 20 points with a segment, according to the following rules:\r\n 1. a segment can appear only once during a game;\r\n 2. no two segments can intersect, except on the endpoints;\r\n 3. the player who has no choice loses.\r\nAssuming both players use their best strategy, which one (first or second) is guaranteed to win?", "Solution_1": "Well...\n\n\n\n[hide]At the end of the game, there will be 18 triangles formed. There will be 37 segments in total. Therefore, player 1 will make the last move.[/hide] - edited because I can't count\n\n\n\nDid I understand the question right?", "Solution_2": "Yes, except there are 37 segments drawn.", "Solution_3": "Wheeps, I subtracted 1 instead. Yeah, it should be 37.", "Solution_4": "Or another way:\n\n[hide]\n\nJoin point 1 to point 11 (counting them in order). Then play symmetrically.\n\n[/hide]", "Solution_5": "TripleM wrote:Or another way:\n[hide]\nJoin point 1 to point 11 (counting them in order). Then play symmetrically.\n[/hide]\n\n\n\n[hide]Actually, it doesn't matter how they play. If a chimp went first, he would always win. The question was to *prove* that player 1 will always win.[/hide]", "Solution_6": "Does anyone have a proof to support their answer?", "Solution_7": "It was to prove it using the best strategies -- it just happens that any strategy is the best strategy. So TripleM was just as correct as the others. (I'm not sure that you were saying he was wrong, but it sounded like it.)", "Solution_8": "[quote=\"JBL\"]It was to prove it using the best strategies -- it just happens that any strategy is the best strategy. So TripleM was just as correct as the others.[/quote]\r\n\r\nYou're right, JBL. TripleM, your strategy works and the proof that player 1 always wins is implied by your strategy. Sorry if it sounded like I was claiming otherwise.\r\n\r\nBut I think smiley wants a proof that *every* strategy provides a win for Player 1.", "Solution_9": "[quote=\"rcv\"][quote=\"JBL\"]It was to prove it using the best strategies -- it just happens that any strategy is the best strategy. So TripleM was just as correct as the others.[/quote]\n\nYou're right, JBL. TripleM, your strategy works and the proof that player 1 always wins is implied by your strategy. Sorry if it sounded like I was claiming otherwise.\n\nBut I think smiley wants a proof that *every* strategy provides a win for Player 1.[/quote]\r\n\r\nWell, the question did say \"Assuming both players use their best strategy\". In which case all you need to do is find a strategy for the winning player. But I suppose if someone *did* want to show that the first player always won, then they'd be happy :)", "Solution_10": "Well, to prove that it always ended in a first player victory I suppose you would have to prove a few things:\r\n\r\n1) In the finishing position, the original 20-gon is disected into triangles. Put a different way, if the polygon is not yet triangluated, then it is possible to draw a segment somewhere under the given rules.\r\n\r\n2) The polygon is triangulated if and only if there are 17 segments drawn.\r\n\r\nFrom those 2 propositions you should be able to conclude that player 1 will be the person who puts in the 17 (and final) segment.", "Solution_11": "Those were the points of my first post - however shouldn't there be 37 segments in total, since the original 20-gon was not drawn?", "Solution_12": "You can do that easily with [hide]strong induction -- assume that it's true for any n-gon up to (and including) a k-gon. First player draws any diagonal. We then have 2 less-than-or-equal-to-k-gons, so apply the inductive hypothesis and we're done.[/hide]\n\n\n\nAlso, it hardly matters in this case whether the outside segments are drawn since they always can be drawn, no matter what the state of the game. So Osiris is right, but it doesn't affect the outcome at all, or even the game-play.\n\n\n\n\n\nHere is a (probably much messier) problem: what if the segments can't intersect anywhere, including endpoints? (This does not mean they have to be parallel -- only the segments must not intersect each other, not the lines they are part of.)", "Solution_13": "Ya, ya'll are right... i forgot about the edges of the 20-gon.\n\n\n\nJBL wrote:Here is a (probably much messier) problem: what if the segments can't intersect anywhere, including endpoints? (This does not mean they have to be parallel -- only the segments must not intersect each other, not the lines they are part of.)\n\n\n\nOh, I know this one! The wonderful [hide]catalan numbers[/hide]!", "Solution_14": "A few variations on this problem and its proofs have been bugging me for a few days - then I couldn't find the problem! \r\n\r\n1. Does N even versus N odd matter?\r\n2. Does including vs. excluding the N edges of the N-gon matter?\r\n\r\nThis yields four cases. Here's my analysis of \r\n1. N even, edges included. P1 wins\r\n Both TripleM's symmetry argument and Joel's induction work (but see note below). Note that the dissection segment becomes the first move in both new games. \r\n\r\n2. N odd, edges included. P1 wins\r\n Though there is no 'middle' diagonal, I believe that the symmetry argument be modified (how?) Induction proof is valid. (Note that even if N even, the induction proof required that P1 wins for both odd and even N upon dissection.)\r\n\r\n3. N odd, edges not included. P2 wins\r\n Since N-3 is even, Player 2 always wins.\r\n Symmetry argument works: Player 2 matches each move of P1.\r\n Induction is more 'interesting' since the dissection is into an odd game and an even game.\r\n\r\n4. N even, edges not included. P1 wins\r\n Original symmetry argument is valid.\r\n Induction again more involved since the first dissection can be into odd,odd games or even,even games. \r\n\r\nConclusion: With minor modifications, I believe that the symmetry argument is valid throughout. Induction requires care. We don't have a response to Zabelman/s challenge of two items to prove. (Basically, the 'simplest' argument for all 4 cases relies on \"exactly N-3 non-edges can always be drawn\" - but why?)\r\n\r\nNote: The induction proof proves that P1 wins for all N when edges included. But the proof does not seem to mention or rely on 'edges' issue. Hence why wouldn't same proof 'prove' that P1 wins for all N when edges not included? Is there a gap in the proof?" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "Sketch the graph of x^x.", "Solution_1": "Weird question. Just take out your graphic calculator? Or think about it a bit. Goes to $ 1$ when $ x$ goes to zero. Then, it takes off sky-rocket with increasing values of $ x$, reaching for example 10 billion at $ x \\equal{} 10$.", "Solution_2": "Haven't got a graphical calculator.\r\n\r\nIt seems clear that for x > 0, x^x is always real and that the graph is continuous, going down from (0,1) to (1,1) and then up quickly. \r\n\r\nHowever, for x < 0, x^x may not exist.\r\n\r\nFor x = -m/n, x^x > 0 when m is even. When m and n are both odd, x^x < 0 (I think) and when m is odd and n is even, x^x seems to be imaginary. Therefore to the left of the y-axis the graph has two branches which are not continuous but are everywhere dense.\r\n\r\nIs this correct?\r\n\r\nAny thoughts on my post about cubic curves?", "Solution_3": "If you want something for $ x<0,$ the interesting graph would be $ y\\equal{}|x|^x.$ Or, more fully,\r\n\\[ y\\equal{}f(x)\\equal{}\\begin{cases}|x|^x&x\\ne0\\\\1&x\\equal{}0\\end{cases}\\]\r\nThis function would have a local minimum at $ x\\equal{}e^{\\minus{}1},$ a local maximum at $ x\\equal{}\\minus{}e^{\\minus{}1},$ and a vertical tangent line at $ x\\equal{}0.$", "Solution_4": "Thanks, Kent." } { "Tag": [ "function", "Putnam", "calculus", "algebra unsolved", "algebra" ], "Problem": "The function $f: R\\to R$ satisfies $f'''(x)$ is a continuos function in $R$.\r\nProve that $\\exists a\\in R$ such that\r\n$f(a)f'(a)f''(a)f'''(a)\\geq 0$", "Solution_1": "[quote=\"phamduyhiep\"]The funtion $f: R\\to R$ satisfies $f'''(x)$ is a continuos function in $R$.\nProve that $\\exists a\\in R$ such that\n$f(a)f'(a)f''(a)f'''(a)\\geq 0$[/quote]\r\n\r\nWhat about $f(x)=e^{-x}$ ?\r\n\r\n :cool:", "Solution_2": "[quote=\"Diogene\"][quote=\"phamduyhiep\"]The funtion $f: R\\to R$ satisfies $f'''(x)$ is a continuos function in $R$.\nProve that $\\exists a\\in R$ such that\n$f(a)f'(a)f''(a)f'''(a)\\geq 0$[/quote]\n\nWhat about $f(x)=e^{-x}$ ?\n\n :cool:[/quote]\r\n\r\n :huh:\r\n\r\nWhat about it ??? :)", "Solution_3": "I didn't see the contracdiction here!\r\nWith $f(x)=e^{-x}$,we have $f(x)f'(x)f''(x)f'''(x)=(e^{-x})^4\\geq 0\\forall x\\in R$", "Solution_4": "if not,we have for $\\any x \\in R$,then $f(x)f'(x)f\\\"(x)f\\\"'(x) <0$\r\nso $f\\\"'(x) \\neq 0$because $f\\\"'(x)$ is a continuos function,\r\nso f(x),f'(x),f\"(x),f\"'(x) must have the only one kind sign when $x \\in R$\r\nCase $f\\\"'(x) >0$,so $f\\\"(x)$ is increasing about $x$.\r\ncase $f\\\"(0)>0$,so $f'(x) \\ge f'(0)+xf\\\"(0)>0$,when x is large enough.\r\nso $f'(x)>0,f\\\"(x)>0$ when $x \\in R$.\r\nsame reason we can prove:$f(x)>0$ when $x \\in R$.\r\ncontradictions,\r\n\r\nWe can also find the contradictions when \r\n$f\\\"'(x) <0$", "Solution_5": "[quote=\"zhaobin\"]if not,we have for $\\any x \\in R$,then $f(x)f'(x)f\\\"(x)f\\\"'(x) <0$\nso $f\\\"'(x) \\neq 0$because $f\\\"'(x)$ is a continuos function,\nso f(x),f'(x),f\"(x),f\"'(x) must have the only one kind sign when $x \\in R$\nCase $f\\\"'(x) >0$,so $f\\\"(x)$ is increasing about $x$.\ncase $f\\\"(0)>0$,so $f'(x) \\ge f'(0)+xf\\\"(0)>0$,when x is large enough.\nso $f'(x)>0,f\\\"(x)>0$ when $x \\in R$.\nsame reason we can prove:$f(x)>0$ when $x \\in R$.\ncontradictions,\n\nWe can also find the contradictions when \n$f\\\"'(x) <0$[/quote]\r\n well done!zhao bin!now i am interested in alberage.", "Solution_6": "This is [url=http://www.kalva.demon.co.uk/putnam/psoln/psol983.html]Putnam 1998/A3[/url]", "Solution_7": "This probably belongs in the \"College Playground\" section - either in the Calculus-Analysis problems, or in the Putnam section." } { "Tag": [ "function" ], "Problem": "Dear group members, \r\n\r\n\r\nDoes anyone have more info about the death of Serge Lang ? -announced \r\nat the following link :( -\r\n\r\nhttp://www.math.yale.edu/public_html/index.html", "Solution_1": "Pray for the souls of the dead!\r\n\r\nSerge Lang was the first person whose book in written English encountered with me.\r\n\r\nkunny", "Solution_2": "RIP.\r\n\r\nI attended a lecture given by him. He put me on the spot (just pointed to me at random) and asked me to come up with a differentiable real function which has an inverse that is not differentiable. I said $sgn(x)x^2$, but then he called on someone else. :roll: :roll: (the other guy said $x^3$). Maybe he thought I said $sin(x)x^2$ instead. Maybe he meant $C^\\infty$ instead of differentiable. I guess we'll never know now.\r\n\r\nAnyway, it amazes me how one guy could know so much (and write so many books)! A great loss to mathematics." } { "Tag": [ "inequalities", "probability" ], "Problem": "I think I got 2\r\n[hide](it just had to be something as beutiful as a 3-4-5, didn't it? A + C = 90, so B was a right angle)[/hide]\nand 4\n[hide](by showing that x = a^5 - a^2 - 1, y = ax gave a solution for all intger a >= 2)[/hide]\r\nplus a load of random workings on the other questions (and a brief sorry to the markers - I didn't think to remove my working on the problems I solved, so there's a lot of paper under \"rough work\" ;).\r\n\r\nHow did people find it?", "Solution_1": "Can u post the the questions???", "Solution_2": "They are at:\r\n\r\nhttp://www.bmoc.maths.org.uk/home/bmo2-2008.pdf", "Solution_3": "Did you ever do the Intermediate Maths Olympiad?", "Solution_4": "Hmmm for #1 [hide]$ \\left( \\sum x^2 \\right)^3 \\ge (x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 2xy \\plus{} 2yz \\plus{} 2zx)(x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} xy \\minus{} yz \\minus{}zx)^2 \\equal{} 1$[/hide]", "Solution_5": "[quote=\"probability1.01\"]Hmmm for #1 [hide]$ \\left( \\sum x^2 \\right)^3 \\ge (x^2 \\plus{} y^2 \\plus{} z^2 \\plus{} 2xy \\plus{} 2yz \\plus{} 2zx)(x^2 \\plus{} y^2 \\plus{} z^2 \\minus{} xy \\minus{} yz \\minus{} zx)^2 \\equal{} 1$[/hide][/quote]\r\nHow do you get any of this?", "Solution_6": "@dgreenb801,\r\n$ (\\sum a^2)^3 \\equal{} (\\sum a^2)(\\sum a^2)^2$\r\nBut we know from Power MEan inequality that $ \\sum x^2 \\ge \\frac {(\\sum x)^2}{3}$.\r\nAlso we know that $ \\sum x^2 \\ge \\sum xy$.\r\nThus we get probability 1.01's lemma.", "Solution_7": "@dgreenb801, \r\n \r\nBut we know from Power MEan inequality that . \r\nAlso we know that . \r\nThus we get probability 1.01's lemma.\r\n__________________________________________________________________ :huh: :rotfl: :rotfl: \r\n[i]||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||[/i]", "Solution_8": "I just did #1 really fast\r\n[hide]Let $ S\\equal{}x^2\\plus{}y^2\\plus{}z^2$ and $ w\\equal{}x\\plus{}y\\plus{}z$\nYou can factor the equation they give as a condition, and then write it in terms of $ S$ and $ w$:\n$ w(S\\minus{}\\dfrac{w^2\\minus{}S}{2})\\equal{}1$.\nSolving for $ S$, you get\n$ S\\equal{}\\dfrac{w^2\\plus{}\\frac{2}{w} }{3} \\ge 1$ by AM-GM.[/hide]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$a>0,b>0,c>0$ and $\\frac{a^3}{b}+\\frac{a^2b}{c}+\\frac{ab^2}{c}+\\frac{c^3}{b}\\leq6.$\r\nProve that $(1+a+a^2+a^3+a^4)a^9bc\\leq100.$", "Solution_1": "Unsolved in 2005. ;)", "Solution_2": "Hi,arqady,can you post the solution now? :blush:", "Solution_3": "$a>0,b>0,c>0$ and $\\frac{a^3}{b}+\\frac{a^2b}{c}+\\frac{ab^2}{c}+\\frac{c^3}{b}\\leq6.$\r\nProve that $(1+a+a^2+a^3+a^4)a^9bc\\leq100.$\r\nProof.\r\n$6\\geq\\frac{a^3}{b}+\\frac{a^2b}{c}+\\frac{ab^2}{c}+\\frac{c^3}{b}=$\r\n$=\\frac{a^3}{3b}+\\frac{a^3}{3b}+\\frac{a^3}{3b}+\\frac{a^2b}{2c}+\\frac{a^2b}{2c}+\\frac{ab^2}{c}+\\frac{c^3}{b}\\geq7\\sqrt[7]{\\frac{a^{14}}{108}}.$\r\nHence, $a\\leq\\sqrt{\\frac{6}{7}}\\cdot\\sqrt[14]{108}.$\r\nBut $6\\geq\\frac{a^3}{b}+\\frac{a^2b}{c}+\\frac{ab^2}{c}+\\frac{c^3}{b}\\geq4\\sqrt[4]{a^6bc}.$ Id est, $a^6bc\\leq\\frac{81}{16}.$\r\nLet $\\sqrt{\\frac{6}{7}}\\cdot\\sqrt[14]{108}=k.$\r\nHence, $(1+a+a^2+a^3+a^4)a^9bc\\leq\\frac{81}{16}(k^3+k^4+k^5+k^6+k^7)=97.8485...<100.$ :)\r\nThank you for interest.", "Solution_4": "Thanks. :)", "Solution_5": "[quote]$(1+a+a^2+a^3+a^4)a^9bc\\leq\\frac{81}{16}(k^3+k^4+k^5+k^6+k^7)=97.8485...<100.$[/quote][quote]$(1+a+a^2+a^3+a^4)a^9bc\\leq100.$[/quote] \r\n:D :D" } { "Tag": [ "rotation" ], "Problem": "I saw a lab tool that I can't remember the name of. \r\n\r\nInfo on the tool:\r\n- used to grind up tissues that were contained in small test tubes (for the one I saw, it was for grinding up mouse lung tissue that was treated earlier with radiation)\r\n- has a disposable tip that was used to grind up tissue\r\n\r\nWhat's this tool called?", "Solution_1": "i think it must be a centrifuge attachment.... :maybe:", "Solution_2": "If it was called that, it had another name...\r\n\r\nIt did not look like it was attached to the centrifuge. You could stick part of the tool into a small tube to mash, not really grind, up small bits of tissue. \r\n\r\nOnce the tissue was mashed up, you could extract the dna and rna using a column in another tube to filter and a centrifuge.", "Solution_3": "If for DNA/RNA extraction, as was mentioned in the third post, I believe the proper tool is a sonicator.\r\n\r\n\"In biological applications, sonication may be sufficient to disrupt or deactivate a biological material. For example, sonication is often used to disrupt cell membranes and release cellular contents.\"\r\n-[url]http://en.wikipedia.org/wiki/Sonicator[/url]", "Solution_4": "oh right yeah and it vibrates and rotates and the actual vibrating and rotating touching the cell mashes it up. \r\n\r\nso I mean its more of a mechanical thing... Just to mash up soft tissue in a tube. \r\n\r\nthink of sticking your finger into a tube and rotating and wiggling your finger to mash up something in the tube.... yeah so its does that." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Suppose that $ \\{a_1,a_2,...,a_n\\}$ and $ \\{b_1,b_2,...,b_k\\}$ are two sets of positive integers with the same sum of elements, and that sum is less than $ nk$. Prove that it's possible to omit some (but not all) summands from both sides of the equality $ a_1\\plus{}a_2\\plus{}...\\plus{}a_n\\equal{}b_1\\plus{}b_2\\plus{}...\\plus{}b_k$ in such a way that the equality remains valid.", "Solution_1": "an idea is to use induction...if you want a full solution ask me and I will post it ( the solution is from a text ,is NOT mine)\r\n\r\nDaniel", "Solution_2": "[hide=\"how about this?\"]We are searching for some set $ A$ of $ a_i$ and some set $ B$ of $ b_i$ such that $ \\displaystyle\\sum_{a_i\\in A}a_i\\minus{}\\displaystyle\\sum_{b_i\\in B}b_i\\equal{}0$, with not all of $ a_i$ in $ A$ or vice versa. So, we prove that such sets exist.\n\nLet $ A_i\\equal{}\\displaystyle\\sum_{j\\equal{}1}^ia_i$ and let $ B_i\\equal{}\\displaystyle\\sum_{j\\equal{}1}^ib_i$ and let $ A_n\\equal{}B_k\\equal{}s$. Consider the set $ X\\equal{}\\{A_i\\plus{}B_j\\}$ which has cardinality $ nk$. First, we prove that modulo $ s$, this set has two equal elements. There are $ nk$ elements and $ s$ remainder classes. Since $ s\\le nk\\minus{}1$, by pigeon hole principle, there are two equal elements modulo $ s$. So, we let $ A_i\\plus{}B_j\\equiv A_p\\plus{}B_q$. Obviously, we cannot have $ i\\equal{}p$ since this implies $ p\\equal{}q$ since $ 0p$. We have two cases:\n\nCase 1: $ j>q$\n\nWe have\n\n$ f\\equal{}\\displaystyle\\sum_{d\\equal{}p\\plus{}1}^i a_d\\plus{}\\displaystyle\\sum_{d\\equal{}q\\plus{}1}^j b_d\\equiv0\\pmod{s}$\n\nSince $ 0\\le fj$\n\nWe have\n\n$ f\\equal{}\\displaystyle\\sum_{d\\equal{}p\\plus{}1}^i a_d\\minus{}\\displaystyle\\sum_{d\\equal{}j\\plus{}1}^q b_d\\equiv0\\pmod{s}$\n\nSince $ f\\minus{}B_q\\ge B_k\\equal{}\\minus{}s$ we have that $ f\\equal{}0$ and we're done.\n[/hide]" } { "Tag": [ "LaTeX" ], "Problem": "SOrry edited....\r\n\r\ni didnt read the question properly...", "Solution_1": "now as per doing by my method i.e put x=0... so a0=1....\r\n\r\ndifferentiate and then put x=0 and we get a1=n*(n+1)/2...........\r\n\r\n\r\nso by applying this method we get Ar=(n+r-1)Cr........\r\n\r\ni am lost after this step...", "Solution_2": "The answer is (2n+1)Cn.....", "Solution_3": "Answer is Wrong... \r\n\r\n[hide]It is $ \\binom{2n}{n}$. Method is right. Please $ \\text{\\LaTeX}$ your post above and write proper steps.... I'll post a more elegant method after that. [/hide]", "Solution_4": "ok answer is $ \\binom{2n}{n}$\r\n\r\nmade a small mistake in the last term..", "Solution_5": "$ a_r\\equal{}\\binom{n\\plus{}r\\minus{}1}{r}$\r\n\r\nthen use the rule $ \\binom{n}{r\\minus{}1} \\plus{}\\binom{n}{r} \\equal{} \\binom{n\\plus{}1}{r}$\r\nso it evaluvates to ur answer shreyas :D" } { "Tag": [ "floor function", "algebra", "polynomial", "quadratics", "function", "trigonometry", "AMC" ], "Problem": "Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005$. Find $\\lfloor P\\rfloor$.", "Solution_1": "[hide=\"Solution\"]We start off by noticing that the polynomial looks suspiciously like $(x-1)^4$.\n\nAdding 1 to both sides, we get $x^4-4x^3+6x^2-4x+1=(x-1)^4=2006$\nThus, $(x-1)^2=-\\sqrt{2006}=x^2-2x+1$, since we only want the nonreal roots.\n\nTh sum of the roots is thus $1+ \\sqrt{2006}$, since the product of the roots of a quadratic is the constant term divided by the leading coefficient.\n\nWhen we apply the greatest integer function, our answer is $45$.[/hide]", "Solution_2": "[hide=\"Answer\"]$x^4-4x^3+6x^2-4x=0\\Rightarrow x^4-4x^3+6x^2-4x+1=1=(x-1)^4$, so we have $(x-1)^4-1=0$. $y=x+1\\Rightarrow y^4-1=0=(y^2-1)(y^2+1)=(y-1)(y+1)(y^2+1)$, so the complex roots are $y=\\pm i\\Rightarrow x-1=\\pm i\\Rightarrow x=1\\pm i$, and the product is $(1+i)(1-i)=\\boxed{2}$.[/hide]\r\n\r\nErr... I'm confused. Did you leave something out of the original equation? Where did the $2005$ come from? And what did you do in your second solution? :?", "Solution_3": "[quote=\"JesusFreak197\"]Err... I'm confused. Did you leave something out of the original equation? Where did the $2005$ come from? And what did you do in your second solution? :?[/quote]\r\n\r\nI'm sorry, I posted the problem incorrectly, hehe. :fool:\r\n\r\nOur two solutions would have been correct if I hadn't misread the problem. :)", "Solution_4": "Okay, I figured there was some problem. I was a little curious as to why the greatest integer function was necessary. :roll:\r\n\r\n[hide=\"Answer\"]$x^4-4x^3+6x^2-4x=2005\\Rightarrow x^4-4x^3+6x^2-4x+1=2006=(x-1)^4$\n$y=x-1\\Rightarrow y^4-2006=0=(y^2-\\sqrt{2006})(y^2+\\sqrt{2006})$\nThe two roots of the first part are both real, so we focus on the second part: $y=\\pm i\\sqrt[4]{2006}=x-1\\Rightarrow x=1\\pm i\\sqrt[4]{2006}$, so the product of the roots is $(1+i\\sqrt[4]{2006})(1-i\\sqrt[4]{2006})=1+\\sqrt{2006}=P\\Rightarrow \\lfloor P\\rfloor=\\boxed{45}$.[/hide]", "Solution_5": "[hide]Adding 1 to both sides, we get $(x-1)^4 = 2005.$ The roots of 2005 are $2006^{1/4} \\text {cis}\\ 0$, $2006^{1/4} \\text {cis}\\ \\pi/2$, $2006^{1/4} \\text {cis}\\ \\pi$, and $2006^{1/4} \\text {cis}\\ 3\\pi/2$. The second and fourth ones are complex. Rewriting in rectangular form, we have $-2006^{1/4}i+1$ and $2006^{1/4}i+1$. Their product is a difference of squares, $P=1^2-(2006^{1/4}i)^2 = 1+\\sqrt{2006}$. Thus $\\lfloor P \\rfloor = 1+44=45$.[/hide]\r\n\r\n(Go interm trig class :D)", "Solution_6": "Uhh... that's hugely unnecessarily complicated. :?", "Solution_7": "[quote=\"JesusFreak197\"]Uhh... that's hugely unnecessarily complicated. :?[/quote]\r\n\r\nIt's not really complicated :huh: It's just using trigonometry instead of the normal way. Unless you don't know DeMoivre's theorem, it shouldn't be too confusing. ;)", "Solution_8": "It was a pretty simple application of Demoivre's theorem, I agree. That seemed like the simplest solution to me at the time. :D \r\n\r\nWas it just me, or were there alot of fast solutions to alot of the 2005 AIME A?", "Solution_9": "For one, I don't know DeMoivre's Theorem (I don't think I do... at least, not by that name), and also, the other method is much simpler than going into the whole $\\text{cis}$ thing...", "Solution_10": "[quote=\"4everwise\"][hide=\"Solution\"]We start off by noticing that the polynomial looks suspiciously like $(x-1)^4$.\n\nAdding 1 to both sides, we get $x^4-4x^3+6x^2-4x+1=(x-1)^4=2006$\nThus, $(x-1)^2=-\\sqrt{2006}=x^2-2x+1$, since we only want the nonreal roots.\n\nTh sum of the roots is thus $1+ \\sqrt{2006}$, since the product of the roots of a quadratic is the constant term divided by the leading coefficient.\n\nWhen we apply the greatest integer function, our answer is $45$.[/hide][/quote]\nhow did u think of that? :D", "Solution_11": "The coefficients of the given polynomial look like the fourth row of Pascal's triangle, i.e. the coefficients of $(x-1)^4$, so we add $1$ and factor to get $(x-1)^4=2006$. Then $x$ can be $1\\pm i\\sqrt[4]{2006}$ or $1\\pm\\sqrt[4]{2006}$. So $$P=(1-i\\sqrt[4]{2006})(1+i\\sqrt[4]{2006})=1+\\sqrt{2006}.$$ After some computation, we find that $\\lfloor \\sqrt{2006}\\rfloor=44$, so the answer is $\\boxed{45}$.", "Solution_12": "[b]Solution:[/b] Exploit the symmetry. Notice that $x^4-4x^3+6x^2-4x$ looks oddly like $x^4-4x^3+6x^2-4x+1 = (x-1)^4$. So making that adjustment, we have that $(x-1)^4 = 2006 \\Rightarrow x = 1 \\pm \\sqrt[4]{2006}$. Clearly, these will be the only real roots. So using Vieta's, we have that $P = \\frac{2005}{\\sqrt{2006} - 1} \\Rightarrow \\lfloor{P}\\rfloor = \\boxed{045}$." } { "Tag": [], "Problem": "How many non-negative integers are less than $ 10\\pi$?", "Solution_1": "That's just 31.4159...\r\n\r\nso ans is 32, because we include 0." } { "Tag": [ "geometry", "geometric transformation", "homothety", "rotation", "geometry unsolved" ], "Problem": "P. point of the triangle abc, so measuring the inside corner PAC equal measure angle PBC. D is the middle of [AB]. L is projected on the vertical told P. (BC), M is projected on the vertical told P. (AC). See chart below. The DM=DL.", "Solution_1": "[quote=\"ashrafmod\"]Given are $\\triangle ABC$ and a interior point $P\\ :$ $\\begin{array}{c}\\widehat{PAC}\\equiv\\widehat{PBC}\\\\ D\\in AB\\ ,\\ DA=DB\\\\ L\\in CB\\ ,\\ PL\\perp CB\\\\ M\\in CA\\ ,\\ PM\\perp CA\\end{array}$ $\\implies DM=DL\\ .$[/quote]\r\n\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=61873 applied to the triangle $APB\\ .$", "Solution_2": "let's do some of darij's work. :lol: \r\n\r\nfirst I restate the problem.\r\n\r\n[b]Problem[/b].[color=blue]in cyclic quadrilareal $ABCD$ let $P$ be intersection of $AC$ and $BD$ and let $H_{1},H_{2}$ be projection of $P$ on $BC$ and $AD$.\nprove that $MH_{1}=MH_{2}$ where $M$ is midpoint of $CD$[/color]\r\n\r\n[i]proof.[/i]let $\\angle BCA=\\angle BDA=\\alpha$\r\nand let $\\chi_{\\alpha}^{k}(O)$ denote a spiral homothecy around center $O$ angle $\\alpha$ and $k$ be the ratio.\r\nlet $f(\\mathcal A)=(\\chi_{\\alpha}^{\\frac{DH_{2}}{DP}}\\circ \\chi_{\\alpha}^{\\frac{CP}{CH_{1}}})(O) \\left(\\mathcal A \\right)$\r\n\r\nhence we have $f(H_{1})=H_{2}$ and $f(M)=M$ so it means $M$ is a constant point under this function.\r\n\r\nand cause $\\frac{CP}{CH_{1}}.\\frac{CH_{2}}{DP}=1$ hence $f(\\mathcal{A})$ is a rotation around center $M$.\r\n\r\nhence $MH_{1}=MH_{2}$" } { "Tag": [ "geometry", "superior algebra", "superior algebra unsolved" ], "Problem": "just like the title", "Solution_1": "just like the title", "Solution_2": "Look at Liu: Algebraic Geometry and Arithmetic Curves pages 28-29." } { "Tag": [ "number theory", "prime factorization" ], "Problem": "Can somebody please post some programs that you have made on a TI-83 plus calculator? :)", "Solution_1": "anyone??? :(", "Solution_2": "Well, I suppose I can share a very simple one, that divides a number into primes.\r\nI'm recalling this one out of my memory, so probably it will have a few bugs.\r\n\r\nDisp \"Which number:\"\r\nInput A\r\nB=2\r\nLbl Q\r\nIf A=1\r\nQuit (I am not sure this commando exists, but some equivalent exists for sure)\r\nIf frac(Q/B)=0\r\nGoto W\r\nB+1-->B\r\nGoto Q\r\n\r\nLbl W\r\nDisp B\r\nQ/B-->Q\r\nGoto Q", "Solution_3": "What kinds do you want? I'll post some later for sample (prime factorization, messup your calculator, radical simplifier...).", "Solution_4": "Here is a program that gives the prime factorization of a number:\r\nNOTE: In LAB, the L means the list thingie.\r\n[hide=\"PRIMEFAC\"]\n:1->Y\n: \u201cInput Number? \u201c,A\n:0->dim(LAB)\n:A->B\n:2->X\n:Goto 3\n:Lbl 5\n:1+X X\n:If A=1:Goto 2\n:If B/A>A:Goto 2\n:Lbl 3\n:Lbl 6\n:If X=1: Goto 2\n:If fPart(A/X)=0\n:Then\n:1+dim(LAB)->dim(LAB)\n:X->LAB(Y)\n:Y+1->Y\n:A/X A\n:If fPart(A/X)\u2260 0:Goto 5\n:Goto 6\n:End\n:Goto 5\n:Lbl 2\n:Disp LAB\n:Delvar LAB\n[/hide]\n\nHere is a harmless program that changes the numbers you put in to incorrect answers. Press ON to exit at any time.\n\n[hide=\"MESSUP\"]\n:ClrHome\n:While 1\n:Input \u201c\u201c,A\n:randInt(1,4) X\n:randInt(-10,10) B\n:If X=1:Disp AB\n:If X=2:Disp B+A\n:If X=3:Disp A-B\n:If X=4:Disp A/B\n:End\n[/hide]\r\n\r\nTell me some other types of programs you might want.", "Solution_5": "I have some games that I programmed, but they are too big to post it here(10814 bytes are considered to be big, right?). :D", "Solution_6": "Sorry for this double post. Below is a program that draws a fractal tree. It asks you to input O, which needs to be a positive integer. below 5 is not that interesting, and above 10 takes really long time to run it.\r\n\r\n[hide=\"PROGRAM: TREE\"]\n: ClrHome\n: Output(2,3,\u201dFRACTAL\n: Output(3,3,\u201dTREE\n: Output(4,3,\u201dGENERATOR\n: Output(7,2,\u201dBY DAVID RHEE\n: Output(8,2,\u201d[PRESS ENTER]\n: Pause\n: ClrHome\n: Prompt O\n: Degree\n: GridOff\n: AxesOff\n: ClrDraw\n: FnOff\n: PlotsOff\n: Zstandard\n: Zsquare\n: 0\u2192X\n: -10\u2192Y\n: 4.5\u2192Z\n: 90\u2192\u03b8\n: Prgm BRANCH\n[/hide]\n\n[hide=\"PROGRAM: BRANCH\"]\n: Line(X,Y,X+Zcos(\u03b8),Y+Zsin(\u03b8\n: If O=0\n: Return\n: O-1\u2192O\n: X\u2192L\u2083(dim(L\u2083)+1\n: Y\u2192L\u2084(dim(L\u2084)+1\n: X+Zcos(\u03b8)\u2192X\n: Y+Zsin(\u03b8)\u2192Y\n: Z\u2192L\u2081(dim(L\u2081)+1\n: \u03b8\u2192L\u2082(dim(L\u2082)+1\n: Z(.7+.2rand)\u2192Z\n: \u03b8+20rand+20\u2192\u03b8\n: Prgm BRANCH\n: L\u2081(dim(L\u2081)+1)\u2192Z\n: L\u2082(dim(L\u2082)+1)\u2192\u03b8\n: Z(.7+.2rand)\u2192Z\n: \u03b8-20rand-20\u2192\u03b8\n: Prgm BRANCH\n: L\u2081(dim(L\u2081))\u2192Z\n: L\u2082(dim(L\u2082))\u2192Z\n: L\u2083(dim(L\u2083))\u2192Z\n: L\u2084(dim(L\u2084))\u2192Z\n: dim(L\u2081)-1\u2192dim(L\u2081\n: dim(L\u2082)-1\u2192dim(L\u2082\n: dim(L\u2083)-1\u2192dim(L\u2083\n: dim(L\u2084)-1\u2192dim(L\u2084\n: O+1\u2192O\n[/hide]\r\n\r\n_____________________________________________________\r\nEDITED: Sorry. It was missing about 5 lines of the program. :blush:", "Solution_7": "How about a prime factorization thingie? I always make errors in factoring.", "Solution_8": "I posted one in my previous post that gives the prime factorization of any number. I edited it to make the lists work better. If I have some spare time, I could post my own version of Hobo Hunt 5, but because it is about 10,000 bytes, it would be a huge post.", "Solution_9": "How do you get into assembly language? I see commands in the catalog that relate to it, but nothing else.", "Solution_10": "[quote=\"1234567890\"]How about a prime factorization thingie? I always make errors in factoring.[/quote]\r\n\r\n The one I made uses a picture of primes(?). I generated a picture so that the p th pixel is off iff p is prime or 1. This one picture stores primes up to 5985, which causes the program to run faster.\r\n\r\nThis is the program to generate such picture. It takes about 10 minutes to run it. After running this program once, it can be deleted.\r\n[hide=\"PROGRAM: PRIME\"]\n: GridOff\n: AxesOff\n: ClrDraw\n: FnOff\n: PlotsOff\n: {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71}\u2192L\u2081\n: For(A,1,20,1\n: For(B, L\u2081(A),iPart(5985/ L\u2081(A)),1\n: Pxl-On(iPart((L\u2081(A)B-1)/95),round(95fPart((L\u2081(A)B-1)/95\n: End\n: End\n: UnArchive Pic1\n: StorePic Pic1\n: Archive Pic1\n[/hide]\n\nHere is my program to factor a positive integer using the picture.\n[hide=\"PROGRAM: FACTOR\"]\n: Clrdraw\n: GridOff\n: AxesOff\n: PlotsOff\n: FnOff\n: UnArchive Pic1\n: RecallPic Pic1\n: Archive Pic1\n: Lbl 0\n: ClrHome\n: Prompt A\n: If A<1 or 0\u2260fPart(A\n: Goto 0\n: ClrHome\n: Output(1,1,A\n: If A=1\n: Then\n: Output(2,1,1\n: Pause\n: Goto 0\n: End\n: If A\u22645985\n: Then\n: If not(pxl-Test(iPart((A-1)/95),round(95fPart((A-1)/95\n: Then\n: Ouput(2,1,A\n: Pause\n: Goto 0\n: End\n: End\n: 1\u2192Y\n: 16\u2192X\n: 0\u2192C\n: While 0=fPart(A/2\n: C+1\u2192C\n: A/2\u2192A\n: End\n: If C\n: Then\n: Output(1,16,\u201d*2\n: X+2\u2192X\n: prgm L\n: If C\u22601\n: Then\n: Output(Y,X,\u201d^\n: X+1\u2192X\n: prgm L\n: Output(Y,X,C\n: X+iPart(log(C))+1\u2192X\n: prgm L\n: End\n: End\n: 3\u2192B\n: Lbl 1\n: 0\u2192C\n: While 0=fPart(A/B\n: C+1\u2192C\n; A/B\u2192A\n: End\n: If C\n: Then\n: Output(Y,X,\u201d*\n: X+1\u2192X\n: prgm L\n: Output(Y,X,B\n: X+iPart(log(B))+1\u2192X\n: prgm L\n: If C\u22601\n: Then\n: Output(Y,X,\u201d^\n: X+1\u2192X\n: prgm L\n: Output(Y,X,C\n: X+iPart(log(C))+1\u2192X\n: prgm L\n: End\n: End\n: B+2\u2192B\n: If B>5985\n: Goto 3\n: While pxl-Test(iPart((B-1)/95),round(95fPart((B-1)/95\n: B+2\u2192B\n: If B>5985\n: Goto 3\n: End\n: Lbl 3\n: If B\u00b2>A\n: Goto 2\n: Output (8,1,B\n: Goto 1\n: Lbl 2\n: If A\u22601\n: Then\n: Output(Y,X,\u201d*\n: X+1\u2192X\n: prgm L\n: Output(Y,X,A\n: End\n: Output(1,16,\u201d \u201c\n: Pause\n: Goto 0\n[/hide]\n\nBelow is a program that prgm FACTOR needs. This program changes lines(this task is also used in a lot of other programs, so I made it into a program.)\n[hide=\"PROGRAM: L\"]\nWhile X>16\nX-16\u2192X\nY+1\u2192Y\nEnd\n[/hide]" } { "Tag": [ "AMC", "AMC 12", "AMC 12 A", "AIME", "search", "MATHCOUNTS", "Putnam" ], "Problem": "What did everyone get?\r\nI got 108.\r\nCoudn't do any of the last four.", "Solution_1": "Beat you by one skipped :wink: 109.5", "Solution_2": "I got a 103.5 (15 right, 9 skipped)\r\nNot bad for an 8th grader :D\r\n\r\nMy reaction to the last 4 problems was :stretcher:", "Solution_3": "Yea I didn't even look at 2 of them.\r\nCan't believe I didn't realize you could add the three equations on 17!!!", "Solution_4": "Well...I'd rather not say...let's just say failed...At least I won't have to go eat dinner during the actual AMC12...", "Solution_5": "Well no duh you guys failed.\r\n\r\nLets see here...\r\n\r\n#12 is the exact problem from 1990 AIME #9\r\n#16 is the exact problem from 1984 AIME #11\r\n#19 was an old AIME problem (it was also one of the 10-15 ones)\r\n#20 is the exact problem from 1988 AIME #12\r\n\r\nI know for a fact that #21 was the 1991 AHSME #30\r\nAlso I that #14 was the 1990 AHSME #30\r\n#17 was also an old AHSME #30 (forgot which year-just search it)\r\n\r\n#9 was an old AHSME problem (like 20-30 problem range)\r\n#7 was an old MATHCOUNTS national problem\r\nEven #1-I would consider that like a #10 on a regular AMC 12 \r\n\r\nI dont know where #22-25 came from but if the ones before that were considered to be very hard AIME problem, then obviously this test is not AMC level. Infact its problaby harder than an AIME test (the last 15 questions can easily be used to make an AIME)\r\n\r\nThats why I didnt do this AMC because I couldve just easily done some old AIME problems and old ASHME/AMC 12 #25-30s.\r\n\r\nAlso if you got 120+ on this test, you probably can get over 8 on AIME easily considering this was only 75 mins.", "Solution_6": "I failed very badly, with like 5 stupid mistakes.\r\nI'm in 7th grade though, so I still have time to get better.", "Solution_7": "firecricket91, what you say is true but let me tell you this. This exam was hard, yes, and I did use a lot of AIME problem and one question (one contestant recognized it!) was from Putnam exam. So technically, this exam fails as a mock AMC exam.\r\n\r\nBut I made it like this so that people can learn from taking this exam. This was NEVER meant to be an exam where you can get your approximate scores. Those were my past mock exams (you can check some of them out on AoPS wiki). This one, on the other hand, meant to provide challenges, and if you broke 100, you probably can get at least 120 on real AMC.\r\n\r\nKnowing this, let's not say that we \"failed\" this test since it was meant to be very difficult. I'm going to post solutions to all problems eventually (hopefully before real AMC) so make sure you read them! They'll be good review as I tried to be pretty thorough.", "Solution_8": "oh ok..\r\n\r\nbut technically you couldve made this like 120 minutes instead of 75 minutes.\r\n\r\nBecause if the point of it was for people to learn how to do knew problems, then time shouldnt be such a constraint.", "Solution_9": "[quote=\"firecricket91\"]oh ok..\n\nbut technically you couldve made this like 120 minutes instead of 75 minutes.\n\nBecause if the point of it was for people to learn how to do knew problems, then time shouldnt be such a constraint.[/quote]\r\n\r\nTrue. But I wanted people to at least attempt as if they were taking real AMC. :D", "Solution_10": "Well, thanks for everything, SilverFalcon.\r\nIt was great practice.", "Solution_11": "I agree. Hopefully the real AMC will be a bit easier :P", "Solution_12": "Very well done... though very difficult.", "Solution_13": "[quote=\"firecricket91\"]Well no duh you guys failed.\n\nLets see here...\n\n#12 is the exact problem from 1990 AIME #9\n#16 is the exact problem from 1984 AIME #11\n#19 was an old AIME problem (it was also one of the 10-15 ones)\n#20 is the exact problem from 1988 AIME #12\n\nI know for a fact that #21 was the 1991 AHSME #30\nAlso I that #14 was the 1990 AHSME #30\n#17 was also an old AHSME #30 (forgot which year-just search it)\n\n#9 was an old AHSME problem (like 20-30 problem range)\n#7 was an old MATHCOUNTS national problem\nEven #1-I would consider that like a #10 on a regular AMC 12 \n\nI dont know where #22-25 came from but if the ones before that were considered to be very hard AIME problem, then obviously this test is not AMC level. Infact its problaby harder than an AIME test (the last 15 questions can easily be used to make an AIME)\n\nThats why I didnt do this AMC because I couldve just easily done some old AIME problems and old ASHME/AMC 12 #25-30s.\n\nAlso if you got 120+ on this test, you probably can get over 8 on AIME easily considering this was only 75 mins.[/quote]\r\n#1 was like #23 on 1993 National Mathcounts (might be wrong).", "Solution_14": "I think the first couple were all MC.", "Solution_15": "[quote=\"Silverfalcon\"]firecricket91, what you say is true but let me tell you this. This exam was hard, yes, and I did use a lot of AIME problem and one question (one contestant recognized it!) was from Putnam exam. So technically, this exam fails as a mock AMC exam.\n\nBut I made it like this so that people can learn from taking this exam. This was NEVER meant to be an exam where you can get your approximate scores. Those were my past mock exams (you can check some of them out on AoPS wiki). This one, on the other hand, meant to provide challenges, and if you broke 100, you probably can get at least 120 on real AMC.\n\nKnowing this, let's not say that we \"failed\" this test since it was meant to be very difficult. I'm going to post solutions to all problems eventually (hopefully before real AMC) so make sure you read them! They'll be good review as I tried to be pretty thorough.[/quote]\r\n\r\nyeah, thanks silverfalcon. now we know to be ready for anything (like a mock amc that's more like a mock aime). it was cool though, with a great variety of problems.", "Solution_16": "Great Problems... I worked on them today and timed out 25 minutes. Here's my breakdown:\r\n\r\n18 Correct\r\n6 Blank\r\nNone Wrong\r\n13 Was a Freebie\r\n\r\nAssuming 13 was 6 points, I got a 123. After I stopped timing myself though, I started to try the last three problems, which I didn't even get to look at. I was able to do the last one pretty easily. I haven't started 23 or 24 yet...\r\n\r\nAnyway, again, great problems. I haven't seen most of these questions (I think I counted 2 of them that I recognized... I haven't worked through too many former AIME problems).", "Solution_17": "If 13 was a freebie I got 114, if its just worth 1.5 as a blank answer I got 109.5\r\n(17-0-8 or 16-0-9)...If I got like 1 more hour I think I mightve gotten like 3 more correct but I am really slow and I have to work on my time (good thing AIME is 3 hours)\r\n\r\nYea this was really hard AMC especially the last 3 or so geometry problems which I didnt even look at, but now you say that they are putnam and AIME probelms I feel better." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "AMC", "AIME", "calculus", "conics" ], "Problem": "A car travels due east at $\\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is 51 miles, moves southeast at $\\frac 12\\sqrt{2}$ mile per minute. At time $t=0,$ the center of the storm is 110 miles due north of the car. At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle. Find $\\frac 12(t_1+t_2).$", "Solution_1": "[hide]\nJust for convenience, I will solve using the center of the storm as the origin of cartesian coordinate system (in the storm's reference point)\nThe horizontal distance between car and storm = x\nThe vertical distance between car and storm = y\n$x_0 = 0 \\text{ mi}$, $y_0=110 \\text{ mi}$\nassuming that the storm moves $45^o$ below the x-axis,\nthe horizontal component of storm's speed = $\\frac{1}{2}$\nthe vertical component of storm's speed = $\\frac{1}{2}$\n\n$\\frac{dy}{dt}=\\frac{1}{2}-0$\n$\\text{(rate of change of y)}=\\text{(difference between vertical components of velocities)}$\n$\\frac{dx}{dt}=\\frac{2}{3}-\\frac{1}{2}$\n$\\text{(rate of change of x)}=\\text{(difference between horizontal components of velocities)}$\n$\\frac{dy}{dx}=3$\n$\\text{(slope of path of car relative to the storm)}$\n\nSince the car is 110 mi below the storm originally, the equations to solve are:\n$x^2+y^2=51^2$\n$y=3x-110$\n(solving these equations for x will give us the horizontal distances at which the path traveled by the car intersects the storm's circumference)\n\nSubstituting, we get $10x^2-660x+7499=0$\nLuckily we are not asked for the exact times, because that would be ugly..\nInstead, we find the sum of the distances by adding the two roots together.\n$x=\\frac{-(-660)\\pm\\sqrt{(-660)^2-4(10)(7499)}}{2(10)}$\n$\\frac{x_1+x_2}{2}=33\\text{ mi}$\nThe time that it takes to travel this amount of horizontal distance is (using d=rt)\n$\\frac{x_1+x_2}{2}=33=\\frac{1}{6}\\frac{t_1+t_2}{2}$\n$\\frac{t_1+t_2}{2}=\\boxed{198}\\text{ min}$\n\n...I hope I did not make any mistakes this time...Well, at least it is a 3 digit integer, since it's AIME :| \n\n[/hide]", "Solution_2": "this can be solved completly without using any calculus (i guess u could just say $\\Delta$ instead of $d$), like every problem in this forum.\r\n\r\ni used the starting point of the car as the orgin and east/ south etc. normally configured.\r\n\r\nthat way the point $C$ at which the car is at the time $t$ is $C(\\frac23 t|0)$ and the equation of the circle is $(x-\\frac12t)^2+(y-(51-\\frac12t))^2 = 51^2$.\r\nby plugging in $x$ and $y$ one gets:\r\n\\[ 10t^2-3960t+341964=10(t-198)^2-50076=0 \\]\r\n\r\ntherefore the center of that parabola is at $t=198$ ...", "Solution_3": "[hide=\"a simpler solution\"]\nUsing coordinates, we see the location of the car is $(\\frac 2 3 t, 0)$ and the location of the storm is $(\\frac 1 2 t, 110 - \\frac 1 2 t)$ at time $t$. Thus, by distance formula, when the car is in the storm, we have $\\left( \\frac 1 6 t \\right)^2 + \\left( 110 - \\frac 1 2 t \\right) ^2 \\le 51^2$. Since we want the times where the car enters and leaves, we just want the roots of $t$. And the answer wants the sum of the roots, which by expanding and using vieta, gives $t_1 + t_2 = 396 \\implies \\frac{t_1 + t_2}{2} = \\boxed{198}$\nInterestingly enough, this value would be the same for any radius of the storm.\n[/hide]", "Solution_4": "I let the storm be a circle of radius $51$ centered at the origin, and let the car's path be a line starting at $(0, 110)$, with slope 3.", "Solution_5": "A car travels due east at $ 2/3$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $ 51$ miles, moves southeast at $ 1/2 \\sqrt2$ mile per minute. At time $ t \\equal{} 0$, the center of the storm is $ 110$ miles due north of the car. At time $ t \\equal{} tsub1$ minutes, the car enters the storm circle, and at time $ t \\equal{} tsub2$ minutes, the car leaves the storm circle. Find $ 1/2(tsub1 \\plus{} tsub2)$\r\n\r\nPlease bear with my inability to use Latex, or something. I'm still fairly new to this site.\r\n\r\nAll right then, so what I thought of doing was creating a diagram in which the storm is $ 110$ miles directly above the car. Therefore, the edge of the storm is only $ 59$ miles above the car at this point. Because its moving $ 1/2 \\sqrt2$ mph, it must be moving $ 1/2$ miles southward every minute.\r\n\r\nOkay, so if its $ 59$ miles above, then it should take $ 118$ minutes to reach the car right?\r\n\r\nIn that time, the storm will have also moved $ 59$ miles east, and the car will have moved $ 78 \\plus{} 2/3$ miles east. So, then I just calculated how much time it would take for the car to move atleast $ 51$ miles east of the storm so that it cannot be inside it (because the radius of the storm is 51).\r\n\r\nWell, the overtake speed of the car is $ 1/6$ miles per minute ($ 2/3 \\minus{} 1/2$). When the car enters the storm, the car will have already progressed $ 78 \\plus{} 2/3 \\minus{} 59 \\equal{} 19 1/3$ miles east of the storm's center. It needs only to go $ 51 \\minus{} 19 \\plus{} 1/3 \\equal{} 31 \\plus{} 1/3$ miles more. At the overtake speed, this will take $ 188$ minutes. Therefore, if the car stays in the storm for $ 188$ minutes, I find that the answer is $ 94$\r\n\r\nWell, if I did anything wrong or assumed something (and I'm sure I did), I hope someone will tell me. Thanks for the help in advance", "Solution_6": "lots of mistakes...\r\n[quote]\nOkay, so if its 59 miles above, then it should take 118 minutes to reach the car right?[/quote]\r\n\r\nNo. Keep in mind that the storm is circular, not square. The southern edge of the storm reaches the x-axis after 118 minutes. But the car has a larger x-velocity than the storm, and thus does not remain directly below the storm.\r\n\r\nLooking at the rest of your math, it seems you assumed a square storm. The storm is a circle - as it moves south, the distance along the line y=0 varies. The car does not have to end up 51 miles east of the center (i.e., all 51 miles in the x-direction); rather, the car just has to be 51 miles from the center of the storm - a component of this 51 will be in the y direction.\r\n\r\nbtw, quick latex tutorial:\r\na_b produces $ a_b$\r\na^b produces $ a^b$\r\n\\frac{a}{b} produces $ \\frac{a}{b}$\r\n\r\none last example: to write a sub 22 we would do a_{22} which produces $ a_{22}$, not a_22 which produces $ a_22$ - latex doesn't recognize anything past the first character after a _ or ^ as in the exponent/subscript unless you use the braces to group them in." } { "Tag": [], "Problem": "n th term of an AP series is 1/2 n(n+1). find sum of n terms of the series.", "Solution_1": "[hide]The $n$th term is $n\\triangle$, so the sum of $n$ terms would be $\\frac{n(n+1)(n+2)}{6}$.[/hide]" } { "Tag": [ "geometry", "3D geometry", "sphere", "Putnam", "linear algebra", "matrix", "induction" ], "Problem": "This problem is inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244052]Putnam 2008 B3.[/url]\r\n\r\nFor $ 1\\le m\\le n$ suppose that $ A$ is an $ n\\times m$ real matrix such that $ A^TA \\equal{} r^2I$ for some real $ r$ and also each diagonal entry of $ AA^T$ is less than or equal to $ 1.$\r\n\r\na) Prove that $ r^2\\le\\frac {n}{m}.$\r\n\r\nb) Conjecture: is it true that for all $ m$ and $ n,$ there exists such a matrix with $ r^2 \\equal{} \\frac {n}{m}\\,?$\r\n\r\n(The problem asked on the Putnam is equivalent to this for $ n\\equal{}4,m\\equal{}2.$ We're convinced that the conjecture is true for $ m\\equal{}2$ and all $ n,$ so the question is now really about $ m\\ge 3.$)", "Solution_1": "I'll throw in the proof of part a), which is quite short.\r\n\r\nSince $ A^TA \\equal{} r^2I$ and $ A^TA$ is $ m\\times m,$ then $ \\text{trace}\\,A^TA \\equal{} mr^2.$\r\n\r\nIf every diagonal element of $ AA^T$ is smaller than $ 1$ and $ AA^T$ is $ n\\times n,$ then $ \\text{trace}\\,AA^T\\le n.$\r\n\r\n$ \\text{trace}\\,A^TA \\equal{} \\text{trace}\\,AA^T$ so $ mr^2\\le n$ or $ r^2\\le\\frac {n}{m}.$", "Solution_2": "This is something I found online from Noam Elkies:\r\n\r\n[quote]\nWhat's the largest circle that will fit into a unit cube? A tough question, suggested by the recent Putnam problem that asked the same question in a hypercube. It turns out that there's a nice answer even for the general question of the largest k-dimensional ball that will fit into an n-dimensional hypercube of unit side --- in each case the maximal diameter is the square root of n/k --- but the proof is beyond our scope.\n[/quote]\r\n\r\nSo your conjecture does indeed appear to be correct, although that doesn't help at all with the proof.\r\n\r\n(This is from near the bottom of http://www.math.harvard.edu/~elkies/FS24i.08/)", "Solution_3": "I have no proof, but found an equivalent question that might be easier to solve. I don't know if it helps, but I guess won't hurt either. \r\n\r\nDoes there exist an orthonormal basis $ u_1,...,u_m$ of an m-dimensional subspace of $ \\mathbb{R}^n$ such that $ \\sum_{k \\equal{} 1}^{n}(u_k)_i^2 \\equal{} \\frac {m}{n}$ for all $ i \\equal{} 1,...,n$?\r\n(where $ (u_k)_i$ denotes the $ i$th component of $ u_k$)\r\n\r\nKent Merryfield, would you mind posting the general example you found for m=2?", "Solution_4": "In their [url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2008s.pdf]solutions document[/url], Kedlaya and Ng present an argument due to Dani Kane. See the remark after the solution to B3.", "Solution_5": "[quote=\"Peter\"]Kent Merryfield, would you mind posting the general example you found for m=2?[/quote]\r\nFor $ m\\equal{}2,n\\equal{}2,$ let $ A\\equal{}I.$ This works for all $ m\\equal{}n$ cases.\r\n\r\nFor $ m\\equal{}2,n\\equal{}3,$ let $ A\\equal{}\\begin{bmatrix}\\frac{\\sqrt{3}}2&\\frac12\\\\ \\minus{}\\frac{\\sqrt{3}}2&\\frac12\\\\ 0&\\minus{}1\\end{bmatrix}.$\r\n\r\nThen $ A^TA\\equal{}\\begin{bmatrix}\\frac32&0\\\\ 0&\\frac32\\end{bmatrix}$ and $ AA^T\\equal{}\\begin{bmatrix}1&\\minus{}\\frac12&\\minus{}\\frac12\\\\ \\minus{}\\frac12&1&\\minus{}\\frac12\\\\ \\minus{}\\frac12&\\minus{}\\frac12&1\\end{bmatrix}.$\r\n\r\nGeometrically, this corresponds to placing a circle in a cube in the plane orthogonal to $ (1,1,1)$ through the center. If you slice the cube with that plane, you get a regular hexagon, and a circle inscribed in that regular hexagon has radius $ \\sqrt{\\frac32}.$\r\n\r\nNow, suppose we have an $ A$ that is maximal for this theorem for some $ n$ and $ m\\equal{}2.$ That is, $ A$ is $ n\\times 2,$ $ A^TA\\equal{}\\frac n2\\,I,$ and every diagonal entry fo $ AA^T$ equals $ 1.$\r\n\r\nLet $ B$ be the $ (n\\plus{}2)\\times 2$ matrix $ B\\equal{}\\begin{pmatrix}A\\\\I\\end{pmatrix}.$\r\n\r\nThen $ B^TB\\equal{}A^TA\\plus{}I\\equal{}\\frac{n\\plus{}2}2\\,I$ and $ BB^T\\equal{}\\begin{pmatrix}AA^T&A\\\\ A^T&I\\end{pmatrix},$ which satisfies our conditions.\r\n\r\nThe sets up an induction in which we eventually come down to the $ 2\\times 2$ identity or the $ 3\\times 2$ case.\r\n\r\nThis was all laid out in the exchange between me and Albanian Eagle on the original topic, although not this explicitly." } { "Tag": [ "geometry", "incenter", "circumcircle", "trigonometry", "geometric transformation", "reflection", "geometry proposed" ], "Problem": "Let $ ABC$ be triangle. Let incenter,circumcenter of triangle$ ABC$ be $ I$,$ O$. line $ l$ is perpendicular to $ IO$ and pass$ I$. $ l$ meet external bisect of angle$ A$ at $ P$ and $ BC$ at $ Q$. Prove that $ PI: IQ=2: 1$", "Solution_1": "[quote=\"Leonhard Euler\"][color=darkred]Let $ I,O$ be incenter and circumcenter of $ \\triangle BC$. The line $ d$ for which $ d\\perp IO$ , $ I\\in d$ meet external bisector of angle $ A$ at $ P$ and $ BC$ at $ Q$.\nProve that $ IP=2\\cdot IQ$.[/color][/quote]\r\n[color=darkred][size=134][b]Nice problem ![/b][/size][/color]\r\n\r\n[color=darkblue] [b]Proof I[/b] ([u]metric[/u]). Suppose w.l.o.g. $ b>c$. Denote $ \\{\\begin{array}{c}D\\in (BC)\\ ,\\ ID\\perp BC\\\\\\ M\\in (BC)\\ ,\\ MB=MC\\\\\\ A'\\in AI\\cap OM\\ ,\\ m(\\widehat{AIP})=\\phi\\end{array}$. Observe that $ \\{\\begin{array}{c}IA\\cdot\\sin\\frac{A}{2}=r\\\\\\\\ 2\\cdot DM=b-c\\end{array}$.\n\n$ 1\\blacktriangleright$ Apply the Sinus' theorem in $ \\triangle AIO$ : $ \\frac{AO}{\\sin\\widehat{AIO}}=\\frac{IO}{\\sin\\widehat{IAO}}$ $ \\implies$ $ \\frac{R}{\\sin (90^{\\circ}\\pm \\phi )}=\\frac{IO}{\\sin\\frac{B-C}{2}}=\\frac{2\\cdot IO\\cdot\\cos\\frac{B+C}{2}}{\\sin B-\\sin C}=\\frac{2\\cdot IO\\cdot\\sin\\frac{A}{2}}{b-c}\\cdot 2R$ $ \\implies$\n$ \\cos\\phi =\\frac{b-c}{4\\cdot IO\\cdot \\sin\\frac{A}{2}}$. Observe that $ IP=\\frac{IA}{\\cos\\phi}$. Thus, $ IP=\\frac{4\\cdot IA\\cdot IO\\cdot \\sin\\frac{A}{2}}{b-c}$ $ \\implies$ $ \\implies$ $ \\boxed{IP=\\frac{4r}{b-c}\\cdot IO}\\ \\ (1)$.\n\n$ 2\\blacktriangleright$ Observe that $ \\tan\\widehat{DMI}=\\frac{ID}{DM}=\\frac{2r}{b-c}$ and the quadrilateral $ IOMQ$ is cyclically, i.e. $ \\widehat{QOI}\\equiv\\widehat{QMI}\\equiv\\ (180^{\\circ}-)\\widehat{DMI}$. Thus,\nfrom $ \\triangle QOI$ ($ IO\\perp IQ$) obtain $ IQ=IO\\cdot\\tan |\\widehat{QOI}|=IO\\cdot\\frac{2r}{b-c}$, i.e. $ \\boxed{IQ=\\frac{2r}{b-c}\\cdot IO}\\ \\ (2)$. From the relations $ (1)$ and $ (2)$ obtain $ IP=2\\cdot IQ$.\n\n[b]Proof II[/b] ([u]synthetic - [b]April's idea[/b][/u]). Denote the second intersections $ A'$, $ B'$, $ C'$ of the circumcircle with the lines $ IA$, $ IB$, $ IC$ respectively.\n\nIs well-known or prove easily that the point $ A$ is the reflection of the point $ I$ w.r.t. the line $ B'C'$ ($ B'I=B'A$, $ C'I=C'A$ and $ B'C'\\perp AI$).\n\nThus, $ B'C'\\parallel AP$ and $ RP=RI$. Apply the [b]theorem of the butterfly[/b] : $ \\{\\begin{array}{c}d\\perp IO\\ ,\\ I\\in d\\\\\\ I\\in BB'\\cap CC'\\\\\\ R\\in B'C'\\cap d\\ ,\\ Q\\in BC\\cap d\\end{array}\\|$ $ \\implies$ $ IR=IQ$.\n\nTherefore, $ PR=RI=IQ$, i.e. $ \\boxed{IP=2\\cdot IQ}$.[/color]", "Solution_2": "[hide=\"Hint.\"] Use Butterfly Theorem :wink: [/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "In the plane there is a close broken line with $ 2006$ vertices(It is a continuous line with $ 2006$ vertices which ends at the same point it starts), no $ 3$ collinear and no $ 3$ sides concurrent; find the maximum number of intersection points of the line.", "Solution_1": "can you post a solution, please(because i cant see your archive). I think the answer is $ \\frac {2(2003) \\plus{} 2004(2002)}{2}$ because i found an example with that but i fail in prove that is the maximum", "Solution_2": "I feel I do not understand the question completely.Also I am not aware of Erdos's result relating to this.I would like to clarify whether my understanding of the problem is correct. Does [i]a[/i] broken line with 4 vertices is this?(pic)\r\nhttp://www.mathlinks.ro/viewtopic.php?mode=attach&id=11451\r\n\r\nIn case my understanding is correct, I have the solution.\r\nObserve that a broken line with 4 vertices can have atmost one intersection. A broken line with $ n$ vertices can also be viewed as end to end join of $ n\\minus{}1$ line segments. If I add another vertex so as to maximize the number of intersection points, the line segment joining $ n$th point and the $ n\\plus{}1$th point cuts through all the previous line segments except the line joining $ n\\minus{}1$th point and $ n$th point, thus generating $ n\\minus{}2$ points.\r\n\r\n$ f(n)$ denotes the max number of intersections of a broken lines with n vertices.\r\n$ f(n\\plus{}1)$=$ f(n)$+$ (n\\minus{}2)$ with $ f(3)\\equal{}0$\r\n\r\nClearly, $ f(n)$= $ \\sum (n\\minus{}3)$\r\n$ f(2006)\\equal{}2003.2004/2$", "Solution_3": "srikanth, the example you gave is correct(is a broken line, but no a closed broken line) but the problem say a closed broken line, in you example if you join v4 and v1, that is a close broken line", "Solution_4": "Thank you for the clarification Cuenca. Here is the solution. \r\n\r\nContinuing the argument from my previous post\r\nLet $ f(n)$ be the maximum number of intersections of a open broken line with n vertices.\r\nLet $ g(n)$ be the maximum number of intersections of a closed broken line with n vertices.\r\n\r\nThen we have the relation,\r\n$ g(n)$ = $ f(n)$ + $ n\\minus{}3$\r\nWhen $ n\\plus{}1$ th point is introduced (in a open broken line of n vertices) , it coincides with point $ 1$. It cuts all the line segments except $ 1\\minus{}\\minus{}2$ , $ (n\\minus{}1)\\minus{}\\minus{}n$.\r\n\r\n$ g(n)$= $ (n\\minus{}3)(n\\minus{}2)/2$ + $ n\\minus{}3$\r\n\r\n(qed?)" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "While working on something I stumbled across a certain problem dealing with Combinatorics. I have been unable to solve it myself.\r\nIt is the following:\r\n\r\nFind the general solution of the following recursion:\r\n\r\n$ F(n,k) \\equal{}\\begin{cases} &F(n\\minus{}1,k\\minus{}1) \\plus{} k \\cdot F(n\\minus{}1,k) \\, \\, \\qquad \\text{if } 2 \\leq k \\leq n\\plus{}1\\\\\r\n&0 \\quad \\quad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\text{if } k > n\\plus{}1\r\n\\end{cases}$\r\nwhere:\r\n$ F(n,1) \\equal{} 1$,\r\n$ F(n,n\\plus{}1) \\equal{} 1$.\r\n\r\nCould anyone help me out with this?", "Solution_1": "These are the [url=http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Recurrence_relation]Stirling numbers of the second kind[/url].", "Solution_2": "Interesting. I sort of know those, but not that they satisfied this recursion.\r\n\r\n\r\nThen we have:\r\n\r\nLet $ 0 < p < 1$ be given.\r\nIf $ U_n : \\equal{} \\sum\\limits_{k \\equal{} 0}^{\\infty} k^n p^k$, then\r\n\r\n$ U_{n\\plus{}1} \\equal{} \\sum_{k \\equal{} 1}^{n \\plus{} 1} S(n,k) \\frac {k! p^k}{(1 \\minus{} p)^{k \\plus{} 1}}$,\r\n\r\nif I am not mistaken with indices.", "Solution_3": "Correct. This identity is equivalent to the identity $ k^n \\equal{} \\sum_{r \\equal{} 0}^{k} S(n, r) {k \\choose r} r!$ together with the identity\r\n\r\n$ \\sum_{k\\equal{}0}^{\\infty} {k \\choose r} p^k \\equal{} \\frac{p^r}{(1 \\minus{} p)^{r\\plus{}1}}$." } { "Tag": [ "modular arithmetic", "number theory", "relatively prime", "Diophantine Equations" ], "Problem": "Find a pair of relatively prime four digit natural numbers $A$ and $B$ such that for all natural numbers $m$ and $n$, $\\vert A^m -B^n \\vert \\ge 400$.", "Solution_1": "$ A\\equal{}3125\\equal{}5^5$\r\n$ B\\equal{}9376$\r\n\r\nIt works because $ B^2\\equiv B (10000)$\r\nAnd $ A\\equiv 3125 (10000)$\r\n$ A^2\\equiv 5625 (10000)$\r\n$ A^3\\equiv 625 (10000)$\r\n$ A^4\\equiv 625 (10000)$\r\n\r\nIs it ok?", "Solution_2": "We can also use $ A\\equal{}2001,B\\equal{}1000$ since $ A^m\\minus{}B^n\\equiv 1001\\pmod{3000}$." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let $ P(x)$ be a polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x)$ and $ R(x)$, again with integer coefficients, such that\r\n[b](i)[/b] $ P(x) \\cdot Q(x)$ is a polynomial in $ x^2$ , and\r\n[b](ii)[/b] $ P(x) \\cdot R(x)$ is a polynomial in $ x^3$.", "Solution_1": "[quote=\"Rijul saini\"]Let $ P(x)$ be a polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x)$ and $ R(x)$, again with integer coefficients, such that\n[b](i)[/b] $ P(x) \\cdot Q(x)$ is a polynomial in $ x^2$ , and\n[b](ii)[/b] $ P(x) \\cdot R(x)$ is a polynomial in $ x^3$.[/quote]\r\n\r\nLet $ P(x)\\equal{}\\sum a_mx^m$\r\n\r\n[b](i)[/b] Let $ Q(x)\\equal{}\\sum a_{2j}x^{2j} \\minus{} \\sum a_{2k\\plus{}1}x^{2k\\plus{}1}$\r\nThen $ P(x) \\cdot Q(x)\\equal{}(\\sum a_{2j}x^{2j})^2 \\minus{} (\\sum a_{2k\\plus{}1}x^{2k\\plus{}1})^2$ is a polynomial in $ x^2$\r\n\r\n[b](ii)[/b] We have $ a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc\\equal{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca)$\r\nNow put $ a\\equal{}\\sum a_{3i}x^{3i}, b\\equal{}\\sum a_{3j\\plus{}1}x^{3j\\plus{}1}, c\\equal{}\\sum a_{3k\\plus{}2}x^{3k\\plus{}2}$ we have $ P(x)\\equal{}a\\plus{}b\\plus{}c$\r\nand $ a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc$ is a polynomial in $ x^3$. So put $ R(x)\\equal{}a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca$ and we are done.", "Solution_2": "here is officiall solution!" } { "Tag": [ "USAMTS" ], "Problem": "I just signed up for an account after I learned about this contest from a couple of my friends. However, I did not even do rounds 1,2, or 3. Will there be any benefit for doing round 4? I know that it will definently help me and is going to be fun, but is it possible to win any prizes? \r\n\r\nThanks", "Solution_1": "[quote=\"IndianaMan\"]I just signed up for an account after I learned about this contest from a couple of my friends. However, I did not even do rounds 1,2, or 3. Will there be any benefit for doing round 4? I know that it will definently help me and is going to be fun, but is it possible to win any prizes? \n\nThanks[/quote]\r\nhttp://www.usamts.org/Rules/U_RulesPrizes.php\r\n\r\nYou need 40 points to get an Honorable Mention, so one round isn't enough.", "Solution_2": "[quote=\"IndianaMan\"]I just signed up for an account after I learned about this contest from a couple of my friends. However, I did not even do rounds 1,2, or 3. Will there be any benefit for doing round 4? I know that it will definently help me and is going to be fun, but is it possible to win any prizes? \n\nThanks[/quote]\r\n\r\nUnfortunately, no. The minimum score to recieve a prize is 40 points, which cannot be achieved in one round, which has a maximum of 25 points.\r\n\r\nBut do the problems anyway! It will be a great experience and practice for next year.", "Solution_3": "Doing the problems just for fun would definitely be a good experience for you. It'll help you get in the good mood for doing well next year, simply to familiarize yourself with USAMTS, how to write rigorous solutions, and how to be creative in writing them. It's really a good contest, in my opinion, because it like stresses creative thinking and problem-solving strategies. Even if you don't win, you'd still enjoy the experience of at least doing the problems, because most importantly, they are fun! :)" } { "Tag": [ "conics", "parabola", "integration", "function", "calculus", "calculus computations" ], "Problem": "Dear all, \r\n\r\nI am having a hard time in solving this problem. Would you please help? \r\n\r\n\"Consider the region R bounded by y = 4x - x^2 and the x-axis. Find the equation of the line that intersects the origin and divides R into two regions of equal area.\" \r\n\r\nThanks a lot! Enjoy the weekend.", "Solution_1": "Let the parabola be $ f$ and let the line be $ y \\equal{} ax$. The intersection of $ y$ and $ f$ is found as follows:\r\n\r\n$ ax \\equal{} \\minus{} x^2 \\plus{} 4x$\r\n$ a \\equal{} \\minus{} x \\plus{} 4$\r\n$ x \\equal{} 4 \\minus{} a$\r\n\r\nIf you draw a graph, you'll see that $ y$ divides $ R$ into an upper and a lower region. The upper region goes from $ x \\equal{} 0$ to $ x \\equal{} 4 \\minus{} a$ and the lower goes from $ x \\equal{} 0$ to $ x \\equal{} 4$. The resulting equation is as follows:\r\n\r\n$ \\int_0 ^{4 \\minus{} a} ( \\minus{} x^2 \\plus{} 4x \\minus{} ax)\\,dx \\equal{} \\int_0 ^{4 \\minus{} a}ax \\, dx \\plus{} \\int_{4 \\minus{} a}^4( \\minus{} x^2 \\plus{} 4x)\\,dx$\r\n\r\nThe right-hand side uses the bounds it does because of which function bounds the lower region on certain intervals. This is easier to see if you draw a graph. To find $ a$, we evaluate these integrals:\r\n\r\n$ \\minus{} \\frac {1}{3}x^3 \\plus{} 2x^2 \\minus{} \\frac {1}{2}ax^2 \\Big]^{4 \\minus{} a}_0 \\equal{} \\frac {1}{2}ax^2 \\Big]^{4 \\minus{} a}_0 \\plus{} \\Big[ \\minus{} \\frac {1}{3}x^3 \\plus{} 2x^2 \\Big]^4_{4 \\minus{} a}$\r\n\r\n$ \\minus{} \\frac {1}{3}(4 \\minus{} a)^3 \\plus{} \\left(2 \\minus{} \\frac {1}{2}a\\right)(4 \\minus{} a)^2 \\equal{} \\frac {1}{2}a(4 \\minus{} a)^2 \\minus{} \\frac {64}{3} \\plus{} 32 \\minus{} \\left[ \\minus{} \\frac {1}{3}(4 \\minus{} a)^3 \\plus{} 2(4 \\minus{} a)^2\\right]$\r\n\r\nNote that the terms on the left are also on the right (except they are negative on the right). So we combine them:\r\n\r\n$ \\minus{} \\frac {2}{3}(4 \\minus{} a)^3 \\plus{} 2\\left(2 \\minus{} \\frac {1}{2}a\\right)(4 \\minus{} a)^2 \\equal{} \\frac {32}{3}$\r\n$ \\minus{} \\frac {2}{3}(4 \\minus{} a)^3 \\plus{} (4 \\minus{} a)(4 \\minus{} a)^2 \\equal{} \\frac {32}{3}$\r\n$ \\minus{} \\frac {2}{3}(4 \\minus{} a)^3 \\plus{} (4 \\minus{} a)^3 \\equal{} \\frac {32}{3}$\r\n$ \\frac {1}{3}(4 \\minus{} a)^3 \\equal{} \\frac {32}{3}$\r\n$ (4 \\minus{} a)^3 \\equal{} 32$\r\n$ 4 \\minus{} a \\equal{} \\sqrt [3]{32}$\r\n$ a \\equal{} 4 \\minus{} \\sqrt [3]{32}$\r\n\r\nSo, the desired line is $ \\boxed{y \\equal{} (4 \\minus{} \\sqrt [3]{32})x}$." } { "Tag": [ "function", "calculus", "integration", "symmetry", "real analysis", "real analysis unsolved" ], "Problem": "let f be reimann integrable function from [a,b] to R and g be function from[a,b] to R which agrees with f on all but finitely many points of[a,b].show g is reimann integrable and integral of f from a to b=integral of g from a to b.does the same result hold when finitely many is replaced by countably many?", "Solution_1": "Every upper and lower step function for $f$ is also an upper and lower step function for\u00a0$g$, except that you may possibly have to modify it in the finitely many points. However, that does not change the values of the integrals of the step functions. By symmetry you can replace $f$ and $g$ in the above statement. So, since the riemann integral definition involves supremum and infimums of such integrals, the statement immediately follows.\r\n\r\nThe statement is not true for countably many points. Let $f(x) = 0$ for $x\\in [0,1]$ and let $g(x) = 1$ if $x$ is rational, and let $g$ agree with $f$ otherwise. Since the rationals are dense in the real numbers it is clear that any lower step function must be $\\le 0$ everywhere except for on finitely many points, while any upper step function must be $\\ge 1$ everywhere except on finitely many points. So the inf of upper functions is not equal to sup of lower functions and hence the function is not riemann integrable." } { "Tag": [], "Problem": "Given any $n+1$ integers between $1$ and $2n$, show that two of them are relatively prime. Is this result best possible, i.e., is the conclusion still true for $n$ integers between $1$ and $2n$?", "Solution_1": "[hide]\nWe just need to show there must be two consecutive numbers, which we know are relatively prime. The most numbers we can get that aren't consecutive would be:\n1, 3, 5, ..., 2n-3, 2n-1\nWhich is a total of n numbers. Since there are n+1 numbers, one of the numbers in between must be filled, thus there will be a pair of consecutive integers.\n[/hide]", "Solution_2": "And the same conclusion is not true for $n$ (as we could choose $n$ even numbers)." } { "Tag": [ "search", "IMO", "IMO Shortlist" ], "Problem": "Can anyone find the longlists of the IMO 2003, 2004 ? If yes caould you give some details about this? Thank you .", "Solution_1": "Some people, e.g. people on the IMO Problem Selection Committee, still may have them though the policy is to return the non-ISL problems to the respective proposer countries and not to collect or to keep track of them. Well that means [b]NO[/b] to your question. The rule is partially based on the lack of good competition problems world-wide and to some extent to the lower quality of some of the proposed problems, lots of \"well-known\", wrong, standard, boring, too easy, too hard problems, too long problem descriptions etc.", "Solution_2": "Thank you Orl and something else. What is the latest IMO in which we have Longlist ?\r\n\r\n1988 ?", "Solution_3": "[quote=\"silouan\"]Thank you Orl and something else. What is the latest IMO in which we have Longlist ?\n\n1988 ?[/quote]\r\n\r\nActually 1989 should have been the latest IMO which made the ILL publically available. But due to a lot of chaos in Moscow's IMO 1992 it was handed out again. But on the other hand a main member (from New Zealand) of the IMO 1992 Problem Selection Committee denied the fact that there was an ILL. Lots of problem were sent in to late back then and were not considered for the contest so the ILL might have been produced after the actual IMO. But anyway I got the ILL 1992 from the extremely helpful American IMO team leader back then. Well I got the ISL and ILL 1992 from different people and it seems not every ISL problem can be matched with an ILL problem, e.g. the ISL is not a subset of the ILL. BTW did you know that the German Democratic Republic (GDR) actually was supposed to administrate the IMO in 1992 but it was too late. At this time GDR's reunification already took place and the event moved to Russia and they obviously were not prepared to handle all the issues which may have to be addressed at an IMO. :)", "Solution_4": "Will you post the ILL from 1989 and 1992 to the forum ?\r\n\r\nBTW thank you for all the informations.", "Solution_5": "[quote=\"silouan\"]Will you post the ILL from 1989 and 1992 to the forum? [/quote]\n\nILL 1989 is partially already on the forum. Search for it! If you are looking for the ILL 1989 and ILL 1992 as distributed to the IMO team leaders check right this forum for the .pdf-files. BTW ILL 1992 is acknowledged to be an ILL of high quality. Eventually I will have posted all available IMO, ISL and ILL problems on ML/AoPS. Check the Resources section frequently! :) \n\n[quote=\"silouan\"]BTW thank you for all the information.[/quote]\r\n\r\nYou are welcome! :)", "Solution_6": "[quote=\"silouan\"]Will you post the ILL from 1989 and 1992 to the forum ?\n[/quote]\n\nSee [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18&year=1989]here[/url] and [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18&year=1992]here[/url]." } { "Tag": [], "Problem": "Hap, the crossing guard, is very popular with the students at Springdale Elementary School. Along with his usual morning smile, Hap has a daily riddle. One of the favorites was this one: How do 5 and 9 more make 2?", "Solution_1": "umm...9 hours after 5 o'clock is 2 o'clock?", "Solution_2": "im not very sure this is a math problem.", "Solution_3": "uh.... sounds more like a riddle than a math problem like it says in the description but...... otherwise, I think the time one is right. I can't think of any other way." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Find $(k1,k2,k3,k4)$ real positives, such that, the solutions $a, b, c, d,$ and $e$ of follow matrix have the same sign.\r\n\r\n$\\begin{pmatrix}(k1-0.112) & (k1-0.127) & (k1-0.089) & (k1-0.186) & (k1-.191) & 0\\\\ (k2-0.031) & (k2-0.017) & (k2-0.039) & (k2-0.044) & (k2-0.595) & 0\\\\ (k3-0.7155) & (k3-0.7) & (k3-0.712) & (k3-0.597) & (k3-0.129) & 0\\\\ (k4-3.77) & (k4-3.64) & (k4-3.73) & (k4-3.71) & (k4-6.87) & 0 \\end{pmatrix}$", "Solution_1": "Is this homework? :maybe: :o \r\n\r\nWhat do you mean the solutions? Have same sign? The question is far from clear." } { "Tag": [ "limit", "inequalities", "probability and stats" ], "Problem": "Suppose that $ P_{n}$ and $ P$ are prob. measures on $ \\mathbb{N}_{0}\\equal{}\\{0,1,2,\\,. \\,. \\,.\\}$ such that $ \\liminf_{n\\to \\infty}P_{n}(j)\\ge P(j)$ for every $ j$. How can we conclude from this that $ \\lim_{n\\to \\infty}P_{n}(j)\\equal{}P(j)$ for every $ j$?", "Solution_1": "Just note that $ P(j)\\equal{}1\\minus{}\\sum_{i: i\\ne j}P(i)$ and similarly for $ P_n$. Now fix $ \\varepsilon>0$ and choose a finite set $ S\\subset\\mathbb N\\setminus\\{i\\}$ such that $ 1\\minus{}\\sum_{i\\in S}P(i)\\le P(j)\\plus{}\\varepsilon$. Then $ \\limsup_{n\\to\\infty}P_n(j)\\le 1\\minus{}\\sum_{i\\in S}\\liminf_{n\\to\\infty}P_n(i)\\le 1\\minus{}\\sum_{i\\in S}P(i)\\le P(j)\\plus{}\\varepsilon$. The rest should be clear.\r\n\r\nThis argument can be shortened a bit by using the Fatou lemma, of course.", "Solution_2": "Yes I got it. Thanks fedja. I tried to use Fatou's lemma but I couldn't do it. If possible, can you give some hints to do by Fatou's lemma?", "Solution_3": "Fatou's lemma says that\r\n$ 1 \\equal{} \\sum_j P(j) \\leq \\sum_j \\liminf_n P_n(j) \\leq \\liminf_n \\sum_j P_n(j) \\equal{} 1$\r\n\r\nHence the equality must hold. However, as far as I can see, fedja is doing exactly this (invoking Fatou's lemma in the leftmost inequality), so the only difference is that he didn't dare do it for an infinite sum.", "Solution_4": "All you have proved now is that $ \\liminf_{n\\to\\infty}P_{n}(j)\\equal{}P(j)$ for every $ j$. But how to prove that $ \\liminf_{n\\to\\infty}P_{n}(j)\\equal{}\\limsup_{n\\to\\infty}P_{n}(j)$", "Solution_5": "Whoops, wrote my response too hastily. I was just trying to say that fedja's solution already uses Fatou's lemma. The calculation he did is\r\n$ \\varlimsup_n P_n(i) \\equal{} 1 \\minus{} \\varliminf_n \\sum_{j \\neq i} P_n(j) \\leq 1 \\minus{} \\sum_{j \\neq i} \\varliminf_n P_n(j) \\leq 1 \\minus{} \\sum_{j \\neq i} P(j) \\equal{} P(i)$" } { "Tag": [ "calculus", "integration", "trigonometry", "induction", "abstract algebra", "calculus computations" ], "Problem": "Evaluate\r\n\r\n\\[{\\int_0^\\frac{\\pi}{2}} \\frac{\\sin 2005x}{\\sin x}dx\\]", "Solution_1": "Again, I noticed a pattern:\r\n\r\nFor odd integers $n$, $\\int_0^{\\frac{\\pi}{2}} \\frac{\\sin(nx)}{\\sin x} \\ dx = \\frac{\\pi}{2}$.\r\n\r\nSo the answer is $\\frac{\\pi}{2}$.", "Solution_2": "[quote=\"Jimmy\"]Again, I noticed a pattern:\n\nFor odd integers $n$, $\\int_0^{\\frac{\\pi}{2}} \\frac{\\sin(nx)}{\\sin x} \\ dx = \\frac{\\pi}{2}$.\n\nSo the answer is $\\frac{\\pi}{2}$.[/quote]\r\n\r\nLet's prove that by induction:\r\n\r\nTo begin, $\\sin(nx+2x)=\\sin nx\\cos 2x+\\sin 2x\\cos nx$\r\n\r\nSince $\\frac {\\sin 2x\\cos nx}{\\sin x}=2\\cos x\\cos nx$, and since $\\frac {\\cos 2x}{\\sin x}=\\csc x-2\\sin x$, we are left with\r\n\r\n$\\sin nx\\csc x+2(\\cos nx\\cos x-\\sin nx\\sin x)=\\sin nx\\csc x+2\\cos((n+1)x)$\r\n\r\nThe latter half of this is zero when integrated, so this establishes the induction.", "Solution_3": "This is the Dirichlet kernel from Fourier analysis.\r\n\r\nClaim:\r\n\r\n$\\sum_{k=-n}^ne^{2ikx}=1+2\\sum_{k=1}^n\\cos(2kx)= \\frac{\\sin((2n+1)x)}{\\sin x}.$\r\n\r\nThis can be proved either by appropriately manipulating geometric sums or by other trigonometric identites.\r\n\r\nThen\r\n\r\n$\\int_0^{\\frac{\\pi}2}\\frac{\\sin((2n+1)x)}{\\sin x}\\,dx = \\int_0^{\\frac{\\pi}2} 1+2\\sum_{k=1}^n\\cos(2kx)\\,dx$\r\n\r\nThe constant term integrates to $\\frac{\\pi}2$ and each of the other terms integrates to zero.", "Solution_4": "[quote=\"Jimmy\"]Again, I noticed a pattern:\n\nFor odd integers $n$, $\\int_0^{\\frac{\\pi}{2}} \\frac{\\sin(nx)}{\\sin x} \\ dx = \\frac{\\pi}{2}$.\n\nSo the answer is $\\frac{\\pi}{2}$.[/quote]\r\n\r\nYes, you are right.\r\n\r\nLet $I_n=\\int_0^{\\frac{\\pi}{2}} \\frac{\\sin (2n+1)x}{\\sin x}dx\\ (n=0,1,2\\cdots)$, $I_n$ satisfies the following equality...\r\n\r\nProbably this integral is classical. It is said that this integral was made an examination at Tokyo University in 1950." } { "Tag": [ "vector" ], "Problem": "While following a treasure map, you start at an old oak tree. You first walk 825 meters directly south, then turn and walk 1.25km at 30.0 west of north, and finally walk 1.00 km at 40.0 north of east, where you find the treasure: a trophy!!!\r\n\r\nTo return to the old oak tree, in what direction should you head ?\r\n\r\nTo return to the old oak tree, how far will you walk?", "Solution_1": "its just vector addition\r\n[hide]\nx component: 0 + 1250 cos 120 + 1000 cos 40\n=-625+766=141\ny component: -825+1250sin(120)+1000sin(40)\n=1080+643-825=9.0*10^2\n\\theta=arctan(y/x)=81.1 north of east = 8.90 east of north.\nr=\\sqrt{x^2+y^2}=911 meters.\n[/hide]" } { "Tag": [ "function", "combinatorics proposed", "combinatorics" ], "Problem": "Find a function $f:N\\rightarrow N$ such that for all $k\\in N$ every graph of average degree at leas $f(k)$ has a bipartite subgraph of minimum degree at least $k$", "Solution_1": "Must we find the least possible $f(k)$? \r\n\r\nIf not, it is clear that some such function exists: at first, we may choose a bipartite subgraph of our graph $G$ without loss more then a half of all edges. Indeed, divided it into two groups such that the number of edges, which ljoin vertices of different groups is maximal. Then each vertex $v$ of degree $d$ is joined with at most $d/2$ vertices of the same group (else move it to other group). \r\n\r\nAfter that, we get a bipartite subgraph with $n$ vertices and at least $nf(k)/4$ edges. \r\n\r\nDelete vertices of degree at most $k-1$ step by step. After some step, we either get a graph, in which all degrees are at least $k$, or an empty graph. The first case is desired, assume that the second case holds. After each step, we deleted at most $k-1$ edges, so the number of edges in initial garph does not exceed $(k-1)n$, which contradicts to an upper estimate above for $f(k)>4(k-1)$.", "Solution_2": "yes least possible" } { "Tag": [ "vector", "articles" ], "Problem": "What is vector notation?", "Solution_1": "hmm... can't help much but here is a link to the mathworld's site [url]http://mathworld.wolfram.com/topics/VectorAlgebra.html[/url]\r\n\r\nAlso, vectors are not getting started but I'll leave that to the moderators to deal with.", "Solution_2": "An informational question like this can usually stay in whatever forum it starts out in. The threads that I \"promote\" to other forums are challenge problems that are clearly at too high a level for participants who are just getting started. A set of problems at the end of an expository article will also usually be left alone." } { "Tag": [ "geometry solved", "geometry" ], "Problem": "Suppse that $ABCDEF$ be the convex hexagon. Let $P=(AB\\cap EF), Q=(EF\\cap CD), R=(CD\\cap AB)$ and $S=(BC\\cap DE), T=(DE\\cap FA), U=(FA\\cap BC).$ Prove that if \\[ \\frac{AB}{PR}=\\frac{CD}{RQ}=\\frac{EF}{QP} \\] then \\[ \\frac{BC}{US}=\\frac{DE}{ST}=\\frac{FA}{TU}. \\]", "Solution_1": "Posted at\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=4822\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=55028\r\n\r\n darij" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $a,b>1$. Solve in $\\mathbb{R}$\r\n$a^{bx}+b^\\frac{a}{x}=a^b+b^a$.", "Solution_1": "Let $f(x)$ be the function $f(x) = a^{cx} + b^\\frac{d}{x} $ where $a,b>1$ and $ c,d > 0 $ . It's trivial that $f(x)=Q > 1$ has only positive solutions.\r\n\r\nFor $x\\in ]0, +\\infty[$, we have $f''(x)>0$, so $f'(x)$ increasing, so $f'(x)=0$ has only one root, so $f(x)=Q$ has at most 2 solutions.\r\n\r\nIt's easy to verify that $f(u)=f( v)$ , where $v= \\frac{d* ln(b)}{u* c* ln(a)}$. In our case $c=a,d=b$ and $u=1$ is a solution, so the oher one is $v= \\frac{a*ln(b)}{b* ln(a)}$ .\r\n :cool: I have already seen/solved this kind of problem :cool:" } { "Tag": [], "Problem": "How many ways can change be made for a quarter using standard U.S. coins?", "Solution_1": "Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1396323#1396323]this[/url]; the answer is $ \\boxed{13}$. Unless you can't make change for a quarter with a quarter, in which case it would be $ \\boxed{12}$.", "Solution_2": "I don't think that a quarter can be changed for a quarter because it wouldn't \"change\" ;). Also, I did this in a sprint round and it said $\\boxed{12}$ was correct.", "Solution_3": "The number of ways to arrange pennies with nickles to from a quarter is: $25/5 = 5$ \nThe number of ways for pennies with dimes is: $25/10 = 2$\nThe # of ways to arrange pennies to form a quarter + # of nickles: $2$\nThe number of ways for dimes and nickles: $2$\nThe number of ways for all of them is $1$\nTherefore $1+2+2+2+5 = \\boxed {12}$" } { "Tag": [ "linear algebra" ], "Problem": "Assuming that $ D_n\\in\\text{M}_n{(p)}$ such that\r\n$ D_n\\equal{}\\left| \\begin{array}{ccccc}\r\na & b \\\\\r\nc & a & b \\\\\r\n & c & a & \\ddots\\\\\r\n & & \\ddots &\\ddots& b\\\\\r\n & & &c &a\r\n\\end{array}\r\n\\right|$\r\nThen calculate $ D_n$.", "Solution_1": "We have : $ D_{n\\plus{}1}\\equal{}aD_{n}\\minus{}bcD_{n\\minus{}1},n \\geq 2$,$ D_1\\equal{}a,D_2\\equal{}a^2\\minus{}bc$, easy calculate $ D_n$ !" } { "Tag": [ "algebra", "polynomial", "search", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Is there somebody who knows how to write a code in kash to find the galois group of a given polynomail?", "Solution_1": "A search directly tells my that you want \"Galois(f);\" :wink:", "Solution_2": "No it doesnot work, it gives an error that f must have a value.\r\n\r\nDo you know any computer program which can help me to find Galois group of an irreducible polynomail of degree 12.", "Solution_3": "It's quite obvious that you need to set $ f$ to you specific polynomial :wink: \r\nTry Galois(X^3-2);\r\nOr in short: RTFM!", "Solution_4": "[quote=\"peteryellow\"]\nDo you know any computer program which can help me to find Galois group of an irreducible polynomail of degree 12.[/quote]format C:", "Solution_5": "ZetaX, i am not taht stupid definitely I have tried with a polynomail insted of f, but that is not working, and they have a pdf-file which tells that how can I use command but the problem is that there is nothing about Galois group.\r\n\r\nBut otherwise thanks for your help.", "Solution_6": "I [b]tried[/b] that command [b]myself[/b].\r\nMaybe you still use kash version 2, then it should be GaloisGroup(x^3-2); or something like that. And still I'm pretty sure it's stated in one of the manuals." } { "Tag": [], "Problem": "Okay. We have this number that is in form of:\r\n\r\n$3^{1001} 7^{1002} 13^{1003}$\r\n\r\nAnd what is the unit digit of this huge number? :D \r\n\r\n :P I changed the wording to make it less boring.", "Solution_1": "[hide]\n\nFirst, note the patterns:\n\n\\[3 \\rightarrow 3, 9, 7, 1\\]\n\\[7 \\rightarrow 7, 9, 3, 1\\]\n\\[13 \\rightarrow 3, 9, 7, 1\\]\n\nWe notice that each of the four integers repeat their units digit every 4 terms. \n\n\\[1001 \\equiv 1 (\\displaystyle{mod} 4)\\]\n\\[1002 \\equiv 2 (\\displaystyle{mod} 4)\\]\n\\[1003 \\equiv 3 (\\displaystyle{mod} 4)\\]\n\nThus, $3^{1001}$ ends in $3$, $7^{1002}$ ends in $9$, and $13^{1003}$ ends in $7$. Hence, the units digit of their product is the units digit of\n\n\\[3 \\cdot 7 \\cdot 9=189\\]\n\nTherefore the units digit of $3^{1001} \\cdot 7^{1002} \\cdot 13^{1003}$ is $9$.\n\n[/hide]", "Solution_2": "Let's change up the question. What is the leftmost digit of $3^{1001} \\cdot 7^{1002} \\cdot 13^{1003}$? :P :P :P\r\n\r\nHint:\r\n[hide]\n$30 prove\r\n\r\n1/(x\u00b2+yx+y\u00b2) + 1/(x\u00b2+xz+z\u00b2)+1/(y\u00b2+yz+z\u00b2) \u2265 1/(x+y+z)\u00b2", "Solution_1": "Actually there is a stronger inequality, for ex : $ \\displaystyle \\sum _{cyc} \\frac 1{x^2 \\plus{} xy \\plus{}y^2} > \\displaystyle \\sum _{cyc} \\frac 1{x^2 \\plus{} 4xy \\plus{}y^2} \\geq \\frac 9{2(x\\plus{}y\\plus{}z)^2}$ :D\r\nBecause of $ \\frac 1t$ is convexe you can write $ \\displaystyle \\sum _{cyc} \\frac 1{x^2 \\plus{} 4xy \\plus{}y^2} \\geq 9/( \\displaystyle 2\\sum_{cyc} x^2 \\plus{}4\\sum_{cyc} xy) \\equal{} \\frac 9{2(x\\plus{}y\\plus{}z)^2}$ \r\nAnd more general for $ n$ positive numbers : $ \\displaystyle \\sum _{1\\leq i 2$). Then the solution can be generated as follows:\r\n\r\nSet $a_1 = b_1, a_2 = b_2, ... a_k = b_k$. At this point, without any other terms set ($k = n$), we have $S < P$. Since we are working $\\in \\mathbb{N}$, we know that $P - S$ must be an integer, $d$. The solution can be generated by setting $a_{k+1} = a_{k+2} = ... = a_{k+d} = 1$. This increases $S$ by $d$ while keeping $P$ the same, so we have $S = P$ and a solution. Q.E.D. \r\n\r\nDemonstration: Suppose we want $S = P = 12$. We arbitrarily choose $b_1 = 2, b_2 = 2, b_3 = 3$, which gives us $2 + 2 + 3 = 7 < 2 \\times 2 \\times 3 = 12$. We have a difference of $5$, so our arbitrary solution is $1 + 1 + 1 + 1 + 1 + 2 + 2 + 3 = 1 \\times 1 \\times 1 \\times 1 \\times 1 \\times 2 \\times 2 \\times 3$, which gives us a solution. \r\n\r\n :)", "Solution_5": "You didn't prove anything significant. You could have just taken $1,1,1,...,2,n$.", "Solution_6": "Just a question: what exactly do you want to hear\u00bf A list of all solutions\u00bf Then good luck...", "Solution_7": "Some more trivia: The number of $k$ such that $a_k=1$ is at last $n-\\mbox{ln_2} (n)-1$.", "Solution_8": "I could just describe the solutions; for $\\mathbb{N}$, the solutions for $n > 1$ consist of all sequences $a_1, a_2, ... a_n$ such that there are at least two terms $\\neq 1$, with a suitable number of $1$s thereafter.\r\n\r\n(I realize the proof was not particularly significant, but that's because the problem is not particularly significant. My method will, however, generate all nontrivial solutions. :))\r\n\r\nFor $\\mathbb{Z}$, the solutions can be generated like so:\r\n\r\nAs before, take any wanted product, $P$, and choose a series of integers $b_1, b_2, ... b_k$ such that $b_i \\neq 1$ for any $1 \\le i \\le k$, and such that $| \\prod_{i=1}^{k} b_i | = |P|$ and $| \\sum_{i=1}^{k} b_i | \\le \\sum_{i=1}^{k} |b_i| < |P|$. As before, we set $a_1 = b_1, ... a_k = b_k$. \r\n\r\nIf $S$ and $P$ have opposite sign, add the terms $a_{k+1} = 1, a_{k+2} = -1$, which reverses the sign of $P$ (this step is optional; see next step). \r\n\r\nFrom there (regardless of whether $S$ and $P$ originally had opposite sign or not), if $S < P$, then add the term $a_{k+3} = 1$ until $S = P$. If $S > P$, then add the terms $a_{k+3} = -1, a_{k+4} = -1, a_{k+5} = 1$ until $S = P$.\r\n\r\nThis will generate all solutions, so I think it counts as a valid answer to the problem, even if it does seem trivial. :)", "Solution_9": "[quote=\"ZetaX\"]For the solutions in $\\mathbb{N}$: apply AM-GM inequality.\nAnd for the general case: there are a lot of 'trivial' solutions...[/quote]...................apply AM-GM inequality?" } { "Tag": [], "Problem": "Who do you think will win it all???\r\n\r\nMost think the Gators will win, but im going with the Buckeyes.", "Solution_1": "osu, obviously.", "Solution_2": "[img]http://shelby.osu.edu/ag/Ohio%20State.jpg[/img]", "Solution_3": "LOL @ 7-0 OSU", "Solution_4": "[size=75]Florida[/size]", "Solution_5": "Let's go OSU!!! :first: :thumbup: \r\n\r\nfor Flordia :winner_second:", "Solution_6": "35-24\r\n\r\nWhat now, all you doubters, what now?", "Solution_7": "NOOOOOOOO! 19.8 seconds left and OSU is down by 10! IT CAN'T BE! IT CAN'T BE!\r\n\r\nEDIT: WTF? I HATE THE FLORIDA GATORS! They beat [b]THE OHIO STATE UNIVERSITY[/b] in both Football and Basketball. WTF. WTF.", "Solution_8": "Huh. Guess my awesomeness is proven once again.", "Solution_9": "[quote=\"footballrocks41237\"]NOOOOOOOO! 19.8 seconds left and OSU is down by 10! IT CAN'T BE! IT CAN'T BE!\n\nEDIT: WTF? I HATE THE FLORIDA GATORS! They beat [b]THE OHIO STATE UNIVERSITY[/b] in both Football and Basketball. WTF. WTF.[/quote]\r\nOMG NOOOOOO\r\nITS ALL A CONSPIRACY TO OVERTHROW THE GOVERNMENT", "Solution_10": "lol, i was going for the gators, but my dad was going for osu\r\ni told him that osu was hopeless by halftime, but he wouldn't listen to me =)", "Solution_11": "i think that florida will win", "Solution_12": "my bracket owned it had a 1510/1920 score\r\n\r\njorian" } { "Tag": [ "trigonometry", "videos" ], "Problem": "Ok it says:\r\n\r\nIn the video game duck shoot, the duck moves in a straight line from point A to B at a rate of 7cm/sec. Bullets fired from point O travel at 25 cm / sec. If a player shoots as soon as a duck appears at A, at which angle [i]a[/i] should the gun be aimed in order to score a direct hit?\r\n\r\nThis is my attempt at drawing the thing how it looks in the exercise (pretty bad drawing, but is for u to understand where is point O and all those things):\r\n\r\nhttp://img171.imageshack.us/img171/7639/duckshootme3.png", "Solution_1": "Erm, correct me if I'm wrong, but isn't the distance that the gun is away from the duck important? \r\n\r\n(unrelated, silly, note: if it's a video game, why are there actual bullets?)", "Solution_2": "It doesnt show the distance from the gun to the thing... and i dont know if there are real bullets are all :P well whatever how would I know.", "Solution_3": "Wait, darn it, never mind, ignore the above post. I was treating the speeds as lengths for some odd reason.\r\n\r\n[hide=\"solution\"]\nLet the distance the duck travels be $ t$, the distance the bullet travels be $ p$.\nWe want $ a$ such that $ \\sin a\\equal{}\\frac{t}{p}$\nNow, $ \\frac{p}{25}\\equal{}\\frac{t}{7}$ due to the d=rt thing. Thus, $ \\frac{t}{p}\\equal{}\\frac{7}{25}$, so $ a\\equal{}\\arcsin \\frac{7}{25}\\equal{}\\text{I'm too lazy to get out a calculator right now}$[/hide]", "Solution_4": "Hey thanks very much xD =]" } { "Tag": [ "geometry", "parallelogram", "ratio", "geometric transformation", "reflection", "3D geometry", "pyramid" ], "Problem": "This is for the hardcore geometry fans out there. Arranged roughly in order of difficulty. \r\n\r\n1. . Let P be an interior point of the parallelogram ABCD. Prove that\r\nangleAPB + angleCPD = 180\u25e6 if and only if anglePDC = anglePBC.\r\n\r\n2. Points K and L are chosen on the sides AB and AC of an equilateral\r\ntriangle ABC such that BK = AL. Segments BL and CK intersect at\r\nP. Determine the ratio AK : KB for which the segments AP and CK\r\nare perpendicular.\r\n\r\n3. The diagonals of a convex quadrilateral ABCD are orthogonal and inter-\r\nsect at point E. Prove that the reflections of E in the sides of quadrilateral\r\nABCD lie on a circle.\r\n\r\n\r\n4. The base of a regular quadrilateral pyramid \u03c0 is a square with side length\r\n2a and its lateral edge has length a(17^.5). Let M be a point inside the\r\npyramid. Consider the five pyramids which are similar to \u03c0, whose top\r\nvertex is at M and whose bases lie in the planes of the faces of \u03c0. Show\r\nthat the sum of the surface areas of these five pyramids is greater or equal\r\nto one fifth the surface of \u03c0, and find for which M equality holds.\r\n\r\n(I don't know how to do #3 or #4, so if someone could post solutions for those I would very much appreciate them.)", "Solution_1": "[quote=\"soltixnj\"]17^1/2[/quote]You must put grouping symbols around the\r\nfractional exponent, because what you have is equivalent to 17/2.", "Solution_2": "[quote=\"Arrange your tan\"][quote=\"soltixnj\"]17^1/2[/quote]You must put grouping symbols around the\nfractional exponent, because what you have is equivalent to 17/2.[/quote]\r\n\r\nI'd recommend just using latex. Saves trouble.", "Solution_3": "[quote=\"soltixnj\"]This is for the hardcore geometry fans out there. Arranged roughly in order of difficulty. \n\n1. . Let P be an interior point of the parallelogram ABCD. Prove that\n$ \\angle{APB} + \\angle{CPD} = 180 ^{\\circ}$ if and only if $ \\angle{PDC} = \\angle{PBC}$.\n\n2. Points K and L are chosen on the sides $ \\overline{AB}$ and $ \\overline{AC}$ of an equilateral\ntriangle $ \\triangle{ABC}$ such that $ \\overline{BK} = \\overline{AL}.$ Segments $ \\overline{BL}$ and $ \\overline{CK}$ intersect at\nP. Determine the ratio $ \\frac{\\overline{AK}}{\\overline{KB}}$ for which the segments $ \\overline{AP}$ and $ \\overline{CK}$\nare perpendicular.\n\n3. The diagonals of a convex quadrilateral ABCD are orthogonal and inter-\nsect at point E. Prove that the reflections of E in the sides of quadrilateral\nABCD lie on a circle.\n\n\n4. The base of a regular quadrilateral pyramid $ \\pi$ is a square with side length\n$ 2a$ and its lateral edge has length $ a\\sqrt{17}$. Let M be a point inside the\npyramid. Consider the five pyramids which are similar to $ \\pi$, whose top\nvertex is at M and whose bases lie in the planes of the faces of $ \\pi$. Show\nthat the sum of the surface areas of these five pyramids is greater or equal\nto $ \\frac{1}{5}$ the surface of $ \\pi$, and find for which M equality holds.\n\n(I don't know how to do #3 or #4, so if someone could post solutions for those I would very much appreciate them.)\n\n[color=red][size=59]~LaTeXed[/size][/color] [/quote]" } { "Tag": [ "limit" ], "Problem": "As usual, if you know how to solve this type of problem, leave this to beginners so they can learn from it.\r\n\r\nThe number represented by the infinite sequence $\\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\cdots}}}}$ is equal to what rational number? \r\n\r\nHint: The answer won't be an integer.", "Solution_1": "[hide]let $A=\\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\cdots }}}}$\nTherefore $A^2=\\frac 94+ \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\cdots }}}}$[/hide]\n[hide=\"Click here to see rest of solution\"]so then $A^2-\\frac 94=A$\n\n\nso $A=\\frac{1+\\sqrt 10}{2}$[/hide]", "Solution_2": "[quote=\"Silverfalcon\"]As usual, if you know how to solve this type of problem, leave this to beginners so they can learn from it.\n\nThe number represented by the infinite sequence $\\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\cdots}}}}$ is equal to what rational number? \n\nHint: The answer won't be an integer.[/quote]\r\n\r\n[hide]\nwe can set it equal to S.\n$S=\\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\sqrt{\\frac{9}{4} + \\cdots\\\\}}}}$\nWe can replace it with itself in this statement:\n$S=\\sqrt{\\frac{9}{4}+S\\\\}$\nSquaring...\n$\\displaystyle S^2=\\frac{9}{4}+S\\\\\nS^2-S-\\frac{9}{4}=0\\\\\nS=\\frac{1 \\pm \\sqrt{1+4*1*\\frac{9}{4}}}{2}=\\frac{1\\pm\\sqrt{10}}{2}$\nThe negative solution is illegal, so the answer is $\\frac{1+\\sqrt{10}}{2}$\n[/hide]", "Solution_3": "I know most people on the forum don't like to have their terminology or notation corrected (I have no idea why), but this is actually a nested (or infinite) radical. A repeated infinite fraction is generally written in the form\\[a_0+\\frac{1}{a_1+\\frac{1}{a_2+\\ldots}}\\]Also, it's not really an infinite sequence, which is something that looks like $\\{a_n\\}^\\infty$. For instance, $\\{2n\\}_{n=1}^\\infty=\\{2,4,6,\\ldots\\}$. However, it can be interpreted as the limiting value of an infinite sequence.\r\n\r\nLet $s_1=\\sqrt\\frac94$ and $s_{n+1}=\\sqrt{\\frac94+s_n}$ for $s\\ge1$.\r\n\r\nThen the number we want to find is $\\lim{s}_n$." } { "Tag": [ "modular arithmetic", "number theory" ], "Problem": "Let $ a_{1},a_{2},\\ldots,a_{100}$ and $ b_{1},b_{2},\\ldots,b_{100}$ be permutations of the positive integers from 1 to 100. Show that for some positive integers $ 1\\le i,j\\le100$, with $ i\\neq j$, we have $ a_{i}b_{i}\\equiv a_{j}b_{j}\\pmod{100}$.", "Solution_1": "Is the following correct?\r\n\r\n[hide]We have 100 products and 100 slots to put them in, so we need exactly one product per slot. In particular, this means that exactly 50 of our products must be even. For any even $ a_{i}$, the product $ a_{i}\\cdot b_{j}$ will be even regardless of the parity of $ b_{j}$, and likewise if $ b_{j}$ is even. Thus, the only way we can have 50 even products is if each even $ a_{i}$ is paired up with an even $ b_{j}$. However, we then have that each of the even products is divisble by 4 (and this is true even mod 100 because 100 is divisible by 4), so in particular we won't achive a product equal to 2, contradiction.[/hide]\r\n\r\nThat proves (with obvious modifications) the following stronger result: if $ p^{2}$ divides $ n$ for some prime $ p$, then for any permutation $ \\sigma$ of the set $ \\{0, 1, \\ldots, n-1\\}$, the set $ \\{0\\cdot \\sigma(0), 1\\cdot \\sigma(1), \\ldots, (n-1)\\cdot \\sigma(n-1)\\}$ contains two elements which are congruent $ \\pmod n$.\r\n\r\nNow, if $ n = p$ is prime, the result still holds, from an entirely different argument using Wilson's Theorem (and possibly other arguments as well). \r\n\r\nThis leaves open the question of what happens in the case of squarefree non-primes. Anyone else want to give it a go?" } { "Tag": [ "\\/closed" ], "Problem": "[url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=13162]this person's[/url] avatar is irritating me. moderators or admin, please do something with this user's avatar.", "Solution_1": "Well there's nothing there anymore so I would assume it was removed.", "Solution_2": "it seems he might mean his signature or profile\r\nit seems like it could...be... you know", "Solution_3": "His interests and signature are both quite racist." } { "Tag": [ "number theory", "Diophantine equation", "combinatorics unsolved", "combinatorics" ], "Problem": "Let A = (a1,a2,...,an) be a permutation of (1,2,...,n). A pair (ai,aj) is said to correspond to an inversion of A, if i aj. How many permutations of (1,2,...,n), n>=3 have exactly two inversions?", "Solution_1": "[quote=\"Sathej\"]Let A = (a1,a2,...,an) be a permutation of (1,2,...,n). A pair (ai,aj) is said to correspond to an inversion of A, if i aj. How many permutations of (1,2,...,n), n>=3 have exactly two inversions?[/quote]\r\n\r\nA little more general. If in the $\\sigma$ permutation in lower row before $1$ stand $t_{1}$ elements grater than $1$, $0\\le t_{1}\\le n-1$, before $2$ stand $t_{2}$ elements greater than $2$, $0\\le t_{2}\\le n-2$,..., before $k$ stand $t_{k}$ elements greater than $k$, $0\\le t_{k}\\le n-k$,..., before $n-1$ stand $t_{n-1}$ elements grater than $n-1$, $0\\le t_{n-1}\\le 1$, then the number of inversions is in the permutation $\\sigma$ is given by the sum $t_{1}+t_{2}+\\cdots+t_{k}+\\cdots+t_{n-1}$. So the number of permutation wich has $p$ inversion is given by the number of solutions of $t_{1}+t_{2}+\\cdots+t_{k}+\\cdots+t_{n-1}=p$ diophantine equation where $0\\le t_{i}\\le n-i, \\forall i=%Error. \"upperline\" is a bad command.\n{1,n}.$ Its very easy to observe that if we note with $S_{p}^{n}$ the number of permutations of rage $n$ wich has $p$ inversions than $S_{p}^{n+1}=\\sum_{i=0}^{p}S_{i}^{n}$.\r\nSo the answer for ur problem is: $S_{2}^{n+1}=S_{0}^{n}+S_{1}^{n}+S_{2}^{n}=1+n+S_{2}^{n}\\Rightarrow S_{2}^{n+1}-S_{2}^{n}=n+1.$ Calcultaing term by term and summed we got $S_{2}^{n+1}-S_{2}^{3}=n-2+\\frac{n(n+1)}{2}-1-2=\\frac{(n-2)(n+5)}{2}$, so $S_{2}^{n}=\\frac{(n+4)(n-3)}{2}+2.$", "Solution_2": "@ cckek : Are you sure that is right ?\r\n\r\nI did 3 cases a1=1 or a2=1 or a3=1 (Clearly a4,a5,....an cannot be equal to one for then we have more than two inversions)\r\n\r\nWe get a recurrence relation that gives n(n-1)/2 -1 after solving.\r\n\r\nNote: (3,1,2), (2,3,1) give two inversions for n=3\r\n(1,4,2,3), (1,3,4,2), (2,1,4,3), (3,1,2,4), (2,3,1,4) are the 5 inversions for n=4\r\n\r\nCorrect me if I am wrong.\r\nI'll post the complete solution in a couple of hours.", "Solution_3": "[quote=\"cckek\"][quote=\"Sathej\"]Let A = (a1,a2,...,an) be a permutation of (1,2,...,n). A pair (ai,aj) is said to correspond to an inversion of A, if i aj. How many permutations of (1,2,...,n), n>=3 have exactly two inversions?[/quote]\n\nA little more general. If in the $\\sigma$ permutation in lower row before $1$ stand $t_{1}$ elements grater than $1$, $0\\le t_{1}\\le n-1$, before $2$ stand $t_{2}$ elements greater than $2$, $0\\le t_{2}\\le n-2$,..., before $k$ stand $t_{k}$ elements greater than $k$, $0\\le t_{k}\\le n-k$,..., before $n-1$ stand $t_{n-1}$ elements grater than $n-1$, $0\\le t_{n-1}\\le 1$, then the number of inversions is in the permutation $\\sigma$ is given by the sum $t_{1}+t_{2}+\\cdots+t_{k}+\\cdots+t_{n-1}$. So the number of permutation wich has $p$ inversion is given by the number of solutions of $t_{1}+t_{2}+\\cdots+t_{k}+\\cdots+t_{n-1}=p$ diophantine equation where $0\\le t_{i}\\le n-i, \\forall i=%Error. \"upperline\" is a bad command.\n{1,n}.$ Its very easy to observe that if we note with $S_{p}^{n}$ the number of permutations of rage $n$ wich has $p$ inversions than $S_{p}^{n+1}=\\sum_{i=0}^{p}S_{i}^{n}$.\nSo the answer for ur problem is: $S_{2}^{n+1}=S_{0}^{n}+S_{1}^{n}+S_{2}^{n}=1+n+S_{2}^{n}\\Rightarrow S_{2}^{n+1}-S_{2}^{n}=n+1.$ Calcultaing term by term and summed we got $S_{2}^{n+1}-S_{2}^{3}=n-2+\\frac{n(n+1)}{2}-1-2=\\frac{(n-2)(n+5)}{2}$, so $S_{2}^{n}=\\frac{(n+4)(n-3)}{2}+2.$[/quote]\r\n\r\na little mistake:)\r\n\r\n$S_{1}^{n}=n-1$ so we got $S_{2}^{n+1}-S_{2}^{n}=n.$ therefore \r\n$S_{2}^{n+1}-S_{2}^{3}=\\frac{n(n+1)}{2}-1-2$ so $S_{2}^{n}=\\frac{(n-1)n}{2}-1.$", "Solution_4": "This is what I did though it basically amounts to just what you did\r\n\r\nLet $I_{n}$ denote number of permutations of $(1,2,3,...n)$ with two inversions.\r\n\r\nWe see that $a_{i}\\ne 1 \\forall i \\ge 4$\r\n\r\nCase 1] $a_{1}= 1$\r\nThen we must have two inversions in $(a_{2},a_{3},...a_{n})$\r\nThus it can be done in $I_{n-1}$ ways.\r\n\r\nCase 2] $a_{2}= 1$\r\nAgain $a_{1}\\ne i \\forall i \\ge 4$ else we will have atleast 3 inversions\r\n\r\nThus a) $a_{1}= 2$\r\nThen we want one inversion in $(a_{3},a_{4},...,a_{n})$\r\nThus n-3 ways to do this.\r\n\r\n b) $a_{1}= 3$\r\nThen we are forced to put $(a_{1},a_{2},...a_{n}) = (3,1,2,4,5,...,n)$\r\nThus one way to do this\r\n\r\nCase 3] $a_{3}= 1$\r\nThen we are forced to put $(a_{1},a_{2},...a_{n}) = (2,3,1,4,5,...,n)$\r\nThus one way to do this\r\n\r\nThus $I_{n}= I_{n-1}+(n-3)+1+1$\r\nThus $I_{n}= I_{n-1}+n-1$\r\n\r\nWe see that $I_{3}= 2$\r\n\r\nThus we have $I_{n}= \\frac{(n-1)n}{2}-1.$", "Solution_5": "Hello Vihang,\r\n How many problems did you get correct this year?\r\nSathej" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "We have a sequence $ x_{n\\plus{}3} \\equal{} x_{n} \\plus{} x_{n\\plus{}1}x_{n\\plus{}2}$ for positive integers $ x_{j}$ .Given \r\n$ x_{1} \\equal{} x_{2} \\equal{} x_{3}\\equal{}1$...Is it possible to convert this into a linear recursive relation ...?", "Solution_1": "no replies???? :(", "Solution_2": "Because $ x_n$ supexponential, increase as $ c^{2^n}$, all lineal recurent sequense $ O(c^n)$.", "Solution_3": ":( sorry can you explain it detailed and express $ x_{n}$ in the form $ a_{1}x_{k} \\plus{}......$ or some similar linear form .... :?:", "Solution_4": "any replies...or explanation for Rust's statement ...?..Is it that dumb that people dont answer it..?If that is the case..I am sorry :blush:", "Solution_5": "What Rust said you is that all linear recurrences $ x_{n\\plus{}k}\\equal{}\\sum_{i\\equal{}1}^ka_ix_{n\\plus{}k\\minus{}i}$ imply solutions $ x_n\\equal{}\\sum_{i\\equal{}1}^k\\alpha_iz_i^n$ whose behaviour when $ n\\to\\plus{}\\infty$ is $ \\approx aA^n$ \r\n\r\nAnd obviously your sequence does not have this behaviour when $ n\\to\\plus{}\\infty$\r\n\r\nSo, we'll surely be unable to find a formula $ x_{n\\plus{}k}\\equal{}\\sum_{i\\equal{}1}^ka_ix_{n\\plus{}k\\minus{}i}$ for your sequence.\r\n\r\nBut maybe this is not what you call a \"linear recurrence\" ? Maybe you accept non fixed summations ? Try to be more clear about the problem.\r\nIn what test / contest / class did you find it ?", "Solution_6": "oh..ok....it was not asked in any contest ...you can say it is a kind of my own problem.I was trying some another problem where I wanted to get a solution by finding the $ n'th$ term of this sequence. What do you mean by non fixed summations ? Anything from which I will be able to find the $ n'th$ term is fine ? I wanted something like a linear recurrence (homogeneous or inhomogeneous) as we know the general way to solve such recurrences .So just find a general expression for $ x_{n}$" } { "Tag": [ "geometry", "geometric transformation", "rotation", "reflection", "trigonometry", "parallelogram", "vector" ], "Problem": "Due to earth's rotation, a plumb bob may not hang exactly along the direction of the Earth's gravitational force on the plumb bob but may deviate slightly from that direction. (a) show that the deflection $\\theta$is given by $\\theta = (\\frac{2\\pi^2R}{gT^2})sin(2L)$, where R is the radius of the Earth and T is the period of the Earth's rotation.\r\n\r\nThis problem is very hard. I try to get a = $\\frac{2\\pi(RcosL)^2}{T^2RcosL}$ and then must get F so I try to separate the thingie into components byut then Fx and Fy get into a mess and I'm confused. Help!\r\n\r\nThanks", "Solution_1": "Look at the attached figure. The circle with the center E and radius R is the Earth. The bob is at the point B with the lattitude angle $\\angle L$. The gravitational force on B is $\\overrightarrow F_{BG} = \\overrightarrow{BG}$ with the magnitude $F_{BG} = mg$. The radius of rotation at the lattitude L is the perpendicular distance $r = R \\cos \\widehat L$ of B from the axis of rotation EN (N is the north pole). The centrifugal force on B is $\\overrightarrow F_{BC} = \\overrightarrow{BC}$ with the magnitude\n\n$F_{BC} = m \\omega^2 r = \\frac{4 \\pi^2 mr}{T^2} = \\frac{4 \\pi^2 mR \\cos \\widehat L}{T^2}$\n\nwhere T is the rotation period of the bob around the axis EN and $\\omega = \\frac{2\\pi}{T}$ the angular velocity. You can get the magnitude of the sum of these these 2 forces as the diagonal BR of the parallelogram BGRC or as the 3rd side BR of the triangle $\\triangle BGR$ by placing the vector $\\overrightarrow F_{BC} = \\overrightarrow {BC}$ tail to the head of the vector $\\overrightarrow F_{BG} = \\overrightarrow{BG}$, i.e., into $\\overrightarrow F_{BC} = \\overrightarrow{GR}$. In either case, you have the triangle $\\triangle BGR$ with 2 known sides BG, GR = BC and the known angle between them $\\angle BGR = \\widehat L$. Using the cosine theorem, you can calculate the side BR, and using the sine theorem next, you can calculate sine of the angle $\\angle GBR = \\theta$ (using formulas, not numbers). You can simplify the task by expressing the centrifugal force $\\overrightarrow F_{BC} = \\overrightarrow{BC}$ as the sum of its horizontal component $\\overrightarrow F_{BH} = \\overrightarrow{BH}$ and vertical component $\\overrightarrow F_{BV} = \\overrightarrow{BV}$ with the magnitudes $F_{BH} = F_{BC} \\sin \\widehat L$ and $F_{BV} = F_{BC} \\cos \\widehat L$. The vectors $\\overrightarrow F_{BG}, \\overrightarrow F_{BV}$ are (anti)parallel and the magnitude of their resultant $\\overrightarrow F_{BD} = \\overrightarrow {BD} = \\overrightarrow F_{BC} + \\overrightarrow F_{BV}$ can be found algebraically by subtraction:\n\n$F_{BD} = F_{BG} - F_{BV} = F_{BG} - F_{BC} \\cos \\widehat L = mg - \\frac{4 \\pi^2 mR \\cos^2 \\widehat L}{T^2}$\n\nThe resultant $\\overrightarrow F_{BR} = \\overrightarrow{BR}$ is now the hypotenuse of the right angle triangle $\\triangle BDR$ and it can be calculated by the Pythagorean theorem. But it is not necessary. The tangent of the angle $\\angle DBR = \\theta$ is the ratio of the known legs BD, BH of the right angle triangle $\\triangle BDR$:\n\n$\\tan \\theta = \\frac{DR}{BD} = \\frac{BH}{BC - BV} = \\frac{BC \\sin \\widehat L}{BG - BC \\cos \\widehat L} =$\n\n$= \\frac{\\dfrac{4\\pi^2 mR \\cos \\widehat L \\sin \\widehat L}{T^2}}{mg - \\dfrac{4\\pi^2 mR \\cos^2\\widehat L}{T^2}} = \\frac{2\\pi^2 R \\sin 2\\widehat L}{gT^2} \\cdot \\frac{1}{1 - \\dfrac{4\\pi^2 R \\cos^2\\widehat L}{gT^2}}$\n\nNow comes the 1st approximation. In the 2nd factor $\\frac{1}{1 - q}$ in the expression for $\\tan \\theta$\n \n${{q = \\frac{4\\pi^2 R \\cos^2\\widehat L}{gT^2}} \\le \\frac{4\\pi^2 R}{gT^2}} \\doteq \\frac{40 \\cdot 6.4 \\times 10^6\\ \\text m}{10\\ \\text{m/sec}^2 \\cdot 86400^2\\ \\text {sec}^2} \\doteq \\frac{25.6}{86.4^2} \\doteq 0.0034$\n\nIf we are content to calculate the small angle $\\theta$ with accuracy 0.5 %, the 2nd factor is so close to 1 that it can be neglected. Hence,\n\n$\\tan \\theta \\doteq \\frac{2\\pi^2 R \\sin 2\\widehat L}{gT^2}$\n\nThe 2nd approximation is that for small angles, $\\tan \\theta \\doteq \\theta$, where $\\theta$ is in radians. This is easily seen if you draw a vertical tangent to a unit circle at its intersection U with the positive x-axis and a ray from the origin at an angle $\\theta$ with the positive x-axis. This ray intersects the unit circle at a point W and the tangent at a point T. The segment UT on the tangent is $\\tan \\theta$ and the arc UW of the unit circle is the angle $\\theta$ in radians. For small angles, the segment UT and the arc UW are about the same, but I would have to use calculus to calculate how close they are. In our case, $\\tan \\theta$ is really small (about half of the above q):\n\n$\\theta \\doteq \\tan \\theta \\doteq \\frac{2\\pi^2 R \\sin 2\\widehat L}{gT^2} \\le \\frac{2\\pi^2 R}{gT^2} \\doteq 0.0017\\ \\text{rad} \\doteq 0.1^\\circ = 6'$", "Solution_2": "hey, thank you so much for the detailed reply, yetti. But where is the diagram?", "Solution_3": "The diagram is at the bottom of my reply, in from of my eyes. The mathlinks web site does not display attachments to messages, if you just browse as a guest, you have to log in using your member name and password. But you certainly have to log in to post messages, how come you do not see the attached drawing ?", "Solution_4": "[quote=\"yetti\"]But you certainly have to log in to post messages, how come you do not see the attached drawing ?[/quote]I think what happened was that he was browsing as a guest, didn't see the diagram, logged in to reply and say that he does not see the diagram, without noticing that once logged in he is able to see it :)" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "An infinite road has traffic lights at intervals of 1500m. The lights are all synchronised and are alternately green for $\\frac 32$ minutes and red for 1 minute. For which $v$ can a car travel at a constant speed of $v$m/s without ever going through a red light?", "Solution_1": "$v=v_0/k, \\ v_0=1500/75=20(m/s), k$ any natural.", "Solution_2": "I think $v_0$ should be 10m/s.", "Solution_3": "you are wrong. Let t(sek) time needed for one interval, then\r\n$\\{\\frac{tm}{150}\\}<\\frac 35$ for all natural m. It give $t=75k,v=v_0/k,v_0=20(m/sek).$", "Solution_4": "I think v is any of the set 1, 2, 5, 10, 20.\nEvidently v needs to be less or equal to 25. Secondly it must be an integer divisor of 10 or be a multiple of 10, but less than 25.", "Solution_5": "The answer depends on the starting point of the car. For example, if the car starts at point $x_0=0$, the velocity should be of the form $v=\\dfrac{20}{2n+1}$ m/s for some non-negative integer $n$. So, answers like $20$ m/s, $20/3$ m/s, $4$ m/s would do.\n\nI write my answer here: https://nivotko.wordpress.com/2022/12/21/avoiding-traffic-lights/\n\n" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ x,y,z$ be real numbers, find the greatest value of $ a$ such that \r\n\\[ x^6\\plus{}y^6\\plus{}z^6\\plus{}a(x^5y\\plus{}y^5z\\plus{}z^5x)\\geq0.\\]", "Solution_1": "Have you a solution to it, pvthuan ? I think it is not easy.", "Solution_2": "[quote=\"pvthuan\"]Let $ x,y,z$ be real numbers, find the greatest value of $ a$ such that\n\\[ x^6 \\plus{} y^6 \\plus{} z^6 \\plus{} a(x^5y \\plus{} y^5z \\plus{} z^5x)\\geq0.\n\\]\n[/quote]\r\nAre you sure there isnt a typo,like the third \"$ \\plus{}$\" sign or there is some condition on $ x,y,z,a$", "Solution_3": "[quote=\"Akashnil\"][quote=\"pvthuan\"]Let $ x,y,z$ be real numbers, find the greatest value of $ a$ such that\n\\[ x^6 \\plus{} y^6 \\plus{} z^6 \\plus{} a(x^5y \\plus{} y^5z \\plus{} z^5x)\\geq0.\n\\]\n[/quote]\nAre you sure there isnt a typo,like the third \"$ \\plus{}$\" sign or there is some condition on $ x,y,z,a$[/quote]\r\nwhy i think the quetsion is right see it readsreal numbers and not positive real numbers if that is ur point" } { "Tag": [ "function", "calculus", "derivative", "real analysis", "real analysis unsolved" ], "Problem": "If a continuous function on [0,1] has zero derivative almost everywhere on the interval (note that we do not require it to be differentiable on [0,1]), what can we say about it in general (existence, for example)?\r\n\r\nAside from being nowhere strictly monotonic. :) In fact, you can add the property that this function is nowhere monotonic.", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=118219]No you can't[/url].", "Solution_2": "That certainly works. :)\r\n\r\nI was hoping there was a non-constructive proof that there isn't such a function which is also nowhere monotonic (assuming that is what 'no you can't' means). I want to exercise my Baire Category muscle.", "Solution_3": "I was providing a counterexample to your implicit statement \"such a function can't be strictly monotonic\". We apparently misunderstood each other.\r\n\r\nNow, if you want to ask whether there are any such functions which are nowhere monotonic, it shouldn't be too hard to modify one of those constructions to create one. Even subtracting two different strictly monotonic solutions is likely to work.", "Solution_4": "I see now. I wasn't clear before. What I meant originally in the first post was that the function is differentiable almost everywhere on [0,1] but the derivative is always zero at any point of differentiability (it is never positive or negative). Moreover, the function can be assumed to be nowhere monotonic. And the question is does such a function exist?" } { "Tag": [], "Problem": "(1/0)*0=?\r\nIs it 0, 1, or infinity?", "Solution_1": "This was discussed in another topic:\r\n[quote=\"K81o7\"]$z=\\frac{1}{0}$\n\nThe flaw comes in saying this. Equality is something we only can assign to certain mathematical objects, ie numbers, vectors, matrices, etc. This mathematical object does not fall under any of thoe categories, so we can't define equailty for it.[/quote]", "Solution_2": "[quote=\"jclarkemathL314159\"](1/0)*0=?\nIs it 0, 1, or infinity?[/quote]\r\nI believe it's indeterminate, meaning we don't know what it is.", "Solution_3": "[quote=\"jclarkemathL314159\"](1/0)*0=?\nIs it 0, 1, or infinity?[/quote]\r\n\r\nThe answer is....\r\n\r\n\r\nUNDEFINED!!!!", "Solution_4": "I am pretty sure it is\r\n[hide]$UNDEFINED!$[/hide] :D" } { "Tag": [ "Miscellaneous Problems" ], "Problem": "The decimal expression of the natural number $a$ consists of $n$ digits, while that of $a^3$ consists of $m$ digits. Can $n+m$ be equal to $2001$?", "Solution_1": "[quote=\"Robert Gerbicz\"]No it can't be equal to 2002. Because:\nif $ a\\geq{10}^{500}$ then $ {a}^{3}\\geq{10}^{1500}$ so $ n+m\\geq 501+1501>2001$\nif $ a<{10}^{500}$ then $ a<{10}^{1500}$ so $ n+m\\leq 500+1500<2001$\nThis proves that $ n+m$ can't be $ 2001$.[/quote]" } { "Tag": [ "geometry", "trapezoid", "Pythagorean Theorem" ], "Problem": "A circle of diameter 10 is inscribed in a trapezoid ABCD where angles A and B are right angles and AD=8. Find the length of longest side of this trapezoid. Express your answer as an improper fraction reduced to lowest terms.", "Solution_1": "[hide=\"hint\"]\nUse the fact that AB+CD=AD+BC[/hide]\n\n[hide=\"Solution\"]Call the P point of tangency to side CD.\n\nLet CP=x. CD can be expressed as $ \\sqrt {10^2 \\plus{} (x \\minus{} 3)^2}$\n\n$ 10 \\plus{} \\sqrt {10^2 \\plus{} (x \\minus{} 3)^2} \\equal{} 8 \\plus{} 5 \\plus{} x$\n$ \\sqrt {10^2 \\plus{} (x \\minus{} 3)^2} \\equal{} x \\plus{} 3$\n$ 100 \\plus{} x^2 \\minus{} 6x \\plus{} 9 \\equal{} x^2 \\plus{} 6x \\plus{} 9$\n$ 12x \\equal{} 100$\n$ x \\equal{} \\frac {25}{3}$\n\nThus, the answer is $ \\frac {25}{3}\\plus{}5\\equal{}\\frac{40}{3}$[/hide]", "Solution_2": "[hide]\n[asy]pathpen = black + linewidth(0.62); pointpen = black;\npair A=(0,0),B=(0,8),D=(10,0),C=(10,40/3),O=(5,5);\nD(MP(\"A\",A)--MP(\"B\",B,NW)--MP(\"C\",C,NE)--MP(\"D\",D)--cycle);\npath p = Circle(O,5); D(p); pair F= IP(p,B--C);\nD(B--MP(\"B'\",(C.x,B.y),E)--C--cycle, blue+linewidth(0.8));\nD(MP(\"E\",D(F),NW)--D(O)); \nD(D(IP(p,C--D))--D(IP(p,A--B)));\nMP(\"10\",(A+D)/2); MP(\"5\",(A+IP(p,A--B))/2,W); MP(\"3\",(B+IP(p,A--B))/2,W); MP(\"3\",(B+F)/2,NW); MP(\"x\",(C+F)/2,NW); MP(\"x-3\",(C+(D.x,5))/2,NE); MP(\"10\",(B+(10,B.y))/2,NE);[/asy]\nLet $ E$ be the point of tangency between the inscribed circle and $ \\overline{BC}$, and $ B'$ be the foot of the altitude from $ B$ to $ \\overline{CD}$. By two-tangent, we have $ BE \\equal{} 8 \\minus{} 5 \\equal{} 3$. Let $ EC \\equal{} x$, by two-tangent again $ B'C \\equal{} x\\minus{}3$. The Pythagorean Theorem on $ \\triangle BB'C$ yields\n\\[ (x\\minus{}3)^2 \\plus{} 10^2 \\equal{} (x\\plus{}3)^2 \\Longrightarrow x \\equal{} \\frac{25}{3}\\]Hence the longest side is $ CD \\equal{} 5 \\plus{} x \\equal{} \\boxed{\\frac{40}{3}}$. \n[/hide]" } { "Tag": [ "ratio", "geometry", "perimeter" ], "Problem": "A janitor has to wash two round floors. If the time to wash the floor is proportional to its size, and the time required to wash the first floor is 4 hours, the time required to watch the second floor is 1 hour, what is the diamater of the first floor if the diamater of the second floor is 20 m.", "Solution_1": "The ratio of the areas is the square of the ratio of the diameters, and thus the ratio of the times is the square of the ratio of the diameters as well, so \\[ \\frac{4}{1}\\equal{}\\left(\\frac{x}{20}\\right)^2 \\Rightarrow x\\equal{}\\boxed{40}\\]", "Solution_2": "[quote=\"5849206328x\"]The ratio of the areas is the square of the ratio of the diameters, and thus the ratio of the times is the square of the ratio of the diameters as well, so\n\\[ \\frac {4}{1} \\equal{} \\left(\\frac {x}{20}\\right)^2 \\Rightarrow x \\equal{} \\boxed{40}\n\\]\n[/quote]\r\n\r\nHow do you know its the areas and not the circumference that its talking about?\r\n\r\nBy the way, shouldnt it be x/20= (1/4)^2 and not the other way around? I'm really confused :(", "Solution_3": "Hi helpmeplz,\r\n\r\nI think 5849206328x did the problem incorrectly. In fact, since the time required to wash is proportional to its SIZE, that implies circumference. Therefore, if we denote the diameter of the first circle with x, then:\r\n\r\n(Time to wash the first floor)/(Time to wash the second floor) = (x)/(20, diameter of second circle)\r\n\r\n4/1 = x/20 which implies that x = 80. Therefore the diameter of the first circle is 80. Notice how there is nothing raised \r\n\r\nto a power. This is only true when if constructing proportions with area and side length (similar polygons).\r\n\r\nHope it makes sense,\r\n\r\nrts2007", "Solution_4": "If size meant circumference then the circumference is what the janitor washes. That usually is not very helpful :wink:", "Solution_5": "[quote=\"rts2007\"]Hi helpmeplz,\n\nI think 5849206328x did the problem incorrectly. In fact, since the time required to wash is proportional to its SIZE, that implies circumference. Therefore, if we denote the diameter of the first circle with x, then:\n\n(Time to wash the first floor)/(Time to wash the second floor) = (x)/(20, diameter of second circle)\n\n4/1 = x/20 which implies that x = 80. Therefore the diameter of the first circle is 80. Notice how there is nothing raised \n\nto a power. This is only true when if constructing proportions with area and side length (similar polygons).\n\n[b]Hope it makes sense,[/b]\n\nrts2007[/quote]\r\n\r\nUh... no.\r\n\r\nIf someone asks you your size, would you measure your circumference and give it to them?\r\n\r\nWell then, how big is that book over there?\r\n\r\nWould you answer the perimeter?", "Solution_6": "So BOGTRO and 5849206328x,\r\n\r\nHow would you interpret the problem? And, by the way, when someone asks me my size I don't say \"blah blah\" square inches. You can't interpret size as being area. And also, that book is huge! Would you say that the book's size is, for example, 45 square inches? \r\n\r\nThanks,\r\n\r\nrts2007", "Solution_7": "area. Pretty obvious.", "Solution_8": "Would you say that the book's size is, for example, 45 square inches?", "Solution_9": "Yes, you would.\r\nYou wouldn't say like $ 15$ inches, referring to its perimeter.\r\nSomething's size is a measure of how much space it takes up,\r\na measure that in this case is two-dimensional and is therefore area.", "Solution_10": "You wouldn't say 45 square inches nor would you say 15 inches, referring to its perimeter. This is a weird problem." } { "Tag": [ "geometry", "trigonometry", "articles", "LaTeX", "trig identities", "Pythagorean Theorem", "Law of Cosines" ], "Problem": "1) The perpendicular bisectors of two of the sides of triangle $ ABC$ intersect the third side at the same point. Prove that the triangle is right-angled.\r\n\r\n2) Segments $ AD$ and $ BE$ are medians of right triangle $ ABC$, and $ AB$ is its hypotenuse. If a right triangle is constructed with legs $ AD$ and $ BE$, what will be the length of its hypotenuse in terms of $ AB$?\r\n\r\n3) If $ \\tan{x} \\equal{} \\frac {2ab}{a^{2} \\minus{} b^{2}}$, where $ a > b > 0$ and $ 0^\\circ < x < 90^\\circ$, then find $ \\sin{x}$ in terms of $ a$ and $ b$.", "Solution_1": "1\r\n[hide]Let the penpendicular bisectors bisect AB and AC. That implies thatt their intersection point O is equidistant from B, C, and A, so triangles ABO and AOC are isoceles, so A = B+C. since A+B+C = 180, A=90.[/hide]", "Solution_2": "[quote=\"elcric\"]1) The perpendicular bisectors of two of the sides of triangle $ ABC$ intersect the third side at the same point. Prove that the triangle is right-angled.\n\n2) Segments $ AD$ and $ BE$ are medians of right triangle $ ABC$, and $ AB$ is its hypotenuse. If a right triangle is constructed with legs $ AD$ and $ BE$, what will be the length of its hypotenuse in terms of $ AB$?\n\n2) If $ \\tan{x} \\equal{} \\frac {2ab}{a^{2} \\minus{} b^{2}}$, where $ a > b > 0$ and $ 0^\\circ < x < 90^\\circ$, then find $ \\sin{x}$ in terms of $ a$ and $ b$.[/quote]\r\n\r\nHint\r\n[hide](1) Answered above.\n\n(2) Let $ AC \\equal{} 2a$ and $ BC \\equal{} 2b$. Use Pythagoras to find $ AD$ and $ BE$.\n\n(3) Construct a right triangle with sides opposite and adjacent to angle $ x$ be $ 2ab$ and $ a^2 \\minus{} b^2$ respectively. [/hide]", "Solution_3": "I still couldn't get numbers 2 and 3. Could someone please help? Thanks a lot!", "Solution_4": "[quote=\"elcric\"]I still couldn't get numbers 2 and 3. Could someone please help? Thanks a lot![/quote]\r\nfor 3, after you construct a right triangle with sides opposite and adjacent to angle $ x$ be $ 2ab$ and $ a^2\\minus{}b^2$ respectively.\r\nwe get $ tanx\\equal{}2ab/{a^2\\minus{}b^2}$ and $ sinx$ is easy to determine.", "Solution_5": "for the three the hint is to use the known: $ \\sin^2x \\equal{} \\frac{\\tan^2x}{1 \\plus{} \\tan^2x}$\r\nthe solution is here though:\r\n[hide]\n$ \\tan^2x\\equal{} \\frac{(2ab)^2}{a^4 \\plus{} b^4 \\minus{}2a^2b^2}$\n$ < \\equal{} >$\n$ \\frac{\\tan^2x}{1\\plus{}\\tan^2x} \\equal{} (\\frac{2ab}{a^2 \\plus{} b^2})^2$\n$ < \\equal{} >$\n$ \\sin x\\equal{}\\frac{2ab}{a^2 \\plus{} b^2}$ because sinx>0[/hide]", "Solution_6": "[quote=\"Athinaios\"]for the three the hint is to use the known: $ \\sin^2x \\equal{} \\frac {\\tan^2x}{1 \\plus{} \\tan^2x}$\nthe solution is here though:\n[hide]\n$ \\tan^2x \\equal{} \\frac {(2ab)^2}{a^4 \\plus{} b^4 \\minus{} 2a^2b^2}$\n$ < \\equal{} >$\n$ \\frac {\\tan^2x}{1 \\plus{} \\tan^2x} \\equal{} (\\frac {2ab}{a^2 \\plus{} b^2})^2$\n$ < \\equal{} >$\n$ \\sin x \\equal{} \\frac {2ab}{a^2 \\plus{} b^2}$ because sinx>0[/hide][/quote]\r\n\r\nHow did you get $ \\sin^2x \\equal{} \\frac{\\tan^2x}{1 \\plus{} \\tan^2x}$? Thanks!", "Solution_7": "[quote]How did you get $ \\sin^2x \\equal{} \\frac{\\tan^2x}{1 \\plus{} \\tan^2x}$? Thanks![/quote]\r\n\r\nOne way is just to use trig identities\r\n\r\n$ \\sin^{2}x \\plus{} \\cos^{2}x \\equal{} 1$; divide by $ \\sin^{2}$ to get\r\n\r\n$ 1 \\plus{} \\frac{1}{\\tan^{2}x} \\equal{} \\frac{1}{\\sin^{2}x}$, so \r\n\r\n$ \\sin^2x \\equal{} \\frac{\\tan^2x}{1 \\plus{} \\tan^2x}$", "Solution_8": "[quote=\"Athinaios\"]for the three the hint is to use the known: $ \\sin^2x \\equal{} \\frac {\\tan^2x}{1 \\plus{} \\tan^2x}$\nthe solution is here though:\n[hide]\n$ \\tan^2x \\equal{} \\frac {(2ab)^2}{a^4 \\plus{} b^4 \\minus{} 2a^2b^2}$\n$ < \\equal{} >$\n$ \\frac {\\tan^2x}{1 \\plus{} \\tan^2x} \\equal{} (\\frac {2ab}{a^2 \\plus{} b^2})^2$\n$ < \\equal{} >$\n$ \\sin x \\equal{} \\frac {2ab}{a^2 \\plus{} b^2}$ because sinx>0[/hide][/quote]\r\n\r\nI don't get how $ \\frac {\\tan^2x}{1 \\plus{} \\tan^2x} \\equal{} (\\frac {2ab}{a^2 \\plus{} b^2})^2$ was gotten.\r\n\r\nThanks!", "Solution_9": "there is a known id that says: if $ \\frac{a}{b} \\equal{}\\frac{c}{d}$ then it is also true that\r\n$ \\frac{a}{b \\plus{} a} \\equal{}\\frac{c}{d \\plus{}c}$(easy to prove) so using this you know\r\n$ \\frac{tan^2x}{1} \\equal{}\\frac{(2ab)^2}{a^2 \\plus{} b^2 \\minus{}2ab}$\r\n$ < \\equal{} >$\r\n$ \\frac{tan^2x}{1 \\plus{} tan^2x} \\equal{}\\frac{(2ab)^2}{a^4 \\plus{} b^4 \\minus{}2a^2b^2 \\plus{} 4a^2b^2}$\r\n$ \\equal{} \\frac{(2ab)^2}{a^4 \\plus{} b^4 \\plus{} 2a^2b^2}\\equal{}$\r\n$ \\equal{}(\\frac{2ab}{a^2 \\plus{} b^2})^2$ i think now it is clear", "Solution_10": "[quote=\"Athinaios\"]there is a known id that says: if $ \\frac {a}{b} \\equal{} \\frac {c}{d}$ then it is also true that\n$ \\frac {a}{b \\plus{} a} \\equal{} \\frac {c}{d \\plus{} c}$(easy to prove) so using this you know\n$ \\frac {tan^2x}{1} \\equal{} \\frac {(2ab)^2}{a^2 \\plus{} b^2 \\minus{} 2ab}$\n$ < \\equal{} >$\n$ \\frac {tan^2x}{1 \\plus{} tan^2x} \\equal{} \\frac {(2ab)^2}{a^4 \\plus{} b^4 \\minus{} 2a^2b^2 \\plus{} 4a^2b^2}$\n$ \\equal{} \\frac {(2ab)^2}{a^4 \\plus{} b^4 \\plus{} 2a^2b^2} \\equal{}$\n$ \\equal{} (\\frac {2ab}{a^2 \\plus{} b^2})^2$ i think now it is clear[/quote]\r\n\r\nThanks, I got number 3 now. Does anyone have a solution to problem 2?", "Solution_11": "For #2,\r\n\r\nUse Stewart's Theorem\r\n$ (dad\\plus{}man\\equal{}bmb\\plus{}cnc)$\r\n\r\nTo obtain\r\n\r\n$ AD^{2}\\equal{}\\frac{1}{2}AC^{2}\\plus{}\\frac{1}{2}AB^{2}\\minus{}\\frac{1}{4}BC^{2}$,\r\n$ BE^{2}\\equal{}\\frac{1}{2}BC^{2}\\plus{}\\frac{1}{2}AB^{2}\\minus{}\\frac{1}{4}AC^{2}$,\r\nSo\r\n$ AD^{2}\\plus{}BE^{2}\\equal{}HYPOTENUSE^{2}\\equal{}\\frac{1}{4}AB^{2}\\plus{}AB^{2}\\equal{}\\frac{5}{4}AB^{2}$\r\nSo\r\n$ HYPOTENUSE \\equal{} \\frac{\\sqrt{5}}{2}AB$", "Solution_12": "[quote=\"iniquitus\"]For #2,\n\nUse Stewart's Theorem\n$ (dad \\plus{} man \\equal{} bmb \\plus{} cnc)$\n\nTo obtain\n\n$ AD^{2} \\equal{} \\frac {1}{2}AC^{2} \\plus{} \\frac {1}{2}AB^{2} \\minus{} \\frac {1}{4}BC^{2}$,\n$ BE^{2} \\equal{} \\frac {1}{2}BC^{2} \\plus{} \\frac {1}{2}AB^{2} \\minus{} \\frac {1}{4}AC^{2}$,\nSo\n$ AD^{2} \\plus{} BE^{2} \\equal{} HYPOTENUSE^{2} \\equal{} \\frac {1}{4}AB^{2} \\plus{} AB^{2} \\equal{} \\frac {5}{4}AB^{2}$\nSo\n$ HYPOTENUSE \\equal{} \\frac {\\sqrt {5}}{2}AB$[/quote]\r\n\r\nWhat's Stewart's Theorem? How does it work and what does it do? Thanks a lot!", "Solution_13": "Wikipedia article\r\nhttp://en.wikipedia.org/wiki/Stewart's_theorem\r\n(URL tags dont work :S) copy and paste into browser", "Solution_14": "umm....i dont really understand anything above.... >.< BUT I HAVE A SOLUTION for 3 except its sort of differentish\r\n\r\nand i just learned that trig stuff so i might have taken some long ways when there was a shorter way thats obvious\r\n\r\n[hide=\"click here for a solution without using that crazy formula!! :P \"]\nmake a right triangle and since that tangent stuff is opposite over adjacent (yay for sohcahtoa!! :D ) you label the leg opposite the angle with a measure of x 2ab and the adjacent leg a^2-b^2 (sorry i cant use latex so the rest will be hard to read.....)\n\nUsing the Pythagorean Theorem we can say that the length of the hypotenuse is \nsqrt((a^2-b^2)^2+(2ab)^2)\n=sqrt(a^4-2(a^2)(b^2)+b^4+4(a^2)(b^2)\n=sqrt(a^4+2(a^2)(b^2)+b^4) \n=sqrt((a^2+b^2)^2)\n=a^2+b^2\n\nsin(x)=opposite/hypotenuse ( :D )\n=(2ab)/(a^2+b^2)\n\nyay!! this was more algebra than geometry :lol: [/hide]", "Solution_15": "[quote=\"elcric\"]\n\nWhat's Stewart's Theorem? How does it work and what does it do? Thanks a lot![/quote]\n\nHere you have [b]Stewart's theorem [/b]and also a proof:\n[quote][color=green]Let the triangle $ ABC$ and $ M$ a point on $ BC$.Prove that\n\\[ AB^2\\cdot MC \\plus{} AC^2\\cdot BM \\equal{} AM^2\\cdot BC \\plus{} BM\\cdot MC\\cdot BC\n\\]\n[/color][/quote]\r\n[hide=\"Proof\"]Use the law of cosines:\n$ AM^2 \\equal{} AB^2 \\plus{} BM^2 \\minus{} 2AB\\cdot BM \\cos B$ \n$ AC^2 \\equal{} AB^2 \\plus{} BC^2 \\minus{} 2AB\\cdot BC \\cos B$ \nThen multiply with $ BC$ the first relation and with $ \\minus{} BM$ the second relation,and by summing will obtain the proof[/hide]." } { "Tag": [], "Problem": "If $ \\minus{}2x\\plus{}1<3$, what is the least integer in the solution set?", "Solution_1": "$ \\minus{}2x\\plus{}1<3\\implies\\minus{}2x<2\\implies x>\\minus{}1\\implies x\\ge\\boxed0$." } { "Tag": [ "geometry", "geometric transformation", "reflection", "parallelogram", "rectangle", "rotation", "circumcircle" ], "Problem": "Let $ABC$ be a right-angled triangle ($AB\\perp AC$) and the mobile points $M\\in AB,\\ N\\in AC$ such that $BN=CM$. Prove that one from the bisectors of the angles formed between the lines $BN,\\ CM$ passes through a fixed point.\r\n\r\n[u]Remark.[/u] I shall be pleased to see a syntetical solution.\r\nThis problem, with analytical geometry, has a easy solution.\r\n\r\n[u]Indication.[/u] The fixed point $F$ is the reflection of the point $A$ w.r.t. the middlepoint of the side $[BC]$ and the distance of the point $F$ to the lines $KB,KC$ is equal to $\\frac{AB\\cdot AC}{l}$ where $K\\in BN\\cap CM,\\ BN=CM=l$.\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=57850", "Solution_1": "[color=blue][b]Lemma.[/b][/color] Let PQRS be a parallelogram inscribed in another parallelogram ABCD, such that $P \\in AB, Q \\in BC, R \\in CD, S \\in DA$. Let the lines AQ, CP meet at a point M and the lines AR, CS at a point N. Then the lines $BN \\parallel DM$ are parallel to the lines $SP \\parallel QR$.\r\n\r\nUsing parallel projection, project the parallelogram PQRS inscribed in the parallelogram ABCD into a rectangle P'Q'R'S' inscribed in a rectangle A'B'C'D' and skip the prime notation. The right angle triangles $\\triangle PQB \\cong \\triangle RSD$ are congruent and the right angle triangles $\\triangle QRC \\cong \\triangle SPA$ are also congruent. In addition, the the 2 triangle pairs are similar. Let $QU, SV$ be lines parallel to the rectangle sides $AB \\parallel CD$ and $RW$ a line parallel to the rectangle sides $BC \\parallel DA$. Let the points $K \\equiv QU \\cap RW$ and $L \\equiv SV \\cap RW$. The rectangles $BVLW \\cong QCRK$ are congruent and they have parallel diagonals $BL \\parallel QR$. Let the line AR, BL, CS meet the lines BC, CA, AB at points E, F, G, respectively. Put a cross-hair through the point F, intersecting the rectangle sides AB, BC, DA at points W', V', U'. Denote $AB = CD = h_0,\\ BC = DA = w_0$ and $BW = QK = VL = CR = h_1$, $BV = WL = AS = QC = KR = UD = w_1$ the heights and widths of the appropriate rectangles. We have\r\n\r\n$\\frac{GA}{GB} = \\frac{AS}{BC} = \\frac{w_1}{w_0}$\r\n \r\n$\\frac{EB}{EC} = \\frac{AB}{CR} = \\frac{h_0}{h_1}$\r\n \r\nThe rectangles $BV'FW' \\sim BVLW \\cong QCRK$ are similar, hence,\r\n\r\n$\\frac{FV'}{FW'} = \\frac{LV}{LW} = \\frac{h_1}{w_1}$.\r\n\r\nSince the point F is on the line CA, we also have\r\n\r\n$\\frac{FW'}{FU'} = \\frac{BC}{AB} = \\frac{w_0}{h_0}$\r\n\r\nConsequently,\r\n\r\n$\\frac{FC}{FA} = \\frac{FV'}{FU'} = \\frac{FV'}{FW'} \\cdot \\frac{FW'}{FU'} = \\frac{h_1 w_0}{w_1 h_0}$\r\n\r\n$\\frac{GA}{GB} \\cdot \\frac{EB}{EC} \\cdot \\frac{FC}{FA} = \\frac{w_1}{w_0} \\cdot \\frac{h_0}{h_1} \\cdot \\frac{h_1 w_0}{w_1 h_0} = 1$\r\n\r\nThus the lines $AE \\equiv AR,\\ CS \\equiv CG,\\ BL \\equiv BF$ are concurrent. But since $N \\equiv AR \\cap CS$, this is the concurency point and as a result, $QR \\parallel BL \\equiv BN$.\r\n\r\n\r\nLet $\\triangle ABC$ be a right angle triangle with the angle $\\angle A = 90^\\circ$. Let BN, CM be arbitrary cevians of this triangle intersecting at a point G in its interior, $M \\in AB, N \\in AC$. Let O be the midpoint of the hypotenuse BC. Reflect the figure in the point O (i.e., rotate it by $180^\\circ$ around the point O). The result is a rectangle $ABA'C$ with the inscribed parallelogram MNM'N', where the lines BN, CM meet at the point G and the lines CM', BN' meet at a point G'. According to the above lemma, the lines $AG' \\parallel A'G \\parallel MN' \\parallel NM'$ are parallel. From a number of similar triangles formed by the lines BC, BN, CM, MN cut by the parallels $AG' \\parallel A'G \\parallel MN' \\parallel NM'$, it quickly follows these lines are cut in the same ratios. For example, let the parallel AG' cut the lines BC, BN, CM, MN at points P, U, V, X and let the parallel A'G cut the lines BC, MN at a point Q, Y. Then form BP = CQ, BQ = CP and MX = NY, MY = NX, we quickly find that\r\n\r\n$\\frac{BU}{BN} = \\frac{CG}{CM},\\ \\ \\frac{UG}{BN} = \\frac{GV}{CM},\\ \\ \\frac{GN}{BN} = \\frac{VM}{CM}$\r\n\r\nand it is clear that UG = VG iff BN = CM. In other words, the triangle $\\triangle GUV$ is isosceles iff BN = CM. This triangle is isosceles iff its base line $AG' \\equiv UV$ is perpendicular to the bisector of the angle $\\angle UGV$ and parallel to the bisector of the complementary angle $\\angle BGC$, which is then identical with the line A'G, where A' is a reflection of the vertex A in the midpoint O of the hypotenuse BC, i.e., the diametrally opposite point of the vertex A on the triangle circumcircle (O). Even simpler is to consider one of the congruent triangles $\\triangle BNM' \\cong \\triangle CN'M$ with the sides BN, BM' = CM and CM, CN' = BN. The 1st triangle is isosceles with BN = BM' = CM iff its base line NM' is perpendicular to the bisector of the angle $\\angle NBM' = \\angle NGC$ and parallel to the bisector of the complementary angle $\\angle BGB = 180^\\circ - \\angle NGC$, i.e., iff the lines A'G and the bisector of the angle $\\angle BGC$ are identical.", "Solution_2": "It is useless to suppose that ABC is right-angled. \r\nIf P is the reflection of A wrt the midpoint of BC, clearly, the triangles PBN, PCM and ABC have the same area.\r\nSo, the distance from P to the line BN is 2.Area(ABC)/BN and the distance from P to the line CM is 2.Area(ABC)/CM \r\nHence, if BN = CM, P lies on a bisector of (BN,CM)", "Solution_3": "The problem was also solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=38363 .\r\n\r\n darij", "Solution_4": "Jpe, you try to solve the my \"dear\" problem mentioned at \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=57850", "Solution_5": "Ph-An introduced the right angle here in an attempt to camouflage the fact that \r\n\r\n BN and CM are common internal tangents of 2 circles. Just move one of them \r\n\r\n parallel to the other towards the midpoint. Angle bisectors direction will not change.\r\n\r\n Now it is not hard to see fixed points.\r\n\r\n\r\n Thank you.\r\n\r\n\r\n M.T." } { "Tag": [], "Problem": "list three times in which a properly working clock's minute and hour hand makes an angle of 180 degrees.", "Solution_1": "[hide]Minute hand moves six degrees per minute. Hour hand moves 1/2 of a degree per minute. Minute hand catches up 5.5 degrees every minute. \n\nIn the second hour, the minute hand needs to catch up 60 degrees then move ahead 180 degrees for a total of 240 degrees. The number of minutes needed to do this is\n\n$\\frac {240}{5.5} = 43 \\frac 7{11}$\n\nSo at $2:43$ and $\\frac 7{11}$ minutes, the hour hand and the minute hand will form a 180 degree angle.\n\nSix o'clock is another time when they form a 180 degree angle.\n\nIn the first hour, the hour hand needs to catch up 30 and move ahead 180 for a total of 210. \n\n$\\frac {210}{5.5} = 38 \\frac 2{11}$\n\nAnother time, then, is one o'clock and thirty-eight and two-elevenths minutes.[/hide]", "Solution_2": "[hide]\n6:00\n7:05:27.27\n8:10:54.54\n[/hide]" } { "Tag": [ "6th edition" ], "Problem": "Hi all,\r\n\r\nPlease click on \"Watch this topic link\" on the right upper (or lower) side of the topic so that you will be informed when updates about this contest appear. \r\n\r\nFor now problems of Round 1 have been posted.\r\n\r\nThe forum for discussions (asking questions about notations, or whatever, is [url=http://www.mathlinks.ro/Forum/index.php?f=382]MathLinks Contest 6th Edition, Fall 2005[/url].", "Solution_1": "Second round problems have been posted.", "Solution_2": "Very good. I was waiting too long. :D", "Solution_3": "in problem 2, $a_k = x_k$, right?", "Solution_4": "[quote=\"perfect_radio\"]in problem 2, $a_k = x_k$, right?[/quote]Yeah, that's correct :blush: The file was modified.", "Solution_5": "Where can I find the problems of the second round ?Please reply. Thanks", "Solution_6": "[quote=\"silouan\"]Where can I find the problems of the second round ?Please reply. Thanks[/quote]\r\nU can find it here:", "Solution_7": "[quote=\"perfect_radio\"]in problem 2, $a_k = x_k$, right?[/quote]\n[quote=\"Valentin Vornicu\"]Yeah, that's correct The file was modified[/quote]\r\n\r\nI didn't see what was modified. What about the case when $k=n$. We have to determine $a_n$ . I'm sorry but i have a problem with english, can me explane the mean of $consecutive$ :blush: Does it mean for example $a_1=p,a_2=p+1$ etc. ?", "Solution_8": "Should we give the possible values of $a_1$ in terms of $n$, or the possible values $a_1$ can take over all values of $n$?", "Solution_9": "[quote=\"Genrih\"]I didn't see what was modified. What about the case when $k=n$. We have to determine $a_n$ . I'm sorry but i have a problem with english, can me explane the mean of $consecutive$ :blush: Does it mean for example $a_1=p,a_2=p+1$ etc. ?[/quote]Yes, that's what consecutive means. If you determine $a_1$ of course you have determinated all the others :D\r\n\r\n@Saul: all possible values of $a_1$. I did not say anothing similar with \"in terms of $n$\" ;)", "Solution_10": "[quote=\"amir2\"][quote=\"silouan\"]Where can I find the problems of the second round ?Please reply. Thanks[/quote]\nU can find it here:[/quote]\r\n\r\nI cannot find it. Could you please post the adreese of the topic. Please mr Valentin.", "Solution_11": "The link is http://www.mathlinks.ro/Forum/contest.php", "Solution_12": "[quote=\"Valentin Vornicu\"]Yes, that's what consecutive means. If you determine $a_1$ of course you have determinated all the others :D\n\n[/quote]\r\nThank you", "Solution_13": "Ok, a few more typos in problem 2. Nothing big, as I assume everyone has understood the problem as they should have. Namely the new statement is \r\n\r\n[quote]Let $a_1,a_2,\\ldots,a_{n-1}$ be $n-1$ consecutive positive integers [b]in increasing order[/b] such that $\\displaystyle k { n \\choose k } \\equiv 0 \\pmod {a_k}$, for all $k\\in\\{1,2,\\ldots,n-1\\}$. Find the possible values of $a_1$.[/quote]", "Solution_14": "Round 3 problems have been posted.", "Solution_15": "Round 4 problems have been posted.", "Solution_16": "Does K have 5 elements, or do the subsets in F have 5 elements?", "Solution_17": "$K$ has 5 elements.", "Solution_18": "Round 5 has been posted. You can discuss about Round 4 now.", "Solution_19": "A small typo (which I am sure many of you noticed anyway) which was signaled by Jin: in problem 2, $x_1,x_2,\\ldots, x_n$ are integers, not real numbers. The PDF is corrected now.", "Solution_20": "Round 6 has been posted :)", "Solution_21": "The statement of problem 2 was updated as follows: \r\n\r\n[quote=\"New statement\"]A $n\\times n$ matrix is filled with non-negative real numbers such that on each line and column the sum of the elements is 1. Prove that one can choose $n$ positive entries from the matrix, such that each of them lies on a different line and different column.[/quote].\n\nAnd also problem 3, a small typo (thanks perfect_radio) \n\n[quote=\"Problem 3\"]Let $\\mathcal{C}_1$, $\\mathcal{C}_2$ and $\\mathcal{C}_3$ be three circles, of radii 2, 4 and 6 respectively. It is known that each of them are tangent exteriorly with the other two circles. Let $\\Omega_1$ and $\\Omega_2$ be two more circles, each of them tangent to all of the 3 circles above, of radius $\\omega_1$ and $\\omega_2$ respectively.\n\nProve that $\\omega_1 + \\omega_2 = 2\\omega_1 \\omega_2$.[/quote]", "Solution_22": "Round 7 has been posted. Good luck!", "Solution_23": "A new edition will start soon. Check the main portal page for details!", "Solution_24": "[url=https://artofproblemsolving.com/community/c1997017_mathlinks_contest_6th]here [/url] is the related post collection" } { "Tag": [], "Problem": "How many positive integer factors of $ 2^2 \\times 3^2 \\times 5$ are multiples of 12?", "Solution_1": "12 = 2^2*3.\r\n\r\nThere are 4 factors of 15. So, 15 is our answer.", "Solution_2": "The correct answer is 4 on FTW.", "Solution_3": "sorry to revive this old thread but can someone tell me what is the right answer ?", "Solution_4": "Hello,\n\n$ 12=2^{2}3$. Therefore we get as possible solutions:\n$2^{2}3=12$\n$2^{2}*3*5=60$\n$2^{2}3^{2}=36$ and\n$2^{2}3^{2}*5=180$.\n\nThus we have $4$ positive integer factors, which are multiplies of $12$.\n\nBest wishes from Leipzig/Germany.", "Solution_5": "Besides counting, actually, there is a faster solution. 12 requires a 2, another 2, and a 3. Thus there is a 3 and 5 left. Using the method to find the factor pairs we get $(1+1)(1+1) = \\boxed {4}.$" } { "Tag": [], "Problem": "solve\r\n\r\nFind the remainder when 2^{1990} is divided by 1990.", "Solution_1": "Google is your friend. This question has been asked at least $ 5$ times:\r\n\r\n[url=http://www.mathhelpforum.com/math-help/number-theory/15484-find-remainder.html]Here[/url], [url=http://answers.yahoo.com/question/index?qid=20081110084811AA640xQ]here[/url], [url=http://answers.yahoo.com/question/index?qid=20081211041845AAZCdHD]here[/url], [url=http://www.goiit.com/posts/list/algebra-find-the-remainder-94670.htm]here[/url], and [url=http://www.pagalguy.com/forum/quantitative-questions-and-answers/852-number-system-402.html]here[/url].", "Solution_2": "[quote=\"AIME15\"]Google is your friend. This question has been asked at least $ 5$ times:\n\n[url=http://www.mathhelpforum.com/math-help/number-theory/15484-find-remainder.html]Here[/url], [url=http://answers.yahoo.com/question/index?qid=20081110084811AA640xQ]here[/url], [url=http://answers.yahoo.com/question/index?qid=20081211041845AAZCdHD]here[/url], [url=http://www.goiit.com/posts/list/algebra-find-the-remainder-94670.htm]here[/url], and [url=http://www.pagalguy.com/forum/quantitative-questions-and-answers/852-number-system-402.html]here[/url].[/quote]\r\n\r\n\r\n\r\nummmmmmmmmmmmmmmmmmm actually I got it from contests on forum.", "Solution_3": "doesn't matter, it's already been posted then if it was from a contest on the forum :wink:", "Solution_4": "It's from the 1990 RMO.", "Solution_5": "anyway I didn't see the solution.\r\n\r\nis the answer 1024.", "Solution_6": "is this a statement or question :huh: :huh:", "Solution_7": "a question", "Solution_8": "seems like it", "Solution_9": "I have a B- in english", "Solution_10": "[quote]saszs I have a B- in english\n [/quote]\r\nIsn't this spam??\r\nEnglish has nothing to do with math, espicially if ur talking about ur own grade.", "Solution_11": "[quote=\"saszs\"]anyway I didn't see the solution.\n\nis the answer 1024.[/quote]\r\n\r\nMaybe clicking the links may help?\r\n\r\nTry the first link...if you're willing to read more than two sentences?", "Solution_12": "If you're asking for the answer and you already know it, then what's the use? You don't need to share these stuff. AoPS is mainly for posted problems. And there's an emphasis on problems. Cause you don't have to post if you don't have a problem. Unless you're sharing something that is productive and helps us learn.", "Solution_13": "[quote=\"jonathanchou711\"]AoPS is mainly for posted problems. [/quote]\r\n\r\nEr. No." } { "Tag": [ "geometry" ], "Problem": "In this figure, $ \\overline{AFB}$ is a semicircle, $ AE$ = 4 cm, $ BC\\equal{}CD$, and $ A$, $ B$, $ C$, and $ E$ are the vertices of a square. Find the area, in square centimeters, of this figure, Express your answer in terms of $ \\pi$. \n\n[asy]pair A,B,C,D,E,F;\nA=(0,0);\nE=(10,0);\nD=(20,10);\nC=(10,10);\nB=(0,10);\nF=(-5,5);\ndraw(Circle((0,5),5));\nfill((10,0)--(10,10)--(0,10)--(0,0)--cycle,white);\ndraw(A--E--D--C--B);\ndraw((0,0)--(0,0.5));\ndraw((10,0)--(10,0.5));\ndraw((10,9.5)--(10,10));\ndraw((0,9.5)--(0,10));\ndraw((-5,5)--(-4.5,5));\nlabel(\"$A$\",A,S);\nlabel(\"$B$\",B,N);\nlabel(\"$C$\",C,N);\nlabel(\"$D$\",D,N);\nlabel(\"$E$\",E,S);\nlabel(\"$F$\",F,W);[/asy]", "Solution_1": "The radius of semicircle $ AFB$ is $ \\frac {AB}{2} \\equal{} 2$, implying that its area is $ 2\\pi$. Since $ ABCE$ is a square, its area is $ 4^2 \\equal{} 16$. Lastly, the area of triangle $ CDE$ is $ \\frac {4\\times{4}}{2} \\equal{} 8$. Adding these up, we have $ 16 \\plus{} 8 \\plus{} 2\\pi \\equal{} \\boxed{24 \\plus{} 2\\pi}$." } { "Tag": [ "LaTeX", "complementary counting" ], "Problem": "How many four-digit numbers greater than 2999 are there such that the product of the two middle digits exceeds 5?", "Solution_1": "hmm set it up this way\r\n\r\n[hide=\"hint\"]\nABCD \n\nwhere BC must be greater than 5\nand the number ABCD must be greater than 2999.\n\nSorry for not using LaTeX I feel really sleepy now :oops: \n[/hide]", "Solution_2": "[hide]100 possibilities for middle digits of numbers.\n00\n01\n02\n03\n04\n05\n06\n07\n08\n09\n10\n11\n12\n13\n14\n15\n21\n22\n31\n41\n51\n20\n30\n40\n50\n60\n70\n80\n90\nare the only possibilities that the product would be equal to or less, and this, i believe, are 29 possibilities,\nso 100-29=71, and since 9999 is the last 4-digit number, there are 71 per 1000 starting from 3000.\n3\n4\n5\n6\n7\n8\n9 so 71*7, 497, so 497 possibilities, but there are 10 possibilities for each ones place, so 4997*10=4970. \n$\\boxed{4970}$ [/hide]\r\n\r\nedited: is it right now?", "Solution_3": "[quote=\"turboturtle\"][hide]100 possibilities for middle digits of numbers.\n00\n01\n02\n03\n04\n05\n06\n07\n08\n09\n10\n11\n12\n13\n14\n15\n21\n22\n31\n41\n51\nare the only possibilities that the product would be equal to or less, and this, i believe, are 21 possibilities,\nso 100-21=79, and since 9999 is the last 4-digit number, there are 79 per 1000 starting from 3000.\n3\n4\n5\n6\n7\n8\n9 so 79*7, 553, so 553 possibilities[/hide][/quote]\n[hide=\"correction\"]\nYou should rethink your answer a bit. If 79/100 possibilities are valid, then how come only 553 of the 7000 numbers are valid?\nFor a four-digit number ABCD, A must be in [3, 9], while D must be in [0, 9]. So there are 7 possible values for A, 10 possible values for D, and 79 possible values for BC.\n7 * 10 * 79 = 5530.\n[/hide]", "Solution_4": "Both of you are incorrect.", "Solution_5": "you missed some Turboturtle", "Solution_6": "Is the answer[hide]\n497[/hide]?\r\n-jorian", "Solution_7": "nope, they didn't miss that many(that don't work)", "Solution_8": "how is the answer not 497?\r\n-jorian[/hide]", "Solution_9": "The total number of numbers greater than 2999 is 7*10*10*10=7000. The ones that don't work are the numbers with the middle digits whose product is five or less. They are:\r\n00-09: 10\r\n10-15: 6\r\n20-22: 3\r\n30-31: 2\r\n40-41: 2\r\n50-51: 2\r\n60-90: 4\r\nThere are 29 for each of the 7 thousands. 29*7=203. 7000-203= [hide]6797 numbers total.[/hide]", "Solution_10": "i think u counted wrong\r\n-jorian", "Solution_11": "What did I do wrong?", "Solution_12": "davidli, you are VERY, VERY close. You forgot about the last digit, for your penultimate step.", "Solution_13": "[quote=\"davidli\"]What did I do wrong?[/quote]\r\nTry adding 10+6+3+2+2+2+4 up again. I don't think you'll get 29. :wink:", "Solution_14": "[quote=\"easyas3.14159...\"][quote=\"davidli\"]What did I do wrong?[/quote]\nTry adding 10+6+3+2+2+2+4 up again. I don't think you'll get 29. :wink:[/quote]\r\n\r\nYeah...before making a post like that check YOUR math first. :wink:", "Solution_15": "is it 4970?\r\n-jorian", "Solution_16": "He already said that was wrong.", "Solution_17": "no, he said [hide]497[/hide]was wrong...i asked if[hide]4970[/hide]was right\r\n-jorian", "Solution_18": "[quote=\"13375P34K43V312\"]davidli, you are VERY, VERY close. You forgot about the last digit, for your penultimate step.[/quote]\r\nBUt there are only 7000 four digit numbers greater than 2999...\r\nI got [hide]4970[/hide] but it has already been said that that is wrong... I used [hide]complementary counting[/hide]...", "Solution_19": "it has to be about [hide]4970[/hide] because [hide]29%[/hide] of the time, u get a number with middle digits having a product equal to or less than 5\n\nlets check....\n\n[hide]$7000*.29=2030$\n\n$7000-2030=4970$[/hide]\r\n-jorian", "Solution_20": "Yes, that answer is correct.", "Solution_21": "[quote=\"13375P34K43V312\"]How many four-digit numbers greater than 2999 are there such that the product of the two middle digits exceeds 5?[/quote]\r\n[hide=\"Solution\"]\nproducts > 5:\n1: 4\n2: 7\n3: 8\n4: 8\n5: 8\n6: 9\n7: 9\n8: 9\n9: 9\n71 possible numbers for the 2 digits\n71*10=710 for every thousand\n710*7=$4970$ Numbers\n[/hide]", "Solution_22": "[quote=\"jhredsox\"]it has to be about [hide]4970[/hide] because [hide]29%[/hide] of the time, u get a number with middle digits having a product equal to or less than 5\n\nlets check....\n\n[hide]$7000*.29=2030$\n\n$7000-2030=4970$[/hide]\n-jorian[/quote]\r\n\r\nHow did you get 29%? Also, you have to factor in the first digit.", "Solution_23": "29 possibilities 29/100 29%\r\n-jorian", "Solution_24": "[hide]4970 is correct[/hide]", "Solution_25": "[hide]First we have to find the number of middle digits with product less than or equal to 5.\n\nWe have\n\n00, 01, 02, 03, 04, 05, 06 , 07, 08 ,09\n10, 11, 12, 13, 14, 15\n20, 21, 22\n30, 31\n40, 41\n50, 51\n60\n70\n80\n90\n\nSo, we have 29 possibilities that aren't allowed so there are 71 possibilities that are allowed.\n\nAlso, since it must be greater than 2999, the thousands digit is greater than or equal to 3, and the units digit can be anything.\n\nSo the answer is 7*71*10=4970\n\n$4970$[/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "how many zeros at the end of 1000! ?\r\nexlanation+answer please.\r\ni got 249, but my teacher says its wrong.", "Solution_1": "I don't see why it should be wrong...\r\nBut you can look here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1000%21&t=17157", "Solution_2": "thanks my teacher is probbly doing more druigs" } { "Tag": [ "linear algebra", "matrix", "induction", "superior algebra", "superior algebra solved" ], "Problem": "A=\r\n(1 -1)\r\n(1 3) compute A^n \r\n----------------------------------------------------------------\r\nA is trigonalisable \r\nA=\r\n(-1 1)(2 1)(0 1)\r\n(1 0) (0 2)(1 1) \r\ndenote T the matrix in the midle\r\nT^n = 2I+N where \r\nN=\r\n(0 1)\r\n(0 0) is nilpotent of order 2\r\nT^n =2^nI+n2^(n-1)N \r\nA^n=\r\n((-n2^(n-1)+2^n) (-n2^(n-1))\r\n((n2^(n-1)) (n2^(n-1))", "Solution_1": "You can also do this using Hamilton-Cayley's relation! :D :D\r\nOr if you want to waste time and try guessing, you can try induction! :D \r\ncheers!", "Solution_2": "Or like this:\r\nA=2I_2+B\r\nwhere B=\r\n-1 -1\r\n1 1 \r\n\r\nThen observe that B^2=O_2 so B^k=O_2\r\nnow apply Newton's formula and get that A^n=2^n*sum(k=0,n) (n,k)B^k=2^n(I_2+nB) and get what Moubinool found! :D :D\r\n\r\n(n,k)=n!/[k!(n-k)!]\r\n\r\ncheers! :D :D" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Define an infinite sequence $ x_n$ by setting $ x_1 \\equal{} 1$ and $ x_{n\\plus{}1}\\equal{}\\frac{x_n}{1\\plus{} \\sqrt{x_n}}$ for all $ n\\geqslant 1$. Show that $ x_n$ converges, and determine its limit.", "Solution_1": "[quote=\"Pythagorean12\"]Define an infinite sequence $ x_n$ by setting $ x_1 \\equal{} 1$ and $ x_{n \\plus{} 1} \\equal{} \\frac {x_n}{1 \\plus{} \\sqrt {x_n}}$ for all $ n\\geqslant 1$. Show that $ x_n$ converges, and determine its limit.[/quote]\r\n\r\n$ x_1 > 0 \\Rightarrow x_2 > 0 \\Rightarrow ... \\Rightarrow x_n > 0 \\forall n \\in N$ \r\n$ x_{n} \\minus{} x_{n \\plus{} 1} \\equal{} x_{n \\plus{} 1} \\sqrt {x_n} > 0$ , so $ x_n$ is decreasing and bounded below $ \\Rightarrow$ $ x_n$ converges , the limit exists and $ \\ge 0$ .\r\n\r\nSo let the limit $ \\equal{} L^2$ with $ L \\ge 0$, we have $ L^2 \\equal{} \\frac {L^2}{1 \\plus{} L} \\Rightarrow L^2(1 \\minus{} \\frac {1}{1 \\plus{} L}) \\equal{} 0 \\Rightarrow L \\equal{} 0$\r\n\r\nSo the limit $ \\equal{} 0$" } { "Tag": [ "trigonometry", "algebra unsolved", "algebra" ], "Problem": "this problem involves the unit circle i think. does anyone know how to solve this???\r\n\r\n\u221a3tan3x= -1\r\n\r\n\r\n\r\nonly the 3 is square rooted", "Solution_1": "Not the correct forum. You want\r\n\r\nhttp://www.artofproblemsolving.com/Forum/index.php?f=149\r\n\r\nAnyway,\r\n\r\n$\\tan 3x =-\\frac{1}{ \\sqrt{3}}$\r\n$3x = 150^{\\circ}, 330^{\\circ}$ (or the corresponding radian measures)\r\n$x = 50, 110, 170, 230, 290, 350$ (or the corresponding radian measures; I've repeatedly added $120^{\\circ}= \\frac{360^{\\circ}}{3}$ to get all the solutions in the circle.)" } { "Tag": [], "Problem": "\u0399\u03c3\u03c9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03b9 \u03bd\u03b1 \u03bc\u03b5 \u03ba\u03b1\u03c4\u03b7\u03b3\u03bf\u03c1\u03ae\u03c3\u03bf\u03c5\u03bd \u03cc\u03c4\u03b9 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03c1\u03ad\u03c6\u03c9 \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03b5 \u03ac\u03c3\u03c7\u03b5\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03c4\u03cc\u03c0\u03b9\u03ba \u03b1\u03bb\u03bb\u03ac \u03b7 \u03c0\u03b5\u03c1\u03af\u03c3\u03c4\u03b1\u03c3\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03b9\u03b4\u03b9\u03ba\u03ae \u03ba\u03b9 \u03b1\u03c2 \u03bc\u03b5 \u03c3\u03c5\u03b3\u03c7\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bd \u03cc\u03bb\u03bf\u03b9 \r\n\r\n\u03b1\u03c2 \u03b3\u03c1\u03ac\u03c8\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03c4\u03b1 \u03c3\u03c5\u03bd\u03b1\u03b9\u03c3\u03b8\u03ae\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b9 \u03c4\u03b7\u03bd \u03c3\u03c5\u03bc\u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03bc\u03b1\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 65 \u03bd\u03b5\u03ba\u03c1\u03bf\u03cd\u03c2 \u03ba\u03b9 \u03c4\u03b1 3,5 \u03b5\u03ba\u03b1\u03c4\u03bf\u03bc\u03bc\u03cd\u03c1\u03b9\u03b1 \u03c3\u03c4\u03c1\u03ad\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b1\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b3\u03ae\u03c2\r\n\r\n\r\n\u03b5\u03b9\u03bc\u03b1\u03c3\u03c4\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03bc\u03b1\u03b6\u03af \u03c4\u03bf\u03c5\u03c2", "Solution_1": "[quote=\"ambros\"]\u0399\u03c3\u03c9\u03c2 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03b9 \u03bd\u03b1 \u03bc\u03b5 \u03ba\u03b1\u03c4\u03b7\u03b3\u03bf\u03c1\u03ae\u03c3\u03bf\u03c5\u03bd \u03cc\u03c4\u03b9 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03c1\u03ad\u03c6\u03c9 \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03bc\u03b5 \u03ac\u03c3\u03c7\u03b5\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03c4\u03cc\u03c0\u03b9\u03ba \u03b1\u03bb\u03bb\u03ac \u03b7 \u03c0\u03b5\u03c1\u03af\u03c3\u03c4\u03b1\u03c3\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03b5\u03b9\u03b4\u03b9\u03ba\u03ae \u03ba\u03b9 \u03b1\u03c2 \u03bc\u03b5 \u03c3\u03c5\u03b3\u03c7\u03c9\u03c1\u03ae\u03c3\u03bf\u03c5\u03bd \u03cc\u03bb\u03bf\u03b9 \n\n\u03b1\u03c2 \u03b3\u03c1\u03ac\u03c8\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03c4\u03b1 \u03c3\u03c5\u03bd\u03b1\u03b9\u03c3\u03b8\u03ae\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b9 \u03c4\u03b7\u03bd \u03c3\u03c5\u03bc\u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7 \u03bc\u03b1\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 65 \u03bd\u03b5\u03ba\u03c1\u03bf\u03cd\u03c2 \u03ba\u03b9 \u03c4\u03b1 3,5 \u03b5\u03ba\u03b1\u03c4\u03bf\u03bc\u03bc\u03cd\u03c1\u03b9\u03b1 \u03c3\u03c4\u03c1\u03ad\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b1\u03bc\u03ad\u03bd\u03b7\u03c2 \u03b3\u03ae\u03c2\n\n\n\u03b5\u03b9\u03bc\u03b1\u03c3\u03c4\u03b5 \u03cc\u03bb\u03bf\u03b9 \u03bc\u03b1\u03b6\u03af \u03c4\u03bf\u03c5\u03c2[/quote]\r\n\r\n \u0391\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c3\u03c4\u03b7\u03bd \u03ba\u03b1\u03bc\u03ad\u03bd\u03b7 \u03c0\u03b5\u03c1\u03b9\u03bf\u03c7\u03ae \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c4\u03b7\u03c2 \u0393\u0384\u03bb\u03c5\u03ba\u03b5\u03af\u03bf\u03c5 \u03ae \u03c0\u03bf\u03c5 \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03bf\u03cd\u03bd \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc , \u03b1\u03c2 \u03bc\u03b1\u03c2 \u03c3\u03c4\u03b5\u03af\u03bb\u03bb\u03bf\u03c5\u03bd PM \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c7\u03b1\u03c1\u03af\u03c3\u03bf\u03c5\u03bc\u03b5 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03bc\u03b5\u03bb\u03ad\u03c4\u03b7 \u03c4\u03bf\u03c5\u03c2 \u03ae \u03c4\u03b9\u03c2 \u03b5\u03be\u03b5\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2. \u0391\u03bb\u03bb\u03ac \u03ba\u03b1\u03b9 \u03cc\u03c0\u03bf\u03b9\u03b1 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1 \u03c0\u03b5\u03c1\u03bd\u03ac\u03b5\u03b9 \u03b1\u03c0\u03cc \u03c4\u03bf \u03c7\u03ad\u03c1\u03b9 \u03bc\u03b1\u03c2 , \u03b8\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c3\u03c6\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03c7\u03b1\u03c1\u03ac.\r\n\r\n [b]\u03a3\u03b7\u03bc\u03b5\u03af\u03c9\u03c3\u03b7[/b]\r\n\r\n \u0391\u03bc\u03b2\u03c1\u03cc\u03c3\u03b9\u03b5 , \u03c4\u03ad\u03c4\u03bf\u03b9\u03b5\u03c2 \u03c0\u03b1\u03c1\u03b5\u03bc\u03b2\u03ac\u03c3\u03b5\u03b9\u03c2 \u03cc\u03c7\u03b9 \u03bc\u03cc\u03bd\u03bf \u03b4\u03b5 \u03c7\u03b1\u03bb\u03ac\u03bd\u03b5 \u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03b1\u03bb\u03bb\u03ac \u03b1\u03bd\u03c4\u03af\u03b8\u03b5\u03c4\u03b1 \u03bc\u03b1\u03c2 \u03b4\u03af\u03bd\u03bf\u03c5\u03bd \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bc\u03b1\u03c3\u03c4\u03b5 \u03bc\u03b5 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03cc\u03c4\u03b5\u03c1\u03bf \u03c0\u03ac\u03b8\u03bf\u03c2. \u0394\u03b9\u03cc\u03c4\u03b9 \u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03cc\u03bb\u03b1 \u03b1\u03bd\u03b8\u03c1\u03c9\u03c0\u03b9\u03c3\u03bc\u03cc\u03c2 !!!" } { "Tag": [ "geometry", "3D geometry", "number theory unsolved", "number theory" ], "Problem": "find the smallest number expressible as the sum of two cubes in two different ways", "Solution_1": "1. Bad statement: can a cube be negative.\r\n2. _not_ hard, just trying in worst case.\r\n3. The Ramanujan-Hardy-story is well known, and so is the taxi number.\r\n4. I lock this thead for the enormous lack of a sensefull title (and I will lock others if it happens again).\r\n5. I need a prime number of points." } { "Tag": [ "function", "search", "induction", "algebra unsolved", "algebra" ], "Problem": "Find all functions $f: \\mathbb{N} \\longrightarrow \\mathbb{N}$ such that for all $m,n \\in \\mathbb{N}$,\r\n\r\n$(2^m+1)f(n)f(2^mn)=2^m f(n)^2 +f(2^mn)^2+(2^m-1)^2n$\r\n\r\n[hide]\nIt is easy to prove that for all primes $p$ we have $f(p)=p+1$, also $f(2^m)=2^m+1$ and $f(1)=2$, so intuition tells me ( :D ) that function $f(n)=n+1$ is the only one, though I don't know how to \"make the final blow\" having informations above :? \n[/hide]", "Solution_1": "I believe the answer is:\r\n\\[f(2^mn)=p+2^mq,\\] where $m$ is any nonnegative integer, $n$ is any positive odd integer, and $p,q$ are any positive integers that satisfy $pq=n$.\r\n\r\nThus the function is not unique; i.e., it can vary according to $p$ and $q$.", "Solution_2": "I know that [tex]2f(n)>f(2n)[/tex] implies [tex]f(n)=n+1[/tex]. Homever I was unable to prove that [tex]2f(n)>f(2n)[/tex] ...", "Solution_3": "I believe that $f(2n)<2f(n)$ does hold. However, I do not see how it implies $f(n)=n+1$, as my solution above also satisfies the relation.", "Solution_4": "Pity the fool ! My proof is wrong. I'll try to fix it.\r\n(Don't botter with the useless things I said before)", "Solution_5": "[b]k2c901_1[/b]: do you have full proof?", "Solution_6": "Yes, I do. I post it a little bit later...", "Solution_7": "After manipulating the original equation we see that \\[(2^mf(n)-f(2^mn))(f(2^mn)-f(n))=(2^m-1)^2n.\\] Note that the two factors in the LHS have a sum that is a positive integer, and hence must both be positive. Setting $n=pq$ for any odd integer $n$ and positive integers $p,q$, setting $m=1$ above yields \\[(2f(n)-f(2n))(f(2n)-f(n))=pq.\\] We see that it is possible for $f(n)=p+q$ for any $p,q$ by simply setting $(2f(n)-f(2n))=p$ and $(f(2n)-f(n))=q$. Now replacing $f(n)=p+q$ in the first equation, and setting $a=(2^mf(n)-f(2^mn))$ and $b=(f(2^mn)-f(n))$, it is easy to see that either $a=(2^m-1)p$ or $a=(2^m-1)q$ for all $m$. Substitution then gives $f(2^mn)=p+2^mq, q+2^mp$. However, replacing $n$ by $2n$ in the original equation, and setting $f(2n)=p+2q$ (since $p$ and $q$ are symmetric), we see that only $f(2^mn)=p+2^mq$ is possible. It is easy to check that this solution is consistent with the initial condition.", "Solution_8": "It has been posted before. You can search it in the in the forum.", "Solution_9": "[quote=\"k2c901_1\"] it is easy to see that either $ a \\equal{} (2^m \\minus{} 1)p$ or $ a \\equal{} (2^m \\minus{} 1)q$ for all $ m$. [/quote]\r\n\r\nHow is it easy? I don't see...", "Solution_10": "I told you that a solution is posted [url=http://www.mathlinks.ro/viewtopic.php?search_id=2146709638&t=6268]here[/url]:\r\n\r\nNotice that:\r\n\\[ (2^mf(n) \\minus{} f(2^mn))(f(2^mn) \\minus{} f(n)) \\equal{} n(2^m \\minus{} 1)^2\r\n\\]\r\nSo the value of $ f(2^mn)$ can be computed by the value of $ f(n)$. Now by setting $ m \\equal{} 1$ and $ n$ to be an odd number we see that we can set\r\n\\[ f(n) \\equal{} g(n) \\plus{} \\frac n{g(n)}\r\n\\]\r\nwhere $ g(n)$ is an arbitrary divisor of $ n$. So we get by induction on $ m$ that\r\n\\[ f(2^nm) \\equal{} 2^mg(n) \\plus{} \\frac n{g(n)}\r\n\\]\r\nand the problem is solved because every natural number can be written as product $ 2^mn$ where $ n$ is an odd number." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "The$ 2^n$ rows of a$ 2^n . n$ table are filled with all the different n-tuples of the numbers+1,-1. Then some numbers were replaced by 0-s.Prove that one can choose a nonemptyset of rows so that the sum of all the numbers written in them is 0", "Solution_1": "I think this was proven by dzeta a while back.", "Solution_2": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6196&highlight=]Here[/url]'s the link." } { "Tag": [ "Support" ], "Problem": "which $ \\ce{H}$ of Slicylic Acid is most acidic knowing that this reaction does take place\r\nSalicylic acid + acetyl chloride -> Aspirin", "Solution_1": "I think you would probably look at the resonance stabilized structures when each hydrogen is deprotonated.", "Solution_2": "yes that gives H attached to the OH group i smore acidic sinve the EWG attached to benzene ring stabilizes it but the case is not the same for -COOH but\r\nMe-COOH is more acidic than Phenol because of greater stabilization of Carboxylate ion by Resonance\r\nso does that play a role here", "Solution_3": "carboxylic acids are more acidic than phenols as the negative charge is delocalised on the oxygens.\r\nBenzoic acid is more acidic than phenol.\r\nSo, the hydrogen of -COOH must be more acidic.", "Solution_4": "obviously the h of cooh is more acidic. \r\nregarding the reaction\r\n[quote]\nSalicylic acid + acetyl chloride -> Aspirin\n[/quote]\r\nthere is no way you can predict the acidities of the hydrogens using this reaction as this is not an acid base reaction", "Solution_5": "yes this is not a acid base reaction.\r\nbut is -COOH is more acidic than -OH then why does the oxygen of -COOH attack the C of acetyl chloride?", "Solution_6": "well , the solvent over here(water usually) acts as a base and so cooh mainly exists as coo-( i know it is not very acidic, but it is more acidic than oh) which makes it better nucleophile than oh (which exists mainly as oh). so cooh reacts with c of acetyl chloride. also it is less sterically hindered than oh\r\n\r\nsorry this is wrong.\r\n oye pardesi you confused me in fact oh group reacts with acetyl chloride( also esterification is usually, not always tho, acid catalyzed so the above explanation is wrong).", "Solution_7": "Wait, I believe aspirin formation is an esterification process so shouldn't it be the oxygen atom of the hydroxyl group rather than the carboxy group?\r\n\r\nP.S. This seems to support the view that the $ \\minus{}COOH$ proton is more acidic:\r\n\r\nhttp://www.chemheritage.org/EducationalServices/pharm/asp/acid03.htm", "Solution_8": "neways wat does this have to do with acidity. it is based on the nucleophilicity of oh or cooh", "Solution_9": "The bottom of the link I posted indicates an acid base reaction where $ \\minus{}\\ce{COOH}$ is deprotonated.", "Solution_10": "[quote=\"Pardesi\"]but is -COOH is more acidic than -OH then why does the oxygen of -COOH attack the C of acetyl chloride?[/quote]\r\n\r\nThat's precisely the reason! If $ \\minus{}CO_2H$ is more acidic than $ \\minus{}OH$, that means $ \\minus{}CO_2^\\minus{}$ is a weaker base than $ \\minus{}O^\\minus{}$, and this also means that $ \\minus{}O^\\minus{}$ is a better nucleophile than $ \\minus{}CO_2^\\minus{}$.", "Solution_11": "yes that is ...\r\ni was looking more from a probablistic point of view", "Solution_12": "By the way an intramolecular hydrogen bond forms in Salicylic Acid so that might explain why the $ \\ce{\\minus{} OH}$ is harder to deprotonate.", "Solution_13": "Why an intramolecular hydrogen bond implies that deprotonation is harder?" } { "Tag": [ "algebra", "polynomial", "geometry", "3D geometry", "search", "algebra open" ], "Problem": "For arbitrary $ x$ being integer polynomial $ P(x)\\equal{}ax^2 \\plus{} bx \\plus{}c$ is a perfect square.\r\n\r\nProve that there exist positive integers $ d$ and $ e$ for which $ P(x)\\equal{}(dx \\plus{} e)^2$", "Solution_1": "1) How can this problem appear in the open question section???? :wink: \r\nPlease look of what \"open question\" means!\r\n\r\n2) There are many problems in this forum where it is proved that if $ P(x)$ is a square, a cube, etc... for all integers, or for some judiciously chosen integers, etc... then $ P(x)$ is itself a square, a cube, etc...\r\nTry a search!\r\n\r\nPierre." } { "Tag": [ "AMC" ], "Problem": "One can holds $ 12$ ounces of soda. What is the minimum number of cans to provide a gallon ($ 128$ ounces) of soda?\r\n\r\n$ \\textbf{(A)}\\ 7 \\qquad\r\n\\textbf{(B)}\\ 8 \\qquad\r\n\\textbf{(C)}\\ 9 \\qquad\r\n\\textbf{(D)}\\ 10 \\qquad\r\n\\textbf{(E)}\\ 11$", "Solution_1": "128/12=10 2/3. rounding up, we get 11." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "x, y and z are positive reals such that xyz>=1\r\nProve that ((x^5-x^2)/(x^5+y^2+z^2))+((y^5-y^2)/(y^5+z^2+x^2))+((z^5-z^2)/(z^5+x^2+y^2))>=0\r\nFrom imo 2005", "Solution_1": "see here http://www.mathlinks.ro/Forum/viewtopic.php?p=281573#p281573" } { "Tag": [ "geometry", "ratio", "USAMTS", "AMC", "AIME" ], "Problem": "A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=609[/img]\r\n\r\n$ \\textbf{(A) } 15 \\qquad \\textbf{(B) } 17 \\qquad \\textbf{(C) } \\frac {35}{2} \\qquad \\textbf{(D) } 18 \\qquad \\textbf{(E) } \\frac {55}{3}$", "Solution_1": "[hide]\n\n*HINT* Use ratio of triangles, knowing that the triangles have the same height. Therefore, the ratio of the area of the triangles is equal to the ratio of the area. I'll elaborate more when I have time to redraw their diagram.\n\n[/hide]", "Solution_2": "Can someone post a solution to this problem? It was the only problem that I missed :(", "Solution_3": "This was a USAMTS problem a few years back. It surprises me it is on the 10.\r\n\r\nAnyway,\r\n[hide]\n$(D)$\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=623[/img]\n\nRearrange the image as shown. The ratio of the areas of two triangles who share an altitude is equal to the ratio of their bases. Thus,\n\\[ \\frac{x}{y} = \\frac{7}{7} = 1 \\implies x = y \\]\nNow, since $\\triangle{DFE}$ and $\\triangle{FEC}$ share an altitude, the ratio of their areas is also equal to 1 so $[DEF] = 3$.\nUsing the same concept, $\\triangle{DEB}$ shares an altitude with $\\triangle{DCE}$ and also $\\triangle{AEB}$ shares an altitude with $\\triangle{ACE}$. Thus,\n\\[ \\frac{[BED]}{[EDC]} = \\frac{BE}{EC} = \\frac{[AEB]}{[ACE]} \\]\nSubstituting,\n\\[ \\frac{10+[BED]}{10} = \\frac{[BED]}{6} \\]\n\\[ 60 = 4[BED] \\]\n\\[ 15 = [BED] \\]\n\nThe area of the quadrilateral is then $15 + 3 = \\boxed{18}$.[/hide]\r\n\r\nCould someone verify this? I don't have the answer, so I just want to be sure.", "Solution_4": "[quote=\"SnowStorm\"]Could someone verify this? I don't have the answer, so I just want to be sure.[/quote]\r\n[hide=\"I get the same answer\"]\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=632[/img]\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=630[/img]\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=631[/img]\n\\begin{eqnarray*} u+3 &=& v \\\\ 7u &=& 3v+21 \\end{eqnarray*}\nSolving these two simultaneous equations:\n\\begin{eqnarray*} u &=& 15/2 \\\\ v &=& 21/2 \\\\ u+v &=& 18 \\end{eqnarray*}\n[/hide]", "Solution_5": "During the competition I solved it this way:\r\n\r\n[hide]Using SnowStorm's diagram, connect $BF$. Let the area of $\\triangle {BEF}$ be $X$ square units and let the area of $\\triangle {BFD}$ be $Y$ units. \n\nNow $\\triangle {DFA}$ and $\\triangle {FCA}$ both share the same altitude, so the ratio of their bases is the same as the ratio of their bases.\n\nThus $DF = FC = x$ for some positive integer $x$. Similarly we have $EF = 3y$ and $FA = 7y$ for some positive integer $y$. \n\n$\\triangle {BCF}$ and $\\triangle {BFD}$ share the same altitude so the ratio of their areas is the ratio of their bases. Therefore $\\frac{X + 3}{Y} = \\frac{x}{x} = 1 \\Rightarrow Y = X + 3$.\n\n$\\triangle {BFA}$ and $\\triangle {BEF}$ also share the same altitude, so the ratio of their areas is also the ratio of their bases. So $\\frac{X}{Y + 7} = \\frac{3y}{7y} = \\frac{3}{7} \\Rightarrow 7X = 3Y + 21$.\n\nSubstituting $Y = X + 3$ in the last equation, $7X = 3(X + 3) + 21$, $4X = 30$, $X = 7.5$. \n\nSo the area of quadrilateral $BEFD = X + Y = X + (X + 3) = 2X + 3 = 2(7.5) + 3 = 15 + 3 = 18$, \n\nhence $(D)$.[/hide]", "Solution_6": "Lol it's exactly the same is rcv's solution... I didn't see his. :blush:", "Solution_7": "[quote=\"aidan\"]During the competition I solved it this way:\n\n[hide]\nNow $\\triangle {DFA}$ and $\\triangle {FCA}$ both share the same altitude, so the ratio of their bases is the same as the ratio of their bases.\n[/hide]\n[/quote]\r\n\r\nTypo? lol :P", "Solution_8": "[quote=\"SnowStorm\"]This was a USAMTS problem a few years back.[/quote]\r\n\r\nAlso see [url=http://www.kalva.demon.co.uk/aime/aime85.html]AIME 1985, Problem 6[/url], which I believe was the inspiration for that USAMTS problem. ;)", "Solution_9": "a really easy solution lies in mass points chess64 :)", "Solution_10": "[hide]\nLet the biggest triangle be $\\triangle ABC$, going around clockwise with $A$ in the lower-left hand corner. Let the important points on $\\overline{AB}$ and $\\overline{BC}$ be $D$ and $E$ respectively, and let the intersection of $\\overline{AE}$ and $\\overline{CD}$ be $X$. Last, let $AB=c$, $AC=b$, $BC=a$, $AD=d$, $CE=e$, and let the area of polygon $PQR$ be $[PQR]$ (for any number of points $P,Q,R$).\n\nConsider $\\triangle ACX$ and $\\triangle XCE$. With respect to $\\overline{AE}$, they have the same height and area, so we have $AX=XE$. Drawing in $\\overline{DE}$, we also have $[DEX]=[ADX]=3$. Now, using $\\triangle AEC$ and $\\triangle ABC$ we have $\\frac ae=\\frac{[ABC]}{[AEC]}=\\frac{[ABC]}{14}$. From $\\triangle ADC$ and $\\triangle ADE$, we also have $\\frac{[ADE]}{[ADC]}=\\frac{6}{10}=\\frac{a-e}a=1-\\frac ea=1-\\frac{14}{[ABC]}$, so $\\frac{14}{[ABC]}=1-\\frac6{10}=\\frac25$, which means that $[ABC]=14\\frac52=35$. Finally, $[DBEX]=[ABC]-[ADX]-[AXC]-[XEC]=35-3-7-7=\\fbox{18}$.[/hide]", "Solution_11": "Spreading the joy of mass points :D \r\n\r\n[hide]Draw triangle $ABC$ with cevians $CD$ and $BE$. Let the intersection point of this cevian be $P$. Assign the following mass points:\n\n$B_m = 1$\n\nSince $[BCP] = [ECP]$, $EB$ is bisected by cevian $CD$. So, \n\n$E_m = 1$\n$P_m = B_m + E_m = 2$\n\nWe try to use the ratio of $DP$ to $PC$ by comparing the areas of $BDP$ and $CPB$. Let the mass point of $D$ be $7x$ and the mass point of $C$ be $3x$. Since $7x + 3x = 2$ because of point $P$, we have $x = \\frac{1}{5}$. So, the mass points of:\n\n$C_m = \\frac{3}{5}$\n$D_m = \\frac{2}{5}$\n\nWe will now draw cevian $AF$ which goes through point $P$ so as to split the quadrilateral into two triangles. The mass point of $F$ is $B_m + C_m = \\frac{8}{5}$. As a result, the mass point of $A$ is $F_m - P_m = \\frac{2}{5}$. The area of $ADP$ is solved by:\n\n$1 * 3 = \\frac{2}{5} * [ADP]$\n\nand $AEP$ by:\n\n$\\frac{3}{5} * 7 = \\frac{2}{5} * [AEP]$\n\nSolving for them, we get the areas to be $\\frac{15}{2}$ and $\\frac{21}{2}$, respectively. By adding, we get:\n\n$\\boxed{18 \\implies D}$[/hide]", "Solution_12": "can someone explain how rcv made the relationships he did about the areas of the various triangles? there's probably some theorem I don't know.", "Solution_13": "It's all of the same-base, same-altitude stuff from AoPS. Basically, just give variables to a few of the lengths that matter (as bases) and the altitudes, then write the equations for areas and you'll see the relationship.", "Solution_14": "[quote=\"crptone\"]Can someone post a solution to this problem? It was the only problem that I missed :([/quote]\r\n\r\nMe too. \r\n\r\nI did it using rcv's method at home in about 3 mins. If only I had seen it earlier.", "Solution_15": "Mod Edit: Please keep this kind of stuff to PM.", "Solution_16": "What's a \"mod edit?\"", "Solution_17": "It's when a moderator edits another person's post. The previous poster quoted nearly the entire thread to say something that should've been in a PM, therefore his post got deleted." } { "Tag": [], "Problem": "Each double desk is for 2 students. There are 30 students and 15 desks in a class. Given that half of the girls sit with the boys. Is it possible to rearrange the sitting plan so that half of the boys sit with the girls?", "Solution_1": ":wink: I didn't get the problem", "Solution_2": "[hide]\nSince [i]half[/i] the girls are sitting with boys, there must be an even number of girls. There must be either $ 4n$ or $ 4n \\plus{} 2$ girls. Let there be $ 4n \\plus{} 2$ girls. Then $ 2n \\plus{} 1$ girls sit with the boys, and $ 2n \\plus{} 1$ girls sit together. However, it's impossible for $ 2n \\plus{} 1$ girls to sit in pairs. Then there are $ 4n$ girls. Similarly, there must be $ 4m$ boys for it to be possible for the rearrangement to be possible. However, there are $ 30 \\minus{} 4n$ boys, which is congruent to $ 2\\mod 4$. Therefore, it is impossible to rearrange it so that half the boys are sitting with girls.\n[/hide]" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "There are $ n$ people and there are $ 6$ comissions. Each comission has at least $ n/4$ members. Prove that there are two comissions and a group at least $ n/30$ members belonging to this two comissions.", "Solution_1": "It has been post before and the solution is:\r\nSuppose that any two comissions has less than $ \\frac{n}{30}$ common members.By using the principle of inclusion and exclusion we have ; $ 6 \\frac n4\\minus{}\\binom{6}{2} \\frac {n}{30}\\plus{}...>n$ and it's contradiction." } { "Tag": [ "induction", "algebra unsolved", "algebra" ], "Problem": "$f: [0,1]\\rightarrow\\mathbb{R}$ is continous.and:\r\n1)$\\forall{x}\\in[0,1], f(x)\\geq0$\r\n2)$f(1)=1$\r\n3)if $x,y,x+y\\in[0,1]$ then $f(x)+f(y)\\leq{f(x+y)}$\r\nfind the minimum c such that if $x\\in[0,1]$ then $f(x)\\leq{cx}$", "Solution_1": "c=2,Prove the following lemma by indution:\r\n$\\forall n,\\forall x\\in[\\frac{1}{2^{n}},\\frac{1}{2^{m-1}}],we have f(x)\\leq 2x$( for n=1,it is triavial)\r\nfor c<2,it not true ,just let $f(x)-\\to \\frac12,x\\geq \\frac12,f(x)=0,x<\\frac12$", "Solution_2": "thank hawk tiger", "Solution_3": "[quote=\"Hawk Tiger\"]for c<2,it not true ,just let $f(x)-\\to \\frac12,x\\geq \\frac12,f(x)=0,x<\\frac12$[/quote]\r\nI dont understand this step.", "Solution_4": "Sorry,I am wrong.\r\nIt should be $f(x)=1,x>\\frac12,f(x)=0,x\\leq \\frac12$\r\nI have a bad memory...", "Solution_5": "[quote=\"Hawk Tiger\"]Sorry,I am wrong.\nIt should be $f(x)=1,x>\\frac12,f(x)=0,x\\leq \\frac12$\nI have a bad memory...[/quote]\r\nf shoud be continous.", "Solution_6": "[quote=\"ali666\"][quote=\"Hawk Tiger\"]Sorry,I am wrong.\nIt should be $f(x)=1,x>\\frac12,f(x)=0,x\\leq \\frac12$\nI have a bad memory...[/quote]\nf shoud be continous.[/quote]\r\n\r\nI think he means $f(x)=0 \\,\\forall\\, x\\leq \\frac{1}{2}-\\epsilon$, $f(x)=1 \\,\\forall\\, x\\geq \\frac{1}{2}+\\epsilon$ and $f(x)=\\frac{1}{2}+\\frac{1}{2\\epsilon}\\left(x-\\frac{1}{2}\\right) \\,\\forall\\, \\frac{1}{2}-\\epsilon\\leq x\\leq \\frac{1}{2}+\\epsilon$ for some arbitrarily small $\\epsilon$", "Solution_7": "[quote=\"weiquan\"][quote=\"ali666\"][quote=\"Hawk Tiger\"]Sorry,I am wrong.\nIt should be $f(x)=1,x>\\frac12,f(x)=0,x\\leq \\frac12$\nI have a bad memory...[/quote]\nf shoud be continous.[/quote]\n\nI think he means $f(x)=0 \\,\\forall\\, x\\leq \\frac{1}{2}-\\epsilon$, $f(x)=1 \\,\\forall\\, x\\geq \\frac{1}{2}+\\epsilon$ and $f(x)=\\frac{1}{2}+\\frac{1}{2\\epsilon}\\left(x-\\frac{1}{2}\\right) \\,\\forall\\, \\frac{1}{2}-\\epsilon\\leq x\\leq \\frac{1}{2}+\\epsilon$ for some arbitrarily small $\\epsilon$[/quote]\r\nI don't mean that(what I mean is wrong :D ).But your example is vailid." } { "Tag": [ "geometry", "circumcircle", "modular arithmetic", "geometric transformation", "homothety", "geometry proposed" ], "Problem": "Dear everyone.\r\n\r\nprove the following problem.\r\n\r\nif from any point O on the circumcircle of triangle ABC, straight lines OA', OB', OC' be inflected to meet\r\nBC, CA, AB in A', B', C' respectively, and to make with them equal angles,\r\nthen the three points A', B', C' are collinear.\r\n\r\nI think it is easy but not bad.\r\nthanks.", "Solution_1": "[quote=\"77ant\"]\n[b]and to make with them equal angles[/b]\n[/quote]\r\n\r\nI cant understand what you mean...?\r\nCan you tell it more clearly???\r\n\r\nThank you \r\nDimitris", "Solution_2": "The normal projections of O on the sides determine the [b] Simson Line[/b]; if, instead of perpendicular we draw lines making the same $ \\angle \\alpha$ with the sides, the property is still valid. The proof is quite similar as for Simson line.\r\n\r\nBest regards,\r\nsunken rock", "Solution_3": "We present two proofs for the problem.\r\n\r\n[b]First solution.[/b]\r\nUsing directed angles, we have:\r\n$ (A'B',A'C')\\equiv (A'B',A'O) \\plus{} (A'O,A'C')$\r\n$ \\equiv (CA,CO) \\plus{} (BO,BA)$\r\n$ \\equiv (CA,CO) \\plus{} (OC,OB) \\plus{} (BO,BA) \\plus{} (OB,OC)$\r\n$ \\equiv (AC,AB) \\plus{} (OB,OC)$\r\n$ \\equiv (AC,AB) \\plus{} (AB,AC)$\r\n$ \\equiv 0 \\pmod{\\pi}$\r\nTherefore $ A',B',C'$ are collinear, QED.\r\n\r\n[b]Second solution.[/b]\r\nUsing the fact about Simson's line. Let $ A_0,B_0,C_0$ be the orthogonal projection of $ O$ on $ BC,CA,AB$ then $ A_0,B_0,C_0$ are collinear (Simson's line). Since $ (OA',BC) \\equiv (OB',CA) \\equiv (OC',AB) \\pmod{\\pi}$, the three triangles $ \\triangle OA_0A',\\triangle OB_0B',\\triangle OC_0C'$ are similar in the same direction. Thus, consider the similar $ \\mathcal{S}$ takes $ A_0\\mapsto A',B_0\\mapsto B',C_0\\mapsto C'$, we have the desired result.\r\n\r\nWith the second solution, we can still point out the line which is the image of line $ \\overline{A'B'C'}$ through the homothety $ \\mathcal{H}(O,k \\equal{} 2)$ passes through $ H\\equiv \\mathcal{S}(H)$ where $ H$ is the orthocenter of $ \\triangle ABC$. Moreover, $ H'$ is the point satisfies that $ (AH',BC)\\equiv (BH',CA)\\equiv (CH',AB) \\pmod{\\pi}$. This is an extension of Steiner's line." } { "Tag": [ "geometry", "3D geometry", "algebra solved", "algebra" ], "Problem": "The sequence {a_n,k} = 1, 2,..., 2^n; n = 0,1,2,... is defined by \r\n\r\n(1) a_0,1 = 2,\r\n(2) a_n,k = 2 * (a_(n-1),k)^3 and\r\n(3) a_n,(k+2^(n-1)) = 1/2 * (a_(n-1),k)^3.\r\n\r\nProve that all a_n,k are distinct.\r\n\r\n[i]German IMO selection test 1979 problem 6.[/i]", "Solution_1": "I didn't check if my solution is wrong, but I think that the idea is the following:\r\n Suppose that there exists m,n,k,p such that a_nk=a_mp. Let us take m and n such that m+n is minimum with the property that there exists k<=2^n and p<=2^m with a_nk=a_mp. We have three cases:\r\n Case 1\r\n If k<=2^n-1 and p<=2^m-1, then it follows that 2a^3_n-1,k=2a^3_m-1,p and so there exits k<=2^n-1 and p<=2^m-1 with a_n-1,k=a_m-1,p, contradicting the minimality of m+n.\r\n Case 2\r\n If k>2^n-1 and p>2^n-1, then write k=k_1+2^n-1 and p=p_1+2^m-1. From the recursive relation it follows that a_n-1,k_1=a_m-1,p_1 and k_1<=2^n-1, p_1<=2^m-1, contradiction.\r\n Case 3\r\n If for example k>2^n-1 and p<=2^m-1 then it follows that 4 is a perfect cube, false." } { "Tag": [], "Problem": "Examine the following list of events that occur during the propagation of a nerve impulse:\r\n\r\nI - Neurotransmitters reach the dendrites\r\nII - Neurotransmitters are released by the end of the axon.\r\nIII - The impulse spreads the axon.\r\nIV - The impulse spreads by dendrites.\r\nV - The impulse reaches the cell body.\r\n\r\nAlternative that presents the chronology of events correct ?\r\n\r\na) V-III-I-IV-II\r\nb) I-IV-V-III-II\r\nc) I-IV-III-II-V\r\nd) II-I-IV-III-V\r\ne) II-III-I-IV-V\r\n\r\n\r\nThanks.", "Solution_1": "Anyone knows?", "Solution_2": "the answer shud b (e)\r\n\r\nbut first check out if ur (III) is gramatically correct :)", "Solution_3": "The answer is B.\r\n\r\nThanks ." } { "Tag": [ "geometry", "parallelogram" ], "Problem": "A point is taken on each side of a parallelogram and the points on adjacent sides are joined by line segments so as to form a quadrilateral.\r\nIf the area of this quadrilateral is half the area of the parallelogram prove that at least one diagonal of the quadrilateral is parallel to a side of the parallelogram", "Solution_1": "[quote=\"Michael Niland\"]A point is taken on each side of a parallelogram and the points on adjacent sides are joined by line segments so as to form a quadrilateral.\nIf the area of this quadrilateral is half the area of the parallelogram prove that at least one diagonal of the quadrilateral is parallel to a side of the parallelogram[/quote]\r\n\r\n[b]HINT[/b]\r\n\r\n If $LN$ is not parallel to $AD,BC$, then we draw $MR \\parallel AB$ , so $( RLMN) = \\frac{1}{2}(ABCD) =(KLMN)$.\r\n This means that $(KLN) = (KRN)$ and hence $KR \\parallel LN$ , contradiction!. So points $K,R$ coincide , that is $MK \\parallel AB$" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c > 0$. Prove that:\r\n$ \\sqrt {(a^2 b \\plus{} b^2 c \\plus{} c^2 a)(b^2 a \\plus{} c^2 b \\plus{} a^2 c)} \\ge abc \\plus{} \\sqrt [3]{(a^3 \\plus{} abc) \\plus{} (b^3 \\plus{} abc) \\plus{} (c^3 \\plus{} abc)}$", "Solution_1": "[quote=\"Shinraehee\"]Let $ a,b,c > 0$. Prove that:\n$ \\sqrt {(a^2 b \\plus{} b^2 c \\plus{} c^2 a)(b^2 a \\plus{} c^2 b \\plus{} a^2 c)} \\ge abc \\plus{} \\sqrt [3]{(a^3 \\plus{} abc) \\plus{} (b^3 \\plus{} abc) \\plus{} (c^3 \\plus{} abc)}$[/quote]\r\nI think it's should be \\[ \\sqrt {(a^2 b \\plus{} b^2 c \\plus{} c^2 a)(b^2 a \\plus{} c^2 b \\plus{} a^2 c)} \\ge abc \\plus{} \\sqrt [3]{(a^3 \\plus{} abc)(b^3 \\plus{} abc)(c^3 \\plus{} abc)}\\]\r\nSee here: http://www.mathlinks.ro/viewtopic.php?t=166168" } { "Tag": [], "Problem": "The molecular weight of an unknown substance was determined to be approximately 94 g/mol. Knowing that each molecule of the substance contains a single oxygen atom, identify the substance.", "Solution_1": "phenol in the monomer state among others", "Solution_2": "And what was the method you used to arrive at that conclusion?", "Solution_3": "I think VARUNARASU guessed since the molar mass of benzene is 78..............................................................", "Solution_4": "Actually phenol is the only possible answer, but the purpose was not to guess.", "Solution_5": "Carcul,can you post more of those ABCD type questions in organic and inorganic?? (please)", "Solution_6": "subtract 16 from 94 to get 78. now consider all compounds with the weight 78 (C6H6 being one of them) divide by 12 to get 6 which corr to the no of c atmos and and the rest are hydrogen and you get the the formula c6h6. Now other compounds could include nitrogen, but the no of nitrogen atoms cannot be odd as that would lead to a fractional du and hence impossible compound(for eg. C5NOH4) presence of two or more nitrogen would lead to a compound with more hydrogen atoms than permissible and also a fractional DU(for eg C4N2OH2 or C3N2OH14). Similarly you can test the presence of other elements. This Is done of course on the assumption that only normal isotopes are present in the compound(for eg, you can have a compound with formula C5NOH2D or C3N2OD7 )so only molecules with formula C6H6O have mass 94", "Solution_7": "wats your method??", "Solution_8": "Well, this is in fact a basic problem of spectroscopy. For a given molar mass M of an organic compound, we start by writting M in the form\r\n\r\n$ \\frac{M}{13} \\equal{} n \\plus{} \\frac{m}{13}$,\r\n\r\nand we then consider our \"base formula\" as $ C_nH_{n \\plus{} m}$. In our case we get n = 7 and m = 3, and so our base formula will be $ C_7H_{10}$. As we have an oxygen, then for adding it we must subtract for the base formula a combination of C and H with the same mass: $ CH_4$. After doing this, we get the formula of our compound $ C_6H_6O$. The most stable compound with this formula is phenol.", "Solution_9": "give us more sums like this carcul", "Solution_10": "The reasoning I gave above is known as the \"Rule of Thirteen\". Here's one more example:\r\n\r\n-Find the molecular formulas for possible organic compounds with molecular masses of 136, in the following cases:\r\n\r\n(a) A compound with two oxygen atoms.\r\n(b) A compound with two nitrogen atoms.\r\n(c) A compound with two nitrogen atoms and one oxygen atom.\r\n(d) A compound with five carbon atoms and four oxygen atoms.\r\n\r\n(For those interested, there are still 3 more interesting problems like this.)", "Solution_11": "is the answer for a part C8H8O2", "Solution_12": "Yes, that's right. Some possible structures?" } { "Tag": [ "ratio", "geometry", "perimeter", "LaTeX", "Pythagorean Theorem" ], "Problem": "You guys will probably need a diagram for this...\r\n\r\nA square ABCD has line segmends drawn from vertex B to the midpoints N and M of sides AD and DC respectively. Find the ratio of the perimeter of quadrilateral BMDN to the perimeter of square ABCD.\r\n\r\nWant a hint? :)\r\n\r\n[hide]Use the Pythagorean Theorem to find some side lengths of quadrilateral BMDN.[/hide]", "Solution_1": "Funny how you can draw a diagram and realize it probably didn't help you very much at all...\r\n[hide=\"solution\"]let the side length of the square be 2x. the perimeter is 8x\nND and MD are x, and by pythagorean theorem BN and BM are xsqrt5.\n(x+x+xsqrt5+xsqrt5)/8x simplifies to (1+sqrt5)/4\n[/hide]", "Solution_2": "[hide]Let the side length be 2. $BN=BM=\\sqrt{5}$.\nPerimeter of BMDN is $2+2\\sqrt{5}$.\nPerimeter of ABCD is 8.\nThe ratio is $\\frac{2+2\\sqrt{5}}{8}=\\frac{1+\\sqrt{5}}{4}$.[/hide]", "Solution_3": "[quote]Funny how you can draw a diagram and realize it probably didn't help you very much at all... [/quote]\r\n\r\nWell, when I drew one, it wuz like my life depended on it...lol :rotfl:", "Solution_4": "[hide]It works out to be$(x(sq5)+x)/4x=(1+sq(5))/4$[/hide]", "Solution_5": "diagram helped me a lot >_>\r\n[hide]\n$\\frac{1 + \\sqrt{5}}{4}$\n[/hide]", "Solution_6": "Let the side be 4. [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/8edac681582cee5938f72fe0bcb232a4.gif[/img], therefore[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/3e1015a8640b40a4abccacf8fc2bc7ab.gif[/img]. Because BN=BM, perimeter ratio would be[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f50d415280176822a4014a3a2843e591.gif[/img]", "Solution_7": "[quote=\"Knuth's_check\"]Let the side be 4. [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/8edac681582cee5938f72fe0bcb232a4.gif[/img], therefore[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/3e1015a8640b40a4abccacf8fc2bc7ab.gif[/img]. Because BN=BM, perimeter ratio would be[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f50d415280176822a4014a3a2843e591.gif[/img][/quote]\r\n[hide]$\\frac{1+\\sqrt{5}}{4}$[/hide]", "Solution_8": "[quote=\"Knuth's_check\"][hide]Let the side be 4. [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/8edac681582cee5938f72fe0bcb232a4.gif[/img], therefore[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/3e1015a8640b40a4abccacf8fc2bc7ab.gif[/img]. Because BN=BM, perimeter ratio would be[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f50d415280176822a4014a3a2843e591.gif[/img] [/hide][/quote]\r\n\r\n1. it can be simplified ;) \r\n2. Please hide your answers and solutions so others will have a chance to do the problems.\r\n3. I think you are using the TeXeR to type up things in LaTeX. If you want to use LaTeX on the forums, just put your code in between two dollar signs.", "Solution_9": "[hide]$\\frac{\\sqrt5+1}{4}$[/hide]" } { "Tag": [ "algebra", "IMO Shortlist", "functional equation" ], "Problem": "Find all functions $ f: \\mathbb{R}\\to\\mathbb{R}$ such that $ f(x+y)+f(x)f(y)=f(xy)+2xy+1$ for all real numbers $ x$ and $ y$.\n\n[i]Proposed by B.J. Venkatachala, India[/i]", "Solution_1": "Given $x=y=0 \\Rightarrow (f(0))^2=1$\r\n$*)f(0)=1$\r\nGiven$y=0\\Rightarrow f(x)\\equiv 1 \\forall x$\r\n$)f(0)=-1$\r\nGiven $x=-y$we have:\r\n $f(x)f(-x)-1=f(-x^2)-2x^2+1$(*)\r\nGiven $x=1$we have\r\n$f(-1)(1-f(1))=0\\Rightarrow f(-1)=0$or$f(1)=1$\r\n$+)f(1)=1$\r\nGiven y=1,we have:\r\n $f(x+1)=2x+1$\r\n $\\Rightarrow f(x)=2x-1 \\forall x$\r\n$+)f(-1)=0$\r\nGiven $y=-1$,we have:\r\n$f(-x)=f(x-1)+2x-1$\r\nGiven$y=-1-x$we have:\r\n$f(x)(f(x)+2x+1)=f(x^2+x-1)$\r\nGiven$x=1$we have:\r\n$f(1)=f(1)(f(1)+3)$\r\n$\\Rightarrow f(1)=0$or$f(1)=-2$\r\n.)$f(1)=0$\r\nGiven$y=1$we have\r\n$f(x+1)=f(x)+2x+1$\r\n$\\Rightarrow f(2)=3$\r\nGiven$y=2$we have:\r\n$f(x+2)+3f(x)=f(2x)+4x+1$\r\n$f(x+2)=f(x)+4x+1 \\Rightarrow f(2x)=4f(x)+3$\r\nGiven $x=x+1,y=x-1$we have:\r\n$f(2x)+f(x-1)f(x+1)=f(x^2-1)+2x^2-1$\r\nBut $f(x^2-1)=f(x^2)\\2x^2+1)$\r\n$\\Rightarrow 4f(x)+3+(f(x)-2x+1)(f(x)+2x+1)=f(x^2)$\r\n$\\Rightarrow (f(x))^2+6f(x)+4-4x^2=f(x^2)$(1)\r\nGiven $x=y$we have:\r\n$f(2x)+(f(x))^2=f(x^2)+2x^2+1$\r\n$\\Rightarrow f(x^2)=(f(x))^2+4f(x)+2-2x^2$(2)\r\n(1)(2)$\\Rightarrow 2f(x)+2-2x^2=0$\r\n$\\Rightarrow f(x)=x^2-1\\forall x$\r\n.)Similar with $f(1)=-2$ we have:$f(x)=-x-1$\r\nThe function satisfly:\r\n$f(x)=2x-1\\forall x$\r\n$f(x)=x^2-1\\forall x$\r\n$f(x)=-x-1\\forall x$", "Solution_2": "[quote=\"vietnamesegauss89\"]\n$\\Rightarrow 4f(x)+3+(f(x)-2x+1)(f(x)+2x+1)=f(x^2)$\n[/quote]\r\nHow you get that ? \r\nI get $4f(x)+3+(f(x)-2x+1)(f(x)+2x+1)=f(-x^2)$\r\n\r\nAnyone help me please!", "Solution_3": "Is there a shorter solution?", "Solution_4": "can anyone explain how he could get :\r\n\r\nGiven $y=-1-x$ we have , $f(x)(f(x)+2x+1)=f(x^2+x-1)$\r\n\r\nthanks", "Solution_5": "Hi Here you have found as follows\r\n[quote=\"vietnamesegauss89\"]Given$y=-1-x$we have:\n$f(x)(f(x)+2x+1)=f(x^2+x-1)$\nGiven$x=1$we have:\n$f(1)=f(1)(f(1)+3)$\n$\\Rightarrow f(1)=0$or$f(1)=-2$[/quote]\r\n$\\Rightarrow f(1)=0$or$f(1)=-2$. you used \"or\" but it could be the same or I am wrong. explain it how to discover that they are different \r\nAbdurashid", "Solution_6": "They are allowed to be the same; the or is not exclusive in that scenario. \r\n\r\nHere's my solution: \r\n\r\nLet $ P(x,y)$ be the assertion that $ f(x \\plus{} y) \\plus{} f(x)f(y) \\equal{} f(xy) \\plus{} 2xy \\plus{} 1$. \r\n\r\n$ P(0,0)$ yields $ f(0) \\equal{} \\pm 1$. If $ f(0) \\equal{} 1$, then $ P(x,0)$ yields $ f(x) \\equal{} 1$ for all $ x$, which $ P(3,3)$ shows is impossible, so $ f(0) \\equal{} \\minus{} 1$. \r\n\r\n$ P(x, \\minus{} x)$ now gives $ f(x)f( \\minus{} x) \\equal{} f( \\minus{} x^2) \\minus{} 2x^2 \\plus{} 2$. Setting $ x \\equal{} 1$ yields $ f(1)f( \\minus{} 1) \\equal{} f( \\minus{} 1)$, so $ f(1) \\equal{} 1$ or $ f( \\minus{} 1) \\equal{} 0$. \r\n\r\nIf $ f(1) \\equal{} 1$, then $ P(x,1)$ yields $ f(x \\plus{} 1) \\plus{} f(x) \\equal{} f(x) \\plus{} 2x \\plus{} 1$, so $ f(x \\plus{} 1) \\equal{} 2x \\plus{} 1$, so $ f(x) \\equal{} 2x \\minus{} 1$ for all real $ x$. It can easily be seen that $ f(x) \\equal{} 2x \\minus{} 1$ indeed satisfies this functional equation. \r\n\r\nSuppose now that $ f(1) \\neq 1$; then $ f( \\minus{} 1) \\equal{} 0$. $ P(1, \\minus{} 2)$ yields $ f( \\minus{} 1) \\plus{} f(1)f( \\minus{} 2) \\equal{} f( \\minus{} 2) \\minus{} 3$, that is, $ f(1)f( \\minus{} 2) \\equal{} f( \\minus{} 2) \\minus{} 3$. $ P( \\minus{} 1,2)$ gives $ f(1) \\plus{} f( \\minus{} 1)f(2) \\equal{} f( \\minus{} 2) \\minus{} 3$, that is, $ f(1) \\equal{} f( \\minus{} 2) \\minus{} 3$. Therefore, $ f(1) \\equal{} f(1)f( \\minus{} 2)$, so $ f(1) \\equal{} 0$ or $ f( \\minus{} 2) \\equal{} 1$. \r\n\r\n$ f(1) \\equal{} 0$ will give $ f(x) \\equal{} x^2 \\minus{} 1$. $ P(x, \\minus{} 1)$ and $ P( \\minus{} x,1)$ yield $ f(x \\minus{} 1) \\equal{} f(1 \\minus{} x) \\equal{} f( \\minus{} x) \\minus{} 2x \\plus{} 1$, so $ f$ is even. $ P(x, \\minus{} x)$, combined with the fact that $ f$ is even, yields $ f(x)^2 \\equal{} f(x^2) \\minus{} 2x^2 \\plus{} 2$, while $ P(x,x)$ gives $ f(2x) \\plus{} f(x)^2 \\equal{} f(x^2) \\plus{} 2x^2 \\plus{} 1$. Since $ f(x)^2 \\equal{} f(x^2) \\minus{} 2x^2 \\plus{} 2$, we see that $ f(2x) \\equal{} 4x^2 \\minus{} 1$, so $ f(x) \\equal{} x^2 \\minus{} 1$ for all $ x$; it can easily be verified that $ f(x) \\equal{} x^2 \\minus{} 1$ satisfies the functional equation. \r\n\r\nIf $ f( \\minus{} 2) \\equal{} 1$, then $ f(1) \\equal{} f( \\minus{} 2) \\minus{} 3 \\equal{} \\minus{} 2$. $ P(1, \\minus{} x)$ yields $ f(1 \\minus{} x) \\equal{} 3f( \\minus{} x) \\minus{} 2x \\plus{} 1$, while $ P( \\minus{} 1,x)$ yields $ f(x \\minus{} 1) \\equal{} f( \\minus{} x) \\minus{} 2x \\plus{} 1$. Subtracting these two equations gives $ 2f( \\minus{} x) \\equal{} f(1 \\minus{} x) \\minus{} f(x \\minus{} 1)$. Substituting $ x \\equal{} 1 \\minus{} y$ yields $ 2f( \\minus{} 1 \\plus{} y) \\equal{} f(y) \\minus{} f( \\minus{} y)$. Replacing $ y$ with $ \\minus{} y$ yields $ 2f( \\minus{} 1 \\minus{} y) \\equal{} f( \\minus{} y) \\minus{} f(y) \\equal{} \\minus{} 2f( \\minus{} 1 \\plus{} y)$, so $ f( \\minus{} 1 \\plus{} y) \\equal{} \\minus{} f( \\minus{} 1 \\minus{} y)$ for all real $ y$. \r\n\r\n$ P(x, \\minus{} 1)$ yields $ f(x \\minus{} 1) \\equal{} f( \\minus{} x) \\minus{} 2x \\plus{} 1$, and $ P( \\minus{} x, \\minus{} 1)$ yields $ f( \\minus{} x \\minus{} 1) \\equal{} f(x) \\plus{} 2x \\plus{} 1$. Since $ f( \\minus{} x \\minus{} 1) \\equal{} \\minus{} f(x \\minus{} 1)$, we have that $ f(x \\minus{} 1) \\equal{} \\minus{} f(x) \\minus{} 2x \\minus{} 1 \\equal{} f( \\minus{} x) \\minus{} 2x \\plus{} 1$, so $ f(x) \\plus{} f( \\minus{} x) \\equal{} \\minus{} 2$ for all real $ x$. \r\n\r\n$ P(0,0)$ yields $ f(x)f( \\minus{} x) \\equal{} f( \\minus{} x^2) \\minus{} 2x^2 \\plus{} 2$, so $ f(x)( \\minus{} 2 \\minus{} f(x)) \\equal{} \\minus{} 2 \\minus{} f(x^2) \\minus{} 2x^2 \\plus{} 2$. Thus, $ 2f(x) \\plus{} f(x)^2 \\equal{} f(x^2) \\plus{} 2x^2$. On the other hand, $ P(x,x)$ yields $ f(2x) \\plus{} f(x)^2 \\equal{} f(x^2) \\plus{} 2x^2 \\plus{} 1$. Subtracting $ 2f(x) \\plus{} f(x)^2 \\equal{} f(x)^2 \\plus{} 2x^2$ gives $ f(2x) \\minus{} 2f(x) \\equal{} 1$. \r\n\r\n$ f(2) \\equal{} \\minus{} 2 \\minus{} f( \\minus{} 2) \\equal{} \\minus{} 3$, so $ P(x,2)$ yields $ f(x \\plus{} 2) \\minus{} 3f(x) \\equal{} f(2x) \\plus{} 4x \\plus{} 1$. But $ f(2x) \\equal{} 2f(x) \\plus{} 1$, so $ f(x \\plus{} 2) \\equal{} 5f(x) \\plus{} 4x \\plus{} 2$. On the other hand, $ P(x,1)$ gives $ f(x \\plus{} 1) \\equal{} 3f(x) \\plus{} 2x \\plus{} 1$; substituting $ x \\equal{} y \\plus{} 1$ here yields $ f(y \\plus{} 2) \\equal{} 3f(y \\plus{} 1) \\plus{} 2y \\plus{} 3$. But $ f(y \\plus{} 1) \\equal{} 3f(y) \\plus{} 2y \\plus{} 1$, so $ f(y \\plus{} 2) \\equal{} 9f(y) \\plus{} 8y \\plus{} 6$. However, $ f(y \\plus{} 2) \\equal{} 5f(y) \\plus{} 4y \\plus{} 2$ as well, so $ 4f(y) \\equal{} \\minus{} 4y \\minus{} 4$, yielding $ f(y) \\equal{} \\minus{} y \\minus{} 1$ for all real $ y$. It can easily be verified that $ f(x) \\equal{} \\minus{} x \\minus{} 1$ satisfies the functional equation. \r\n\r\nHence, our three solutions to this functional equation are $ \\boxed{f(x) \\equal{} 2x \\minus{} 1, x^2 \\minus{} 1, \\minus{} x \\minus{} 1}$.", "Solution_7": "This is my solution from WOOT some time ago...\n\nLet $P(x,y)\\implies f(x+y)+f(x)f(y)=f(xy)+2xy+1$. First, note that $P(1,-1)\\implies f(-1)[f(1)-1]=0$.\n\n[b]Case 1:[/b] $f(1)=1$. Then\n\\[P(x,1)\\implies f(x+1)=2x+1\\implies f(x)=2x-1\\]for all $x$.\n\n[b]Case 2:[/b] $f(1)\\ne1\\implies f(-1)=0$. Note that $P(x,0)\\implies f(0)=-1$ (otherwise, $f(x)$ is constant, which is clearly impossible since $2xy$ is surjective). Now let $c=1-f(1)$ (by assumption, $c\\ne0$). Then\n\\[P(x,1)\\implies f(x+1)=cf(x)+2x+1.\\]The second order difference gives us\n\\[f(x+3)-(c+2)f(x+2)+(2c+1)f(x+1)-cf(x)=0\\]for all $x$.\n\n[b]Subcase 2.1:[/b] If $c=1$ (i.e. $f(1)=0$), so $f(x+1)=f(x)+2x+1$ for all $x$ and we find by simple induction that $f(n)=n^2-1$ and $f(x+n)=f(x)+(x+n)^2-x^2$ for all integers $n$, whence\n\\[P(x,\\pm n)\\implies f(-xn)=f(xn)+1=n^2[f(x)+1].\\]Thus $f$ is even, and\n\\[P(x,x)\\implies f(x^2)=[f(x)]^2+4f(x)-2x^2+2\\]while\n\\[P(x,-x)\\implies f(x^2)=f(-x^2)=[f(x)]^2+2x^2-2.\\]Equating, we arrive at\n\\[f(x)=x^2-1,\\]which indeed satisfies the original equation.\n\n[b]Subcase 2.2:[/b] If $c\\ne1$ (i.e. $f(1)\\ne0$), then the characteristic polynomial of the sequence $f(n)$ has roots $1,1,c$ and\n\\[f(n)=\\alpha+\\beta n+\\gamma c^n\\]for all integers $n$ (for some real constants $\\alpha,\\beta,\\gamma$). Considering $f(-1),f(0),f(1)$, we find that\n\\[0=\\alpha-\\beta+\\frac{\\gamma}{c},\\quad -1=\\alpha+\\gamma,\\quad 1-c=\\alpha+\\beta+\\gamma c.\\]Solving this system,\n\\[f(n)=-\\frac{c+1}{(c-1)^2}+\\frac{2n}{1-c}-\\frac{c(c-3)}{(c-1)^2}c^n.\\]Now,\n\\[P(2,2)\\implies f(2)=\\pm3.\\]If $f(2)=3$, then\n\\[3=f(2)=cf(1)+2(1)+1=cf(1)+3\\implies cf(1)=0.\\]But we have $c\\ne0$ and $f(1)\\ne0$. So $f(2)=-3$, and\n\\[-3=f(2)=cf(1)+2(1)+1\\implies [f(1)-3][f(1)+2]=0.\\]\n[b]Subcase 2.2.1:[/b] $f(1)=3\\implies c=-2$. Then\n\\[f(n)=\\frac{1}{9}+\\frac{2n}{3}-\\frac{10}{9}(-2)^n\\]for all integers $n$, so $f(2)=-3$, $f(-2)=-3/2$, and $f(-4)=-21/8$. This contradicts $P(2,-2)\\implies f(2)f(-2)=f(-2^2)$.\n\n[b]Subcase 2.2.2:[/b] $f(1)=-2\\implies c=3$. Then\n\\[f(n)=-1-n\\]for all integers $n$. By induction, we find $f(x+n)=3^nf(x)+(3^n-1)x+(3^n-n-1)$. Thus\n\\[P(x,n)\\implies f(xn)+xn+1=(3^n-n-1)[f(x)+x+1].\\]So\n\\begin{align*}\n(3^n-n-1)^2[f(x)+x+1]&=(3^n-n-1)[f(xn)+xn+1]\\\\\n&=f(xn^2)+xn^2+1=(3^{n^2}-n^2-1)[f(x)+x+1]\n\\end{align*}for all real $x$ and integers $n$. Take $n=-1$. We find\n\\[f(x)=-x-1.\\]\nFinally, our solutions are $f(x)=2x-1\\forall x$, $f(x)=x^2-1\\forall x$, and $f(x)=-x-1\\forall x$ (all three work).", "Solution_8": "[quote=\"andre.l\"]Find all functions $ f: \\mathbb{R}\\to\\mathbb{R}$ such that $ f\\left(x+y\\right)+f\\left(x\\right)f\\left(y\\right)=f\\left(xy\\right)+2xy+1$ for all real numbers $ x$ and $ y$.[/quote]\nLet $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$\nLet $f(0)=a$\nLet $f(1)=b$\nLet $f(-1)=c$\n\n(e1) : $P(x,-1)$ $\\implies$ $f(x-1)+cf(x)=f(-x)-2x+1$\n(e2) : $P(x-1,1)$ $\\implies$ $f(x)+(b-1)f(x-1)=2x-1$\n(1-b)e1 + e2 $\\implies$ $(1+c-bc)f(x)+(b-1)f(-x)=-2bx-b$\nchanging $x\\to -x$, we get $(1+c-bc)f(-x)+(b-1)f(x)=2bx-b$\n\nAnd so the system :\n$(1+c-bc)f(x)+(b-1)f(-x)=-2bx-b$\n$(b-1)f(x)+(1+c-bc)f(-x)=2bx-b$\n\nIf the determinant of the system is non zero, this gives $f(x)=ux+v$ for some real $u,v$ and plugging this in original equation, \nwe get two solutions $\\boxed{f(x)=-x-1}$ and $\\boxed{f(x)=2x-1}$\n\nIf the determinant of the system is zero, this means :\neither $1+c-bc=b-1$ and so $-2bx-b=2bx-b$ and so $b=0$\neither $1+c-bc=1-b$ and so $-2bx-b=-(2bx-b)$ and so $b=0$\n\nSo $b=0$ and :\n$P(0,1)$ $\\implies$ $b+ab=a+1$ and so $a=-1$\n$P(-1,1)$ $\\implies$ $a+bc=c-1$ and so $c=0$\nThe system above becomes then $f(x)=f(-x)$ and the function is even.\n\n$P(\\frac x2,\\frac x2)$ $\\implies$ $f(x)+f(\\frac x2)^2=f(\\frac {x^2}4)+\\frac{x^2}2+1$\n\n$P(\\frac x2,-\\frac x2)$ $\\implies$ $-1+f(\\frac x2)^2=f(\\frac {x^2}4)-\\frac{x^2}2+1$\n\nSubtracting these two lines implies $\\boxed{f(x)=x^2-1}$ which indeed is a solution", "Solution_9": "Let $P(x,y)$ the assertion : $f(x+y)+f(x)f(y)=f(xy)+2xy+1$\n$P(x,0) : (f(0)+1)(f(x)-1)=0$, we can see easily that the constant function $f \\equiv 1$ is not a solution then $f(0)=-1$\n$P(1,-1): f(-1)(f(1)-1)=0$ \n- If $f(1)=1$ then :\n$P(x,1): f(x+1)=2x+1$ which give us $\\forall x \\in \\mathbb{R} : f(x)=2x-1$ which is indeed a solution.\n- If $f(-1)=0$ : \n$P(2,-1) : f(-2)=f(1)+3$\n$P(-2,1): f(-2)(1-f(1))=3$ hence either $f(1)=0$ either $f(1)=2$\n-- If $f(1)=0$\n$P(x,1): f(x+1)=f(x)+2x+1$\n$P(x,-1):f(x-1)=f(-x)-2x+1$\n$P(-x,1):f(-x+1)=f(-x)-2x+1$ hence we get $f(-x+1)=f(x-1)$ which mean that f is even.\n$P(x,-x): (f(x))^{2}=f(x^{2})+2-2x^{2}$\n$P(x,x): f(2x)+(f(x))^{2}=f(x^{2})+2x^{2}+1$ hence $ \\forall x \\in \\mathbb{R} : f(2x)=4x^{2}-1=(2x)^{2}-1$ hence we get $\\forall x \\in \\mathbb{R}: f(x)=x^{2}-1$\n-- If $f(1)=-2$:\n$P(x,1):f(x+1)=3f(x)+2x+1$\nHence :$f(x+2)=9f(x)+8x+6 \\forall x$ (*)\nOtherwise : \n$P(x,-1): f(x-1)=f(-x)-2x+1$\n$P(-x,1): f(-x+1)-2f(-x)=f(-x)-2x+1 \\Rightarrow f(-x)=f(x-1)+2x-1=\\frac{f(-x+1)+2x-1}{3}$\nHence : $f(-x+1)=3f(x-1)+4(x-1)+2$ which mean $\\forall x \\in \\mathbb{R} f(-x)=3f(x)+4x+2$\n$P(x,-x): -1+f(x)(3f(x)+4x+2)=3f(x^{2})+4x^{2}+2$ which is equivalent to : $f(x^{2}) = \\frac{-3-4x^{2}+f(x)(3f(x)+4x+2)}{3}$ (1)\n\nFrom (*) we get $f(2)=-9+6=-3$ \n$P(x,2) f(x+2)-3f(x)=f(2x)+4x+1 \\Rightarrow f(2x)=6f(x)+4x+5$ (2)\n$P(x,x) : f(2x)+(f(x))^{2}=f(x^{2})+4x^{2}+1$ (3)\nUsing (1) +(2) +(3) we get :\n$ \\forall x \\in \\mathbb {R} : f(x)=-x-1$\nFinally the F.E have 3 solutions which are :\n$\\forall x \\in \\mathbb{R} : f(x)=-x-1 , f(x)=x^{2}-1, f(x)=2x-1$", "Solution_10": "Let $P(x,y)$ be the assertion.\nFirst see that no constant functions exist.\n$P(x,0)\\implies f(x)+f(x)f(0)=f(0)+1\\implies (f(0)+1)(f(x)-1)=0$\nAs $f$ is non constant $\\exists x$ such that $f(x)\\neq 1\\implies f(0)=-1$\n$P(1,-1)\\implies f(0)+f(1)f(-1)=f(-1)-1\\implies f(-1)(f(1)-1)=0$\nIf $f(1)=1$\n$P(x,1)\\implies f(x+1)=2x+1\\implies \\boxed{f(x)=2x-1\\text{ for all } x\\in\\mathbb{R}}$\nIf $f(-1)=0$.\n$P(-1,-1)\\implies f(-2)+f(-1)^2=f(1)+3\\implies f(-2)=f(1)+3$.\n$P(1,-2)\\implies f(-1)+f(1)f(-2)=f(-2)-3\\implies f(-2)(f(1)-1)=-3$\n$\\implies (f(1)+3)(f(1)-1)=-3$\n$\\implies f(1)^2+2f(1)=0\\implies f(1)=0\\text{ or } f(1)=-2$\n$\\text{ If }f(1)=0$\n$P(x,1)\\implies f(x+1)=f(x)+2x+1\\implies f(x-1+1)=f(x-1)+2x-2+1$\n$\\implies f(x)=f(x-1)+2x-1$\n$P(x,1)\\implies f(x-1)=f(-x)-2x+1$\nbut we have just proved $f(x)=f(x-1)+2x-1$.\nComparing these two equations, we get $f(x)=f(-x)\\text{ for all } x\\in\\mathbb{R}$\n$P(x,-y)\\implies f(x-y)+f(x)f(-y)=f(-x)-2xy+1\\text{ but as }f(x)=f(-x)$\n$f(x-y)=f(xy)-f(x)f(y)-2xy+1$\n$P(x,y)\\implies f(x+y)+f(x)f(y)=f(xy)+2xy+1$\nComparing these two equations, we get \n$f(x+y)-f(x-y)=4xy$\nTake $x=y$ to get $f(2x)=4x^2-1$, \nso, $x\\rightarrow \\frac{x}{2}\\implies \\boxed{f(x)=x^2-1\\text{ for all } x\\in\\mathbb{R}}\\text{ is another solution}$\nIf $f(1)=-2$\n$P(x,1)\\implies f(x+1)=3f(x)+2x+1\\implies f((x-1)+1)=3f(x-1)+2x-1\\implies f(x)=3f(x-1)+2x-1$\n$P(x,-1)\\implies f(x-1)=f(-x)-2x+1$\n$\\implies 3(f(x-1))=3f(-x)-6x+3=f(x)-2x+1$\n$\\implies f(x)+4x=3f(-x)+2$\nNow $x\\rightarrow -x\\implies f(-x)-4x=3f(x)+2$\nAdding these two, we get, $f(x)+f(-x)=-2$\n$P(x,-y)\\implies f(x-y)+f(x)f(-y)=f(-xy)-2xy+1$\n$\\implies f(x-y)+f(x)\\left[-2-f(y)\\right]=-2-f(xy)-2xy+1$\n$\\implies f(x-y)-2f(x)+2xy+1=f(x)f(y)-f(xy)$\nBut $P(x,y)\\implies f(x)f(y)-f(xy)=2xy+1-f(x+y)$\nTherefore, $2xy+1-f(x+y)=f(x-y)-2f(x)+2xy+1$\n$\\implies f(x-y)+f(x+y)=2f(x)$\n$x=y\\implies -1+f(2x)=2f(x)\\implies f(2x)-2f(x)=1$\nWe have already proved $f(x)=3f(x-1)+2x-1\\stackrel{x=-1}{\\implies} f(-2)=1$\nAs $f(x)+f(-x)=-2\\implies f(-2)=3$\n$P(x,2)\\implies f(x+2)+f(x).f(2)=f(2x)+4x+1\\implies f(2x)=f(x+2)-3f(x)-4x-1$\nWe have already proved \n$f(x+1)=3f(x)+2x+1$\n$\\implies f((x+1)+1)=3f(x+1)+2x+3=9f(x)+6x+3+2x+3$\n$\\implies f(x+2)=9f(x)+8x+6$\nTherefore $9f(x)+8x+6-3f(x)-4x-1=f(2x)$\nBut $f(2x)=1+2f(x)\\implies 6f(x)+4x+5-2f(x)=1\\implies 4f(x)=-4x-4$\n$\\implies \\boxed{f(x)=-x-1\\text{ for all }x\\in\\mathbb{R}}$", "Solution_11": "[hide=\"Solution\"]Easy to get $f(0)=-1$ and $(f(1)-1)f(-1)=0$. So if $f(1)=1$, then $P(x-1,1)\\implies f(x)=2x-1$ which is definitely a solution. So assume $f(-1)=0$. Let $f(1)=c$.\n\n$P(-x+1,-1)\\implies f(-x)-f(x-1)=2x-1$\n$P(x-1,1)\\implies f(x)+cf(x-1)-f(x-1)=2x-1$\nCombining these we get $f(-x)=f(x)+cf(x-1)\\quad (1)$\nSo $f(x)=f(-x)+cf(-x-1)$\nAdding them gives $c[f(x-1)+f(-x-1)]=0$ So at least one of them is zero. Suppose $c=0$. Then (1) yields $f(x)=f(-x)$. Now subtracting $P(x,-x)$ from $P(x,x)$ gives $f(2x)=4x^2-1\\implies f(x)=x^2-1$ which is another valid solution.\n\nNow assume $f(x-1)+f(-x-1)=0\\implies f(x)+f(-x-2)=0$. So $f(-2)=-f(0)=1$. And now $P(-1,-1)$ implies $f(1)=-2$.\n$P(x,-1)\\implies f(x-1)=f(-x)-2x+1$\n$P(x,1)\\implies f(x+1)=3f(x)+2x+1$\nSo $-f(x)=f(-x-2)=f(x+1)+2x+3=3f(x)+4x+2$ $\\implies f(x)=-(x+1)$\n\nSo all the functions are $f(x)=2x-1,x^2-1,-x-1$[/hide]", "Solution_12": "Why $f(x)=1$ for all $x$ don't satisfy in this problem?", "Solution_13": "[quote=\"Goblik\"]Why $f(x)=1$ for all $x$ don't satisfy in this problem?[/quote]\nBecause with $f(x)=1$ $\\forall x$ :\n\n$LHS$ of functional equation is $2$\n$RHS$ of functional equation is $2+2xy$\n\nAnd obviously we dont have $LHS=RHS$ $\\forall x,y$", "Solution_14": "I hate casework.\n[hide=Solution]\nThe functions are $2x-1, x^2-1,$ and $-x-1$.\n\nLet $P(x, y)$ be the assertion that $f(x+y) + f(x)f(y) = f(xy) + 2xy + 1$. \n$P(0, 0)$ yields that $f(0)^2 = 1$, so $f(0) = 1$ or $f(0) = -1$. However, if $f(0) = 1$, then $P(x, 0)$ tells us that $f(x)$ is a constant function, which is clearly false. Thus, $f(0) = -1$. $P(1, -1)$ gives us $f(1)f(-1) = f(-1)$, so $f(1) = 1$ or $f(-1) = 0$. \n\n[b]Case 1: $f(1) = 1$[/b]\n$P(x - 1, 1)$ shows that $f(x) = 2x - 1$.\n\n[b]Case 2:$f(-1) = 0$[/b]\n$P(-2, 1)$ shows that $f(-2)f(1) = f(-2) - 3$ and $P(-1, -1)$ implies $f(-2) = f(1) + 3$. Putting together these two observations, we see that $f(1) = f(1)f(-2)$, so $f(1) = 0$ or $f(-2) = 1$.\n\n[b]Subcase 1:$f(1) = 0$[/b]\n$P(x - 1, 1)$ and $P(-x + 1, -1)$ tell us that $f(x) = f(-x)$. $P(-x, x)$ implies $-1 + f(-x)f(x) = f(-x^2) - 2x^2 + 1$, and substituting $f(x)$ for $f(-x)$ and $f(x^2)$ for $f(-x^2)$ gives us $f(x)^2 = f(x^2) - 2x^2 + 2$. $P(x, x)$ shows that $f(2x) + f(x)^2 = f(x^2) + 2x + 1$. Subtracting this equation from the previous one, we get $f(2x) = 4x^2 - 1$, or $f(x) = x^2 - 1$.\n\n[b]Subcase 2: $f(-2) = 1$[/b]\nWe can easily show that $f(-2) = 1$ implies that $f(1) = -2$. \n$P(-x, -1)$ gives us \n$f(-x-1) = f(x) + 2x + 1$. (1)\n $P(x, 1)$ gives us \n$f(x+1) = 3f(x) + 2x + 1$. (2)\nSubtracting, we get\n$f(x+1) - f(-x-1) = 2f(x)$, or equivalently $f(x) - f(-x) = 2f(x-1)$. Substituting $x$ for $-x$, we can easily see that $f(x-1) = -f(-x-1)$. (3)\nSubstituting $x-1$ in (2), multiply (1) by 3, and subtracting those two equations, we get $f(-x-1) = 3f(x-1) + 4x$. Using (3), we immediately get $f(-x-1) = x$, or $f(x) = -x - 1$.\n\nAll those functions work, so QED.\n[/hide]", "Solution_15": "very beatiful function.", "Solution_16": "This problem took a really long while for me... Well, here's my solution. Feedback would be greatly appreciated - it is loooong (4 o's to indicate it's pretty long, but not insane).\n[hide=\"Spoilers!\"]\nWe claim that the only possible functions $f$ are $f(x)=2x-1, f(x)=-x-1, f(x)=x^2-1$. It is easy to see that these values satisfy the problem condition.\n\nUsing $x=0, y\\in\\mathbb{R}$ is any real number, we have $f(y)(f(0)+1)=f(0)+1$.\n\nIf $f(0)\\neq -1, f(y)=1$ for all real number $y$. Choosing $x=y=10$ gives us $1+1=1+200+1$, which is clearly not valid.\nTherefore, $f(0)=-1$.\n\nUsing $x=1, y=-1$, we have $f(0)+f(1)f(-1)=f(-1)-2+1, f(-1)(f(1)-1)=0$. Therefore, $f(1)=1$ or $f(-1)=0$.\n\nIf $f(1)=1$, using $y=1$ gives us $f(x+1)+f(x)f(1)=f(x)+2x+1, f(x+1)=2x+1$ for all real $x$. Therefore, $f(x)=2x-1$ for all real $x$.\n\nIf $f(1)\\neq 1$, $f(-1)=0$ and denote $c=f(1)$.\nUsing $y=1$, we have $f(x+1)+f(x)c=f(x)+2x+1$, or $f(x+1)=(1-c)f(x)+2x+1(\\clubsuit)$ for all real $x$.\nUsing $y=-1$, we have $f(x-1)+f(x)f(-1)=f(-x)-2x+1, f(x-1)=f(-x)-2x+1$ for all real $x$.\nTherefore, $f(-x-1)=f(x)+2x+1(\\star)$ for all real $x$.\n\nMultiplying $(\\star)$ by $1-c$ and subtract $(\\clubsuit)$gives us $(1-c)f(-x-1)-f(x+1)=(2x+1)(-c)$ for any real $x$. Therefore, we also have $(1-c)f(x)-f(-x)=(2x-1)(-c)$, $(1-c)f(-x)-f(x)=(-2x-1)(-c)$ for all real $x$.\n\nIf $c\\neq 0$, Solving for $f(x)$, we get that $f(x)=\\frac{-2xc^2}{c^2-2c}-1$ for all real $x$. Using $x=1$, we get $c(c+2)(c-1)=0$. Since $c\\neq 0,1$, we get $f(x)=-x-1$.\n\nIf $c=0$, then our equation gives us $f(x)=f(-x)$. Using $y=x$, we have $f(2x)+f(x)^2=f(x^2)+2x^2+1$. Furthermore, using $y=-x$, we have $f(0)+f(x)f(-x)=f(-x^2)-2x^2+1$.\nUsing $f(x)=f(-x)$ and subtracting, we have $f(2x)=4x^2-1$ for all real $x$. Therefore, $f(x)=x^2-1$.\n\nHaving exhausted all possible value of $f(1)=c$, the only possible functions $f$ are $f(x)=x^2-1, 2x-1, -x-1$, as desired.\n[/hide]", "Solution_17": "[quote=IMO ShortList 2005 A4]Find all functions $ f: \\mathbb{R}\\to\\mathbb{R}$ such that $ f(x+y)+f(x)f(y)=f(xy)+2xy+1$ for all real numbers $ x$ and $ y$.\n\n[i]Proposed by B.J. Venkatachala, India[/i][/quote]\n\n[hide=solution] Answer: The only functions which work are $f(x)=2x-1$ for all $x$, $f(x)=x^2-1$ for all $x$ and $f(x)=-(x+1)$ for all $x$. It is clear that they indeed satisfy the desired conditions.\n\nLet $P(x,y)$ denote the given assertion and $f$ be a function which works. Let $h(x)=f(x)-x^2+1$ and $g(x)=f(x)+x+1$. I shall use them later on.\n\nNote that $P(0,0) \\Longrightarrow f(0)=\\pm 1$ so if $P(0)=1$ then $P(x, 0) \\Longrightarrow f(x)=1$ for all $x$ which is not a valid function. So $f(0)=-1$. Observe $$P(x, 1) \\Longrightarrow f(x+1)+f(x)\\cdot \\left(f(1)-1\\right)=2x+1$$ and $P(2, 2) \\Longrightarrow f(2)=\\pm 3$. Henceforth, $P(x, 1)$ is called the $(+)$. Plugging $x=1$ in $(+)$ yields $$f(1)(f(1)-1)=3-f(2) \\in \\{0, 6\\} \\Longrightarrow f(1) \\in \\{0, 1, -2, 3\\}.$$ Consider the four cases for different values of $f(1)$ as follow. \n\nFirstly, the easier ones.\n\n[b]Case 1.[/b] $f(1)=1$\n\nOur equation $(+)$ becomes $f(x+1)=2x+1 \\Longrightarrow f(x)=2x-1$ for all reals $x$ which is one of the claimed solutions.\n\n[b]Case 2.[/b] $f(1)=3$\n\nPut $x=-1$ in $(+)$ to get $f(-1)\\cdot \\left(f(1)-1\\right)=0 \\Longrightarrow f(-1)=0$. Observe $P(-1, -1) \\Longrightarrow f(-2)=f(1)+3 \\Longrightarrow f(-2)=6$ but $x=-2$ in $(+)$ gives $f(-2)\\cdot \\left(f(1)-1\\right)=-3 \\Longrightarrow f(-2)=-1.5$ so we have a contradiction! \n\nHarder cases incoming...\n\n[b]Case 3.[/b] $f(1)=0$\n\nNotice that $P(x, y)$ is equivalent to the assertion $$Q(x, y) \\overset{\\text{def}}{:=}h(xy)=h(x+y)+h(x)h(y)+(x^2-1)h(y)+(y^2-1)h(x). $$ Equation $(+)$ boils down to $f(1+x)-f(x)=2x+1 \\Longrightarrow h(1+x)=h(x)$ for all reals $x$. Notice that $$Q(1+x, y)-Q(x, y) \\Longrightarrow h(xy+y)-h(xy)=(2x+1)h(y).$$ Plugging $x=\\frac{1}{y}$ for non zero $y$ in the last equation gives $\\frac{h(y)}{y}=0$ so $h \\equiv 0$ and we get $f(x)=x^2-1$ for all reals $x$, which is again one of the stated solutions.\n\n[b]Case 4.[/b] $f(1)=-2$\n\nEquation $(+)$ is equivalent to $g(1+x)=3g(x)$ and the assertion $P(x, y)$ is equivalent to $$R(x, y) \\overset{\\text{def}}{:=} g(x+y)+g(x)g(y)=g(xy)+(1+x)g(y)+(1+y)g(x).$$ Evidently, $$R(1+x, y)-3R(x, y) \\Longrightarrow g(xy+y)-3g(xy)=(2x+1)g(y).$$ Plugging $x=\\frac{1}{y}$ we conclude $2\\left(\\frac{1}{y}-1\\right)g(y)=0$ so $g(y)=0$ for all $y \\ne 1$, which combined with $g(1)=0$ yield $g \\equiv 0$ or $f(x)=-(x+1)$ for all reals $x$, another one of the mentioned solutions.\n\nBy our rather long analysis, it is clear that these are the only valid functions. \n\nEnd of Story.\n\n[/hide]", "Solution_18": "[hide=Solution]\nAnswer: $f(x)=2x-1, f(x)=x^2-1,$ and $f(x)=-1-x$ are the three solutions.\n\nLet $P(x, y)$ denote the assertion that \\[f(x+y)+f(x)f(y)=f(xy)+2xy+1\\] for all real numbers $x, y$. Then $P(0, 0)\\implies f(0)^2=1\\implies f(0)=1 \\ \\text{or} \\ f(0)=-1$.\n\n[b]Case 1[/b]: $f(0)=1$. Then $P(x, 0)\\implies 2f(x)=2\\implies f(x)=1$, which clearly doesn't satisfy the original FE.\n\n[b]Case 2[/b]: $f(0)=-1$. Then \\begin{align*}P(1, -1)&\\implies -1+f(1)f(-1)=f(-1)-1 \\\\ &\\implies f(-1)=0 \\ \\text{or} \\ f(1)=1.\\end{align*}\n[list]\n[*][b]Subcase 2.1[/b]: $f(1)=1$. Then \\begin{align*}P(x, 1)&\\implies f(x+1)+f(x)=f(x)+2x+1) \\\\ &\\implies f(x+1)=2x+1 \\\\ &\\implies f(x)=2x-1.\\end{align*} We can easily check that $\\boxed{f(x)=2x-1}$ indeed satisfies the original equation, as \\[f(x+y)+f(x)f(y)=2x+2y-1+4xy-2x-2y+1=f(xy)+2xy+1.\\]\n[*][b]Subcase 2.2[/b]: $f(-1)=0$. Let $g(x)=f(x)+1$, and we have that $P(x, y)$ is equivalent to the assertion \\[Q(x, y):=g(x+y)+g(x)g(y)-g(x)-g(y)=g(xy)+2xy\\] for all reals $x, y$. Note that $g(0)=0$ and $g(-1)=1$. \n\n$Q(-1, -1)$ gives $g(-2)=g(1)+3$, and $Q(1, -2)$ gives \\[1+g(1)g(-2)-g(1)-g(-2)=g(-2)-4 \\implies g(1)=1 \\ \\text{or} \\ -1\\] after substituting $g(-2)=g(1)+3$.\n[list]\n[*][b]Subcase 2.2.1[/b]: $g(1)=1$. Then $Q(x, 1)\\implies g(x+1)-1=g(x)+2x$. We also have that $Q(-x, -1)\\implies g(-x-1)-1=g(x)+2x$, so $g(x+1)=g(-(x+1))$ for all $x$, and $g$ is even. But \\begin{align*}Q(x, -x)&\\implies g(x^2)-2g(x)=g(x^2)-2x^2 \\\\ &\\implies g(x)=x^2.\\end{align*} Because $f(x)=g(x)-1$, the $g(x)=x^2$ corresponds to $\\boxed{f(x)=x^2-1}$ for all real $x$. This can indeed be checked to satisfy the original FE, as \\[f(x+y)+f(x)f(y)=x^2+2xy+y^2+(xy)^2-x^2-y^2+1=f(xy)+2xy+1.\\]\n[*][b]Subcase 2.2.2[/b]: $g(1)=-1$. Then $Q(x, 1)\\implies g(x+1)=3g(x)+2x-1 \\ \\ \\ (*)$. Plugging $x=-\\tfrac{1}{2}$ into $(*)$ gives \\[g\\left(\\tfrac{1}{2}\\right)=3g\\left(-\\tfrac{1}{2}\\right)-2. \\ \\ \\ (1)\\] Replacing $x$ with $2x-\\tfrac{1}{2}$ in $(*)$ gives \\[g\\left(2x+\\tfrac{1}{2}\\right)=3g\\left(2x-\\tfrac{1}{2}\\right)+4x-2. \\ \\ \\ (2)\\] Write \\begin{eqnarray*}Q\\left(2x, -\\tfrac{1}{2}\\right)\\implies g\\left(2x-\\tfrac{1}{2}\\right)+g(2x)g\\left(-\\tfrac{1}{2}\\right)-g(2x)-g\\left(-\\tfrac{1}{2}\\right)=g(-x)-2x\\end{eqnarray*} and then use $(1), (2)$ to get \\begin{align*}Q\\left(2x, \\tfrac{1}{2}\\right)&\\implies g\\left(2x+\\tfrac{1}{2}\\right)+g(2x)g\\left(\\tfrac{1}{2}\\right)-g(2x)-g\\left(\\tfrac{1}{2}\\right)=g(x)+2x \\\\ &\\implies 3\\left[g\\left(2x-\\tfrac{1}{2}\\right)+g(2x)g\\left(-\\tfrac{1}{2}\\right)-g(2x)-g\\left(-\\tfrac{1}{2}\\right)\\right]=g(x)-2x \\\\ &\\implies 3\\left[g(-x)-2x\\right] = g(x)-2x\\ \\\\ &\\implies 3g(-x) = g(x) + 4x. \\ \\ \\ (3)\\end{align*} Replacing $x$ with $-x$ in $3g(-x)=g(x)+4x$ yields \\[3g(x)=g(-x)-4x, \\ \\ \\ (4)\\] and solving for $g(x)$ in $(3)$ and $(4)$ returns $g(x)=-x$. The corresponding function in $f$ for $g(x)=-x$ is $\\boxed{f(x)=-1-x}$, and it is easy to check that it works: \\[f(x+y)+f(x)f(y)=-1-x-y+xy+x+y+1=f(xy)+2xy+1.\\] \n\n[/list]\n[/list]\n---------\nWe are done, having exhausted all cases.\n[/hide]", "Solution_19": "Let $P(x,y)$ denote the assertion that $ f(x+y)+f(x)f(y)=f(xy)+2xy+1$.\n$$P(x,0) \\implies (1+f(0))(f(x))=1+f(0) \\implies f(0)=-1$$\nIf $f(1)=1$, then $P(x-1,1) \\implies f(x)=2x-1$. Now suppose that $f(1) \\neq 1$ and let $c=f(1)$.\n$$P(1,-1) \\implies f(1)f(-1)=f(-1) \\implies f(-1)=0$$\n$$P(x,-1) \\implies f(x-1)=f(-x)-2x+1$$\n$$P(x-1,1) \\implies f(x)+cf(x-1)=f(x-1)+2x-1 \\implies f(x)+c(f(-x)-2x+1)=f(-x) \\implies (c-1)f(-x)=2cx-c-f(x) \\ \\ (Q(x))$$\n$$Q(-1) \\implies c(c-1)=-3c \\implies c=0 \\ \\text{or} \\ c=-2$$\nLet's check the case $c=-2$.\n$$(c-1)Q(-x)-Q(x) \\implies (c-1)f(-x)=-(c-1)(2cx+c+(c-1)f(x))=2cx-c-f(x) \\implies 3(-4x-2-3f(x))=-4x+2-f(x) \\implies f(x)=-x-1$$\nNow let's check the case when $c=0$. Note that $Q(x) \\implies f(x)=f(-x)$\n$$P(x,x)-P(x,-x) \\implies f(2x)+1=4x^2 \\implies f(x)=x^2-1$$\nTo conclude, the only solutions are $f(x)=x^2-1$ , $f(x)=-x-1$ and $f(x)=2x-1$.", "Solution_20": "1. y=1\n2. y=y+1 (expand)\n3. use symmetry\nDone. That easy (try it)", "Solution_21": "Help I dont follow", "Solution_22": "Beautiful FE. Worthy of its position as A4. Here's my solution (probably different): \nWe claim that the only solutions that work are $f(x)=2x-1,f(x)=x^2-1,f(x)=-x-1 \\text{ } \\forall x \\in \\mathbb{R}$. One can easily check that all three of these satisfy the given equation. We'll now show that these are the only functions that work.\n\nLet $P(x,y)$ denote the given assertion. Note that no constant function satisfies the given equation. So from now on assume that $f$ is non-constant.\n\n$$P(x,0) \\Rightarrow f(x)+f(x)f(0)=f(0)+1 \\Rightarrow (f(x)-1)(f(0)+1)=0 \\Rightarrow f(0)=-1$$\n\n$$P(1,-1) \\Rightarrow f(1)f(-1)=f(-1) \\Rightarrow f(1)=1 \\text{ OR } f(-1)=0$$\n\n[b]CASE-1[/b] $(f(1)=1):$ $$P(x-1,1) \\Rightarrow f(x)+f(x-1)=f(x-1)+2(x-1)+1 \\Rightarrow \\boxed{f(x)=2x-1}$$\n\n[b]CASE-2[/b] $(f(-1)=0):$ Let $f(1)=k \\neq 1$. Then $P(-1,-1) \\Rightarrow f(-2)=f(1)+2+1=k+3$. Also, $P(2,2) \\Rightarrow f(2)^2=9 \\Rightarrow f(2)= \\pm 3$ \n\n[b]SUBCASE-2.1[/b] $(f(2)=3):$ Note that $P(2,-2) \\Rightarrow -1+f(2)f(-2)=f(-4)-7 \\Rightarrow f(-4)=3(k+3)+6=3k+15$\n\nAnd, $P(2,1) \\Rightarrow f(3)+f(2)f(1)=f(2)+4+1 \\Rightarrow f(3)=3(1-k)+5=8-3k$\n\nBut, $P(-3,-1) \\Rightarrow f(-4)=f(3)+6+1 \\Rightarrow 3k+15=8-3k+7 \\Rightarrow k=0 \\Rightarrow f(1)=0$\n\nNow, $P(x-1,1) \\Rightarrow f(x)=f(x-1)+2(x-1)+1$ and $P(1-x,-1) \\Rightarrow f(-x)=f(x-1)+2(x-1)+1$\n\nTogether, These two statements give that $f$ is an even function. Then $$P(x,x)-P(x,-x) \\Rightarrow f(2x)-f(0)=2x^2-(-2x^2) \\Rightarrow f(2x)=4x^2-1 \\Rightarrow \\boxed{f(x)=x^2-1}$$\n\n[b]SUBCASE-2.2[/b] $(f(2)=-3):$ Note that $P(2,-2) \\Rightarrow -1+f(2)f(-2)=f(-4)-7 \\Rightarrow f(-4)=-3(k+3)+6=-3k-3$\n\nAnd, $P(2,1) \\Rightarrow f(3)+f(2)f(1)=f(2)+4+1 \\Rightarrow f(3)=-3(1-k)+5=3k+2$\n\nBut, $P(-3,-1) \\Rightarrow f(-4)=f(3)+6+1 \\Rightarrow -3k-3=3k+2+7 \\Rightarrow 6k=-12 \\Rightarrow k=-2 \\Rightarrow f(1)=-2$\n\nNow, $P(x,1) \\Rightarrow f(x+1)=3f(x)+2x+1$ and $P(-x,-1) \\Rightarrow f(-x-1)=f(x)+2x+1$\n\nTogether these two statements give that $f(x+1)=2f(x)+f(-x-1)$. Putting $x \\rightarrow x-1$, we have $f(x)-f(-x)=2f(x-1)$\n\nAlso, $P(-x,1) \\Rightarrow f(1-x)=3f(-x)-2x+1=3(f(x)-2f(x-1))-2x+1$. Putting $x \\rightarrow x+1$ in this equation, we get that $$f(-x)=3(f(x+1)-2f(x))-2x-1=3f(x+1)-6f(x)-2x-1$$\n\nAnd, $$P(x,-1) \\Rightarrow f(x-1)=f(-x)-2x+1 =(3f(x+1)-6f(x)-2x-1)-2x+1=3(3f(x)+2x+1)-6f(x)-4x=3f(x)+2x+3$$\n\nThus, $$P(x-1,1) \\Rightarrow f(x)=3f(x-1)+2x-1 \\Rightarrow f(x-1)=3(3f(x-1)+2x-1)+2x+3=9f(x-1)+8x \\Rightarrow f(x-1)=-x$$\n\nPutting $x \\rightarrow x+1$, we get our final solution as $\\boxed{f(x)=-x-1}$", "Solution_23": "Found the source of the problem I was looking for :D\n\nLet $P(x,y)$ be the assertion of the problem statement. $P(0,0)$ gives $f(0)^2 = 1$ or $f(0)=-1$ or $1$. \n\nIf $f(0) = 1$, $P(x,0)$ gives $2f(x) = 2$ or $f(x) = 1$ for all $x$ which is not a solution. Thus $f(0) = -1$. \n\nIf $f(1) = 1$, then $P(x,1)$ gives $f(x+1) +f(x) = f(x) + 2x+1\\implies \\boxed{f(x) = 2x-1}$ which is a solution. \n\nNow assume $f(1) = c\\neq 1$. $P(-1,1)$ gives $f(0) + f(1)f(-1) = f(-1) - 1 \\implies f(-1)(f(1) - 1)=0$ or $f(-1) = 0$. Now $P(x-1,1)$ gives $f(x) + cf(x-1) = f(x-1) + 2x-1$. Also $P(x,-1)$ gives $f(x-1) = f(-x) - 2x+1$.\n\nThus \n\\begin{align*}\nf(x) &= (1-c)f(x-1) + 2x-1\\\\\n& = (1-c)(f(-x)-2x+1) + 2x-1\\\\\n& = f(-x)-2x+1-cf(-x)+c(2x-1) + 2x+1\\\\\n& = f(-x) - cf(-x) + c(2x-1) = (1-c)f(-x) + c(2x-1).\n\\end{align*}\n\nPlugging $x=-1$ in the above equation gives $f(-1) = f(1)(1-c)-3c\\implies c(1-c) = 3c\\implies c=0,-2$.\n\nIf $c=0$ then \\[f(x) = (1-c)f(-x)+c(2x-1) = f(-x).\\] We have $P(x,-x)\\implies -1 + f(x)^2 = f(x^2) - 2x^2 + 1$ and $P(x,x)$ gives $f(2x) + f(x)^2 = f(x^2) +2x^2 + 1$ thus \\[f(2x) + 1 = 4x^2\\implies \\boxed{f(x) = x^2-1}.\\]\n\nIf $c=-2$ then \\[f(x) = (1-c)f(-x)+c(2x-1) = 3f(-x) - 2(2x-1).\\] Solving linear equation by plugging $x=-x$ in the previous equation gives \\[-8f(x) = -4x+2 + 12x+ 6 = 8x+8\\implies \\boxed{f(x) = -(x+1)}.\\]", "Solution_24": "Taking $y=0$ gives $f(x)(1+f(0))=f(0)+1$. Clearly $f(x)=1$ is a not a solution, so we must have $f(0)=-1$. Applying $(x,y)=(2,2)$ yields $f(2)=\\pm 3$. Taking $(x,y)=(1,1)$ gives $f(2)+f(1)^2=f(1)+3$.\n\n[b]Case 1: f(2)=3[/b] Then $f(1)^2=f(1)$, so $f(1)=1$ or $f(1)=0$. Suppose $f(1)=1$. Plugging $y=1$, \n\\begin{align}\nf(x+1)+f(x)f(1)&=f(x)+2x+1\n\\end{align}\nHence, $f(x+1)=2x+1$. Thus, $\\boxed{f(x)=2x-1}$ for all real $x$, and it is easy to verify this works. Now we visit the case when $f(1)=0$. Note $(1)$ transforms to $$f(x+1)=f(x)+2x+1\\implies f(x+1)-(x+1)^2=f(x)-x^2.$$\nA simple double induction proves that $f(x+n)=f(x)+2nx+n^2$ for all integers $n$. Another induction shows that $f(n)=n^2-1$. Taking $y=n$ for any integer $n$, we see $$f(nx)=f(x+n)+f(x)f(n)-2nx-1 = f(x)+2nx+n^2+(n^2-1)f(x)-2nx-1=n^2f(x)+n^2-1.$$\nNote that taking $n=-1$ gives $f(-x)=f(x)$. Plugging $y=x$ gives \n\\begin{align*}\\tag{2}\nf(x^2)+2x^2+1=f(2x)+f(x)^2=f(x)^2+4f(x)+3\n\\end{align*}\nNow taking $y=-x$ grants \\begin{align*}\\tag{3} f(0)+f(x)f(-x)=f(-x^2)-2x^2+1\\implies f(x)^2=f(x^2)-2x^2+2\\end{align*}\nCombining $(2)$ and $(3)$ gives $2x^2+1-(4f(x)+3)=-2x^2+2$. Therefore $\\boxed{f(x)=x^2-1}$. \n\n[b]f(2)=-3[/b] In this case $f(1)^2-f(1)-6=0$, so that $f(1)=3$ or $f(1)=-2$. Suppose $f(1)=3$. Trying $y=1$ grants $$f(x+1)=-2f(x)+2x+1.$$ Computing, we see $f(3)=11$, $f(5)=39$ and $f(6)=-67$. Now taking $(x,y)=(2,3)$ gives $$f(5)+f(2)f(3)=f(6)+13\\implies f(6)=-7,$$ which does not match with $f(6)=-67$. Thus, we must have $f(1)=-2$. Again, we take $y=1$ to get $$f(x+1)=3f(x)+2x+1.$$ A little bashing gives $f(x-1)=(f(x)-2x+1)/3$. Note $(x,-1)$ grants $$f(-x)=2x-1+f(x-1)=2x-1+(f(x)-2x+1)/3.$$\nReplacing $x$ with $-x$ in the above equation and adding gives $f(x)+f(-x)=-2$. Plugging $f(-x)=-2-f(x)$ back into the equation gives $$-2-f(x)=2x-1+(f(x)-2x+1)/3.$$ Solving, we get $\\boxed{f(x)=-x-1}$. We can easily verify all the derived solutions work.\n", "Solution_25": "Let $Q(x)$ be the assertion given in the problem statement. Then, $Q(x, 0)$ gives:\n$f(x)(f(0) + 1) = f(0) + 1$. As $f \\equiv 1$ is not a solution, we have $f(0) = -1$.\nNow, let $g(x) = f(x) + 1$.\nThen $Q(x)$ can be written as $P(x)$:\n$g(x+y) - (g(x) + g(y)) = g(xy) - g(x)g(y) + 2xy$. We have $g(0) = 0$.\n$P(1, -1)$ gives: $g(-1) = 1$ or $g(1) = 2$\nCase 1: $g(1) = 2$\n$P(x,1)$ gives $g(x+1) = 2(x+1)$ so $g(x) \\equiv 2x$. One can see that this is a valid solution.\nCase 2: $g(-1) = 1$\n$P(x, -1)$ gives $g(x-1) - 1 = g(-x) - 2x$. Call this assertion $R(x)$.\n$R(2)$ gives $g(-2) = g(1) + 3$\nConsidering $P(-2, 1)$ with the above relation and solving for $g(1)$ gives $g(1) = 1$ or $-1$.\nSubcase 2.1: $g(1) = 1$\n$P(x-1, 1)$ gives $g(x-1) = g(x) - 2x + 1$, which combines with $R(x)$ to give us that $g$ is even.\nUsing this knowledge and considering $P(x, x)$ and $P(x, -x)$ together gives $g(2x) = 4x^2$, thus $g(x) \\equiv x^2$, which can be checked to be a valid solution.\nSubcase 2.2: $g(1) = -1$\n$P(x-1, 1)$ gives $g(x) = 3g(x - 1) + 2x - 3$ -> $(1)$\nCombining this with $R(x)$ gives $g(x) = 3g(-x) - 4x$\nConsidering $R(-x)$ gives $3g(x+1) + 4x + 3 = g(-x-1) - 1 = g(x) + 2x$ -> $(2)$\nSolving for $g(x)$ using $(1)$ and $(2)$ gives $g(x) = -x$, thus $g(x) \\equiv -x$, which can be checked to be a valid solution.\n\nThus, the solution set is $f(x) \\equiv 2x - 1$, $f(x) \\equiv -x - 1$, and $f(x) \\equiv x^2 - 1$.", "Solution_26": "Probably the same as some other solutions in the thread. \n\nFirst $(0,0)$ yields $f(0)^2=1\\implies f(0)=\\pm 1$. \n\n[b]Case 1:[/b] $f(0)=1$. Then $y=0$ yields $f(x)=1$, which doesn't work. \n\n[b]Case 2:[/b] $f(0)=-1$. Now $(1,-1)$ yields $f(1)f(-1)=f(-1)$ so $f(-1)=0$ or $f(1)=1$. \n\n[i]Case 2a:[/i] $f(1)=1$. Then $y=1$ gives $f(x+1)=2x+1$ so $\\boxed{f(x)=2x-1}$ here. \n\n[i]Case 2b:[/i] $f(-1)=0$ and $f(1)\\neq 1$. First $(2,2)$ yields $f(2)^2=9\\implies f(2)=\\pm 3$. \n\n[u]Case 2ba:[/u] $f(2)=-3$. Now $(1,1)$ yields $f(1)^2-f(1)=6$ so $f(1)$ is $3$ or $-2$. Note from $y=-1$ that $f(x-1) =f(-x)-2x+1$, so letting $x=3,2$ gives $f(-3)=2$ and that $f(-2)$ is either $6$ or $1$ depending on our choice of $f(1)$. But now $(-3,1)$ yields $f(-2)+2f(1)=-3$, from which we find that $f(-2)=1$ and $f(1)=-2$. Now $y=-1$ yields $f(x-1)=f(-x)-2x+1$ and $y=1$ yields $f(x+1) =3f(x)+2x+1$. It follows that $f(-x)-2x+1=f(x-1) = \\dfrac{f(x)-2x+1}{3}$, which rearranges as $3f(-x)+2=f(x)+4x$.This also implies $3f(x)+2=f(-x)-4x$, and now we solve to find $\\boxed{f(x)=-1-x}$ for all $x$. \n\n[u]Case 2bb:[/u] $f(2)=3$. Now $(1,1)$ yields $f(1)^2=f(1)$ so $f(1)\\neq 1$ implies $f(1)=0$. Now $y=1$ gives $f(x+1)=f(x)+2x+1$ while $y=-1$ gives $f(x-1)=f(-x)-2x+1$. This last equation implies $f(-x-1)=f(x)+2x+1=f(x+1)$, so $f$ must be even. Now $y\\to -y$ gives $f(x-y)+f(x)f(y)=f(xy)-2xy+1$, so subtracting from the original equation gives $f(x+y)-f(x-y) = 4xy$. It follows that $f(x+y)-(x+y)^2 = f(x-y) - (x-y)^2$ so $f(u)-u^2$ is constant, hence $u=0$ implies the constant is $-1$ and $\\boxed{f(x)=x^2-1}$. \n\nAll cases are done so we have $\\boxed{f(x) = x^2-1, -1-x, 2x-1}$. ", "Solution_27": "Arghh, why so long BJV...\nAgain, I'll to make my solution motivated, but this becomes hard to do with FEs, so... well, here's my solution-\n\nFirst we get the basics substitutions over with- \n\n$P(x,0): f(x)(1+f(0))=(1+f(0))$. This immediately gives us that if $f(0) \\neq -1 => f \\equiv 1$ which of course doesn't work. \nHence $f(0)=-1$.\n\nSo we've just gotten some information about $f(0)$- the next natural substitution seems to be $P(x,-x)$:\n$$P(x,-x): -1+f(x)f(-x)=f(-x^2)+1-2x^2$$\nPutting in $x=1 =>-1+f(1)f(-1)=f(-1)-1=>f(-1)(f(1)-1)=0$.\n\nCase 1: $f(1)=1$\n$P(x,1): f(x+1)+f(x)f(1)=f(x)+2x+1=>f(x+1)=2x+1=>f(x)=2x-1$. \n\nOk, easiest solution found- time to deal with the other, rather more messier case:\n\nCase 2: $f(-1)=0$\nLet's try and extract as much as we can from such a nice zero:\n$$P(x,-1): f(x-1)=f(-x)-2x+1=>f(-x-1)=f(x)+2x+1$$\n$$P(x,-x-1): f(x)f(-x-1)=f(-x(x+1))-2x(x+1)+1=>f(x)^2+2xf(x)+f(x)=f(-x(x+1))-2x^2-2x+1$$\nPutting in $x=1$, we get $f(1)^2+3f(1)=f(-2)-3$. But by the first equation, putting in $x=2 =>f(1)=f(-2)-3$.\n\nSolving for $f(-2)$, we get $(f(2)-3)^2(f(-2)-1)=0=>f(-2)=3$ or $f(-2)=1$. \n\nSub-case 1: $f(-2)=1 =>f(1)=f(-2)-3=>f(1)=-2$.\nNow the deal with both these cases is to play with $f(x+1),f(-x)$ and similar $f$s and see if we can get something...\nAs we've already gotten $f(-x-1)=f(x)+2x+1$, and $P(x,1)=>f(x+1)=3f(x)+2x+1$\n$$=> f(-x)=f(x)+2f(x-1)=3f(x)+4x+2$$\nNow $P(x-1,1): f(x)-2f(x-1)=f(x-1)+2x-1=>f(x)=3f(x-1)+2x-1=>f(x-1)=\\frac{f(x)+1-2x}3$\n$=>f(-x)-2x+1=\\frac{f(x)+1-2x}3 =>f(-x)=\\frac{4x-2+f(x)}3$.\n\nEquating both expressions for $f(-x)$, we get $(3f(x)+4x+2) \\cdot 3=4x-2+f(x)=>8f(x)=-8x-8 =>f(x)=-x-1$.\nAha, only one case left- we're almost there.\n\nSubcase 2: $f(-2)=3 =>f(1)=\\frac{f(-2)-3}3=>f(1)=0$\nOkay we have another nice zero here- clearly it would be ideal to put in $(x,1)$ and see what we get:\n$$P(x,1): f(x+1)=f(x)+2x+1$$\nWait this expression seems familiar- turning our pages back (or scrolling up :P ) we see that $f(x)+2x+1$ is also equal to $f(-x-1)$- hence $f$ is even.\nNow the natural substitution is $P(x,-x): -1+f(x)^2=f(-x^2)+1-2x^2=>f(x^2)=f(x)^2+2x^2-2$.\n\nBut $P(x,x)$ gives us $f(2x)+f(x)^2=f(x^2)+2x^2+1=f(x)^2+2x^2-2+2x^2+1=>f(2x)=4x^2-1$ (I was so surprised this worked, but it did, so lucky me :) )\n\nThus our 3 solutions are $ f \\equiv 2x-1, f \\equiv -x-1$ and $f \\equiv x^2-1$, and as all 3 work, we're finally done! \n\n", "Solution_28": "Let $P(x,y)$ denote the assertion.\n-----\n[b]Claim 1.[/b] If $f(x)=2x-1$ for all $x$, then $f(0)=-1$ and $f(-1)=0$.\n\n[i]Proof.[/i] $P(0,0)$ immediately gives $f(0)=\\pm1$, but if $f(0)=1$, then $P(x,0)$ gives $f\\equiv1$, contradiction. Thus $f(0)=-1$.\n\n Now $P(1,-1)$ gives us $f(1)f(-1)=f(-1)$, so either $f(-1)=1$ or $f(-1)=0$. Say $f(-1)=1$. Then $P(x-1,1)$ immediately yields $f(x)=2x-1$, as claimed. $\\blacksquare$\n-----\n[b]Claim 2.[/b] The tuple $(f(-2),f(1),f(2))$ is either $(1,-2,-3)$ or $(3,0,3)$.\n\n[i]Proof.[/i] Note that $P(1,-2)$ gives $f(1)f(-2)=f(-2)-3$ and $P(2,-1)$ gives $f(1)=f(-2)-3$. Solving, $(f(-2),f(1))$ is either $(1,-2)$ or $(3,0)$.\n\n However $P(1,1)$ gives $f(2)=-f(1)^2+f(1)+3$, which equals $-3$ and $3$ respectively. $\\blacksquare$\n-----\n[b]Claim 3.[/b] If $(f(-2),f(1),f(2))=(1,-2,-3)$ then $f(x)=-x-1$ for all $x$.\n\n[i]Proof.[/i] Note that $P(1-x,-1)$ and $P(1+x,-1)$ give the two equations\n \\begin{align*}\n f(-x)&=f(x-1)+2x-1,\\\\\n f(-x-1)&=f(x)+2x+1.\n \\end{align*}\n respectively. Furthermore $P(x-1,1)$ and $P(-x-1,1)$ give the two equations\n \\begin{align*}\n f(x)=3f(x-1)+2x-1,\\\\\n f(-x)=3f(-x-1)-2x-1.\n \\end{align*}\n Solving the system of equations, we find $f(x)=-x-1$, as desired. $\\blacksquare$\n-----\n[b]Claim 4.[/b] If $(f(-2),f(1),f(2))=(3,0,3)$ then $f(x)=x^2-1$ for all $x$.\n\n[i]Proof.[/i] First $P(x-1,1)$ and $P(-x+1,-1)$ give \\[f(x)=f(x-1)+2x-1=f(-x),\\]\n so $f$ is even. With this, $P(x/2,x/2)$ and $P(x/2,-x/2)$ give the two equations\n \\begin{align*}\n f(x)+f(x/2)^2&=f(x^2/4)+x^2/2+1,\\\\\n -1+f(x/2)^2&=f(x^2/4)-x^2/2+1.\n \\end{align*}\n Subtracting, $f(x)=x^2-1$, as desired. $\\blacksquare$\n-----\nThis exhausts all possible cases, so we conclude the only answers are $f(x)=2x-1$, $f(x)=-x-1$, $f(x)=x^2-1$, which clearly work.", "Solution_29": "The solutions are $f(x)\\equiv\\boxed{2x-1,-x-1,x^2-1}$, which can all be checked to work. Now let $P(x,y)$ be the given FE, and suppose $f$ is a solution. Note that \\[P(x,0)\\implies f(x)[1+f(0)]=f(0)+1,\\] and since $f\\equiv 1$ isn't a solution, we have $\\boxed{f(0)=-1}$. Now, we have \\[P(1,-1)\\implies -1+f(1)f(-1)=f(-1)-1\\implies f(1)f(-1)=f(-1).\\] If $f(-1)\\ne 0$, then we have $f(1)=1$, so \\[P(x,1)\\implies f(x+1)+f(x)=f(x)+2x+1,\\] so $f(x)\\equiv 2x-1$, which is a solution. Thus, we may now assume $\\boxed{f(-1)=0}$ and $f(1)\\ne 1$.\n\nWe see that \\[P(x,1)\\implies \\boxed{f(x+1)=2x+1+f(x)[1-f(1)]}\\] and \\[P(x+1,-1)\\implies f(x)=f(-(x+1))-2x-1.\\] Adding the two equations, we get \\[\\boxed{f(x+1)-f(-(x+1))=-f(1)f(x)}.\\] Call the first boxed equation $Q(x)$ and the second one $R(x)$.\n\nWe have \\[Q(-2)\\implies 0=-3+f(-2)[1-f(1)]\\implies f(-2)=\\frac{3}{1-f(1)},\\] and \\[Q(1)\\implies f(2)=3+f(1)[1-f(1)].\\] Thus, \\[R(1)\\implies 3+f(1)[1-f(1)]-\\frac{3}{1-f(1)}=-f(1)^2,\\] so solving for $f(1)$ gives either $f(1)=0$ or $f(1)=-2$. We thus have two cases.\n\n[b]Case 1[/b]: Suppose $f(1)=0$. Then, $Q(x)$ reduces to \\[f(x+1)-f(x)=2x+1.\\] It is not hard to show by induction that \\[f(x+n)-f(x)=2nx+n^2\\] for any integer $n$. Combined with $f(0)=-1$, we see that $f(n)=n^2-1$ for any integer $n$. Thus, \\[P(x,n)\\implies f(x)+2nx+n^2+f(x)(n^2-1)=f(nx)+2nx+1,\\] so \\[f(nx)=n^2f(x)+(n^2-1).\\] We now have \\begin{align*} P(x,x) &\\implies f(2x)+f(x)^2=f(x^2)+2x^2+1 \\\\ &\\implies 3+4f(x)+f(x)^2=f(x^2)+2x^2+1 \\end{align*} and \\begin{align*} P(x,2x) &\\implies f(3x)+f(x)f(2x)=f(2x^2)+4x^2+1 \\\\ &\\implies 9f(x)+8+f(x)[4f(x)+3]=4f(x^2)+3+4x^2+1 \\\\ &\\implies 4f(x)^2+12f(x)+4=4f(x^2)+4x^2 \\\\ &\\implies 3f(x)+f(x)^2+1=f(x^2)+x^2. \\end{align*} Comparing the value for $f(x^2)$ we get from these two equations, we learn \\[f(x)^2+4f(x)+2-2x^2 = f(x)^2+3f(x)+1-x^2,\\] so $f(x)=x^2-1$, as desired. This completes Case 1.\n\n[b]Case 2[/b]: Suppose $f(1)=-2$. Then, $Q(x)$ gives \\[f(x+1)=2x+1+3f(x).\\] It is not hard to show by induction that \\[f(x+n)=3^nf(x)+(3^n-1)x+(3^n-n-1)\\] for any integer $n\\ge 0$. In particular, $f(n)=-n-1$ for integers $n\\ge 0$. Thus, \\[P(x,n)\\implies 3^nf(x)+(3^n-1)x+(3^n-n-1)-(n+1)f(x)=f(nx)+2nx+1,\\] so \\[f(nx)=(3^n-n-1)f(x)+(3^n-1-2n)x+(3^n-n-2).\\] We now have \\begin{align*} P(x,x) &\\implies f(2x)+f(x)^2=f(x^2)+2x^2+1 \\\\ &\\implies 6f(x)+4x+5+f(x)^2=f(x^2)+2x^2+1 \\\\ &\\implies f(x^2)=f(x)^2+6f(x)-2x^2+4x+4 \\end{align*} and \\begin{align*} P(x,2x) &\\implies f(3x)+f(x)f(2x)=f(2x^2)+4x^2+1 \\\\ &\\implies 23f(x)+20x+22+f(x)[6f(x)+4x+5]=6f(x^2)+4x^2+5+4x^2+1. \\end{align*} Substituting the value for $f(x^2)$ from $P(x,x)$ into the above equation and simplifying, we get \\[(x-2)f(x)=(2-x)(x+1),\\] so $f(x)=-x-1$ when $x\\ne 2$. However, $f(2)=-3$ as $2$ is an integer, so we have $f(x)=-x-1$, as desired.\n\nThis completes the proof.", "Solution_30": "Let $P(x,y) $ denotes the assertion.\nIf $a=0$ then $f(b)+f (0)f(b)=f(b)=f(0)+1\\implies f(b)=-1, f(0)\\ne -1\\quad\\forall b\\in\\mathbb R$\nSimilarly, if $b=0$ we obtain $f(a)=f(b)=-1, f(a)=-1\\quad\\forall a,b\\in\\mathbb R$\n\nIf $a=1, P(x-1,1)\\implies f(x)=2x-1, a=-2\\implies f(x)=-x-1\\forall x\\ne (0,1), f(x)=-x-1\\forall x$", "Solution_31": "[hide=Correction in #24][quote=\"TheDarkPrince\"]\n\\begin{align*}\nf(x) &= (1-c)f(x-1) + 2x-1\\\\\n& = (1-c)(f(-x)-2x+1) + 2x-1\\\\\n& = f(-x)-2x+1-cf(-x)+c(2x-1) + 2x+1\\\\\n& = f(-x) - cf(-x) + c(2x-1) = (1-c)f(-x) + c(2x-1).\n\\end{align*}\n[/quote]\n\nNice solution TDP, just make it $2x-1$ in the third line :)[/hide]", "Solution_32": "[i]My solution[/i]\n$$P(x,y) \\implies f(x+y)+f(x)f(y)=f(xy)+2xy+1$$\n[b]Answers are:[/b] $f(t)=2t-1,$ $f(t)=t^2-1,$ $f(t)=-t-1.$ It all works.\\\\\n\n[color=#f00][b]Claim 1:[/b][/color] $f(0) = -1.$\\\\\n[color=#00f][b]Proof:[/b][/color] $P(0,0) \\implies f(0)^2=1$, if $f(0)=1,$ by putting $P(x,0)$ we get $f(x)=1,$ which is wrong.\\\\\n\n[color=#f00][b]Claim 2:[/b][/color] $f(1)=1$ or $f(-1)=0.$\\\\\n[color=#00f][b]Proof:[/b][/color] $P(1,-1) \\implies f(1)f(-1)=f(-1).$\\\\\n\nIf $f(1)=1$, $P(x-1,1) \\implies f(x)=2x-1.$\\\\\n\nIf $f(-1)=0,$ then by putting [color=#0f0](1)[/color] $P(x,-1) \\implies f(x-1)=f(-x)-2x+1$ and [color=#0f0](2)[/color]$P(x,-x) \\implies f(x)f(-x)=f(-x^2)-2x^2+2$\\\\\nSince [color=#0f0](1)[/color] $f(x^2-1)+1 = f(-x^2)-2x^2+2 = f(x)f(-x) = f(x)(f(x-1)+2x-1)$, we get that [color=#0f0](3)[/color] $f(x^2-1) = f(x)(f(x-1)+2x-1)-1$\\\\\n\nBy (1), $f(-2) = f(1)+3$, and by putting $P(-2,1)$ we get $f(1)^2+3f(1)=f(1) \\implies f(1)=0 \\text{ or } f(1)=-2.$\\\\\nIf $f(1)=0$, by putting $P(x,1) \\implies f(x+1)=f(x)+2x+1 \\implies f(2)=3$\\\\\n\nBy $P(x,2) \\implies f(x+2)+3f(x)=f(2x)+4x+1$. Since $f(x+2) = f(x+1)+2x+3 = f(x)+4x+4.$ we get that $f(2x) = 4f(x)+3.$\\\\\n\nBy $P(x-1,x+1) \\implies f(2x)+f(x-1)f(x+1)=f(x^2-1)+2x^2-1,$ expanding all this by (3), and $f(x+1)=f(x)+2x+1$ and by $f(2x)=4f(x)+3,$ we get a desired solution $f(x)=x^2-1.$\\\\\n\nThe last case $f(1)=-2$. By putting $P(x,1) \\implies f(x+1) = 3f(x)+2x+1 \\implies f(2)=-3$\\\\\nWorking similarly on $P(x,2)$ and $P(x-1,x+1)$ gives a last solution namely $f(x)=-x-1,$ We are done!\\\\", "Solution_33": "The solutions are $f\\equiv 2x-1$, $f\\equiv x^2-1$, and $f\\equiv -x-1$. It is easy to check that each works.\n\nSet $x=y=0$ to yield $f(0)^2=1$. Set $y=0$ to yield $f(x)(f(0)+1)=f(x)+f(x)f(0)=f(0)+1$. Thus either $f(0)=-1$ or $f\\equiv1$, which is clearly absurd. Then $f(0)=-1$. Let $y=-x$ to yield\n\\[-1+f(x)f(-x)=f(-x^2)-2x^2+1.\\]\nIn particular, for $x=1$ this yields $f(1)f(-1)=f(-1)$, so either $f(1)=1$ or $f(-1)=0$. In the case of $f(1)=1$, we have $f(x+1)=f(x)=f(x)+2x+1$ so $f(x+1)=2[x+1]-1$, yielding the solution $f\\equiv 2x-1$. \n\nNow suppose $f(-1)=0$. Then we write the equation $f(-x-1)=f(x)+2x+1$. Take $y=-x-1$ so \n\\[f(-x^2-x)-2x^2-2x+1=f(x)f(-1-x)=f(x)f(x)+2xf(x)+f(x)\\qquad (\\diamondsuit).\\]\nSo for $x=1$, this reads $f(-2)+1=f(1)^2+3f(1)$. But we also have $f(-2)=f(1)+3$, so\n\\[f(1)=f(1)^2+3f(1)\\iff f(1)\\in \\{0,-2\\}.\\]\nThe first case yields the equation $f(x+1)=f(x)+2x+1$ so $f$ is even, so then $f(2x)+f(x)f(-x)=f(-x^2)+2x^2+1$, which implies $f(2x)=4x^2-1$ and $f\\equiv x^2-1$. \n\nNow we have $f(1)=-2$, $f(-1)=0$, and $f(0)=-1$. Take $x=1$ to yield $f(1+y)=3f(y)+2y+1$. Now, write\n\\[f(y)=3f(y-1)+2y-1=3[f(-y)-2y+1]+2y-1 = 3f(-y)-4y+2 = 3[3f(y)+4y+2]-4y+2=9f(y)+8y+8\\implies f\\equiv -y-1.\\]", "Solution_34": "Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$.\n\n$$P(x,0)\\implies\\left(1+f(0)\\right)\\left(f(x)-1\\right)=0$$\nIf $f(0)\\ne-1$ then $f(x)=1$, which doesn't work. Therefore, we must have $f(0)=-1$.\n\nAlso note that:\n$$P(1,-1)\\implies f(1)f(-1)=f(-1)$$\nSo either $f(1)=1$ or $f(-1)=0$.\n\n[b]Case 1:[/b] $f(1)=1$\n\nWe have that:\n$P(x-1,1)\\implies\\boxed{f(x)=2x-1}$, which works.\n\n[b]Case 2:[/b] $f(-1)=0$\n\n\\begin{align*}\nP(-x,-1)&\\implies f(-x-1)=f(x)-2x+1\\qquad(1) \\\\\nP(x,1)&\\implies f(x+1)+f(x)f(1)=f(x)+2x+1\\qquad(2) \\\\\n&\\implies f(-x-1)+4x=f(x+1)+f(x)f(1)\\qquad(3)\n\\end{align*}\n\nTaking $x\\mapsto-x-1$ in $(3)$, we get:\n$f(x)=f(-x)+f(-x-1)f(1)-4x$\n\nSubstituting $(1)$, we can get an expression in terms of just $f(x)$, $f(-x)$, $x$, and constants.\n$f(x)=f(-x)+f(x)f(1)+(2f(1)-4)x+f(1)$\n$\\implies f(x)\\left(1-f(1)\\right)=f(-x)+(2f(1)-4)x+f(1)$\n\nNow, taking $x\\mapsto-x$ and adding will cancel out the $2xf(1)$ term:\n$f(-x)\\left(1-f(1)\\right)=f(x)+(4-2f(1))x+f(1)$\n\nAdding both of these:\n$\\left(f(x)+f(-x)\\right)\\left(1-f(1)\\right)=\\left(f(x)+f(-x)\\right)+2f(1)$\n$-f(1)\\left(f(x)+f(-x)\\right)=2f(1)$\nThis means that either $f(1)=0$ or $f(x)+f(-x)=-2$.\n\n[b]Case 2.1:[/b] $f(x)+f(-x)=-2$\n\nWe get $f(-x)=-2-f(x)$. Comparing $P(x,y)$ with $P(-x,-y)$ proves:\n\n\\begin{align*}\n&\\phantom{\\implies}f(-x-y)+f(-x)f(-y)=f(x+y)+f(x)f(y) \\\\\n&\\implies -2-f(x+y)+4+2f(x)+2f(y)+f(x)f(y)=f(x+y)+f(x)f(y) \\\\\n&\\implies f(x+y)=f(x)+f(y)+1\\end{align*}\n\nNote that the substitution $g(x)=f(x)-1$ proves that $g$ is additive, but there's no way to prove that it's linear.\n\nSubstituting this into $P(x,y)$, we get:\n$$f(x)+f(y)+1+f(x)f(y)=f(xy)+2xy+1$$\n\nNow let $h(x)=f(x)+1$. We have:\n$$Q(x,y):h(x)h(y)=h(xy)+2xy$$\nSince $f(-1)=0$ and $f(x)+f(-x)=-2$, we know that $h(-1)=1$ and $h$ is odd. The latter means that $h(1)=-1$.\n\nNow, $Q(x,1)\\implies h(x)=-x$\n\nTherefore, $f(x)=h(x)-1=\\boxed{-x-1}$, which indeed works. There are no other solutions in this case.\n\n[b]Case 2.2:[/b] $f(1)=0$\n\nSubstituting this into $(3)$, we see that $f(-x-1)=f(x+1)\\forall x$, and $x\\mapsto x-1$ shows that $f$ is even.\n\\begin{align*}\nP(x,-y)&\\implies f(x-y)+f(x)f(-y)=f(-xy)-2xy+1 \\\\\n&\\implies f(x-y)+f(x)f(y)=f(xy)-2xy+1\n\\end{align*}\nWe know that $f(x+y)+f(x)f(y)=f(xy)+2xy+1$, so we can say that:\n$$f(x-y)+2xy=f(x+y)-2xy$$\n$$f(x-y)+4xy=f(x+y)$$\nHere, taking $x=y$ gives:\n$$f(2x)=4x^2-1$$\nAnd doing the transformation $x\\mapsto\\frac x2$, we get:\n$$\\boxed{f(x)=x^2-1}$$\nwhich also works.\n\n\nThus, our solutions are $\\boxed{f(x)=2x-1}$, $\\boxed{f(x)=x^2-1}$ and $\\boxed{f(x)=-x-1}$. $\\square$", "Solution_35": "Let $P(x,y)$ denote the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$. First, $P(x,0)$ gives $(f(0)+1)(f(x)-1)=0$, so either $f\\equiv 1$ or $f(0)=-1$. The former fails, so $f(0)=-1$. Next, $P(1,-1)$ gives $f(-1)f(1)=f(-1)$, so $f(-1)=0$ or $f(1)=1$. [list] [*] [color=#00f][b][i][b]Case 1: $f(1)=1$[/i][/b][/b][/color]. Then $P(x,1)$ gives $f(x+1)+f(x)=f(x)+2x+1$, so $f(x+1)=2x+1$, so $f(x)=2x-1$. This is a valid solution for all $x$. [*] [color=#00f][b][i][b]Case 2: $f(-1)=0$[/i][/b][/b][/color]. Firstly, \\[ P(x,-1): \\quad f(x-1)=f(-x)-2x+1. \\quad (\\heartsuit)\\] We can use the cancellation trick on $P(x,y)$, to cancel $f(x+y)$ and $f(xy)$ with the following. $P(x,\\tfrac{x}{x-1})$ gives $f(x)f(\\tfrac{x}{x-1})=\\tfrac{(2x-1)(x+1)}{x-1}$. Now, $x=\\tfrac{x}{x-1}$ when $x=2$, so in particular $P(2,2)$ gives $f(2)^2=9$. Hence $f(2)=\\pm 3$. [list] [*][color=#00f] [i][b]Case 2.1: $f(2)=3$.[/i][/b][/color] We can use this information to find $f(1)$ via $P(1,1)$: $f(1)^2=f(1)$, so $f(1) \\in \\{0,1\\}$. [list] [*] [color=#00f][i][b]Case 2.1.1: $f(1)=0$.[/i][/b] [/color]Then $P(x-1,1)$ gives $f(x)=f(x-1)+2x-1$. Combined with $(\\heartsuit)$, this implies $f$ is even. Then compare $P(x,y)$ and $P(x,-y)$: \\begin{align*} P(x,y)&: \\quad f(x+y)+f(x)f(y)=f(xy)+2xy+1 \\\\ P(x,-y)&: \\quad f(x-y)+f(x)f(y)=f(xy)-2xy+1. \\end{align*} Subtracting, $f(x+y)-f(x-y)=4xy$. Plug in $y=x$ into this: $f(2x)-f(0)=4x^2$, so $f(2x)=4x^2-1$, so $f(x)=x^2-1$. This is a valid solution for all $x$. [*][color=#00f] [i][b]Case 2.1.2: $f(1)=1$.[/i][/b] [/color]Then $P(x-1,1)$ gives $f(x)=2x-1$ for all $x$, a valid solution. [/list] [*] [color=#00f][i][b]Case 2.2: $f(2)=-3$.[/i][/b] [/color]Then $P(1,1)$ gives $f(1)^2=f(1)+6$, so $f(1)\\in \\{-2,3\\}$. [list] [*][color=#00f] [i][b]Case 2.2.1: $f(1)=-2$.[/i][/b][/color] Then $P(x-1,1)$ gives $f(x)=3f(x-1)+2x-1$. Combined with $(\\heartsuit)$, this implies $f(x)=3f(-x)-4x+2$. These two both give us that $f(1-x)=3f(-x)+2(1-x)-1=f(x)+2x-1$. The RHS's of $P(x,1-x)$ and $P(-x,x-1)$ are the same since $x(1-x)=(-x)(x-1)$, so the LHS's are equal too. This gives a relation between all the f's we currently do not know how to relate. This gives \\[ f(1)+f(x)f(1-x) = f(-1)+f(-x)f(x-1). \\] We have \\begin{align*} f(-x) &= \\tfrac13(f(x)+4x-2) \\\\ f(x-1) &= \\tfrac13(f(x)-2x+1) \\\\ f(1-x) &= f(x)+2x-1. \\end{align*} Plugging these into the above equation: \\begin{align*} &\\qquad -2 + f(x)(f(x)+2x-1) = \\tfrac19(f(x)+4x-2)(f(x)-2x+1) \\\\ &\\implies -18+9f(x)^2 + 18xf(x)-9f(x)=f(x)^2+2xf(x)-f(x)-8x^2+8x-2 \\\\ &\\implies 8f(x)^2+16xf(x)-8f(x)+8x^2-8x-16=0 \\\\ &\\implies f(x)^2 + 2xf(x) - f(x)+x^2-x-2 = 0 \\\\ &\\implies (f(x)+x)^2 - (f(x)+x) - 2 = 0 \\\\ &\\implies f(x)+x \\in \\{2,-1\\}. \\end{align*} Therefore, for each $x$, either $f(x)=-x+2$ or $f(x)=-x-1$. Now we have to deal with the pointwise trap. Suppose $f(a)=-a+2$ for some $a$. Then $P(a,2)$ gives $f(a+2)+(-a+2)(-3)=f(2a)+4a+1$, so $f(a+2)=f(2a)+a+7$. But $f(a+2) \\in \\{-a,-a-3\\}$ and $f(2a)\\in \\{-2a+2,-2a-1\\}$, so we cannot have $f(a+2)-f(2a) =a+7$. Therefore, $f(a)\\not = -a+2$ for any $a$, so $f(x)=-x-1$ for all $x$. This is a valid solution. [*][color=#00f] [i][b]Case 2.2.2: $f(1)=3$.[/i][/b] [/color]We show this is impossible. We know $f(0)=-1$ and $f(-1)=0$ and $f(2)=-3$. [list] [*] $P(2,-1)$ gives $f(1)+f(2)f(-1)=f(-2)-3$ so $f(1)=f(-2)-3$, so $f(-2)=6$. [*] $P(-2,1)$ gives $f(-1)+f(-2)f(1)=f(-2)-4+1$, so $f(-2)(f(1)-1)=-3$, so $f(-2)=-3/2$. [/list] This is a contradiction. [/list] [/list] [/list] Finally, the solutions we recovered are $f(x)=2x-1 \\forall x$, and $f(x)=x^2-1\\forall x$, and $f(x)-x-1\\forall x$.", "Solution_36": "Another solution since I re-did this problem without realizing it.\n\nLet $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$.\n$P(x,0)\\Rightarrow f(x)+f(x)f(0)=f(0)+1$ so $f(0)=-1$ (else $f(x)=1$, which doesn't work).\n$P(1,-1)\\Rightarrow f(1)f(-1)=f(-1)\\Rightarrow f(-1)=0\\vee f(1)=1$\n\n[b]Case 1:[/b] $f(-1)=0$\n$P(1,-2)\\Rightarrow f(1)f(-2)+3=f(-2)$\n$P(-1,-1)\\Rightarrow f(-2)=f(1)+3$\nSo $f(1)^2+2f(1)=0$, thus either $f(1)\\in\\{0,-2\\}$.\n\n[b]Case 1.1:[/b] $f(1)=0$\nLet $g(x)=f(x)-x^2+1$, so that $g$ has zeroes at all elements of the set $\\{-1,0,1\\}$. We get the assertion\n$$Q(x,y):g(x+y)+g(x)g(y)-g(x)-g(y)+x^2g(y)+y^2g(x)=g(xy)$$\n$Q(x,1)\\Rightarrow g(x+1)=g(x)$\n$Q(-x,-1)\\Rightarrow g(-x-1)=g(x)$\nthus $g$ is even, since $g(x+1)=g(-x-1)$.\n$Q(x,-y)\\Rightarrow g(x-y)+g(x)g(y)-g(x)-g(y)+x^2g(y)+y^2g(x)=g(xy)$ (using evenness)\nSo $g(x+y)=g(x-y)$, which we obtain when we compare the above with $Q(x,y)$. Setting $x=y$ yields $g(2x)=g(0)=0$, so $g(x)=0$, and then $\\boxed{f(x)=x^2-1}$, and it's easy, but a bit annoying to verify that this works.\n\n[b]Case 1.2:[/b] $f(1)=-2$\n\\begin{align*}\nP(x-1,1)&\\Rightarrow f(x)-2x+1=3f(x-1)\\\\\nP(-x+1,-1)&\\Rightarrow f(-x)=f(x-1)+2x-1\\\\\n&\\Rightarrow f(-x)=\\frac{f(x)+4x-2}3\\\\\n&\\Rightarrow3f(x)=f(-x)-4x-2=\\frac{f(x)+4x-2}3-4x-2\\\\\n&\\Rightarrow8f(x)=-8x-8\\\\\n&\\Rightarrow\\boxed{f(x)=-x-1},\\\\\n\\end{align*}\nwhich works.\n\n[b]Case 2:[/b] $f(1)=1$\n$P(x-1,1)\\Rightarrow\\boxed{f(x)=2x-1}$, which also works.\n\nSolutions, which all work: $\\boxed{f(x)=x^2-1}$, $\\boxed{f(x)=-x-1}$, $\\boxed{f(x)=2x-1}$\n", "Solution_37": "I redid this problem yet again without realizing it.\n\nLet $P(x,y)$ denote this assertion.\nIf $f(0)\\ne-1$, then:\n$P(x,0)\\Rightarrow f(x)=1$ which doesn't work.\nSo $f(0)=-1$.\n$P(1,-1)\\Rightarrow f(1)f(-1)=f(-1)$\nIf $f(1)=1$ we have:\n$P(x-1,1)\\Rightarrow\\boxed{f(x)=2x-1}$, which is a solution.\nOtherwise, assume $f(-1)=0$.\n$P(-2,1)\\Rightarrow f(-2)f(1)=f(-2)-3$\n$P(-1,-1)\\Rightarrow f(-2)=f(1)+3$\nSubstituting $P(-1,-1)$ into $P(-2,1)$, we have $f(1)\\in\\{-2,0\\}$.\n\n[b]Case 1:[/b] $f(1)=0$\n$P(x,1)\\Rightarrow f(x+1)=f(x)+2x+1$\n$P(x,y+1)-P(x,y)\\Rightarrow 2x^2+4xy+2y^2+2yf(x)+f(x)=f(xy+x)-f(xy)+2x-1$\nSetting $y=\\frac1x$ and making $x\\ne0$, we obtain $f(x)=x^2-1$ for $x\\ne0$. Since $f(0)=-1$, it completes the solution $\\boxed{f(x)=x^2-1}$ which fits.\n\n[b]Case 2:[/b] $f(1)=-2$\n$P(x,1)\\Rightarrow f(x+1)=3f(x)+2x+1$\n$P(x,y+1)-3P(x,y)\\Rightarrow 2y+3+2yf(x)+f(x)=f(xy+x)-3f(xy)-4xy$\nWe repeat with $y=\\frac1x,x\\ne0$, we have $f(x)=-x-1$ for $x\\ne0$. Since $f(0)=-1$ this is $\\boxed{f(x)=-x-1}$, which is a solution.", "Solution_38": "I rather dislike casework to be honest.\n\n Let the assertion be $P(x,y).$\n\n$P(0,0)$ gives $f(0)+f(0)^2=f(0)+1$ which implies $f(0)^2=1.$ If $f(0)=1$ then $P(x,0)$ implies $f(x)+f(x)=2$ so $f(x)=1$ for all $x.$ Note that this construction doesn't work when plugging into original. Now, we work under the assumption that $f(0)=-1.$\n\n$P(-1,1)$ gives $-1+f(-1)f(1)=f(-1)-1,$ which implies either $f(-1)=0$ or $f(1)=1.$ If $f(1)=1$ then $P(x-1,1)$ gives $f(x)+f(x-1)=f(x-1)+2x-1$ so $f(x)=2x-1.$ Note that this satisfies the original condition. Now, we assume $f(-1)=0.$\n\n$P(-1,-1)$ gives $f(-2)=f(1)+3.$ On the other hand, $P(-2,1)$ gives $f(-2)f(1)=f(-2)-3$ so $f(-2)f(1)=f(1).$ Thus, $f(1)=0$ or $f(-2)=1.$\n\nIf $f(1)=0$ then $P(x,1)$ implies $f(x+1)=f(x)+2x+1$ and so $f(2)=3.$\n\n$P(x,2)$ implies $f(x+2)+3f(x)=f(2x)+4x+1.$ On the other hand, $P(x+1,1)$ implies $f(x+2)=f(x+1)+2(x+1)+1=f(x+1)+2x+3=f(x)+2x+1+2x+3=f(x)+4x+4.$\n\nThis implies $f(2x)+4x+1-3f(x)=f(x)+4x+4$ so $f(2x)=4f(x)+3.$ $P(x,x+2)$ gives $f(2x+2)+f(x)f(x+2)=f(x^2+2x)+2x(x+2)+1.$ This implies \\[f(2x+2)+f(x)f(x+2)=f((x+1)^2)-2(x+1)^2+1+2x^2+4x+1.\\]\nThe RHS cancels out to $f((x+1)^2)$ so \\[f((x+1)^2)=4f(x+1)+3+(f(x+1)-2(x+1)+1)(f(x+1)+2(x+1)+1)\\]\nThis simplifies down to $f(x^2+2x+1)=f(x+1)^2+6f(x+1)-4(x+1)^2+3$ so $f(x^2)=f(x)^2-4x^2+6f(x)+4.$ However, $P(x,x)$ gives $f(2x)+f(x)^2=f(x^2)+2x^2+1$ so $f(x^2)-f(x)^2=-4x^2+6f(x)+4$ while it also equals $4f(x)+2-2x^2.$ This all implies that $f(x)=x^2-1.$ Note that this is a valid solution.\n\nIf $f(-2)=1$ then $f(1)=-2.$ Now $P(x-1,1)$ gives $f(x)=3f(x-1)+2x-1$ and $P(-x+1,-1)$ gives $f(-x)=f(x-1)+2x-1$ (remember that $f(-1)=0$). Note that equivalently $f(-x-1)=f(x)+2x+1$\n\nThus, $f(x)-f(-x)=2f(x-1).$ Also, $f(-x)-f(x)=-2f(x-1)$ and $2f(-x-1)$ at the same time. Thus $f(x-1)=-f(-x-1).$\n\nWe see that $f(x-1)=-f(x)-2x-1$ so $f(x)=3(-f(x)-2x-1)+2x-1=-3f(x)-4x-4$ so $f(x)=-x-1,$ and this also works so $f(x)=-x-1,x^2-1,2x-1$", "Solution_39": "Let $P(x,y)$ denote the assertion.\n$P(x,1)\\implies f(x+1)=cf(x)+2x+1,~~~~~(**)$\nwhere $c=1-f(1).$ \n$P(x,y+1)\\implies c(f(x+y)+f(x)f(y))+(2y+1)(1+f(c))=f(x(y+1))+2xy+1.$\n$P(2k,-1/2)\\implies c(f(-k)-2k+1)=f(-k)+2k+1.~~~~~(*)$\nCombining yields, $(-c^2)f(k)=2(1-c)^2k+c^2-1.$\nNote that $c\\neq -1.$ \n----------\n\nCase 1: $c\\neq 1$\nThen $1-c^2\\neq 0 \\implies f(k)=2(\\frac{1-c}{1+c})k-1.$\n$k\\to 1\\implies c=0$ or $c=3\\implies \\boxed{f(x)=2x-1}$ or $\\boxed{f(x)=-1-x},$ both work.\n------------\n\nCase 2: $c=1.$\n$(*)\\implies f(k)=f(-k).$\n$P(x,x)-P(x,-x)\\implies f(2x)=4x^2+f(0).$\nSetting $x=0$ in $(**)\\implies f(0)=-1.$\nIt follows that $\\boxed{f(x)=x^2-1},$ which also fits.", "Solution_40": "Really straightforward problem. \n\nLet $P(x;y)$ denotes the assertion of given functional equation. Note that $P(0;0)$ gives us: \n$$ f(0)+f(0)^2 = f(0)+1 \\implies f(0)= 1 \\quad \\text{or} \\quad f(0)=-1 $$ \nSuppose that $f(0)=1$. Then from $P(x;0)$ we have: \n$$ f(x)+f(x) =1 +1 \\implies f(x)=1 $$ \nIt is easy to see that this function does not work; therefore we can assume that $f(0)=-1$. \n\nConsider $P(1;-1)$: \n$$ f(0)+f(1)f(-1)=f(-1)-2+1 \\implies f(1)f(-1)=f(-1) \\implies f(-1)=0 \\quad \\text{or} \\quad f(1)=1 $$ \nSuppose that the latter is the case. Then from $P(x;1)$ we have: \n$$ f(x+1) +f(x)=f(x)+2x+1 \\implies f(x+1)=2x+1 \\implies f(x)=2x-1$$ \nIt is easy to see that this function works; thus we can assume that $f(-1)=0$. \n\nWe would like to determine the value of $f(1)$. To do so we consider $P(2;-1)$, which gives us: \n$$ f(1)=f(-2)-3 $$ \nOn the other hand, $P(1;-2)$ reveals that: \n$$ f(1)f(-2)=f(-2)-3 \\implies f(1)(f(1)+3)=f(1)+3-3 \\implies f(1) =0 \\quad \\text{or} \\quad f(1)=-2 $$ \nLet's consider the case when $f(1)=0$. \n\nFrom $P(x;1)$ we have: \n$$ f(x+1)=f(x)+2x+1 \\implies f(x)=f(x-1)+2x-1 \\implies f(x)-2x+1=f(x-1)$$ \nOn the other hand, $P(x,-1)$ gives us: \n$$ f(x-1) = f(-x)-2x+1 = f(x)-2x+1 \\implies f(x)=f(-x) $$ \nNow it is quite natural to consider $P(x;-x)$: \n$$ -1 +f(x)f(-x)=f(-x^2)-2x^2+1 \\implies f(x)^2=f(x^2)-2x^2 +2 \\qquad (\\star)$$ \nAlso note that $P(x;x)$ gives us: \n$$ f(2x) +f(x)^2 = f(x^2) +2x^2 +1 \\qquad (\\star \\star) $$ \nSubtracting $(\\star)$ from $(\\star \\star )$: \n$$ f(2x) = 4x^2 -1 \\implies f(x) = x^2 -1 $$ \nIt is easy to check that this function indeed works. \n\nNow comes the painful part, when $f(1)=-2$. Since $f(1)=f(-2)-3$, we must have $f(-2)=1$. Also note that from $P(1;1)$ we have $f(2)=-3$. First of all, observe that from $P(x;1)$ we have: \n$$ f(x+1) -2f(x) = f(x)+2x+1 \\implies f(x+1) = 3f(x)+2x+1 $$ \nNow $P(x;2)$ gives us: \n\\begin{align*}\nf(x+2) -3f(x) =f(2x)+4x+1 \\\\ \n3f(x+1)+2x+3 =3f(x)+f(2x)+4x+1 \\\\\n9f(x)+6x+3+2x+3 =3f(x)+f(2x)+4x+1 \\\\ \n6f(x)+4x+5=f(2x) \\implies 6f(-x)-4x+5 = f(-2x) \n\\end{align*} \nIt also makes sense now to consider $P(x;-2)$: \n\\begin{align*} \nf(x-2) +f(x)=f(-2x) -4x+1 \\\\ \n9f(x-2)+9f(x) =9f(-2x)-36x+9 \\\\ \n3(f(x-1)-2(x-2)-1)+9f(x) = 9f(-2x)-36x+9 \\\\ \n3f(x-1)-6x+9+9f(x) =9f(-2x)-36x+9 \\\\ \n3f(x-1)+9f(x) =9f(-2x)-30x \\\\ \nf(x)-2(x-1)-1+9f(x)=9f(-2x)-30x \\\\ \nf(x)-2x+1+9f(x)=9f(-2x)-30x \\\\ \n10f(x) +1 = 9(-2x)-28 x \\\\ \n10f(x)+1 = 9(6f(-x)-4x+5) -28x \\\\ \n10f(x) =54f(-x)-64x+44 \\\\ \n5f(x) =27f(-x)-32x+22 \n\\end{align*} \nOur pain ends here, since now we can relate $f(x)$ and $f(-x)$ using only $x$ and some constants. The final step is to take a look at $P(x;-1)$: \n\\begin{align*} \nf(x-1) =f(-x)-2x+1 \\\\ \n27f(x-1)=27f(-x)-54x+27 \\\\\n27f(x-1)=5f(x)+32x-22-54x+27 \\\\ \n27f(x-1)=5f(x)-22x+5 \\\\ \n9(3f(x-1)) =5f(x)-22x+5 \\\\ \n9(f(x)-2(x-1)-1) = 5f(x)-22x+5 \\\\ \n9f(x) -18x+9 = 5f(x)-22x+5 \\\\ \n4f(x) =-4x-4 \\\\ \nf(x) = -x-1 \n\\end{align*} \nIt is easy to check that this function also satisfies given equations. With all cases being considered, we conclude that only possible functions are $f(x)=x^2-1$; $f(x)=2x-1$ and $f(x)=-x-1$.", "Solution_41": "Let $P(x,y)$ be the assertion.\n\n[hide=Finding f(0)]$P(0,0) \\rightarrow f(0)=1, -1$\nAssume $f(0)=1$.\n$P(x,0) \\rightarrow f(x)=1$\n$P(2,2) \\rightarrow f(2)=\\pm 3$, contradiction.\nThus, $f(0)=-1$.[/hide]\n\n[hide=Cases]$P(1,-1) \\rightarrow f(1)=1$ or $f(-1)=0$.\nFirst case, let $f(1)=1$.\n$P(x,1) \\rightarrow f(x)=2x-1$, which works.\nSecond case, $f(1) \\neq 1, f(-1)=0$.\n$P(-2, 1) \\rightarrow f(1)f(-2)=f(-2)-3$.\n$P(-1,-1) \\rightarrow f(-2)=f(1)+3$.\nCombining gives $f(1)=0, f(-2)=3$ or $f(1)=-2, f(-2)=1$.\nFirst subcase, let $f(1)=0, f(-2)=3$.\n$P(x,-1), P(-x,1)$ gives $f(x)=f(-x)$.\n$P(x,-x) \\rightarrow f(x)^2+2x^2=f(x^2)+2$.\n$P(x,x) \\rightarrow f(x)=x^2-1$, which works.\nSecond subcase, let $f(1)=-2, f(-2)=1$.\n$P(x-1,1) \\rightarrow f(x)+1=3f(x-1)+2x$.\n$P(1-x, -1) \\rightarrow f(-x)+1=f(x-1)+2x$.\nThen $f(x-1)=-f(-x-1)$ and some manipulations yields $f(x)=-x-1$, which works.[/hide]\n\n[hide=Finish/Summary]Thus, the only solutions are $f(x)=2x-1, x^2-1, -x-1$, which all work. $\\blacksquare$[/hide]", "Solution_42": "Let $P(x,y)$ be the assertation that $f(x+y)+f(x)f(y)=f(xy)+2xy+1$\n $P(0,0) \\Rightarrow f(0)=\\pm1$\n If $f(0)=1$ then $P(x,0) \\Rightarrow f(x)=1$\n $P(x,y) \\Rightarrow 0=2xy \\Rightarrow$ contradiction $\\Rightarrow f(0)=-1$\n $P(-1,1) \\Rightarrow f(0)+f(-1)f(1)=f(-1)-1 \\Rightarrow f(-1)(f(1)-1)=0$\n[b][u]Case 1[/u]:[/b] [b]f(1)=1[/b] $\\Rightarrow$\n$ P(x,1) \\Rightarrow f(x+1)+f(x)f(1)=f(x)+2x+1 \\Rightarrow f(x+1)=2x+1 \\Rightarrow \\boxed{f(x)=2x-1}$\n[b][u]Case 2[/u]:[/b] [b]f(-1)=0[/b] $\\Rightarrow P(-1,-1) \\Rightarrow f(-2)=f(1)+3$\n$P(-2,1) \\Rightarrow f(-2)f(1)=f(-2)-3 \\Rightarrow f(-2)=\\frac{-3}{f(1)-1} \\Rightarrow f(1)+3=\\frac{-3}{f(1)-1} \\Leftrightarrow $ $f(1)=0$ or $f(1)=-2$\n[b][u]Subcase 2.1[/u][/b][b]: f(1)=0[/b] $\\Rightarrow P(x,1) \\Rightarrow f(x+1)=f(x)+2x+1$\n$P(x+1,-1) \\Rightarrow f(x)=f(-x-1)-2x-1 \\Rightarrow f(x+1)=f(-x-1) \\Rightarrow f(x)=f(-x)$ for all $x \\in \\mathbb{R}$\n$P(x,-x) \\Rightarrow f(x)^2=f(x^2)-2x^2+2 \\Rightarrow P(x,x) \\Rightarrow f(2x)+f(x)^2=f(x)^2+2x^2-2+2x^2+1 \\Rightarrow \\boxed{f(x)=x^2-1}$\n[b][u]Subcase 2.2[/u]: f(1)=-2 [/b]$\\Rightarrow P(1,1) \\Rightarrow f(2)=-3$\n$P(x,1) \\Rightarrow$ $f(x+1)=3f(x)+2x+1$ [b](1)[/b] $\\Rightarrow f(x+2)=9f(x)+8x+6$ [b](2)[/b]\n$P(x,-1) \\Rightarrow$ $f(x)=f(-x-1)-2x-1 \\Rightarrow f(x)=3f(-x)-4x+2$ [b](3)[/b] \nUsing (1) and (3): $P(x,-x) \\Rightarrow$ $f(x)f(-x)=f(-x^2)-2x^2+2 \\Rightarrow f(x^2)=\\frac{3f(x)^2+4xf(x)+2f(x)-2x^2-4}{3}$[b](4)[/b]\nUsing (1),(3) and (4): $P(x,x) \\Rightarrow f(2x)+f(x)^2=f(x^2)+2x^2+1 \\Rightarrow f(2x)=\\frac{4xf(x)+2f(x)+4x^2-1}{3}$[b](5)[/b]\nUsing (2) and (5): $P(x,2) \\Rightarrow f(x+2)-3f(x)=f(2x)+4x+1\\Rightarrow$ after substituting $\\boxed{f(x)=-x-1}$\nAfter checking we see that all 3 solution work." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove the inequalities\r\n\r\n\r\n1) m_a+m_b+m_c \\geq a+b\r\n\r\n2) m_b+m_c \\leq 3/4(a+b+c)\r\n\r\n\r\nwhere m_a,m_b,m_c are medians of a triangle of sides a,b,c.", "Solution_1": "2) Is easy :\r\nLet B' be the midpoint of the side AC, and B'' be the symmetric of B with respect to B'.\r\nThus BB' = 2m_b < BA + AB'' = c + a.\r\nSimilarly 2m_c < b + a.\r\nAdding, we get the result from the triangle inequality.\r\n\r\nPierre.", "Solution_2": "For the first one:\r\n\r\nLet G be mass-center. Then \r\n GA+GC >= b\r\n GB + GC>= a\r\n (GA+GB)/2 - GC >= 0\r\nSum up these ineqs we have done.\r\n\r\nNamdung" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "show that the set of numbers in the interval [0,1] having decimal expansions using only odd digits is closed. describe this set by a Cantor-set type construction.", "Solution_1": "Well, if they've got only odd digits, then the only possibility is $\\{1\\}$, which is of course closed :lol:\r\n\r\nEdit: I'm joking, of course. The closedness is obvious (show that the complement is open)." } { "Tag": [ "MATHCOUNTS", "AMC", "USA(J)MO", "USAMO", "Putnam", "USAMTS", "\\/closed" ], "Problem": "You can now hide the forums that you don't usually visit, to reduce some of the clutter on the Forum main page.\r\n\r\nThe new option is available in the [url=http://www.mathlinks.ro/Forum/usercp.php]options menu[/url]. Once there click on \"Hide Forums Panel\". \r\n\r\nAs in the pictures below, for each forum you can:\r\n\r\n- Select \"View\" to have that forum appear on the Forum main page.\r\n\r\n- Select \"Hide Forum\" to have that forum be hidden on the Forum main page.\r\n\r\n- Click on the green arrow to copy that forum's setting to all of its subforums.\r\n\r\nYou must click Submit at the bottom of the page for any changes to take effect.\r\n\r\nIf you hide a forum but leave visible any of its subforums, then the subforums will still appear on the Forum main page.\r\n\r\nYou can still visit a forum that you have hidden, by selecting it on the \"Jump to:\" link on the lower-right of the Forum page.\r\n\r\nIf you have any questions about this, please post them in this topic.", "Solution_1": "The options pane doesn't display with the same line spacing as the forum list pane when I view the hide forums with Firefox. I'm not sure what forum I'm choosing for that way, so I don't dare make any changes.", "Solution_2": "[quote=\"tokenadult\"]The options pane doesn't display with the same line spacing as the forum list pane when I view the hide forums with Firefox. [/quote]I cannot reproduce this particular problem on my copy of Firefox (version 1.0). However, there are definitely some issues using this feature in Firefox...in particular, it is [b]very[/b] slow to load. This is due to the way that Firefox processes Javascript. However, once the page is completely loaded (and it takes about 2 minutes :sleeping: on my machine), it does seem to be lined up fine (again, at least on my machine) and the feature [i]does[/i] work.\r\n\r\nWe'll look into it, but I don't think this is going to be any easy fix (if it's fixable at all).", "Solution_3": "YAY!!! Thank you! :clap: I now don't see College or Olympiad or non-North American forums, since I've never been there. Let me reiterate: THANK YOU!\r\n\r\nNo problem on Deer Park Alpha 1, though it takes some time to load; that's why you have multiple tabs loading at one time ;)\r\n\r\n:clap: :clap2: \r\n\r\nBilly", "Solution_4": "[quote=\"tokenadult\"]The options pane doesn't display with the same line spacing as the forum list pane when I view the hide forums with Firefox. I'm not sure what forum I'm choosing for that way, so I don't dare make any changes.[/quote]The quickest and simplest solution would be just to use another browser to modify your hide forums settings. Obviously you won't do this often (it beats the whole purpose of it :P) so I recommend you use IE to set up your options. \r\n\r\nFirefox has some major issues dealing with Java Script. This is a problem of the browser (on some computers, on mine it loads in about 20 seconds and it works fine), but I don't think there are any problems with either Opera or IE :)", "Solution_5": "Firefox actually works fine for me...it didn't take long to load at all! :) Thankyou for this new feature, you guys are doing a great job with this forum! :)", "Solution_6": "[quote=\"chess64\"]Firefox actually works fine for me...it didn't take long to load at all! :)[/quote]Mine work fine too (on both my home computer, my laptop, and my aops computer). However on some computers (especially those not having a high processor speed and a ton of RAM) it might not work. I generally use IE to switch the options, as IE is instant.\n\n[quote=\"chess64\"]Thankyou for this new feature, you guys are doing a great job with this forum! :)[/quote]Thanks! :P We are trying :D", "Solution_7": "Yep, I use Firefox, but you're right, it is loaded with ram and processing speed. By the way, nice avatar/picture: GO YODA! YODA RULES!", "Solution_8": "thank you for all the AOPs people who set up this forum, yall r the best! :first: :clap2: \r\n\r\nand go yoda", "Solution_9": "Firefox works fine here.\r\n\r\nBut there's one little bug, in Firefox and IE. The Middle School green arrow copies the setting to all boars, instead of only the Middle School-subforums. The other green arrows work fine.", "Solution_10": "[quote=\"Stijn\"]Firefox works fine here.\n\nBut there's one little bug, in Firefox and IE. The Middle School green arrow copies the setting to all boars, instead of only the Middle School-subforums. The other green arrows work fine.[/quote]What is \"boars\"? Are you sure you are clicking the middle school arrow and not the whole forum arrow? There's no such bug on my computer(s) ...", "Solution_11": "I meant to write 'boards'/'forum'.\r\n\r\nYes, I'm sure clicking the correct arrow. I'm going to 'digg' into the source code a bit, see if I can find the problem.", "Solution_12": "[quote=\"Stijn\"]I meant to write 'boards'/'forum'.\n\nYes, I'm sure clicking the correct arrow. I'm going to 'digg' into the source code a bit, see if I can find the problem.[/quote]Can you post a print screen of the arrow you are clicking and what it does (before and after the click)? \r\n\r\nPS How are you planning to \"digg into the source code\"?!! :huh:", "Solution_13": "For some reason, he thinks of the MATHCOUNTS forum having both id 132 (correct) as 301 (incorrect/nonexistant).\r\nThe Middle School green arrow does javascript:branch_set('styles',298,301), but since there is no board 301, the keeps changing the View/Hide option to the bottom.\r\nThe same with the MATHCOUNTS green arrows, which does javascript:branch_set('styles',132,301).", "Solution_14": "[quote=\"Valentin Vornicu\"]Can you post a print screen of the arrow you are clicking and what it does (before and after the click)? [/quote]\r\n\r\nIt doesn't work for me either. See attached.", "Solution_15": "[quote=\"Stijn\"]For some reason, he thinks of the MATHCOUNTS forum having both id 132 (correct) as 301 (incorrect/nonexistant).\nThe Middle School green arrow does javascript:branch_set('styles',298,301), but since there is no board 301, the keeps changing the View/Hide option to the bottom.\nThe same with the MATHCOUNTS green arrows, which does javascript:branch_set('styles',132,301).[/quote]Can you post a print screen?!! Thanks. I have not experienced this problem on any computer(s) ...", "Solution_16": "[url=http://download.stijnshome.be/before.png]Before[/url] and [url=http://download.stijnshome.be/after.png]after[/url] I clicked on the Middle School green arrow. Notice everything below changes to 'Hide Forum'. Notice the status bar javascript-message too.", "Solution_17": "I am having no such problems on none of my installed browsers ... this is very weird. What browsers are you guys using?", "Solution_18": "Internet Explorer. He's using Firefox. See another screenshot attached.", "Solution_19": "I have the problem on both Firefox 1.0.4 and IE6. So I doubt this is an browser-related issue...", "Solution_20": "[quote=\"Stijn\"]I have the problem on both Firefox 1.0.4 and IE6. So I doubt this is an browser-related issue...[/quote]Well I have those 2. And I have tried reproducing the bug on 3 different computers, and with 2 different accounts (my account, and a regular user account, just to be on the safe side) and I could not do it ... it might be a problem with your java installation. Do you have the latest java rte?", "Solution_21": "Anyway, besides this (minor) bug, are there any important bugs you guys have noticed? In the end just avoid hiding the middle school forums, or do it manually (there are only like 5 subforums :P) ...", "Solution_22": "I didn't notice the arrow thingy and changed each and every one of the forums I didn't want to see manually (that's all of the forums and subforums in the olympiad and college math sections). Bleh. Oh well...thanks so much for this feature. The forums loading and then collapsing was beginning to drive me crazy.", "Solution_23": "Valentin, if you don't experience this, YOUR java is not working ;)\r\n\r\nWhere the source code writes\r\n\r\njavascript:branch_set('styles', 298, 301);\r\n\r\nit should be \r\n\r\njavascript:branch_set('styles', 298, xxx);\r\n\r\nwith xxx the last forum in that category, I don't know which forums admins have there but for normal users it is 132.", "Solution_24": "[quote=\"Valentin Vornicu\"]Anyway, besides this (minor) bug, are there any important bugs you guys have noticed? In the end just avoid hiding the middle school forums, or do it manually (there are only like 5 subforums :P) ...[/quote]\r\n\r\nI noticed another strange thing, though it may be because of my stupidity.\r\n\r\nIn high school, I hided only:\r\nGetting Started\r\n\r\nAnd the forums that I see there now in high school are:\r\nGrading Issues\r\nOther US Contests & Programs\r\nInternational Contests & Programs\r\n\r\nWhich do not belong there at all. This also prevents me from accessing the forum I am moderating :o which cannot be intended?\r\n\r\n[edit] when I then click on the category, I get the normal selection. It would be nice if I could get it main screen though ;) [/edit]", "Solution_25": "Valentin, could you please make that posts from hidden forums do not appear at \"View posts since last visit\" link?", "Solution_26": "[quote=\"Peter VDD\"]I noticed another strange thing, though it may be because of my stupidity.\n\nIn high school, I hided only:\nGetting Started\n\nAnd the forums that I see there now in high school are:\nGrading Issues\nOther US Contests & Programs\nInternational Contests & Programs\n\nWhich do not belong there at all. This also prevents me from accessing the forum I am moderating :o which cannot be intended?\n\n[edit] when I then click on the category, I get the normal selection. It would be nice if I could get it main screen though ;) [/edit][/quote]If you won't post a print screen it is very hard to understand what is the problem you are experiencing .... \r\n\r\nYou have probably hidden only the mother forums, not all the subforums too ... :)\r\n\r\n@Bojan: no, that will not happen very soon. Partially because the hidden forums thing, is just what it says it is: hiding the forums, so that your index.php pages are not as crowded. But I want everyone to still be able to click on links to the hidden forums or jump into them using the jumpbar.", "Solution_27": "[quote=\"Valentin Vornicu\"]You have probably hidden only the mother forums, not all the subforums too ... :)[/quote]I'm not THAT stupid ;)\r\n\r\nHere are the print screens, the error in the index and the correct display in the category-view.", "Solution_28": "I discovered why it is. There are a couple of forums (grading issues, other US contests, classes information - very annoying forum :?) that don't have buttons, and therefor can't be hidden. (and I still had to roll them out - this is quite confusing)\r\n\r\nCould you get those without a button fixed soon? :) \r\n\r\nThanks for this great system by the way, it is indeed this we wanted when the roll-out system came. :)\r\n\r\n(besides, is it normal that the configuration page takes >1 minute to display? :?)", "Solution_29": "[quote=\"Valentin Vornicu\"]\n@Bojan: no, that will not happen very soon. Partially because the hidden forums thing, is just what it says it is: hiding the forums, so that your index.php pages are not as crowded. But I want everyone to still be able to click on links to the hidden forums or jump into them using the jumpbar.[/quote]\r\nWhen I come to the forum I click on \"\"View posts since last visit\" and read new posts. I think it is logical that if one hid some forums then doesn't want to read new posts from them, but if you have different opinion - no problem, it's your choice. Thank you anyway.", "Solution_30": "Awesom feature! :ninja:", "Solution_31": "[quote=\"Bojan Basic\"][quote=\"Valentin Vornicu\"]\n@Bojan: no, that will not happen very soon. Partially because the hidden forums thing, is just what it says it is: hiding the forums, so that your index.php pages are not as crowded. But I want everyone to still be able to click on links to the hidden forums or jump into them using the jumpbar.[/quote]\nWhen I come to the forum I click on \"\"View posts since last visit\" and read new posts. I think it is logical that if one hid some forums then doesn't want to read new posts from them, but if you have different opinion - no problem, it's your choice. Thank you anyway.[/quote]I said that it will not happen very soon. You guys should understand that it\u00b4s not very easy to add all these features and a lot of effort is put into each of them (besides all the other stuff we have to do like jobs, college, classes and stuff :) ). \r\n\r\nPeter: some of the forums cannot and will not be hidden on purpose. We want people to be able to see those forums all the time (I am saying that it\u00b4s not a bug).", "Solution_32": ":o \r\n\r\nwhy do non-US residents need to see \"other US contests\"?? :huh: we can't participate in any of them?\r\n\r\nSame for \"grading issues\", what's the use of it? The classes I can understand, since it is a source of income to you, but those other 2 forums seem completely useless to me, especially since they focus on US-things in which I cannot participate.\r\n\r\nIt is a sad evolution, and excuse me if I'm wrong, but I get more and more the impression that the site gets mainly designed for americans... and if you're not, too bad. :|", "Solution_33": "[quote=\"Peter VDD\"]:o \n\nwhy do non-US residents need to see \"other US contests\"?? :huh: we can't participate in any of them?\n\nSame for \"grading issues\", what's the use of it? The classes I can understand, since it is a source of income to you, but those other 2 forums seem completely useless to me, especially since they focus on US-things in which I cannot participate.\n\nIt is a sad evolution, and excuse me if I'm wrong, but I get more and more the impression that the site gets mainly designed for americans... and if you're not, too bad. :|[/quote]I think that you are over reacting :dry: . As I told you, most of you [u]have absolutely no idea how much amount of work[/u] is spent such that you could have all this. Of course nobody is ever fully satisfied, but we are doing out best :?\r\n\r\nUp until a couple of days ago you coudn't hide any forums. Now you can hide 95% of them, but I am making modifications for US residents only .... :| Jeez ... \r\n\r\nHas it [b]ever[/b] crossed your mind that some of the forums had simply been forgotten on the setting \"not able to hide\"? Of course not, it's easier to blame it on the Romanians that do work only for the US-residents :no:\r\n\r\nPS Just for the record you need not be a US-resident to participate in many of the US competitions. I know a bunch of Romanians that took AMCs, and there are 3 Canadian winners of the USAMO this year on the forum.\r\n\r\nPPS If you can't participate in a competition that doesn't mean that you can't or shouldn't check that competitions' forum ... or you shouldn't be ever looking in the Putnam forum, right?", "Solution_34": "my screen just doesn't let me have the \"show\" or \"not show\" buttons... do you know why?? I'm using Internet Explorer, and the buttons showed the first time but now they don't show. is there a problem with my computer? Thanks. :)\r\n\r\n\r\n\r\nsorry if that image is too big but I wasn't sure how to make it so you could download the pic. but as you can see it doesn't work :(\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=13[/img]", "Solution_35": "Lol... Valentin, I think you misunderstood me a bit ;)\r\n\r\nI'm very thankful for the modification made here, as I've posted some posts above! I was just replying to what you said, that it is _intended_ that those forums cannot be hidden. \r\n\r\nFirst you say:\r\n[quote=\"Valentin Vornicu\"]Peter: some of the forums cannot and will not be hidden on purpose. We want people to be able to see those forums all the time. (I am saying that it\u00b4s not a bug).[/quote]\nThen you say:\n[quote=\"Valentin Vornicu\"]Has it [b]ever[/b] crossed your mind that some of the forums had simply been forgotten on the setting \"not able to hide\"?[/quote]\r\n-> I don't get which of those it is now :)\r\n\r\nBut again, don't get me wrong. I don't \"blame\" you for anything! I just wonder why \"purely american\" forums should be on the list \"cannot be hidden by anyone\"...\r\n\r\n...And eum, perhaps that reply is a bit overreacting as well ;) I have a lot of respect and thanks to your work here. Sorry if my previous message wasn't clear in that. :oops:", "Solution_36": "Ive also got a similar problem:\r\n\r\n\r\nhttp://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=16", "Solution_37": "Ok, as soon as I am back from the IMO I will take a look to check if all the forums that are supposed to be able to be hidden are actually. For now, I have to much things to do here in Mexico :D :juggle: :play_ball:", "Solution_38": "[quote=\"Peter VDD\"]Same for \"grading issues\", what's the use of it?[/quote]\r\n\r\nI assume it's just a temporary oversight, as it clearly is useful only to the USAMTS participants here.", "Solution_39": "I think this is an excellent feature, I don't get what you guys are talking about with all that javascript error fancy things (I know, awful grammar :P ). \r\n\r\nNow, I barely have to scroll to get to the forums I want. :D \r\n\r\nAnd you can't hide the other US contests and programs? :huh:", "Solution_40": "May you explain your rationale behind prohibiting the Calculus/Olympiad/College links from being hidden? Not be to disrespectful; I appreciate your work, and I'm just curious as to why they are forced when there are easier ways to get to the desired forums.", "Solution_41": "I don't believe they are prohibited form being hidden. I have them hidden on my account. Make sure you hide them them in Profile->Hide Forums." } { "Tag": [ "probability" ], "Problem": "A deli sandwich consists of one type of bread, one type of meat, and one type of sauce. The deli offers wheat, rye, and white bread; ham, turkey, roast beef, and tuna; ranch and southwest chipotle sauce. Javier is allergic to rye bread, tuna, and southwest chipotle sauce. He tells the server to assemble a random deli sandwich. What is the probability that Javier will suffer an allergic reaction?", "Solution_1": "There are 3 types of bread, 4 types of meat, and 2 types of sauce. He is allergic to one kind of bread, one kind of meat, and one kind of sauce. The probability that he doesn't have an allergic reation is thus $ \\frac {2}{3}\\cdot{\\frac {3}{4}}\\cdot{\\frac {1}{2}} \\equal{} \\frac {1}{4}$. Therefore, the probability that he does suffer an allergic reation is $ 1 \\minus{} \\frac {1}{4} \\equal{} \\boxed{\\frac {3}{4}}$", "Solution_2": "To people like me with food allergies, this problem seems partially disrespectful. \n\n[hide=Solution]There are $3\\cdot2\\cdot1=6$ ways for him to choose a sandwich without any of his allergens, and $4\\cdot3\\cdot2=24$ ways to make a sandwich. So, the probability of him suffering an allergic reaction is $1-\\frac{3\\cdot2\\cdot1}{4\\cdot3\\cdot2}=\\boxed{\\frac{3}{4}}$.[/hide]\n\nI hope he's carrying some epinephrine or something else like that.", "Solution_3": "why why why did javier tell the server to make a random deli sandwich\n\n[hide=hhhhhhh]so there are 3*4*2 different deli sandwiches\n2*3*1 deli sandwiches that wont trigger an allergic reaction\n6/24=1/4\nsubtract from 1\nthen there's 3/4 chance that he suffers an allergic reaction[/hide]", "Solution_4": "what was bro thinking, bros literally going to die\n[hide=smh]\n3*4*2 = 24 total sandwiches\n2*3*1 = 6 where he lives\n18 sandwiches where he dies[/hide]", "Solution_5": "bru Javier" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "If $X$ and $Y$ are metric spaces and $f\\;: \\;X\\;\\rightarrow\\;Y$ is a mapping between them, show that the following statements are equivalent:\r\na) $f^{-1}(B)$ is open in $X$ whenever $B$ is open in $Y$.\r\nb) $f^{-1}(B)$ is closed in $X$ whenever $B$ is closed in $Y$.\r\nc) $f(\\overline{A})\\subseteq\\overline{f(A)}$ for every subset $A$ of $X$.", "Solution_1": "[quote=\"towersfreak2006\"]If $X$ and $Y$ are metric spaces and $f\\;: \\;X\\;\\rightarrow\\;Y$ is a mapping between them, show that the following statements are equivalent:\na) $f^{-1}(B)$ is open in $X$ whenever $B$ is open in $Y$.\nb) $f^{-1}(B)$ is closed in $X$ whenever $B$ is closed in $Y$.\nc) $f(\\overline{A})\\subseteq\\overline{f(A)}$ for every subset $A$ of $X$.[/quote]\r\n\r\nIt has been done in one of my very last posts." } { "Tag": [], "Problem": "What is the largest 2 digit prime factor of $ {200\\choose 100}$?", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=328131#328131[/url]" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "geometry unsolved" ], "Problem": "We are given a square $ ABCD$. Let $ P$ be a point not equal to a corner of the square or to its center $ M$. For any such $ P$, we let $ E$ denote the common point of the lines $ PD$ and $ AC$, if such a point exists. Furthermore, we let $ F$ denote the common point of the lines $ PC$ and $ BD$, if such a point exists. All such points $ P$, for which $ E$ and $ F$ exist are called acceptable points. Determine the set of all acceptable points, for which the line $ EF$ is parallel to $ AD$.", "Solution_1": "First, we show that P must lie on the circumcircle of $ DOC$, where O is the center of the square.\r\n\r\nAssume $ EF$ is parallel to $ AD$. Then $ EF \\perp CD$ and $ CE \\perp DF$. Therefore, $ E$ is the orthocenter of $ FDC$, and thus $ DE \\perp CF$. Therefore, $ \\angle DPC\\equal{}90$, and this along with $ \\angle DOC\\equal{}90$ yield the desired result.\r\n\r\nNow, the acceptable points on this circle are merely those points that are [b]not $ O,C,D$[/b] or the reflection of $ O$ about $ CD$. We will show that all of the points on the circle that are not $ O,C,D$ or the reflection of $ O$ about $ CD$ work.\r\n\r\n$ CO \\perp DF$ and $ DP \\perp CF$ together imply that the intersection of $ CO$ and $ DP$, $ E$, is the orthocenter of triangle $ DFC$. Therefore, $ EF\\perp CD$ and we are done.", "Solution_2": "it suffices to prove $\\angle CPD=90$\nbesides,I think that the answer can include O." } { "Tag": [ "geometry", "geometric transformation", "rotation", "analytic geometry" ], "Problem": "The disk of m mass has a rotation axis. Find the lagrangian of the system.\r\nThe generalized coordinate is the angle $\\phi$ \r\n[url=http://img53.imageshack.us/my.php?image=discku6.jpg][img]http://img53.imageshack.us/img53/4086/discku6.th.jpg[/img][/url]", "Solution_1": "$I = \\frac{1}{2}m r^{2}$.\r\n\r\n$T = \\frac{1}{2}I (\\dot \\phi)^{2}= \\frac{1}{4}m r^{2}(\\dot \\phi)^{2}$.\r\n\r\n$V = V_{1}+V_{2}= \\frac{1}{2}C_{1}x^{2}+\\frac{1}{2}C_{2}x^{2}= \\frac{1}{2}C x^{2}$,\r\nwhere $C = C_{1}+C_{2}$.\r\n\r\nBut, $x = r \\phi$.\r\n\r\nThus, $V = \\frac{1}{2}C r^{2}\\phi^{2}$.\r\n\r\nThus, $L = \\frac{1}{4}m r^{2}(\\dot \\phi)^{2}-\\frac{1}{2}C r^{2}\\phi^{2}$.\r\n\r\nBut, we can cancel the $r^{2}$, since it is just a constant factor of everything in $L$.\r\n\r\nTherefore, we have $\\frac{1}{4}m (\\dot \\phi)^{2}-\\frac{1}{2}(C_{1}+C_{2}) \\phi^{2}$." } { "Tag": [ "geometry" ], "Problem": "$ ABCD$ is a quadrilateral such that $ AB \\equal{} AD$. If $ \\angle ACB \\equal{} \\angle ACD$, prove that $ ABCD$ is cycle quadrilateral.\r\n\r\nSorry for my bad English :D .", "Solution_1": "Are you sure this is true?\r\n [geogebra]1836c2422573c312791e2e5bffdabefd8015762f[/geogebra]" } { "Tag": [ "LaTeX" ], "Problem": "In my Latex document for an AoPS class final, I used eqnarray with the equation numbers for one of the problems. If I use eqnarray again for a later problem, how can I get the equation numbers to start back at (1)? Or should I just have everything continuing throughout the document, even though the problems are unrelated?", "Solution_1": "Use the command \\setcounter{equation}{0} before begining the new eqnarray enivronment.", "Solution_2": "yeah that works.\r\n\r\nThanks!" } { "Tag": [], "Problem": "the chickens and pigs in Farmer McCoy's barn have a total of 50 heads and 170 legs. How many pigs are in the barn?", "Solution_1": "Let $ x\\equal{}\\text{chickens}$, $ y\\equal{}\\text{pigs}$.\r\n$ \\left\\{\\begin{array}{l}\r\nx\\plus{}y\\equal{}50 \\\\\r\n2x\\plus{}4y\\equal{}170\r\n\\end{array}\\right.$\r\n$ \\iff (x, y)\\equal{}(15, 35)$\r\n\r\nThus, $ \\boxed{35}$." } { "Tag": [ "Diophantine equation", "number theory unsolved", "number theory" ], "Problem": "Given the Pell equation $ x^2 \\minus{} dy^2 \\equal{} 1$ where $ d$ is not a perfect square and a fixed positive integer $ k$, prove that you can find infinitely many solutions with $ y$ divisible by $ k$.", "Solution_1": "put $ y\\equal{}y'k$ then the pell equation will be:\r\n\r\n$ x^2\\minus{}(dk^2)y'^2\\equal{}1$\r\n\r\nput $ d'\\equal{}dk^2$ then $ d'$ isn't a perfect square and so the pell equation $ x^2\\minus{}d'y'^2\\equal{}1$ has infinitely many solutions." } { "Tag": [ "Putnam", "function", "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Let $ \\mathbb{P}_n$ be the set of subsets of $ \\{1,2, \\cdots, n\\}$. Let $ c(n,m)$ be the number of functions $ f : \\mathbb{P}_n \\to \\{1,2, \\cdots, m\\}$ such that $ f(A \\cap B) \\equal{} \\mbox{min}\\{f(A), f(B)\\}$. Prove that $ c(n,m) \\equal{} \\sum_{j \\equal{} 1}^m j^n.$\r\n\r\nI don't really get the notation used in this question, what does $ f(A \\cap B) \\equal{} \\mbox{min}\\{f(A), f(B)\\}$ mean? And how do you approach this question?\r\n\r\nThank you!", "Solution_1": "The minimum of two numbers is another number. That makes this a meaningful equation, and we want to find all functions on sets satisfying it.\r\n\r\nThe nicest way to prove a counting formula is to construct a bijection. For a typical term $ j^n$, what does $ j^n$ count? Well, the obvious choice is functions from $ \\{1,2,\\cdots,n\\}$ (or some other set of size $ n$) to $ \\{1,2,\\cdots,j\\}$ (or some other set of size $ j$).\r\n\r\nOK, what if we started with a function $ g$ from $ \\{1,2,\\cdots,n\\}$ to $ \\{1,2,\\cdots,j\\}$; we want to turn that into an $ f$ satisfying our property. Clearly*, we should have $ f(\\{k\\})\\equal{}g(k)$. Then $ f(\\{1,2\\})\\equal{}\\dots$ Hmm, maybe this isn't so obvious- but if we look instead for $ f(A\\cup B)\\equal{}\\min\\{f(A),f(B)\\}$, we can build everything easily, and $ f(A)\\equal{}\\min_{x\\in A}g(x)$ for nonempty $ A$. How to go from unions to intersections? DeMorgan's laws; just flip to the complement $ f(\\{1,2,\\cdots,n\\}\\setminus k)\\equal{} g(k)$.\r\n* OK, that was really just a guess- but one-element sets are usually a good place to start when building a set function.\r\n\r\nThat's all very nice, but we've got two problems: the functions we get for $ j$ are a subset of those for $ j\\plus{}1$ rather than being disjoint, and we haven't accounted for the value of $ f$ on the whole set $ S$. How to deal with that? The obvious choice is to just pick $ f(S)$ and see what that does to the rest of the problem. Since $ f(A)\\equal{}f(S\\cap A)\\equal{}\\min\\{f(S),f(A)\\}$ for any $ A$, $ f(S)$ is at least as large as any other $ f(A)$. If $ f(S)\\equal{}j$, all other values are in $ \\{1,2,\\cdots,j\\}$; pick $ n$ of them to be $ f(S\\setminus k)$, and we have $ j^n$ functions with $ f(S)\\equal{}j$. Sum over $ j$, and we get the desired formula $ c(n,m)\\equal{}\\sum_{j\\equal{}1}^m j^n$.\r\n\r\nThat whole post was pretty much stream-of-consciousness; it's the process I just went through to solve this.", "Solution_2": "Thanks for that but what about using induction to prove it?\r\n\r\nThis is what I have so far:\r\n\r\nSay you had a function $ f : \\{1,2,3\\} \\to \\{1,2,3\\}$ then there exist $ 3^3$ different functions.\r\n\r\nIf you had $ f: \\{1,2...n\\} \\to \\{1,2...m\\}$ then there exist $ m^n$ different functions.\r\n\r\nIf you had $ f: \\mathbb{P}_n \\to \\{1,2...m\\}$ then there exist $ m^{2^n}$ different functions.\r\n\r\nSo out of these $ m^{2^n}$ functions we need to find how many satisfy the restriction $ f(A \\cap B) \\equal{} \\mbox{min}(f(A) , f(B))$\r\n\r\nSo then using induction:\r\n\r\nBase Case:\r\n\r\nLet $ m \\equal{} 1$\r\n\r\n$ f: \\mathbb{P}_n \\to \\{1\\}$\r\n\r\n$ c(n,1) \\equal{} 1 and 1^{2^n} \\equal{} 1$\r\n\r\nWhich is true since the only possible function satisfies the restriction $ f(A \\cap B) \\equal{} \\mbox{min}(f(A) , f(B))$\r\n\r\nInductive Hypothesis:\r\n\r\nAssume it is true for $ m \\equal{} k$\r\n\r\nSo $ c(n,k) \\equal{} \\sum_{j\\equal{}1}^k j^n \\equal{} 1\\plus{} 2^n\\plus{}3^n \\plus{} ... \\plus{} k^n$ is true\r\n\r\nProof:\r\n\r\nNeed to prove it is true for $ m \\equal{} k\\plus{}1$\r\n\r\n$ c(n,k\\plus{}1) \\equal{} \\sum_{j\\equal{}1}^{k\\plus{}1} j^n \\equal{} 1\\plus{}2^n\\plus{}3^n\\plus{}...\\plus{}k^n\\plus{}(k\\plus{}1)^n$\r\n\r\n$ \\therefore c(n,k\\plus{}1) \\equal{} c(n,k) \\plus{} (k\\plus{}1)^n$\r\n\r\nThen what?", "Solution_3": "Then you ask which functions are the new ones (they're the ones with the value $ k\\plus{}1$ somewhere), which puts you in pretty much the same place as my solution, only possibly handled in a different order. I don't see a way to do this without that explicit bijection for each $ j^n$." } { "Tag": [ "algebra", "polynomial", "college contests" ], "Problem": "a) Find all polynomials of the form \r\n $ f(x) = x^5 - 3x^4 + 2x^3 + ax^2 + bx + c $\r\nif it is known that $f(x)$ is divided by $(x-1)(x+1)(x-2)$.\r\nb) Let $P(x), Q(x), R(x)$ be plolynomials with real coefficients of degree 3, 2, 3 respectively satisfying condition $(P(x))^2 + Q(x)^2 = R(x)^2 $. Find the minimal possible number of real roots (counted with multiplicity) of polynomial $T(x) = P(x)Q(x)R(x)$.", "Solution_1": "[b]a[/b]: Easy to find out the result\r\n[b]b[/b]: At least 6 ;)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "ARML", "AMC 10" ], "Problem": "Perhaps it is my ignorance of perhaps it is rumor but... \r\n\r\nI have heard that you can qual for USAMO (MOP etc.) as a first yr. undergrad. Is this rite? I was wondering why it said grade 13 on some of those USAMOers on the list...", "Solution_1": "Sorry, but you can not take the USAMO in college. That is because, as you know, you must take the AMC 10 or AMC 12 to qualify. The AMC 12 is only open to people in 12th grade, or under. Also, I checked and there are no people in 13th grade who qualfied to take the USAMO.", "Solution_2": "Were those grade 13ers Canadian?", "Solution_3": "Silly silly me.. I should have checked earlier.. Blasted rumors... Sorry for wastng your time..\r\n\r\nPeace,\r\nIW.OAV", "Solution_4": "Hah! I was right! I think... Look at Eunice Kim here \r\n\r\nhttp://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2001-ua/01usamoqualstate.html\r\n\r\nFrom the 2001 olympiad... Eunice is a pretty tite name...\r\n\r\nOn a completely random aside, I enjoy cheese and my focus is brokus....\r\n\r\nPeace,\r\nIW.OAV :twisted:", "Solution_5": "Hm...That's peculiar. Perhaps there are some exceptions. Also, I have heard people who take the year off after finishing high school, so maybe that's what \"13\" is referring to.\r\n\r\n-interesting_move", "Solution_6": "Yeah take the year off just to do USAMO... very chic indeed.\r\n\r\nIW.OAV", "Solution_7": "No, that might not be the reason for taking a year off. It could be that you want to take a break from all the studying in high school. Or, you might just do the USAMO...lol\r\n\r\n-interesting_move", "Solution_8": "While extremely bored, I was checking out the Exeter Academy website. Apparantly kids go to Exteter after they finished high school for one year. That can also be another thing.", "Solution_9": "some kids in india go for a 13th grade. it doesnt mean theyre held back or anything. i dunno why though. hmmmmmmm. i never did read that arml math jam. i should do that", "Solution_10": "The 13th graders taking the USAMO are typically Canadians who choose to take that additional year of precollege schooling. At least, that's what I always heard.", "Solution_11": "There's also the possibility of a typo/misbubbling/scanning error. The AMC is hardly perfect in these things.", "Solution_12": "JONGMIN BAEK 9 CUPERTINO HS CUPERTINO CA \r\nAARON KLEINMAN 11 CUPERTINO HS CUPERTINO CA \r\n\r\nch-ching", "Solution_13": "[quote=\"Lord Venom\"]JONGMIN BAEK 9 CUPERTINO HS CUPERTINO CA \nAARON KLEINMAN 11 CUPERTINO HS CUPERTINO CA \n\nch-ching[/quote]\r\n\r\nAnd how exactly does this have anything to do with the present topic? Not being mean or anything, just confused. :?", "Solution_14": "Those are the people from his school who qualified for the USAMO 2001, but I don't understand it either.", "Solution_15": "Query: Is he one of them?", "Solution_16": "No.", "Solution_17": "Ooh, I think Simon just posted the shortest message yet (see \"languages\" under Etc. if you don't understand, but since Simon Reads All, he probably does know...)", "Solution_18": "I was thinking about that when I read your other post. It may have been the shortest post, except I believe Chinaboy posted LOL with no period, so that would be tied in terms of the number of characters." } { "Tag": [ "analytic geometry", "geometry", "trapezoid", "algorithm", "area of a triangle" ], "Problem": "Let points A and B have coordinates (0,1) and (4,0) respectively. There is a point C in the first quadrant that lies on the line y = x + 1 and such that the area of triangle ABC is 20 square units. Find the coordinates of C.\r\nhint:\r\n[hide] use a determinant for the area of a triangle...[/hide]", "Solution_1": "[quote=\"Apprentice\"]hint:\n[hide] use a determinant for the area of a triangle...[/hide][/quote]\nThere's a way to solve this problem even if you don't know determinants.\n[hide=\"Hint\"]\n$A (0,1)$ is on the line $y=x+1$.\nSo if we regard $AC$ as the base of the triangle, the height is the distance between $B(4,0)$ and the line $y=x+1$.\n[/hide]", "Solution_2": "[quote=\"Apprentice\"]\n use a determinant for the area of a triangle...[/quote]\r\nHow can you expres the area of a triangle $ABC$ as a determinant with $A(x_{A},y_{A}),B(x_{B},y_{B}),C(x_{C},y_{C})$. Is there a way to obtain it?", "Solution_3": "See my PM. It's kind of long and is in Romanian, so I won't put it here. There are two ways to prove it:\r\n\r\n1. the formula for distance (from $A$ to $BC$, which is an altitude in our case);\r\n\r\n2. consider the trapezium formed by $AB$ and $Ox$ (with a right angle) and so on... sum up the areas of the three trapeziums. (or smth like this)", "Solution_4": "OK. I'm just going to post the solution now.\r\n[hide=\"Answer\"] $\\boxed{C(8,9)}$\nI think. :D [/hide]\n[hide=\"Solution\"] Call $C(x,x+1)$. $AC$ is the base of length $x\\sqrt{2}$. The height of the triangle is the distance from $B$ to the line $y=x+1$, which is $\\frac{5\\sqrt{2}}{2}$. So the area of the triangle is $\\frac{1}{2}\\times\\frac{5\\sqrt{2}}{2}\\times{x\\sqrt{2}}$.\n$\\frac{5x}{2}=20 \\Longrightarrow x=8$. So the coordinates of C are $\\boxed{\\boxed{(8,9)}}$ :D [/hide]", "Solution_5": "[hide]Use Shoesting Algorithm:\nA=(0,1), B=(4,0), C=(x,x+1)\n\n0 1\n4 0\nx x+1\n0 1\n\nFirst sum is 4(x+1)+x=5x+4\nSecond sum is 4\n\nSince x is positive, 5x+4 is positive\n$\\frac{(5x+4)-4}{2}=\\frac{5x}{2}=20$\n$x=8$\n$y=9$\n$\\boxed{(8,9)}$[/hide]", "Solution_6": "Could we use like Shoelace or something?", "Solution_7": "i took mathnerd314's idea, kind of.\r\n[hide]\nwhat i did was graph the two points, and the line. I took random points on the line and found the area of it using the shoelace method. this is not a very good method, theres probably a better way to do it. \n\nanswer: (8,9)[/hide]", "Solution_8": "[quote=\"mathnerd314\"]Could we use like Shoelace or something?[/quote]\r\nI used Shoelace.", "Solution_9": "[quote=\"math92\"]i took mathnerd314's idea, kind of.\n[hide]\nwhat i did was graph the two points, and the line. I took random points on the line and found the area of it using the shoelace method. this is not a very good method, theres probably a better way to do it. \n\nanswer: (8,9)[/hide][/quote]\r\nGuess and check with shoestring? Variables are definately better with larger numbers.", "Solution_10": "Isn't Shoelace using the determinants?", "Solution_11": "Using the determinant...\r\n[hide]\n$x+4(y-1)=20$\n$y=x+1$\n\n$x=4, y=5$\n\n$(4,5)$[/hide]" } { "Tag": [ "limit", "function", "real analysis", "real analysis solved" ], "Problem": "Define \r\n\\[ F(X)=1.875X - 0.625 X^3 + 0.09375X^5\\; \\; ,\\; \\; \\; x_{k+1}=F(x_k) \\; ,\\; k=0,1,...,\\]\r\nwhere $ x_0:=1.5\\; . $\r\nWhat about $ L:=\\lim\\limits_{n\\to \\infty}x_n \\; $ ?\r\nMoreover $ |x_{2004}-L | < ?? $ .", "Solution_1": "$F(x)$ can be rewritten as $(3x^5-20x^3+60x)/32$. Any fixed point for iteration of this function must be a root of $F(x)=x$, or $3x^5-20x^3+28x=0$. This has five roots: $0, \\pm\\sqrt{2}, \\pm\\sqrt{\\frac{14}3}$. Of these, $0$ and $\\pm\\sqrt{\\frac{14}3}$ are unstable and $\\pm\\sqrt{2}$ are stable. (This can be seen by drawing the graph of $y=F(x)$ and seeing where and how it crosses the graph of $y=x$.)\r\n\r\n$F'(x) = \\frac{15}{32}(x^2-2)^2 \\ge 0$. From this it follows that the sequence cannot jump over any of the five roots referred to above.\r\n\r\nIf $x_0 < -\\sqrt{\\frac{14}3}$, then $x_n \\to -\\infty$.\r\nIf $x_0 > \\sqrt{\\frac{14}3}$, then $x_n \\to \\infty$.\r\nIf $-\\sqrt{\\frac{14}3}0", "Solution_1": "Plugging $ x\\equal{}1$ into $ P(\\frac{1}{x}) \\geq \\frac{1}{P(x)}$ we see that $ P(1)^2 \\geq 1$\r\nBut by Cauchy-Schwarz we have that for all $ x>0$ we have $ P(x)P(\\frac{1}{x}) \\geq P(1)^2$ :lol:" } { "Tag": [], "Problem": "Find $ m\\in \\mathbb{R}$ such that the equation $ z^3\\plus{}i(z^2\\minus{}1)\\minus{}mz\\equal{}0$ has real roots .", "Solution_1": "[hide=\"Not sure..\"]$ z \\equal{} a \\plus{} bi$\n\n$ a^3 \\plus{} 3a^2bi \\minus{} 3ab^2 \\minus{} ib^3 \\plus{} i(a^2 \\plus{} 2abi \\minus{} b^2 \\minus{} 1) \\minus{} ma \\minus{} mbi \\equal{} 0$\n\n$ (a^3 \\minus{} 3ab^2 \\minus{} 2ab \\minus{} ma) \\plus{} i(3a^2b \\plus{} a^2 \\minus{} b^3 \\minus{} b^2 \\minus{} 1 \\minus{} mb) \\equal{} 0$\n\n$ 3a^2b \\plus{} a^2 \\minus{} b^3 \\minus{} b^2 \\minus{} 1 \\minus{} mb \\equal{} 0$\n\nSince $ z$ is real, $ b \\equal{} 0$.\n\n$ a^2 \\equal{} 1$\n\n$ a \\equal{} \\pm 1$\n\n$ a^3 \\minus{} 3ab^2 \\minus{} 2ab \\minus{} ma \\equal{} 0$\n\nSince $ b \\equal{} 0$ and $ a \\equal{} \\pm 1$ we get $ \\boxed{m \\equal{} 0}$.[/hide]\n\n[hide=\" Then I got a different solution\"]We want $ z$ to be real. So we must have that $ z^2 \\minus{} 1 \\equal{} 0$ so $ z \\equal{} \\pm 1$. Plugging back in we have $ m \\equal{} \\pm 1$.[/hide]", "Solution_2": "All three roots can never be all real, since their product is $ i$. If we only want at least one root real, be that $ x$, then $ x(x^2 \\minus{} m) \\equal{} i(1 \\minus{} x^2)$ requires $ x \\equal{} \\pm 1$ and so $ m \\equal{} 1$." } { "Tag": [ "geometry", "perimeter" ], "Problem": "One leg of a right triangle has a length of 9 m. The other sides have lengths that are consecutive integers. Find the number of meters in the perimeter.", "Solution_1": "[hide=\"Solution if you didn't memorize Pythagorean Triples\"]Let us call the remaining to sides $ x,y$ with $ x$ being the length of the leg and $ y$ being the length of the hypotenuse. Then,\n\\[ y^2 \\minus{} x^2 \\equal{} 9^2\n\\]\n. From this, we get\n\\[ (y \\minus{} x)(y \\plus{} x) \\equal{} 9\\cdot9\n\\]\nBut we know $ y \\equal{} x \\plus{} 1$ as the integers are consecutive and the hypotenuse, $ y$, must be the largest.\n\\[ (1)(2x \\plus{} 1) \\equal{} 81\n\\]\nBut multiplication by $ 1$ doesn't matter.\n\\[ 2x \\plus{} 1 \\equal{} 81\n\\]\nSolving the equation we get $ x \\equal{} 40$ and $ y \\equal{} 41$. Hence, the perimeter is $ 9 \\plus{} 40 \\plus{} 41 \\equal{} 90$.[/hide]\n\n[hide=\"Solution if you did memorize the Pythagorean Triples\"]We know the measures of the triangle are $ (9,40,41)$. Thus, the perimeter is $ 9 \\plus{} 40 \\plus{} 41 \\equal{} 90$.[/hide]\n\nAnyway, I would highly recommend memorizing Pythagorean Triples upto a certain degree which you choose.[/hide]" } { "Tag": [ "ratio", "geometric series" ], "Problem": "Prove that\r\n \r\n$\\sum_{x=1}^{\\infty}\\left(\\frac{1}{n}\\right)^x=\\frac{1}{n-1}$\r\n\r\nWhere $n>1$ and is an integer\r\n\r\n\r\nI don't think I'm allowed to post Sigma problems in Basics or MOEMS, so this'll have to suffice.", "Solution_1": "[hide]The first term is $1/n$, the ratio is $1/n$, so the sum is $\\frac{1/n}{1-1/n}=\\frac{1/n}{(n-1)/n}=\\frac{1}{n-1} \\ \\ \\ \\ \\blacksquare$[/hide]", "Solution_2": "You didn't prove anything, you just used a formula. You're supposed to prove the formula.\r\n\r\nAs for that, I don't know how to do it. It seems intuitive, though; say $n=2$. You have the first $\\frac{1}{2}$, then the sum of the rest will equal $\\frac{1}{2}$. I can't prove that, but it's just intuitive, and you can extend it out far enough pretty quickly that you see that it must approach $\\frac{1}{2}$.", "Solution_3": "$s=\\left(\\frac1n\\right)^1+\\left(\\frac1n\\right)^2+\\left(\\frac1n\\right)^3+\\left(\\frac1n\\right)^4+\\cdots$\r\n\r\n$ns=1+\\left(\\frac1n\\right)^1+\\left(\\frac1n\\right)^2+\\left(\\frac1n\\right)^3+\\left(\\frac1n\\right)^4+\\cdots$\r\n\r\n$(n-1)s=1$\r\n\r\n$s=\\frac1{n-1}$", "Solution_4": "I thought there weren't supposed to be any infinate series in middle school sections, but I guess these are okay.", "Solution_5": "yea, but it wasn't a very complicated series either.", "Solution_6": "[quote=\"nat mc\"]$s=\\left(\\frac1n\\right)^1+\\left(\\frac1n\\right)^2+\\left(\\frac1n\\right)^3+\\left(\\frac1n\\right)^4+\\cdots$\n\n$ns=1+\\left(\\frac1n\\right)^1+\\left(\\frac1n\\right)^2+\\left(\\frac1n\\right)^3+\\left(\\frac1n\\right)^4+\\cdots$\n\n$(n-1)s=1$\n\n$s=\\frac1{n-1}$[/quote]\r\n\r\nMaybe it's late at night and my brain's not working properly, but\r\n\r\nHow did you get from the 2nd step to the 3rd step?", "Solution_7": "Just subtract the first equation from the second, and all cancel except the 1", "Solution_8": "Is that the formula for all geometric series like this? \r\n\r\nI didn't realize the equation subtracting from the second to the third step either, must be just too early", "Solution_9": "Right, that makes sense. It's like getting a fraction from repeating decimals.", "Solution_10": "yeah, thats pretty much the exact way you do repeating decimals, and it proves the sum for all infinite series" } { "Tag": [ "calculus", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Find: \u222b(dx)/(x(1-e^x))", "Solution_1": "Try develop $\\frac{1}{1-e^x} = \\frac{e^x}{1-e^{-x}}= e^x \\sum_{n=0}^{\\infty}e^{-nx}$ and then integrate each term separately by redeveloping $e^x$" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2=1$. Prove the following inequality \\[ \\sum \\dfrac{a}{a^3+bc} \\geq\\dfrac{23}{7}. \\]", "Solution_1": "That's false : take $a=b$ close to $0$ and $c$ close to $1$ \r\n:cool:" } { "Tag": [], "Problem": "New poll", "Solution_1": "Yeah two people is what we discussed, and it's better anyway. Btw, I accidentally hit 3 in the poll, so consider that.", "Solution_2": "is this ever going to happen?", "Solution_3": "here is a better poll:\r\n\r\nHOW MANY TIMES HAS THIS QUESTION BEEN POLLED??!?!?!??!!?", "Solution_4": "I actually think that two is better, but then we might have a 16 team competition.", "Solution_5": "only one smart person is needed. if you have too many people u have a wide mem expansion" } { "Tag": [ "quadratics", "Gauss", "modular arithmetic", "number theory", "relatively prime", "Divisibility Theory", "pen" ], "Problem": "(Wolstenholme's Theorem) Prove that if \\[1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{p-1}\\] is expressed as a fraction, where $p \\ge 5$ is a prime, then $p^{2}$ divides the numerator.", "Solution_1": "We will treat rational numbers, which have their denominator relatively prime to $p$, as residues mod $p$.\r\nFrom the Fermat's Little Theorem: $\\frac{1}{i} \\equiv i^{p-2} \\mod{p}$\r\nSo we see that:\r\n$\\sum_{i=1}^{p-1} \\frac{1}{i} \\equiv \\sum_{i=1}^{p-1} i^{p-2} = \\sum_{i=1}^{\\frac{p-1}{2}} i^{p-2} + (p-i)^{p-2} \\equiv \\sum_{i=1}^{\\frac{p-1}{2}} p(p-2)i^{p-3} = p(p-2) \\sum_{i=1}^{\\frac{p-1}{2}} i^{p-3} \\mod{p^2}$\r\nThe last congruence comes from the fact that the rest term vanishes mod $p^2$. We are left with proving that:\r\n$\\sum_{i=1}^{\\frac{p-1}{2}} i^{p-3} \\equiv 0 \\mod{p}$\r\nor (one more time from FTL):\r\n$\\sum_{i=1}^{\\frac{p-1}{2}} \\frac{1}{i^2} \\equiv 0 \\mod{p}$\r\nNow if $i, j \\leq \\frac{p-1}{2}$ and $\\frac{1}{i^2} \\equiv \\frac{1}{j^2} \\mod{p}$ then also $(i-j)(i+j) \\equiv 0 \\mod{p}$ but since $i, j \\leq \\frac{p-1}{2}$ we must have $i=j$. So the numbers: $\\frac{1}{i^2}$ for $i=1,2,...,\\frac{p-1}{2}$ are all different quadratic residues (and there are exactly $\\frac{p-1}{2}$ quadratic residues), as well as the numbers $i^2$ for $i=1,2,...,\\frac{p-1}{2}$ so:\r\n$\\sum_{i=1}^{\\frac{p-1}{2}} \\frac{1}{i^2} \\equiv \\sum_{i=1}^{\\frac{p-1}{2}} i^2 = \\frac{p(p-1)(p+1)}{24} \\equiv 0 \\mod{p}$\r\nwhich ends the proof.", "Solution_2": "I love it. This is a beautiful proof.", "Solution_3": "[quote=\"Peter\"](Wolstenholme's Theorem) Prove that if\n\\[ 1 + \\frac {1}{2} + \\frac {1}{3} + \\cdots + \\frac {1}{p - 1}\\]\nis expressed as a fraction, where $ p \\ge 5$ is a prime, then $ p^{2}$ divides the numerator.[/quote]\n\nThis is an updated version of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46121]an older MathLinks post of mine[/url], and I am submitting it here since I will refer to it in a number of other proofs.\n\nI will generalize the problem, but first I explain some preliminaries about primes.\n\n[color=blue][b]1. p-adic valuation[/b][/color]\n\nLet $p$ be an arbitrary prime. We denote by $ \\mathbb{Z}_{p}$ the field of residues of integers modulo $p$.\n\nFor any rational number $x$, we will now define the so-called [i]$p$-adic valuation of $x$[/i]. This valuation will be denoted by $ v_{p}\\left(x\\right)$, and it is defined as follows: If $ x\\neq 0$, then $ v_{p}\\left(x\\right)$ is the greatest integer $k$ such that $\\frac{x}{p^{k}}$ can be written as a ratio of two integers with the denominator not being divisible by $p$. Besides, we set $ v_{p}\\left(0\\right) = + \\infty$, where $ + \\infty$ is just a symbol satisfying $ + \\infty > a$ and $ + \\infty + a = + \\infty$ for any integer $ a$. (The main thing we will need about $+ \\infty$ is that $ v_{p}\\left(0\\right) > v_{p}\\left(x\\right)$ for any nonzero $x$).\n\nWe can easily see how to compute $ v_{p}\\left(x\\right)$ for rationals $ x\\neq 0$:\n\n* If $x$ is a nonzero integer, then $ v_{p}\\left(x\\right)$ is the greatest integer $k$ such that $ p^{k}$ divides $x$ (thus, in particular, we have $ v_{p}\\left(x\\right) = 0$ if $p$ doesn't divide $x$).\n\n* If $x$ is a ratio of two nonzero integers, say $ x = \\frac{a}{b}$ with nonzero integers $a$ and $b$, then $ v_{p}\\left(x\\right) = v_{p}\\left(a\\right) - v_{p}\\left(b\\right)$.\n\nNote that there is a simple way to interpret the sign of the $p$-adic valuation $ v_{p}\\left(x\\right)$ of a rational number $x$: If we write $x$ as a reduced fraction (i.e., a fraction with numerator and denominator coprime; if $x$ is an integer, then just write it in the form $x = \\frac{x}{1}$), and it turns out that the numerator is divisible by $p$, then $ v_{p}\\left(x\\right) > 0$; if neither the numerator nor the denominator is divisible by $p$, then $ v_{p}\\left(x\\right) = 0$; if the denominator is divisible by $p$, then $ v_{p}\\left(x\\right) < 0$.\n\nA [i]$p$-integer[/i] shall mean a rational number $x$ satisfying $v_p\\left(x\\right) \\geq 0$. In particular, every integer is a $p$-integer. Moreover, any fraction of integers is a $p$-integer if its denominator is coprime to $p$.\n\nIt is rather easy to prove that for any two rational numbers $x$ and $y$, we have $ v_{p}\\left(xy\\right) = v_{p}\\left(x\\right) + v_{p}\\left(y\\right)$ and $v_{p}\\left(x + y\\right)\\geq\\min\\left(v_{p}\\left(x\\right);\\;v_{p}\\left(y\\right)\\right)$. Thus, the sum and the product of two $p$-integers are $p$-integers again. Hence, it is not hard to see that any sum or product of finitely many $p$-integers is a $p$-integer, and that a difference of two $p$-integers is always a $p$-integer. (In the language of abstract algebra, this is saying that the $p$-integers form a subring of $\\mathbb{Q}$.)\n\n[color=blue][b]2. Working with fractions modulo $p$[/b][/color]\n\nI will use the notation $\\mathbb{Z}_{p}$ for the ring $\\mathbb{Z} / p\\mathbb{Z}$ (that is, the ring of integers modulo $p$).\n\nWe will also need some familiarity with fractions in $\\mathbb{Z}_{p}$. The main idea is to extend the notion of congruence modulo $p$ from integers to $p$-integers: If $x$ and $y$ are two $p$-integers, then we say that $x\\equiv y\\mod p$ (in words: \"$x$ is congruent to $y$ modulo $p$\") if and only if $v_{p}\\left(x - y\\right) > 0$. Thus we have defined a relation (\"congruence modulo $p$\") on the set of all $p$-integers. This relation is an equivalence relation (this is easy to prove; do it!). Moreover, if $x$ and $y$ are two integers, then $x \\equiv y \\mod p$ with respect to this new equivalence relation if and only if $x \\equiv y \\mod p$ in the usual sense (i.e., we have $p \\mid x-y$ in $\\mathbb{Z}$). This fact allows us to use the same notation for both equivalence relations without worrying about possible ambiguity.\n\nIt is easy to see that this equivalence relation has the same properties as the classical \"congruence modulo $p$\" relation on integers:\n\n* If $a$, $b$, $c$ and $d$ are $p$-integers, and if $a \\equiv b \\mod p$ and $c \\equiv d \\mod p$, then $a+c \\equiv b+d \\mod p$ and $a-c \\equiv b-d \\mod p$ and $ac \\equiv bd \\mod p$. In other words, congruences modulo $p$ between $p$-integers can be added, subtracted and multiplied just as congruences between integers.\n\n* The same holds for larger sums: i.e., if $a_1, a_2, \\ldots, a_j, b_1, b_2, \\ldots, b_j$ are $p$-integers, and if every $i$ satisfies $a_i \\equiv b_i \\mod p$, then $\\sum\\limits_{i=1}^j a_i \\equiv \\sum\\limits_{i=1}^j b_i \\mod p$.\n\n* If $a$, $b$, $c$ and $d$ are $p$-integers such that $v_p\\left(b\\right) = 0$ and $v_p\\left(d\\right) = 0$, and if $a \\equiv b \\mod p$ and $c \\equiv d \\mod p$, then $a/c \\equiv b/d \\mod p$. In other words, congruences modulo $p$ between $p$-integers can be divided by one another, as long as the $p$-integers we are dividing by have $p$-adic valuations equal to $0$. We shall refer to this fact as the [i]division rule[/i].\n\nRecall that congruence modulo $p$ is an equivalence relation on the set of $p$-integers. Let $F_p$ denote the set of equivalence classes of this equivalence relation. One might expect that $F_p$ is larger than $\\mathbb{Z}_p$, seeing that the equivalence classes that comprise $F_p$ consist of $p$-integers whereas the equivalence classes that comprise $\\mathbb{Z}_p$ consist of integers only. But in fact, it turns out that $F_p$ is in bijection with $\\mathbb{Z}_p$. Indeed, it is easy to prove that for every $p$-integer $x$, there is one and only one integer $n$ satisfying $0\\leq n < p$ such that $x\\equiv n\\mod p$. Thus, every equivalence class in $F_p$ contains exactly one of the integers $0, 1, \\ldots, p-1$, and thus corresponds to a unique equivalence class in $\\mathbb{Z}_p$. Equivalence classes in $F_p$ can be added, subtracted and multiplied (just as in $\\mathbb{Z}_p$), and these operations match precisely the analogous operations in $\\mathbb{Z}_p$ (that is, if you add two equivalence classes in $F_p$, then the corresponding class in $\\mathbb{Z}_p$ is obtained by adding the classes in $\\mathbb{Z}_p$ that correspond to the two addends; and similarly for subtraction and multiplication). Furthermore, if $x$ and $y$ are two equivalence classes in $F_p$, then $x/y$ is a well-defined equivalence class in $F_p$ provided that $y$ is the equivalene class of a $p$-integer $u$ with $v_p\\left(u\\right) = 0$. (Again, it is constructed as usual: Pick an element $a \\in x$ and an element $b \\in y$, and form the equivalence class of the rational number $a/b$.) Hence, you can work with $p$-integers modulo $p$ in the same way as you work with integers modulo $p$. This also means that modulo $p$, you can uniquely divide by any integer not divisible by $p$. [A nice application of this is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44573]problem 4 of the IMO 2005[/url] - see Fedor Petrov's proof in post #2.]\n\n[color=blue][b]3. Generalization of Wolstenholme's Theorem[/b][/color]\n\nNow, the problem A 23 asks us to show that if $ p$ is a prime number such that $ p\\geq 5$, then $ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k}\\right)\\geq 2$. This is a particular case (namely, the $ u = 1$ case) of the following result:\n\n[color=blue][b]Theorem 1.[/b] Let $ p$ be a prime number, and $ u$ be an odd integer such that $ p\\geq u + 3$ and $ u\\geq 0$. Then,\n\n$ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right)\\geq 2$.[/color]\n\n[i]Proof of Theorem 1.[/i] We have $ p > 2$ (since $ p\\geq u + 3$ and $ u\\geq 0$). We start with the Gauss trick:\n\n$ 2\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}} = \\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}} + \\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}} = \\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}} + \\sum_{k = 1}^{p - 1}\\frac {1}{\\left(p - k\\right)^{u}}$\n$= \\sum_{k = 1}^{p - 1}\\left(\\frac {1}{k^{u}} + \\frac {1}{\\left(p - k\\right)^{u}}\\right) = \\sum_{k = 1}^{p - 1}\\frac {k^{u} + \\left(p - k\\right)^{u}}{k^{u}\\left(p - k\\right)^{u}}$.\n\nNow, define $ s_k=\\sum_{i=2}^u\\left(-1\\right)^{u-i}\\binom{u}{i}p^{i-2}k^{u-i}$ for every $ k\\in\\left\\{1,2,...,p-1\\right\\}$. Obviously, $ s_k$ is an integer for every $ k$. We can expand $ \\left(p - k\\right)^{u}$ according to the binomial formula:\n\n$ \\left(p - k\\right)^{u}=\\sum_{i=0}^u\\left(-1\\right)^{u-i}\\binom{u}{i}p^ik^{u-i}$\n$ =\\left(-1\\right)^{u-0}\\binom{u}{0}p^0k^{u-0}+\\left(-1\\right)^{u-1}\\binom{u}{1}p^1k^{u-1}+\\sum_{i=2}^u\\left(-1\\right)^{u-i}\\binom{u}{i}p^ik^{u-i}$\n$ =\\left(-1\\right)^uk^u+\\left(-1\\right)^{u-1}upk^{u-1}+p^2\\sum_{i=2}^u\\left(-1\\right)^{u-i}\\binom{u}{i}p^{i-2}k^{u-i}$\n$ =\\left(-1\\right)^uk^u+\\left(-1\\right)^{u-1}upk^{u-1}+p^2s_k$\n$=\\left(-1\\right)k^u+1upk^{u-1}+p^2s_k$ (since $u$ is odd, so that $ \\left(-1\\right)^u=-1$ and $ \\left(-1\\right)^{u-1}=1$)\n$ =-k^u+upk^{u-1}+p^2s_k$.\n\nThus, $ k^u+\\left(p-k\\right)^u = upk^{u-1}+p^2s_k$ for every $ k\\in\\left\\{1,2,...,p-1\\right\\}$, and therefore\n\n$ 2\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}} = \\sum_{k = 1}^{p - 1}\\frac {k^{u} + \\left(p - k\\right)^{u}}{k^{u}\\left(p - k\\right)^{u}} = \\sum_{k = 1}^{p - 1}\\frac {upk^{u-1}+p^2s_k}{k^{u}\\left(p - k\\right)^{u}} = \\sum_{k = 1}^{p - 1} p \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}$\n$ = p \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}$.\n\nHence,\n\n$ v_{p}\\left(2\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right) = v_{p}\\left( p \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right) = \\underbrace{v_{p}\\left(p\\right)}_{=1} + v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right)$\n$ = 1 + v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right)$.\n\nBut since $ 2$ isn't divisible by $ p$ (since $ p > 2$), we have $ v_{p}\\left(2\\right) = 0$, so that\n\n$ v_{p}\\left(2\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right) = \\underbrace{v_{p}\\left(2\\right)}_{=0} + v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right) = v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right)$.\n\nComparing these two equalities, we get\n\n$ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right) = 1 + v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right)$.\n\nHence, instead of showing $ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right)\\geq 2$, it will be enough to prove $ v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right)\\geq 1$. This is equivalent to $ v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right) > 0$, i.e., to $ \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\equiv 0\\mod p$. Here, we are using the fact that for every $ k\\in\\left\\{1,2,...,p-1\\right\\}$, the fraction $ \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}$ is a $p$-integer (because its denominator, $ k^{u}\\left(p - k\\right)^{u}$, is not divisible by $ p$, since $ 1\\leq k\\leq p - 1$ and since $ p$ is a prime), and therefore it makes sense to say that the sum of all these fractions is $\\equiv 0 \\mod p$.\n\nLet us now look more closely at such a fraction $\\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}$ (with $k \\in \\left\\{1,2,\\ldots,p-1\\right\\}$).\nThe numerator $ uk^{u-1}+ps_k$ of this fraction is congruent to $ uk^{u - 1}$ modulo $p$ (since $ p\\equiv 0\\mod p$, and thus all terms containing $ p$ can be omitted). Also, the denominator can be simplified modulo $p$: Since $ p - k\\equiv - k\\mod p$, we have $ k^{u}\\left(p - k\\right)^{u}\\equiv k^{u}\\left( - k\\right)^{u} = \\left( - 1\\right)^{u}k^{2u} \\mod p$. Dividing the congruence $uk^{u-1}+ps_k \\equiv uk^{u-1} \\mod p$ by the latter congruence, we obtain $ \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\equiv \\frac {uk^{u-1}}{\\left(-1\\right)^u k^{2u}}\\mod p$ (here, we have applied the division rule, since the rational numbers $k^{u}\\left(p - k\\right)^{u}$ and $\\left(-1\\right)^u k^{2u}$ have $p$-adic valuation $0$). Therefore,\n\n[color=green][b](1)[/b][/color] $ \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}\\equiv\\sum_{k = 1}^{p - 1}\\frac {uk^{u - 1}}{\\left( - 1\\right)^{u}k^{2u}} = \\sum_{k = 1}^{p - 1}\\frac {uk^{-u - 1}}{\\left( - 1\\right)^{u}} = \\frac {u}{\\left( - 1\\right)^{u}}\\sum_{k = 1}^{p - 1}k^{ - u - 1}\\mod p$.\n\nBut by Fermat's Little Theorem, $ k^{p - 1}\\equiv 1\\mod p$ for every integer $ k$ such that $ 1\\leq k\\leq p - 1$. Thus, every such $k$ satisfies $ k^{ - u - 1}\\equiv k^{ - u - 1}k^{p - 1} = k^{p - u - 2}\\mod p$, and the relation [color=green][b](1)[/b][/color] becomes\n\n$ \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}\\equiv\\frac {u}{\\left( - 1\\right)^{u}}\\sum_{k = 1}^{p - 1}k^{p - u - 2}\\mod p$.\n\nNow, $ p - u - 2\\geq 1$ (since $ p\\geq u + 3$) and $ p - u - 2\\leq p - 2$. Also, $ p$ is an odd prime (since $ p$ is a prime and $ p > 2$). Now, according to http://www.problem-solving.be/pen/viewtopic.php?t=676 part [b]a)[/b] (and also http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 ), we have:\n\n[color=blue][b]Theorem 2.[/b] If $ p$ is an odd prime and $ \\rho$ is an integer such that $ 1\\leq\\rho\\leq p - 2$, then $ p\\mid\\sum_{k = 1}^{p - 1}k^{\\rho}$.[/color]\n\nApplying this Theorem 2 to our prime $ p$ and $ \\rho = p - u - 2$ (we know that $ p$ is an odd prime and $ \\rho = p - u - 2$ satisfies $ 1\\leq p - u - 2\\leq p - 2$), we get $ p\\mid\\sum_{k = 1}^{p - 1}k^{p - u - 2}$, so that $ \\sum_{k = 1}^{p - 1}k^{p - u - 2}\\equiv 0\\mod p$, and thus\n\n$ \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}}\\equiv\\frac {u}{\\left( - 1\\right)^{u}} \\underbrace{\\sum_{k = 1}^{p - 1}k^{p - u - 2}}_{\\equiv 0\\mod p} \\equiv \\frac {u}{\\left( - 1\\right)^{u}}\\cdot 0 = 0\\mod p$.\n\nThis completes the proof of Theorem 1.\n\n[color=blue][b]4. A divisibility by $ p^3$[/b][/color]\n\nWe can use almost the same tactics to prove a similar result:\n\n[color=blue][b]Theorem 3.[/b] Let $ p$ be a prime number such that $ p>3$. Then,\n\n$ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^p}\\right)\\geq 3$.[/color]\n\n[i]Proof of Theorem 3.[/i] It is known that for every integer $ i$ such that $ 00$ (since $ p>3>2$).\n\nSet $ u=p$. Also, denote $ s_k=\\sum_{i=2}^u\\left(-1\\right)^{u-i}\\binom{u}{i}p^{i-2}k^{u-i}$ for every $ k\\in\\left\\{1,2,...,p-1\\right\\}$. Obviously, $ s_k$ is an integer for every $ k$. Then, every $ k\\in\\left\\{1,2,...,p-1\\right\\}$ satisfies\n\n$ s_k=\\sum_{i=2}^u\\left(-1\\right)^{u-i}\\binom{u}{i}p^{i-2}k^{u-i}=\\sum_{i=2}^p\\left(-1\\right)^{p-i}\\binom{p}{i}p^{i-2}k^{p-i}$ (since $ u=p$)\n$ =\\sum_{i=2}^{p-1}\\left(-1\\right)^{p-i}\\underbrace{\\binom{p}{i}}_{\\substack{\\equiv 0\\mod p,\\\\ \\text{ since } 00}}k^{p-p}$\n$ \\equiv \\underbrace{\\sum_{i=2}^{p-1}\\left(-1\\right)^{p-i}0p^{i-2}k^{p-i}}_{=0}+\\underbrace{\\left(-1\\right)^{p-p}\\binom{p}{p}0k^{p-p}}_{=0}=0+0=0\\mod p$.\n\nJust as in the proof of Theorem 1, we can show that\n\n[color=green][b](2)[/b][/color] $ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right) = 1 + v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right)$.\n\nSince $ u=p$, we have\n\n$ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{u}}\\right) = v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^p}\\right)$\n\nand\n\n$ v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {uk^{u-1}+ps_k}{k^{u}\\left(p - k\\right)^{u}} \\right) = v_{p}\\left( \\sum_{k = 1}^{p - 1} \\frac {pk^{p-1}+ps_k}{k^p\\left(p - k\\right)^p} \\right)$\n$ = v_{p}\\left( p\\cdot \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right)$ (since $ \\sum_{k = 1}^{p - 1} \\frac {pk^{p-1}+ps_k}{k^p\\left(p - k\\right)^p} = \\sum_{k = 1}^{p - 1} \\frac {p\\left(k^{p-1}+s_k\\right)}{k^p\\left(p - k\\right)^p} = \\sum_{k = 1}^{p - 1} p\\cdot \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} = p\\cdot \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p}$)\n$ = \\underbrace{v_p\\left(p\\right)}_{=1} + v_p\\left( \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right) = 1 + v_p\\left( \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right)$.\n\nThus, [color=green][b](2)[/b][/color] becomes\n\n$ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^p}\\right) = 1 + \\left(1 + v_p\\left( \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right)\\right)$.\n\nIn other words,\n\n$ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^p}\\right) = 2 + v_p\\left( \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right)$.\n\nHence, instead of showing $ v_{p}\\left(\\sum_{k = 1}^{p - 1}\\frac {1}{k^{p}}\\right)\\geq 3$, it will be enough to prove $ v_p\\left( \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right)\\geq 1$. This is equivalent to $ v_p\\left( \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\right) > 0$, i.e., to $ \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\equiv 0\\mod p$. Here, we are using the fact that for each $k \\in \\left\\{1,2,\\ldots,p-1\\right\\}$, the fraction $\\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p}$ is a $p$-integers (because its denominator, $ k^p\\left(p - k\\right)^p$, is not divisible by $ p$, since $ 1\\leq k\\leq p - 1$ and since $ p$ is a prime), and thus it makes sense to speak of congruence modulo $p$ for the sum of these fractions. Let us study such a fraction $\\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p}$ (with $k \\in \\left\\{1,2,\\ldots,p-1\\right\\}$) more closely: Its numerator $ k^{p-1}+s_k$ is congruent to $ k^{p-1}$ modulo $p$ (since $ s_k\\equiv 0\\mod p$, and thus the term $ s_k$ can be omitted). Also, the denominator can be simplified: Since $ p - k\\equiv - k\\mod p$, we have $ k^p\\left(p - k\\right)^p\\equiv k^p\\left( - k\\right)^p = \\left( - 1\\right)^pk^{2p} \\mod p$. Hence, $ \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\equiv \\frac {k^{p-1}}{\\left( - 1\\right)^pk^{2p}} \\mod p$ (by the division rule). Therefore,\n\n[color=green][b](3)[/b][/color] $ \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\equiv \\sum_{k = 1}^{p - 1}\\frac {k^{p-1}}{\\left( - 1\\right)^pk^{2p}} = \\frac{1}{\\left(-1\\right)^p} \\sum_{k = 1}^{p - 1}\\frac {k^{p-1}}{k^{2p}} \\mod p$.\n\nBut by Fermat's Little Theorem, $ k^p\\equiv k\\mod p$ for every integer $ k$. Thus, $ \\left(k^p\\right)^2\\equiv k^2\\mod p$, so that $ k^{2p}=\\left(k^p\\right)^2\\equiv k^2\\mod p$. Hence, for every $k \\in \\left\\{1,2,\\ldots,p-1\\right\\}$, we have $\\frac{k^{p-1}}{k^{2p}} \\equiv \\frac{k^{p-1}}{k^2} \\mod p$ (here we used the division rule again, because $k^{2p}$ and $k^2$ are rational numbers with $p$-adic valuation $0$). Hence, the relation [color=green][b](3)[/b][/color] becomes\n\n$ \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\equiv \\frac{1}{\\left(-1\\right)^p} \\sum_{k = 1}^{p - 1}\\underbrace{\\frac {k^{p-1}}{k^2}}_{ = k^{\\left(p-1\\right)-2}=k^{p-3}} = \\frac{1}{\\left(-1\\right)^p} \\sum_{k = 1}^{p - 1}k^{p-3} \\mod p$.\n\nNow, $ p-3\\geq 1$ (since $ p>3$ yields $ p-3>0$, and $ p-3\\in\\mathbb{Z}$) and $ p-3\\leq p-2$. Also, $ p$ is an odd prime (since $ p$ is a prime and $ p > 2$). Applying Theorem 2 to our prime $ p$ and $ \\rho = p - 3$ (we know that $ p$ is an odd prime and $ \\rho = p-3$ satisfies $ 1\\leq p-3\\leq p - 2$), we get $ p\\mid\\sum_{k = 1}^{p - 1}k^{p - 3}$, so that $ \\sum_{k = 1}^{p - 1}k^{p - 3}\\equiv 0\\mod p$, and thus\n\n$ \\sum_{k = 1}^{p - 1} \\frac {k^{p-1}+s_k}{k^p\\left(p - k\\right)^p} \\equiv \\frac{1}{\\left(-1\\right)^p} \\underbrace{\\sum_{k = 1}^{p - 1}k^{p-3}}_{\\equiv 0\\mod p} \\equiv \\frac{1}{\\left(-1\\right)^p}\\cdot 0 = 0\\mod p$.\n\nThis completes the proof of Theorem 3.\n\n[EDIT: Made the proof of Theorem 1 slightly more formal, and added section 4. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46121]the other thread[/url] for the old version of this post.\n\nEDIT 2: Improved the exposition of $p$-integers and their congruence significantly.]\n\n Darij", "Solution_4": "[quote=\"TomciO\"]\n$ \\sum_{i \\equal{} 1}^{p \\minus{} 1} \\frac {1}{i} \\equiv \\sum_{i \\equal{} 1}^{p \\minus{} 1} i^{p \\minus{} 2}$\n[/quote]\r\n\r\nWhy is this true mod $ p^2$?", "Solution_5": "Yes, you are right. Thank you for pointing out the mistake. I have corrected the proof accordingly since it's not a big difference.\r\n\r\nWe will treat rational numbers, which have their denominator relatively prime to $ p$, as residues mod $ p$.\r\nFrom the Euler's Theorem: $ \\frac {1}{i} \\equiv i^{p(p\\minus{}1) \\minus{} 1} \\mod{p^2}$\r\nSo we see that:\r\n\\[ \\sum_{i \\equal{} 1}^{p \\minus{} 1} \\frac {1}{i} \\equiv \\sum_{i \\equal{} 1}^{p \\minus{} 1} i^{p(p\\minus{}1) \\minus{} 1} \\equal{} \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} i^{p(p\\minus{}1) \\minus{} 1} \\plus{} (p \\minus{} i)^{p(p\\minus{}1) \\minus{} 1} \\equiv \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} p(p(p\\minus{}1) \\minus{} 1)i^{p(p\\minus{}1) \\minus{} 2} \\equiv \\minus{}p \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} i^{p(p\\minus{}1) \\minus{} 2} \\mod{p^2}\\]\r\nThe congruence come from the fact that the rest term vanishes mod $ p^2$. We are left with proving that:\r\n$ \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} i^{p(p\\minus{}1) \\minus{} 2} \\equiv 0 \\mod{p}$\r\nor from FTL:\r\n$ \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} \\frac {1}{i^2} \\equiv 0 \\mod{p}$\r\nNow if $ i, j \\leq \\frac {p \\minus{} 1}{2}$ and $ \\frac {1}{i^2} \\equiv \\frac {1}{j^2} \\mod{p}$ then also $ (i \\minus{} j)(i \\plus{} j) \\equiv 0 \\mod{p}$ but since $ i, j \\leq \\frac {p \\minus{} 1}{2}$ we must have $ i \\equal{} j$. So the numbers: $ \\frac {1}{i^2}$ for $ i \\equal{} 1,2,...,\\frac {p \\minus{} 1}{2}$ are all different quadratic residues (and there are exactly $ \\frac {p \\minus{} 1}{2}$ quadratic residues), as well as the numbers $ i^2$ for $ i \\equal{} 1,2,...,\\frac {p \\minus{} 1}{2}$ so:\r\n$ \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} \\frac {1}{i^2} \\equiv \\sum_{i \\equal{} 1}^{\\frac {p \\minus{} 1}{2}} i^2 \\equal{} \\frac {p(p \\minus{} 1)(p \\plus{} 1)}{24} \\equiv 0 \\mod{p}$\r\nwhich ends the proof.", "Solution_6": "Maybe we can use fields to solve this in a more efficient way!", "Solution_7": "We begin with a lemma.\n\nLemma. Let $k \\in \\mathbb{Z}$. Then $1^k + 2^k + \\cdots + (p - 1)^k \\equiv 0 \\pmod{p}$ whenever $p - 1$ does not divide $k$.\n\nProof: This is an often used result in number theory. See my post in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=422958[/url] for a proof.\n\nUnless mentioned otherwise, all quantities should be taken as elements of $\\mathbb{Z}/p^2\\mathbb{Z}$.\nNext, observe that if $1 \\leq i \\leq p - 1$, we have\n\\[(p - i)(p + i) = p^2 - i^2 = -i^2,\\]\nwhich, after some rearranging, gives\n\\[\\frac1{p - i} = -\\frac{p + i}{i^2}.\\]\nTherefore,\n\\[1 + \\cdots + \\frac1{p - 1} = -\\left(\\frac{p + 1}{1^2} + \\frac{p + 2}{2^2} + \\cdots + \\frac{p + (p - 1)}{(p - 1)^2}\\right),\\]\nso, after splitting up the fractions on the right side and rearranging, we get\n\\[2\\left(1 + \\cdots + \\frac1{p - 1}\\right) = -p\\left(1 + \\frac 1{2^2} + \\cdots + \\frac1{(p - 1)^2}\\right).\\]\nSince $p \\neq 2$, if we can show that $p | 1 + \\frac1{2^2} + \\cdots + \\frac1{(p - 1)^2}$, then we know that $p^2 | 1 + \\frac12 + \\cdots + \\frac1{p - 1}$, as desired. Indeed, since $p > 3$, $p-1$ does not divide $-2$, so the lemma gives that \n\\[1 + \\frac1{2^2} + \\cdots + \\frac1{(p - 1)^2} = 1^{-2} + 2^{-2} + \\cdots + (p - 1)^{-2} \\equiv 0 \\pmod{p},\\]\ncompleting the proof.", "Solution_8": "is that obvious that we can write fractions mod n that way?", "Solution_9": "[quote=\"Peter\"](Wolstenholme's Theorem) Prove that if \\[1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{p-1}\\] is expressed as a fraction, where $p \\ge 5$ is a prime, then $p^{2}$ divides the numerator.[/quote]\n$1+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{p-1}\\ = \\sum_{i=1}^{\\frac{p-1}{2}} (\\frac{1}{i}+\\frac{1}{p-i}) = \\sum_{i=1}^{\\frac{p-1}{2}} \\frac{p}{i(p-i)}$. So it'a enough to show that $\\sum_{i=1}^{\\frac{p-1}{2}} \\frac{1}{i(p-i)} \\equiv 0 \\mod{p}$. $\\sum_{i=1}^{\\frac{p-1}{2}} \\frac{1}{i^2}=0$. It's obviously that if $i^2 \\equiv j^2 \\mod{p}$. Than $i \\equiv j$, if $0 (a \\minus{} b)^2 \\Longleftrightarrow \\frac {a(b \\plus{} c)}{(a \\minus{} b)^2} \\ge \\frac {a(b \\plus{} c)}{c^2}$.\r\nSo $ \\sum\\frac {a(b \\plus{} c)}{(a \\minus{} b)^2}\\ge \\sum\\frac {a(b \\plus{} c)}{c^2} \\ge 6 > 1$." } { "Tag": [ "inequalities" ], "Problem": "Find all positive reals $a,b,c$ which fulfill the following relation\r\n\\[ 4(ab+bc+ca)-1 \\geq a^2+b^2+c^2 \\geq 3(a^3+b^3+c^3) . \\]\r\n\r\ncreated by Panaitopol Laurentiu.", "Solution_1": "Here the short version of my sol:\r\n\r\nfrom the first piece of inequality we get $a+b+c \\geq 1$:\r\norder inequality, replacing 1 ($ab+bc+ac$ by $a+b+c$ will keep the inequality, then simplify and again do this gives $a+b+c \\geq 1$\r\n\r\nfrom the second part we get $a+b+c \\leq 1$:\r\n$(a+b+c)(a+b+c) \\leq 3(a+b+c)$ but the problem says without $(a+b+c)$ it's $\\geq$ so $a+b+c \\leq 1$\r\n\r\nTherefor, a+b+c = 1\r\nwhich also implies that order inequality in the second case is an equality, so a=b=c\r\n\r\nWhich results in $a=b=c=\\frac{1}{3}$", "Solution_2": "i don't get this solution\r\ncan anybody clear it up to me?", "Solution_3": "what is it exactly you don't get? \r\n\r\nI first prove that a+b+c must be >= 1 (first part)\r\nThen I prove that a+b+c must be <= 1 at the same time, so that means a+b+c = 1 for sure.\r\n\r\nfrom there I call in the order inequality I used to prove the above, which states that it becomes an equality iff a=b=c. So since the sign is reversed, we must have an equality here for the given to be possible, thus a=b=c with a+b+c=1, this results in a=b=c=1/3.\r\n\r\nIs it one of the proofs of the first/second step you don't get? :)", "Solution_4": "actually both of them i don't understand (what is \"order\" ineq?)", "Solution_5": ":) order ineq says that 5 bills of 5 dollar + 2 bills of 1 dollar is worth more than 2 bills of 5 dollar and 5 bills of 1 dollar :)\r\n\r\n(so the greedy law, take most of the highes and fewest of the lowest to get best result)\r\n\r\nthus, order ineq: if a>=b>=c then a^2+b^2+c^2>=ab+bc+ca [ that is a common ineq, which can also be constructed by taking (a-b)^2+(b-c)^2+(c-a)^2>=0 ]\r\n\r\nthat's quite important, a^2+b^2+c^2>=ab+bc+ca for all a,b,c in R.\r\n\r\n(so in each of the steps I could have added: assume without loss of generality that a>=b>=c)\r\n\r\nIs it clear now, or do you want me to explain it in full detail? :)", "Solution_6": "Adding $2(a^2+b^2+c^2)$ to both sides of the left inequality, we obtain\r\n\r\n\\[2(a+b+c)^2-1 \\geq 3(a^2+b^2+c^2).\\]\r\n\r\nClearly, $3(a^2+b^2+c^2) \\geq (a+b+c)^2$ (easily proven by Cauchy or Power-Mean, for example and equality iff $a=b=c$), so we have\r\n\r\n\\[2(a+b+c)^2-1 \\geq (a+b+c)^2 \\iff a+b+c \\geq 1,\\]\r\n\r\nsince $a,b,c$ are positive.\r\n\r\nNow note that $3(a^3+b^3+c^3) \\geq (a+b+c)(a^2+b^2+c^2)$ which easily follows from $(a^3+a^3+b^3)/3 \\geq a^2b$, etc by AM-GM. The right side of the inequality implies that $a+b+c \\leq 1$, so combining this with the above, we conclude that $a+b+c=1$, so equality must hold, or $a=b=c=1/3$.", "Solution_7": "Yes. [quote=\"ThAzN1\"]... so equality must hold, or $a=b=c=1/3$.[/quote]=> here for the case it may be unclear: you substitute the given of equalness in the second step, to find that (like I said) to find another equality, from which you can conclude like I said that a=b=c. :)", "Solution_8": "thanks, i get it now.\r\nI think order ineq is the ineq known in my country as \"rearrangament ineq\":\r\ngiven $a_{1} \\geq \\cdots \\geq a_{n}$ and $b_{1} \\geq \\cdots \\geq b_{n}$ we have:\r\n$a_{1} \\cdot b_{1} + \\cdots + a_{n} \\cdot b_{n} \\geq a_{1} \\cdot b_{ \\tau (1) } \\cdots a_{n} \\cdot b_{ \\tau (n) } \\geq a_{1} \\cdot b_{n} + \\cdots + a_{n} \\cdot b_{1}$\r\nfor any permutation $\\tau$ of $(b_{1}, \\cdots, b_{n})$.", "Solution_9": "yes, that is another (even better) name for it, rearrangement :)", "Solution_10": "$\\frac{4(a+b+c)^2}{3}\\geq4(ab+ac+bc)\\geq a^2+b^2+c^2+1\\geq \\frac{(a+b+c)^2}{3}+1 \\Rightarrow a+b+c\\geq1$\r\n$a^2+b^2+c^2 \\geq 3(a^3+b^3+c^3) \\geq (a+b+c)(a^2+b^2+c^2) \\Rightarrow a+b+c\\leq1$ \r\n Then $ a=b=c=\\frac{1}{3}$", "Solution_11": "[quote=Mamat]$\\frac{4(a+b+c)^2}{3}\\geq4(ab+ac+bc)\\geq a^2+b^2+c^2+1\\geq \\frac{(a+b+c)^2}{3}+1 \\Rightarrow a+b+c\\geq1$\n$a^2+b^2+c^2 \\geq 3(a^3+b^3+c^3) \\geq (a+b+c)(a^2+b^2+c^2) \\Rightarrow a+b+c\\leq1$ \n Then $ a=b=c=\\frac{1}{3}$[/quote]\n\nNice! I did the same thing, after spending a ton of time on it. A bit of an explanation: The first one follows from a^2+b^2+c^2+2ab+2ac+2bc >= 3ab+3ac+3bc, which is true by a^2+b^2-2ab >=0, so adding all cyclic permutations of it and dividing by two, gives a^2+b^2+c^2-ab-ac-bc >= 0. (a^2+b^2+c^2)(1+1+1) >= (a+b+c)^2 by C-S, so the third-fourth part is true. thanks @below for clarifying a final part, I thought you would have to use some inequality but I wasn\u2019t considering just expanding :)", "Solution_12": "[b]Claim:[/b] $3(a^3+b^3+c^3)\\ge (a+b+c)(a^2+b^2+c^2)$\n\n[b]Proof:[/b] Rearrange the inequality to\n\\[2a^3+2b^3+2c^3\\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2.\\]\nIt suffices to prove\n\\[a^3+b^3\\ge a^2b+ab^2\\]\nbut this is equivalent to \n\\[(a-b)^2(a+b)\\ge 0\\]\nafter rearranging, clearly true since $a$, $b$, and $c$ are positive reals.\n$\\blacksquare$" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "$ f$ is a real-valued function on the reals. It satisfies $ f(x^3 \\plus{} y^3) \\equal{} (x \\plus{} y)(f(x)^2 \\minus{} f(x) f(y) \\plus{} f(y)^2)$ for all $ x, y$. Prove that $ f(1996x) \\equal{} 1996 f(x)$ for all $ x$", "Solution_1": "[hide=\"Hint\"] Use $ 2$ lemmas:\n1) If $ f(ax)\\equal{}af(x)\\forall x$ then $ f(\\sqrt[3]{a}x)\\equal{}\\sqrt[3]{a}f(x)\\forall x$ ($ a>0$)\n2) If $ f(ax)\\equal{}af(x);f(bx)\\equal{}bf(x)$ then $ f((a\\plus{}b)x)\\equal{}(a\\plus{}b)f(x)$ ($ a,b>0$)[/hide]", "Solution_2": "Let me complete this argument.\r\n\r\n$ (1)$ $ f(x^3\\plus{}y^3)\\equal{}(x\\plus{}y)(f(x)^2\\minus{}f(x)f(y)\\plus{}f(y)^2)$ for all $ x,y\\in\\mathbb{R}$\r\n\r\n$ x\\equal{}y\\equal{}0$ gives $ f(0)\\equal{}0$.\r\nThen $ y\\equal{}0$ gives $ (2)$ $ f(x^3)\\equal{}xf(x)^2$ for all $ x\\in\\mathbb{R}$.\r\n\r\nThen $ f(x)f(y)\\equal{}\\sqrt[3]{xy}\\left[ f(\\sqrt[3]{x})f(\\sqrt[3]{y}) \\right] ^2$ for all $ x,y\\in\\mathbb{R}$.\r\nSo $ (3)$ $ f(x)f(y)\\ge 0$ for all $ x,y\\in\\mathbb{R},xy\\ge 0$.\r\n\r\nIf $ n\\in\\mathbb{N}$ with $ f(nx)\\equal{}nf(x)$ for all $ x\\in\\mathbb{R}$, then using $ (2)$\r\n$ nxf(x)^2\\equal{}nf(x^3)\\equal{}f(nx^3)\\equal{}(\\sqrt[3]{n}x)f(\\sqrt[3]{n}x)^2$\r\nand using $ (3)$ we have $ \\sqrt[3]{n}f(x)f(\\sqrt[3]{n}x)\\ge 0$.\r\nSo $ f(\\sqrt[3]{n}x)\\equal{}\\sqrt[3]{n}f(x)$ for all $ x\\in\\mathbb{R}$. (using $ f(0)\\equal{}0$ for $ x\\equal{}0$)\r\n\r\nAnd then using $ (1)$ and $ (2)$\r\n$ f((n\\plus{}1)x)\\equal{}f((\\sqrt[3]{nx})^3\\plus{}(\\sqrt[3]{x})^3)\\\\\r\n\\hspace*{0.67in}\\equal{}(\\sqrt[3]{nx}\\plus{}\\sqrt[3]{x})(f(\\sqrt[3]{nx})^2\\minus{}f(\\sqrt[3]{nx})f(\\sqrt[3]{x})\\plus{}f(\\sqrt[3]{x})^2)\r\n\\\\ \\hspace*{0.67in}\\equal{}(\\sqrt[3]{n}\\plus{}1)\\sqrt[3]{x}\\, ((\\sqrt[3]{n})^2\\minus{}\\sqrt[3]{n}\\plus{}1)f(\\sqrt[3]{x})^2\\\\ \\hspace*{0.67in}\\equal{}(n\\plus{}1)f(x)$\r\nfor all $ x\\in\\mathbb{R}$.\r\n\r\n\r\nThen by inuction $ f(nx)\\equal{}nf(x)$ for all $ n\\in\\mathbb{N}_0,x\\in\\mathbb{R}$. \r\n$ (2)$ gives $ f(1)\\equal{}f(1)^2$, so $ c: \\equal{}f(1)\\in\\{0,1\\}$.\r\nThen $ f(n)\\equal{}cn$ for all $ n\\in\\mathbb{N}$ and $ (x,y)\\equal{}(2,\\minus{}1)$ in $ (*)$ gives \r\n$ 7c\\equal{}4c^2\\minus{}2cf(\\minus{}1)\\plus{}f(\\minus{}1)^2\\Rightarrow f(\\minus{}1)\\equal{}\\minus{}c$ or $ f(\\minus{}1)\\equal{}3c$.\r\nBut $ (2)$ gives $ f(\\minus{}1)\\equal{}\\minus{}c$, so $ f(n)\\equal{}cn$ for all $ n\\in\\mathbb{Z}$.\r\n\r\nNow $ nf({m\\over n})\\equal{}f(m)\\equal{}cm$ for all $ m\\in\\mathbb{Z},n\\in\\mathbb{N}$,\r\nso $ f(x)\\equal{}cx$ for all $ x\\in\\mathbb{Q}$.\r\n\r\nFor all $ x\\in\\mathbb{R}$ we find $ x_1,x_2,\\ldots\\in\\mathbb{Q}$ with $ x_n\\mathop{\\longrightarrow}_{n\\to\\infty}x$.\r\nThen $ y_n: \\equal{}\\sqrt[3]{x_n^3\\minus{}x^3}\\mathop{\\longrightarrow}_{n\\to\\infty}0$ with $ x^3\\plus{}y_n^3\\equal{}x_n^3\\in\\mathbb{Q}$ and $ x\\plus{}y_n\\mathop{\\longrightarrow}_{n\\to\\infty}x$.\r\nSo for $ x\\not\\equal{}0$ we have $ x\\plus{}y_n\\not\\equal{}0$ for sufficiently large $ n$, and then\r\n$ c(x^2\\minus{}xy_n\\plus{}y_n^2)\\equal{}{f(x^3\\plus{}y_n^3)\\over x\\plus{}y_n}\\equal{}f(x)^2\\minus{}f(x)f(y_n)\\plus{}f(y_n)^2\\\\ \\hspace*{1in}\\equal{}{3\\over 4}f(x)^2\\plus{}({f(x)\\over 2}\\minus{}f(y_n))^2\\ge {3\\over 4}f(x)^2$\r\nand $ n\\to\\infty$ gives $ cx^2\\ge {3\\over 4}f(x)^2$ for all $ x\\not \\equal{}0$.\r\nSo $ f(x)\\to 0$ for $ x\\to 0$. (using $ f(0)\\equal{}0$)\r\n\r\nAnd now for $ x\\in\\mathbb{R}$ we find $ x_1,x_2,\\ldots\\in\\mathbb{Q}$ with $ x_n\\mathop{\\longrightarrow}_{n\\to\\infty}\\sqrt[3]{x}$.\r\nThen $ y_n: \\equal{}\\sqrt[3]{x\\minus{}x_n^3}\\mathop{\\longrightarrow}_{n\\to\\infty}0$ with $ x\\equal{}x_n^3\\plus{}y_n^3$ and also $ f(y_n)\\mathop{\\longrightarrow}_{n\\to\\infty}0$.\r\nThis gives\r\n$ f(x)\\equal{}f(x_n^3\\plus{}y_n^3)\\equal{}(x_n\\plus{}y_n)(f(x_n)^2\\minus{}f(x_n)f(y_n)\\plus{}f(y_n)^2)\r\n\\\\ \\hspace*{1.11in}\\equal{}(x_n\\plus{}y_n)(cx_n^2\\minus{}cx_nf(y_n)\\plus{}f(y_n)^2)\\mathop{\\longrightarrow}_{n\\to\\infty}cx$\r\nand therefore $ f(x)\\equal{}cx$ for all $ x\\in\\mathbb{R}$.\r\n\r\n\r\nSo we find only $ 2$ solutions $ f\\equiv 0$ and $ f(x)\\equal{}x,x\\in\\mathbb{R}$." } { "Tag": [], "Problem": "If i partially solved a solution, according to the grading scale, I can still receive some points. Should i just send in the solution with the proof to where i got to w/o an answer?", "Solution_1": "It's best to solve the problem, which almost always guarantees a 5/5. However, partially solving a problem can almost always get you a 3, or maybe even a 4, depending on how close you got to the answer.\r\n\r\nHope this helps. :)" } { "Tag": [ "probability", "analytic geometry", "trigonometry", "quadratics", "inequalities" ], "Problem": "I am going to post a string of NYCIML questions.\r\n\r\n#1:\r\nan ordinary pack of playing cards is shuffled, and two cards are delt facce up. Find the probability that one of these is a spade.", "Solution_1": "Exactly one or at least one? Two different questions...", "Solution_2": "[hide=\"Exactly one\"]spade, not-spade = $\\frac{13}{52}\\cdot\\frac{39}{51} = \\frac{507}{2652} = \\frac{13}{68}$\nnot-spade, spade = $\\frac{39}{52}\\cdot\\frac{13}{51} = \\frac{507}{2652} = \\frac{13}{68}$\nSo the answer should be $\\frac{13}{34}$, I believe.[/hide]\n\n[hide=\"At least one\"]1 - (not-spade, not-spade) = $1-\\frac{39}{52}\\cdot\\frac{39}{51} = \\frac{29}{68}$[/hide]\r\n\r\nIt all depends on what you're being asked for.\r\n\r\nI believe these are both correct.", "Solution_3": "Small error there, the second one should be...\r\n[hide]$1-\\frac{39}{52}\\cdot\\frac{38}{51} = \\frac{15}{34}$\nBecause there is one less non-spade card after the first choice.[/hide]", "Solution_4": "[quote=\"lyra\"]an ordinary pack of playing cards is shuffled, and two cards are delt facce up. Find the probability that one of these is a spade. \n\n1/104?[/quote]\r\n\r\nReality check: if I turn over a couple of cards from a deck, I should expect to get a spade between a quarter and a half of the time. If 1/104 were the correct answer, that would be telling me that it's nearly impossible to draw a spade on two tries -- something which common sense should dictate is totally unreasonable.", "Solution_5": "sorry, at least one.", "Solution_6": "As to JBL your right, I was being stupid, I am going to start a string of questions here if no one minds?\r\n\r\nProblem #2\r\nIn a rectangular coordinate system, a tangent from teh point (24,7) to the circle whose equation is x^2+y^2=400 has a point of tangency (a,b) where b>0. Find a.\r\n\r\nSmiles,\r\nlyra", "Solution_7": "I'm not completely sure about this solution so if you think there's something wrong with it, feel free to attack it.\r\n\r\n[hide=\"Solution for problem 2\"]\n\nLet the center of this circle be $O$ and certain point that has $b > 0$ be $A$. Also, call $(24,7)$ as $B$. From this, we know that there is a right triangle $OAB$ where $A$ is the right angle. $OB$ is simply distance from origin to $(24,7)$ and this can be calculated $\\sqrt{24^2+7^2} = 25$. Also, $OA$ is radius so we know that it's 20. This means that $AB = \\sqrt{25^2-20^2} = 15$. So, comparing $A$ And $B,$ we get $(24-a)^2+(7-b)^2 = 225$. But, we also know that $a^2+b^2 = 400$ by the problem. Playing with algebra gives $24a+7b = 400$. At $(12,16)$ this equation is satisfied with $b > 0$. I had to use calculator for this part but yeah, this one works out so maybe it is right. [/hide]", "Solution_8": "[hide]Let $O = (0,0), P = (a,b), Q = (24,7)$ and $R = (24,0)$.\nLet $x = \\angle QOR$ and $y = \\angle QOP$\n$\\sin x = \\frac 7{25}, \\cos x = \\frac{24}{25}, \\sin y = \\frac 35, \\cos y = \\frac 45$\n$a = 20\\cos(x+y) = 20\\left(\\frac{24*4-7*3}{25*5}\\right) = 20\\left(\\frac35\\right) = \\boxed{12}$[/hide]", "Solution_9": "Nice solutions Silver Falcon and alan!!! Silver Falcon: can you explain what you did with the algebra, I kindof see what you did but I am not quite sure.\r\n\r\nProblem #3:\r\nFind the smallest positive value of m such that the solutions (for x) of ((x^2+x+1)/x)=m are real numbers\r\n\r\nis m=3?", "Solution_10": "[quote=\"lyra\"]Problem #3:\nFind the smallest positive value of m such that the solutions (for x) of ((x^2+x+1)/x)=m are real numbers\n\nis m=3?[/quote]\r\n[hide=\"Solution 1\"]Rewrite the equation into a quadratic: $x^2+(1-m)x+1=0$. Since the roots of this equation must be real, its discriminant must be nonnegative so $(1-m)^2-4(1)(1)=m^2-2m-3=(m-3)(m+1)\\ge0$. Solve for $m$ and you get $m\\ge3$ because $m$ is positive.[/hide]\n\n[hide=\"Solution 2\"]This solution requires a bit of an insight: Write the equation as $m=x+1+\\frac1x$. Easily, by AM-GM, we have $x+\\frac1x\\ge2$ because $x$ is a nonzero real number. So, $m=x+1+\\frac1x\\ge3$.[/hide]", "Solution_11": "So you are saying if you had to write a solution it would be m=3 because that is the smallest. That is what I did.\r\n\r\nSmiles,\r\nlyra", "Solution_12": "Problem #4:\r\nFor all real numbers x, $f(x)=\\sqrt{x^2+25} + \\sqrt{(x-1)^2+25}$. If the minimum value of $f(x)$ is $\\sqrt{a}$, find a.\r\n\r\nSmiles,\r\nlyra", "Solution_13": "[hide=\"Cauchy it\"]\nBy Cauchy's inequality, \n\n\\begin{eqnarray*} (1\\cdot \\sqrt{x^2+25}+1\\cdot \\sqrt{(x-1)^2+25})^2 &\\le & [1+1][(x^2+25)+((x-1)^2+25)]\\\\ &\\le & 2(2x^2-2x+51)\\\\ &\\le & (2x-1)^2+101\\\\ &\\le & 101\\\\ \\Rightarrow a &=& \\boxed {101} \\end{eqnarray*}\n[/hide]", "Solution_14": "[quote=\"lyra\"]Nice solutions Silver Falcon and alan!!! Silver Falcon: can you explain what you did with the algebra, I kindof see what you did but I am not quite sure.[/quote]\r\n\r\nWell, you have $(24-a)^2 + (7 - b)^2 = 225$ and $a^2+b^2 = 400$. Expand the first one and subtract to cancel $a^2$ and $b^2$. You should get something as what I have.", "Solution_15": "Oh, yes I see what your saying.\r\n\r\nThe next one is pretty fun, you should try it.\r\n\r\nSmiles,\r\nlyra", "Solution_16": "[quote=\"chess64\"][hide=\"Cauchy it\"]\nBy Cauchy's inequality, \n\n\\begin{eqnarray*} (1\\cdot \\sqrt{x^2+25}+1\\cdot \\sqrt{(x-1)^2+25})^2 &\\le & [1+1][(x^2+25)+((x-1)^2+25)]\\\\ &\\le & 2(2x^2-2x+51)\\\\ &\\le & (2x-1)^2+101\\\\ &\\le & 101\\\\ \\Rightarrow a &=& \\boxed {101} \\end{eqnarray*}\n[/hide][/quote]\r\n\r\nOoh, good solution, I didn't think of that, that takes away some messy stuff!\r\n\r\nSmiles,\r\nlyra" } { "Tag": [ "geometry", "geometry theorems" ], "Problem": "Is there any good source where I can learn everything about barycenters and their applications?\r\nI have seen mr. Luis Geometria solve a lot of problems using that technique hin Gemetry problems section.Has he or someone else have any material on this topic? I am in urgent need of it. Thank you very much in advance.", "Solution_1": "Dear Mathlinkers,\r\nsee for example\r\nhttp://garciacapitan.auna.com/baricentricas/ (in Spain)\r\nSincerely\r\nJean-Louis" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "find all natural numbers $n$ such that $2^{n}+1$is the power of natural number.", "Solution_1": "The only solution is 2^3+1=3^2. Assume 2^n+1=x^a where a>1. Then 2^n=(x-1)(x^(a-1)+x^(a-2)+...+x+1) Then x=2^k+1. As a result a must be even for otherwise the second expression would be a sum of an odd number of odd numbers which can't be a power of two. Set a=2b. Then one has 2^n=(x^b-1)(x^b+1). This results in n=3, x=3, a=2 as the only solution for a>1." } { "Tag": [ "geometry", "trapezoid", "rectangle", "power of a point", "radical axis" ], "Problem": "Circles with centers $O$ and $O'$ are disjoint. A tangent from $O$ to the second circle intersects the first in $A$ and $B$ .\r\nA tangent from $O'$ to the first circle intersects the second circle in $A'$ and $B'$ such that $A$ and $A'$ lie on the same side of $OO'$ .Prove that $AA'B'B$ is a trapezoid .", "Solution_1": "[hide]Note that $\\triangle OAB$ and $\\triangle O'A'B'$ are isosceles with bases $\\overline{AB}$ and $\\overline{A'B'}$ respectively.\nDraw $\\overline{OO'}$. It bisects $\\angle A'O'B'$ and is thus perpendicular to $\\overline{A'B'}$. Similarly it is perpendicular to $\\overline{AB}$. Thus $\\overline{AB}\\parallel\\overline{A'B'}$.[/hide]", "Solution_2": "O,A,B are collinear :maybe: so what triangle $OAB$", "Solution_3": "The way you state it, it sounds like you construct 2 line segments from $O$, one for each of the possible tangents, and each one only intersects circle $O$ once. In that case $AB$ and $A'B'$ are obviously parallel.\r\n\r\nNow that I know $A$ and $B$ are colinear, here's a proof:\r\n\r\n[hide]Look at $A$, the point above $O$ where $AB$ intersects circle $O$. Call the angle $AOO'$ $\\theta$. Call the radius of $O$ $r$, and the radius of $O'$ $r'$. Call the length of $OO'$ $d$. The height, $h$ of $A$ above line $OO'$ is $r$sin($\\theta$). Now look at where line $AB$ is tangent to circle $O'$, and call it point $C$. Triangle $OCO'$ has right angle $C$ and acute angle $\\theta$. Looking at it we can see that sin($\\theta$) $= \\frac{r'}{d}$, and therefore $h = \\frac{r*r'}{d}$. The same argument applies to the other circle, and we can see that $A$ and $A'$ have the same height above $OO'$, as do $B$ and $B'$. Therefore, $AA'$ and $BB'$ are both horizontal, and parallel.[/hide]", "Solution_4": "[quote=\"silouan\"]Circles with centers $O$ and $O'$ are disjoint. A tangent from $O$ to the second circle intersects the first in $A$ and $B$ .\nA tangent from $O'$ to the first circle intersects the second circle in $A'$ and $B'$ such that $A$ and $A'$ lie on the same side of $OO'$ .Prove that $AA'B'B$ is a trapezoid .[/quote][color=darkblue]Denote : the circles $w=C(O)$, $w'=C(O')$ ; the tangent points $T\\in w$, $T'\\in w'$ [u]so that the line[/u] $OO'$ [u]separates these points[/u] $T$, $T'$; the intersection $S\\in \\overline{BOAT'}\\cap\\overline{A'O'B'T}$. Therefore, the quadrilateral $OTO'T'$ is cyclically , i.e. $\\widehat{AOT}\\equiv\\widehat{A'O'T'}$ $\\Longrightarrow$ the isosceles triangles $AOT$, $A'O'T'$ are similarly $\\Longrightarrow$ $\\widehat{OAT}\\equiv\\widehat{T'A'T}\\ ,\\ \\widehat{STB}\\equiv\\widehat{BT'B'}$ $\\Longrightarrow$ the quadrilaterals $ATA'T'$, $BTB'T'$ are cyclically $\\Longrightarrow$ $SA\\cdot ST'=ST\\cdot SA'\\ ,\\ SB\\cdot ST'=ST\\cdot SB'$ $\\Longrightarrow$ $\\frac{SA}{SB}=\\frac{SA'}{SB'}$ $\\Longrightarrow$ $\\boxed{\\ AA'\\parallel BB'\\ }\\ .$\n\n[b]Remarks.[/b] Prove easily that : $BT\\parallel A'T'$, $AT\\parallel B'T'$, i.e. and the quadrilaterals $BTA'T'$, $ATB'T'$ are trapezoids ; if denote the intersections $U\\in AT\\cap A'T'$ and $V\\in BT\\cap B'T'$ then the quadrilateral $TUT'V$ is rectangle and the line $UV$ is radical axis between the circles $w$, $w'$. [/color]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Show that there exist infinitely many natural numbers that are not representable as sum of two different powers of $2$ (with non-nagative exponents) and a prime number.", "Solution_1": "I can give you the classic proof by Crocker, if it is of interest." } { "Tag": [], "Problem": "Does anyone know where can I found the APMO tests? :D", "Solution_1": "Currently at http://www.kalva.demon.co.uk/apmo.html and very soon at http://www.mathlinks.ro/Forum/resources.php?c=1&cid=20" } { "Tag": [ "number theory open", "number theory" ], "Problem": "it's well-known that :\r\n 1+2^3+3^3+...............+n^3=(n(n+1)/2)^2\r\n but i want someone to find it with a good way", "Solution_1": "Good way $n^3=(\\frac{n(n+1)}{2})^2-(\\frac{n(n-1)}{2})^2$.", "Solution_2": "no man you did not understand me \r\n calculate \r\n 1+2^3+3^3+...................+n^3", "Solution_3": "it's true with n=1 or n=2.Then if it's true with n=k, you can easily prove that it's true with n=k+1.I think it's a traditional way to prove the problems like this.What do you think?", "Solution_4": "I think he understood you well (or at least I understand it the same way).\r\nWhat do you mean by 'calculate'\u00bf Rust just gave you one idea how to proof it by induction.\r\nYou could also take a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=59905 , it has a similar content." } { "Tag": [], "Problem": "Two friends A and B receive cups of coffee those are at equal temperature.A adds some cool cream immediately and takes his coffee 10minutes later.B waits for 10 minutes,and adds the same amount of cool cream and begins to take.Decide who takes hotter coffee?\r\n\r\nI cannot understand how should I use Newton's law of cooling here in this example.Please help me to start with.", "Solution_1": "Assuming that the cool cream is cooler than room temperature, friend B takes the hotter coffee.\r\n \r\nNewton's law of cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of its surroundings. \r\n \r\nThe rate of change of temperature when the cream is immediately added has 10 minutes worth of effect on the temperature, combined with the temperature of its surroundings.\r\n \r\nThe rate of change of temperature has no time have much effect on the temperature of friend B's coffee, so only the surroundings' temperature cooled the coffee." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be positive reals with $ ab \\plus{} bc \\plus{} ca \\equal{} 3$. Prove that:\r\n\\[ \\frac {1}{1 \\plus{} a^2(b \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} b^2(a \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} c^2(b \\plus{} a)}\\le \\frac {1}{abc}.\r\n\\]", "Solution_1": "[quote=\"Ahiles\"][color=brown]Let $ a,b,c$ be positive reals with $ ab \\plus{} bc \\plus{} ca \\equal{} 3$. Prove that:\n\\[ \\frac {1}{1 \\plus{} a^2(b \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} b^2(a \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} c^2(b \\plus{} a)}\\le \\frac {1}{abc}\n\\]\n[/color][/quote]\r\n\r\nIt just mathlinks contest but weaker :wink: .", "Solution_2": "Yes it is very similar and here there exist and easy solution .\r\nWe have \r\n\r\n$ \\sum\\frac{1}{1\\plus{}a^2(b\\plus{}c)}\\leq\\sum\\frac{1}{abc\\plus{}a^2(b\\plus{}c)}\\equal{}\\sum\\frac{1}{3a}\\equal{}\\frac{1}{abc}$", "Solution_3": "[quote=\"silouan\"]Yes it is very similar and here there exist and easy solution .\nWe have \n\n$ \\sum\\frac {1}{1 \\plus{} a^2(b \\plus{} c)}\\leq\\sum\\frac {1}{abc \\plus{} a^2(b \\plus{} c)} \\equal{} \\sum\\frac {1}{3a} \\equal{} \\frac {1}{abc}$[/quote]\r\n\r\nNice solution,silouan ! :o", "Solution_4": "[quote=\"akai\"]\nNice solution,silouan ! :o[/quote]\r\n\r\nThank you akai :) .", "Solution_5": "Well the nice solution has already been posted... so I might as well note that there is a lower bound with Jensen's:\r\nLet $ f(x)\\equal{}\\frac{1}{x\\plus{}k}$ which is convex for arbitrary $ k$. In particular set $ k\\equal{}1\\minus{}abc$.\r\n\\[ \\sum\\frac{1}{1\\plus{}a^2(b\\plus{}c)}\\equal{}\\sum f(3a)\\geq\\frac{1}{a\\plus{}b\\plus{}c\\plus{}\\frac13\\minus{}\\frac{abc}{3}}\\]", "Solution_6": "[quote=\"silouan\"]Yes it is very similar and here there exist and easy solution .\nWe have \n\n$ \\sum\\frac{1}{1\\plus{}a^2(b\\plus{}c)}\\leq\\sum\\frac{1}{abc\\plus{}a^2(b\\plus{}c)}\\equal{}\\sum\\frac{1}{3a}\\equal{}\\frac{1}{abc}$[/quote]\n\n\nhow did you write first inequality? i didn't understand why abc <= 1? (abc small or equal = 1)\n\nsorry for my english.", "Solution_7": "Use am_gm\n$\\sum_{cyc} ab \\geq3 \\sqrt[3] {a^2b^2c^2}$", "Solution_8": "From ma>=mg we have (ab+bc+ac)/3>=\u221b(a*a*b*b*c*c) <=> 3/3=1>=abc (ab+bc+ca=3)\na*b*c<=1;\nWe shrink the denominator replaces 1 with a*b*c.\n\u22111/(1+a*a(b+c) )<=\u22111/(a*b*c+a*a(b+c) )=\u22111/(a(ab+ac+bc))= 1/3 \u2211 1/a=1/3(1/a+1/b+1/c)= 1/3 (ab+bc+ac)/abc=1/3*3/abc=1/(a*b*c)", "Solution_9": "[quote=\"Ahiles\"]Let $ a,b,c$ be positive reals with $ ab \\plus{} bc \\plus{} ca \\equal{} 3$. Prove that:\n\\[ \\frac {1}{1 \\plus{} a^2(b \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} b^2(a \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} c^2(b \\plus{} a)}\\le \\frac {1}{abc}.\n\\][/quote]\n\n\n\\[ \\frac {1}{1 \\plus{} a^2(b \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} b^2(a \\plus{} c)} \\plus{} \\frac {1}{1 \\plus{} c^2(b \\plus{} a)}\\le Q \\le \\frac {1}{abc}.\n\\]\n\n\n$Q=\\frac{1}{3}(\\frac{c(1+a)}{1+b}+\\frac{a(1+b)}{c+1}+\\frac{b(c+1)}{1+a})$", "Solution_10": "We have \\[\\sum_{cyc} \\frac{1}{1+a^2(b+c)}\\leq \\frac{1}{abc} \\iff \\sum_{cyc}\\frac{1}{3a}-\\frac{1}{1+a(3-bc)}\\geq 0 \\iff \\sum_{cyc} \\frac{1-abc}{3a(1+3a-abc)} \\geq 0\\iff 1\\geq abc\\] which follows from \\[1=\\frac{ab+bc+ca}{3}\\geq \\sqrt[3]{a^2b^2c^2} \\implies 1\\geq abc.\\]" } { "Tag": [ "geometric series", "real analysis", "real analysis unsolved" ], "Problem": "[img]http://www.picturerack.com/ims/mathquark/64153.gif[/img]", "Solution_1": "The series $1+\\frac12+\\frac13+\\cdots+\\frac18+\\frac1{10}+\\cdots\\frac1{18}+\\frac1{20}+\\cdots$ \"strangely\" converges. The sum of every line \\[1+\\frac12+\\frac13+\\cdots+\\frac18\\] \\[\\frac1{10}+\\cdots+\\frac1{18}+\\cdots+\\frac1{88}\\] \\[\\cdots\\] is $\\le 8^{n}\\frac1{10^{n-1}}$.\r\nThen \\[\\sum_{n=1}^{\\infty}8^{n}\\frac1{10^{n-1}}=400\\]", "Solution_2": "Nene beat me by a minute or two :-)\r\n\r\nNo -- there are $8\\cdot9^{n}$ numbers with $n+1$ digits, none of which is 9, so the sum of their reciprocals is bounded above by $\\frac{8\\cdot9^{n}}{10^{n}}$ and the entire sum is bounded by $\\sum_{n\\geq 0}\\frac{8\\cdot9^{n}}{10^{n}}= 72$.", "Solution_3": "Yes, except I think that your last geometric series evaluates to 80, not 72. Some of us with too much time on our hands figured out that the sum is approximately 22.9. See\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32625[/url].", "Solution_4": "[quote=\"Ravi B\"]Yes, except I think that your last geometric series evaluates to 80, not 72. Some of us with too much time on our hands figured out that the sum is approximately 22.9. See\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32625[/url].[/quote]\r\n\r\nYes, I accidentally started my sum at $n = 1$ instead of $n = 0$. :oops:\r\n\r\n\r\nBy the way, it's not as hard as you suggested in that post to extend your method to grouping by the first two digits. One of the 72-term sum we need is (I think) $h_{88}-h_{9}-\\frac1{19}-\\frac1{29}-\\frac1{39}-\\frac1{49}-\\frac1{59}-\\frac1{69}-\\frac1{79}$ where $h_{n}$ is the $n$th harmonic number, which is easy to approximate. Using Mathematica to get that sum exactly, and following through with the same methods, I get an upper bound of $23.25777\\ldots$.", "Solution_5": "Looks good." } { "Tag": [ "inequalities", "calculus", "inequalities unsolved" ], "Problem": "Find the minimum of:\r\n\r\n$\\frac{a+2c}{b+2c}+\\frac{b+2a}{c+2a}+\\frac{c+2b}{a+2b}$", "Solution_1": "[quote=\"hungkhtn\"]Find the minimum of:\n$\\frac{a+2b}{a+2c}+\\frac{b+2c}{b+2a}+\\frac{c+2a}{c+2b}$[/quote]\r\nI dont know why you cant slove it. It's very easy !\r\nP >= 3.\r\n\\[ P+3= 2(a+b+c)(\\frac{1}{a+2c}+\\frac{1}{b+2a}+\\frac{1}{c+2b}) \\geq 6 \\]\r\n+++++++>>>>>>>> P>=3\r\nand P can be 3 when a=b=c>0", "Solution_2": "[quote=\"imababy\"][quote=\"hungkhtn\"]Find the minimum of:\n$\\frac{a+2b}{a+2c}+\\frac{b+2c}{b+2a}+\\frac{c+2a}{c+2b}$[/quote]\nI dont know why you cant slove it. It's very easy !\nP >= 3.\n\\[ P+3= 2(a+b+c)(\\frac{1}{a+2c}+\\frac{1}{b+2a}+\\frac{1}{c+2b}) \\geq 6 \\]\n+++++++>>>>>>>> P>=3\nand P can be 3 when a=b=c>0[/quote]\r\n\r\n Really,so how about the case in which at least one of three nos is less than 0...\r\nI don't think it can have minimum value.....Because by specializing it we can see there will be alot of set (a,b,c ) that can makes it tend to negative infinite...\r\n Hope i'm not wrong...", "Solution_3": "what are a,b,c?", "Solution_4": "For c tend to -1( but still less than -1)..a=2 and b=0 we have negative infinite....I think there must be at least 1 more conditions for a,b,c...Agree with Imababy,if a,b,c>0 this inequality is very easy...", "Solution_5": "Oh..I'm Sorry.. Sorry alot. My problem ism't it. Here is right:\r\n\r\nFind the minimum of:\r\n\r\n$\\frac{a+2c}{b+2c}+\\frac{b+2a}{c+2a}+\\frac{c+2b}{a+2b}$\r\n\r\nWith $a,b,c \\ge 0$.", "Solution_6": "[quote=\"Kummer\"]For c tend to -1( but still less than -1)..a=2 and b=0 we have negative infinite....I think there must be at least 1 more conditions for a,b,c...Agree with Imababy,if a,b,c>0 this inequality is very easy...[/quote]\r\n\r\n\r\n\r\nUh`, i didnt conside some case! But i think a,b,c>0 must be on!@@@@@@@@", "Solution_7": "Yes. Mo matter. All problems of mine allways have condition $a,b,c \\ge 0$.", "Solution_8": "[quote=\"hungkhtn\"]Find the minimum of:\n\n$\\frac{a+2c}{b+2c}+\\frac{b+2a}{c+2a}+\\frac{c+2b}{a+2b}$\n\nwith $a,b,c\\ge0$[/quote]\r\n\r\nI still don't see the problem in imababy's (trivial) solution? That are the conditions he took right? :?", "Solution_9": "[quote=\"Peter VDD\"][quote=\"hungkhtn\"]Find the minimum of:\n\n$\\frac{a+2c}{b+2c}+\\frac{b+2a}{c+2a}+\\frac{c+2b}{a+2b}$\n\nwith $a,b,c\\ge0$[/quote]\n\nI still don't see the problem in imababy's (trivial) solution? That are the conditions he took right? :?[/quote]\r\n\r\nhungkhtn had a typo originally, and now he fixed his post.", "Solution_10": "I think the classical substitution $b=a+x$, $c=a+x$, $x,y \\geq 0$ works here to show that\r\n$\\frac{a+2c}{b+2c}+\\frac{b+2a}{c+2a}+\\frac{c+2b}{a+2b} \\geq 3$.", "Solution_11": "Hmm... what's wrong with this?\r\n\r\n$\\sum_{cyc}\\frac{a+2c}{b+2c} \\ge 3$\r\n\r\n$\\sum_{cyc}\\frac{a-b}{b+2c} \\ge 0$\r\n\r\n$\\sum_{cyc}\\frac{a}{b+2c} \\ge \\sum_{cyc}\\frac{b}{b+2c}$\r\n\r\nwhich follows from rearrangement?", "Solution_12": "Rearrangement? How?", "Solution_13": "Denote for short $A=\\sum_{cyc}\\frac{a}{b+2c}, B=\\sum_{cyc}\\frac{b}{b+2c}, C=\\sum_{cyc}\\frac{c}{b+2c}$.\r\n\r\nI thought it was quite obvious that $A\\ge B\\ge C$... apparently it is not that easy to prove this. The part $B\\ge C$ is trivial but the part $A\\ge B$ is not. We have:\r\n$A-B = \\frac{2S-3T+3xyz}{(y+2z)(z+2x)(x+2y)}\\ge0$ with $S=x^3+y^3+z^3, T=xy^2+yz^2+zx^2$\r\n\r\nI think this nice ineq has been on before? But it is not easy indeed. :)", "Solution_14": "[quote=\"imababy\"]I dont know why you cant slove it. It's very easy !\nP >= 3.\n\\[ P+3= 2(a+b+c)(\\frac{1}{a+2c}+\\frac{1}{b+2a}+\\frac{1}{c+2b}) \\geq 6 \\]\n+++++++>>>>>>>> P>=3\nand P can be 3 when a=b=c>0[/quote]\r\n\r\nI think imababy solution is nicest and shortest\r\n\r\nIt's interesting that hungkhtn ineq is equivalent to\r\n\r\n2(a^3+b^3+c^3) + 3abc >= 3(ab^2+bc^2+ca^2)\r\n\r\nSo, the natural question is to find the maximum k such that\r\n\r\n a^3 + b^3 + c^3 + kabc >= (1+k/3)(ab^2+bc^2+ca^2)\r\n\r\nNamdung", "Solution_15": "[quote=\"Namdung\"]I think imababy solution is nicest and shortest[/quote]No, imababy works with an old version...\n\n[quote=\"Namdung\"]It's interesting that hungkhtn ineq is equivalent to\n\n2(a^3+b^3+c^3) + 3abc >= 3(ab^2+bc^2+ca^2)\n\nSo, the natural question is to find the maximum k such that\n\n a^3 + b^3 + c^3 + kabc >= (1+k/3)(ab^2+bc^2+ca^2)\n\nNamdung[/quote]isn't that exactly what I said? ;)", "Solution_16": "Oh, yes, so sorry. For long time I've not been here and I did so many misteakes.\r\n\r\nThe ineq 2(a^3 + b^3 + c^3) + 3abc >= 3(ab^2+bc^2+ca^2) (*)\r\n\r\ncan be proven as Vasc comment: put b = a + x and c = a+ y then\r\n\r\n LHS - RHS = 3a(x^2 - xy + y^2) + 2x^3 + 2y^3 - 3xy^2\r\n\r\nMy question about maximal k is still in force: Find the max k such that\r\n\r\n (a^3 + b^3 + c^3) + kabc >= (1+k/3)(ab^2+bc^2+ca^2)\r\nfor all a, b, c >= 0.\r\n\r\nIn (*) k = 3/2, but I believe that it can be replaced by greater constant.\r\n\r\nAlso not that k = 3 not works. Just take a = 0, b = 2, c = 3.", "Solution_17": "[u]A easy generalization[/u]:\r\n\r\n$\\{ x,y,a,b,c\\} \\subset (0,\\infty),\\ x0\\Longrightarrow \\sum \\frac{(a+b)+xb}{c+x(a+c)}\\ge 3\\cdot \\frac{x+2}{2x+1}$ (with equality iff $a=b=c$). \r\n\r\n[u]Remark.[/u] What is the minimum for the sum $\\sum \\frac{a+b}{a+c}\\ ?$", "Solution_18": "$\\sum_{cyc}\\frac{a+b}{c+a}\\geq 3$ by AM-GM. Equality holds when $a=b=c$.", "Solution_19": "Stupid Mistake of mine. I don't know that after expanding We have a easy problem. Because at first I don't think the minimum is $3$.", "Solution_20": "How about this:\r\n\r\nFind the minimum of:\r\n\r\n$\\frac{a+4c}{b+4c}+\\frac{b+4a}{c+4a}+\\frac{c+4b}{a+4b}$\r\n\r\nAt first when I creat it, I intend to find an easy express cyclic inequality, but \r\n\r\nthe equal is not hold if $a=b=c$. But I don't believe that if $k=2$, the equal \r\n\r\neasy hold if $a=b=c$. Notice that if we can't guess the minimum is $3$ ,we can't \r\n\r\nexpand anything.\r\n\r\nSorry and try to help me if $k=4$ in the following:\r\n\r\n$\\frac{a+kc}{b+kc}+\\frac{b+ka}{c+ka}+\\frac{c+kb}{a+kb}$", "Solution_21": "[quote=\"Namdung\"]\nThe ineq 2(a^3 + b^3 + c^3) + 3abc >= 3(ab^2+bc^2+ca^2) (*)\n\ncan be proven as Vasc comment: put b = a + x and c = a+ y then\n\n LHS - RHS = 3a(x^2 - xy + y^2) + 2x^3 + 2y^3 - 3xy^2\n\nMy question about maximal k is still in force: Find the max k such that\n\n (a^3 + b^3 + c^3) + kabc >= (1+k/3)(ab^2+bc^2+ca^2)\nfor all a, b, c >= 0.\n\nIn (*) k = 3/2, but I believe that it can be replaced by greater constant.\n\nAlso not that k = 3 not works. Just take a = 0, b = 2, c = 3.[/quote]\r\nWe have\r\n$(LHS - RHS) = (2-k/3)(x^2 - xy + y^2) + x^3 + y^3 - (1+k/3)xy^2$.\r\nBy the AM-GM inequality, we get\r\n$x^3+y^3/2+y^3/2 \\geq \\frac{3}{\\sqrt[3]{4}}xy^2$. Thus, the best $k$ is $\\frac{9}{\\sqrt[3]{4}}-3$.", "Solution_22": "Yes. Thank you Vasc. Let think about Hung prb.", "Solution_23": "I think without calculus we are forceless, because max ~ 2.8836030805", "Solution_24": "Oh? Why? We haven't any condition of $a,b,c$?Do you calculate it when $a=b$?", "Solution_25": "[quote=\"hungkhtn\"]$ a,b,c>0.$ Find the minimum of:\n\n$ \\frac {a \\plus{} 2c}{b \\plus{} 2c} \\plus{} \\frac {b \\plus{} 2a}{c \\plus{} 2a} \\plus{} \\frac {c \\plus{} 2b}{a \\plus{} 2b}$[/quote]\r\n\r\nThis inequality is [b]Problem 2.23[/b], page 70 of new book (where it is presented with a simple proof by Cauchy-Schwarz):\r\n\r\nVasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009." } { "Tag": [ "search", "linear algebra", "matrix", "induction", "vector", "linear algebra unsolved" ], "Problem": "what is the maximal dimension of a subspace of $M_n(K)$ such that any $2$ matrices of that space commute.", "Solution_1": "I posted this problem not too long ago\r\nI can give you an answer: $[\\frac{n^2}{4}]+1$.Also in that i post i mentioned i gave an example\r\nI'll try to search it", "Solution_2": "I just can't believe this was the real problem. I asked a teacher in muy school who is a real genius and he told me the problem of finding the maximal dimension of a commutative subalgebra is extremely difficult. Alekk, are you generalizing again the problem? ;)", "Solution_3": "No. My friend really got this at the oral examination last week (the examinator was also the director of the departement of mathematic at ENS :) ). Anyway he told me that he received a lot of hints ... Nevertheless I do not know the proof since my friend forgot how he solved the pb :(", "Solution_4": "in fact the field was $\\mathbb{C}$ but I'm not sure it helps very much ..", "Solution_5": "That guy I was talking about was the first at ENS contest at his age and he said this problem is impossible to solve in one hour. And they give it at oral examination. I think the examinator was a little bit drunk. :)", "Solution_6": "Of course as for me it is too difficult problem for an examination\r\nI have a proof a bit less then a page in my book", "Solution_7": "I would be very grateful to you, eugene, if you posted the solution. Thanks. I remember you posted the problem, but nobody made any headway. :(", "Solution_8": "i'm really beginner in linear algebar but if $K=\\mathbb{C}$ i think such a idea maybe work.there exist $P$ which is invertible such that $PSP^{-1}$ for any $S\\in subspace$ such that its form is some lowertriangular matrix with equal diagonal .it is obvious that $PSP^{-1}$ also commute. so we compute maximum independent matrix for one block.if our block is $m*m$ and $m=2k$.they should be $\\begin{pmatrix}xI_k&0\\\\ *&xI_k\\end{pmatrix}$and if $m=2k+1$ ,they should be like $\\begin{pmatrix}xI_{k+1}&0\\\\ *&xI_k\\end{pmatrix}$,so for this block we have $[\\frac{m^2}{4}]+1$ and at last totaly we have $\\sum [\\frac{m_i^2}{4}]+k$and we have $n=\\sum m_i$ so i think we are done any way excuse me if i'm totaly wrong :huh:", "Solution_9": "[quote=\"sam-n\"]i'm really beginner in linear algebar but if $K=\\mathbb{C}$ i think such a idea maybe work.there exist $P$ which is invertible such that $PSP^{-1}$ for any $S\\in subspace$ such that its form is some lowertriangular matrix with equal diagonal [/quote]\r\nSam-n, could you please explain, why all the matrices $PSP^{-1}$ have equal numbers on it's diagonal, i thought they should have it's eigenvalues on main diagonal", "Solution_10": "i just mean in each block .", "Solution_11": "Maybe i don't understand what theorem you are using, but i t seemed to me that that you are using at first this one: if the set of matrices is commuting then we can find an invertible $P$ s.t. $PSP^{-1}$ are lowertriangular(uppertriangular) for all $S$ from our set of matrices\r\nIn this meaning i quite don't undertstand what kind of blocks $\\begin{pmatrix}xI_k&0\\\\ *&xI_k\\end{pmatrix}$ you are meaning (if you are meaning blocks obtained by equal eigenvalues on the diagonal, then as far as i undertsand it isn't necesseraly for them to commute with blocks from other matrices, our lowertriangular matrices are not block-form)\r\nSam-n,could you please clairify", "Solution_12": "[quote=\"harazi\"]I would be very grateful to you, eugene, if you posted the solution. Thanks. I remember you posted the problem, but nobody made any headway. :([/quote]\r\n\r\nWe'll prove our result by induction on $n$.\r\nLet's suppose our dimension $m \\geq [\\frac{n^2}{4}]+2$.\r\nIt is known fact that every set of commuting matrices have common eigenvalue, thus let it be $x$ for our set of matrices.Therefore all the mtarices in our set can be represented in form $A_j=\\left( \\begin{array}{cc} \\lambda_j & * \\\\ 0 & B_j \\end{array} \\right) $. Since all the matrices $A_1,A_2,...,A_m$ commute the matrices $B_1,B_2,...B_m$ also commute. Let $\\mathfrak{R}=_\\mathbb{C}$. By the induction hypothesis $r=dim \\mathfrak{R} \\leq [\\frac{(n-1)^2}{4}]+1$. WLOG we can assume that matrices $B_1,B_2,...,B_r$ are linearly independent, thus $B_i= \\sum_{j=1}^{r} \\alpha_{ij}B_j$. For $i>r$ put $C_i=A_i-\\sum_{j=1}^{r} \\alpha_{ij}A_j$. It is evident that all the matrices $C_j$ are linearly independent and for $i=r+1,...,m$ they are of the form $C_i= \\left( \\begin{array}{c} c_i \\\\ 0 \\end{array} \\right)$, where $c_i$ is $1 \\times n$ matrix. Moreover, vectors $c_i$ must be linearly independent over $\\mathbb{C}$.\r\n\r\nForm now let's do almost the same from another point: Our uppertriangular commuting matrice we can also write in form $A_j=\\left( \\begin{array}{cc} B'_j & u_{1j} \\\\ & . \\\\ & . \\\\ & . \\\\ 0 & u_{nj} \\end{array} \\right)$ with commuting matrices $B'_j$. Using again induction hypothesis we can again get the set of $n \\times 1$ linearly independent matrices $c'_{s+1},...,c'_m$, where $s \\leq [\\frac{(n-1)^2}{4}]+1$. Now $C'_j= \\left( \\begin{array}{c} c'_j \\\\ 0 \\end{array} \\right)$ for $s \\geq s+1$. Since all the matrices $C_i$ and $C'_j$ belong to the same commuting set, $c_ic'_j=0$ for $i\\r+1,...,m$ and $j=s+1,...,m$.\r\n\r\nTo this end, condsider $(m-r) \\times n$ matrix $C$, which has as it's $i$'th row $c_i$, $i=r+1,...,m$. Since all the $c_i$ are linearly independent we have $rank A \\geq m-r$. From the other side $Cc'_j=0$ for $j = s+1 ,...,m$. Since $c'_j$ are linearly independent and $dimKer\\mathfrak{C}+dimIm\\mathfrak{C}=n$ for linear operator $\\mathfrak{C}$ with matrix $C$, taking into account ineaulities with $m,r,s$ we get\r\n$n \\geq (m-r)+(m-s) \\geq 2 \\left( [\\frac{n^2}{4}]-[\\frac{(n-1)^2}{4}]+1 \\right) \\geq 2[\\frac{n}{2}]+2 \\geq n$ - contradiction which proves our hypothesis.", "Solution_13": "I decompose matrix in lowertrianguar block simultenously and maximum nuber of independent matrix for block of $m*m$ which commute with each other is $[\\frac{m^2}{4}]+1$ as i said so maximum in total is $\\sum [\\frac{m_i^2}{4}]+1$ which is lesser than$[\\frac{n^2}{4}]+2$." } { "Tag": [ "function", "inequalities solved", "inequalities" ], "Problem": "Prove that if x is in [0,pi/2] then we have x/sqrt(1+x^2)<=sin(x)<=x.", "Solution_1": "sin x -x is negative on [0,pi/2], you can study the function f(x)=sinx-x\r\n\r\nfor x/ \\sqrt (1+x^2) \\leq sinx take the square \r\nx^2/(1+x^2) \\leq (sinx)^2 \r\n<=>\r\n1 \\leq (sinx)^2 + 1/(1+x^2) \r\n\r\nminimum of RHS is x=0" } { "Tag": [ "SFFT" ], "Problem": "The sum of the 3 different unit fractions is 6/7. What is the sum of the denominator?\r\n\r\nAnswer?\r\n[hide]\n47[/hide]", "Solution_1": "[quote=\"DonkeyKong\"]The sum of the 3 different unit fractions is 6/7. What is the sum of the denominator?\n\nAnswer?\n[hide]\n47[/hide][/quote]\r\n\r\nCouldn't you just have $ \\frac17\\plus{}\\frac27\\plus{}\\frac37\\equal{}\\frac67$?", "Solution_2": "The problem said unit fraction - 1 over some integer.", "Solution_3": "[quote=\"DonkeyKong\"]The sum of the 3 different unit fractions is 6/7. What is the sum of the denominator?\n\nAnswer?\n[hide]\n47[/hide][/quote]\r\n\r\nWe have $ \\frac1a \\plus{} \\frac1b \\plus{} \\frac1c \\equal{} \\frac67$. Assume, WLOG, that $ \\frac1a\\geq\\frac1b\\geq\\frac1c$.\r\n\r\nThen $ \\frac1a\\geq\\frac {\\frac1a \\plus{} \\frac1b \\plus{} \\frac1c}3 \\equal{} \\frac27$, and $ a\\leq3 < \\frac72$. However, $ a\\geq2$ as well, since $ \\frac1a < \\frac67 < 1$.\r\n\r\nIf $ a \\equal{} 2$, then $ \\frac1b \\plus{} \\frac1c \\equal{} \\frac5{14}$, so, similar to above, we have $ \\frac1b\\geq\\frac5{28}\\implies2\\leq b\\leq5$ (remember that $ \\frac1a\\geq\\frac1b\\geq\\frac1c$!), and testing, we find that $ (a,b,c) \\equal{} (2,3,42)$ is the only possibility for this case. Note that SFFT would have worked as well.\r\n\r\nIf $ a \\equal{} 3$, then $ \\frac1b \\plus{} \\frac1c \\equal{} \\frac {11}{21}$. Again, we have $ \\frac1b\\geq{11}{42}\\implies3\\leq b\\leq3$. However, $ b \\equal{} 3$ is not valid.\r\n\r\nWe conclude that the only solution is $ (a,b,c) \\equal{} (2,3,42)$, which gives an answer of $ \\boxed{47}$.\r\n\r\nEdit: Thanks, mewto55555. :) You see, I was typing this at school today while helping emotionless1 in a countdown. :P", "Solution_4": "[quote=\"math154\"][quote=\"DonkeyKong\"]The sum of the 3 different unit fractions is 6/7. What is the sum of the denominator?\n\nAnswer?\n[hide]\n47[/hide][/quote]\n\nWe have $ \\frac1a \\plus{} \\frac1b \\plus{} \\frac1c \\equal{} \\frac67$. Assume, WLOG, that $ a\\geq b\\geq c$.\n\nThen $ \\frac1a\\geq\\frac {\\frac1a \\plus{} \\frac1b \\plus{} \\frac1c}3 \\equal{} \\frac27$, and $ a\\leq3 < \\frac72$. However, $ a\\geq2$ as well, since $ \\frac1a < \\frac67 < 1$.\n\nIf $ a \\equal{} 2$, then $ \\frac1b \\plus{} \\frac1c \\equal{} \\frac5{14}$, so, similar to above, we have $ \\frac1b\\geq\\frac5{28}\\implies2\\leq b\\leq5$ (remember that $ a\\geq b\\geq c$!), and testing, we find that $ (a,b,c) \\equal{} (2,3,42)$ is the only possibility for this case. Note that SFFT would have worked as well.\n\nIf $ a \\equal{} 3$, then $ \\frac1b \\plus{} \\frac1c \\equal{} \\frac {11}{21}$. Again, we have $ \\frac1b\\geq{11}{42}\\implies3\\leq b\\leq3$. However, $ b \\equal{} 3$ is not valid.\n\nWe conclude that the only solution is $ (a,b,c) \\equal{} (2,3,42)$, which gives an answer of $ \\boxed{47}$.[/quote]\r\n\r\nA slight problem there. You say that a>b>c yet your answer is a0$\r\n\r\nthis is probably a very simple problem...i'm actually really ashamed that i got stuck on it...but yea...i got stuck on it... :( \r\n\r\nso can someone please help me...tank you!!!", "Solution_1": "$3\\sqrt[3]{\\frac{1}{4}}$", "Solution_2": "Just AM-GM on $x/2$, $x/2$ and $1/x^{2}$.\r\n\r\nTank you... :rotfl:", "Solution_3": "oooooohh...i see...that's soooo easy...haha...\r\n\r\ntank you!!!! :rotfl:", "Solution_4": "It's also easy with calculus. Call it f(x), take the derivative, set it equal to 0, and plug the value of x you get into the original function. Then verify that it's a minimum with a second derivative test.", "Solution_5": "Julian are you taking calculus in school?", "Solution_6": "o...part of my math course covers calculus...but i sort of skip math...mentally...lol :P \r\nschool is finally over...hurray!!!\r\n1 more provincials though... :|" } { "Tag": [ "algebra", "polynomial", "Galois Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Find \\[ min(\\sqrt{2}+\\sqrt [3] {3},\\mathbb{Q}). \\]", "Solution_1": "I have edited. ;)", "Solution_2": "Cezar i dont understand the problem,could you reformulate and give some examples if possible ?", "Solution_3": "well, the minimal polynomial of an algebraic number $\\alpha$ over the field $K$ is the monic irreducible polynomial $\\mu_\\alpha \\in K[x]$ such that $\\mu_\\alpha(\\alpha)=0$.\r\nfor example, the minimal polynomial $\\mu_{\\sqrt2}$ of $\\sqrt2$ over $\\mathbb{Q}$ is trivially $\\mu_{\\sqrt2}(x) = x^2-2$, and so on...\r\nby the way, the degree of the minimal polynomial equals the degree of the simple extension: $\\partial \\mu_\\alpha = [K(\\alpha): K]$.", "Solution_4": "And when someone is really bored, he could take all the $6$ the conjugates $a_1,...,a_6$ of $\\sqrt{2}+\\sqrt[3]{3}$, the minimal polynomial is $(x-a_1)(x-a_2)...(x-a_6)$ then (the conjugates are $\\pm \\sqrt{2}+\\zeta_3^{k}\\sqrt[3]{3}$).", "Solution_5": "The result is $x^6 - 6 x^4 - 6 x^3 + 12 x^2 - 36 x + 1$. ;)\r\n\r\nIt is easy to see that the conjugates of $\\sqrt{2} + \\sqrt[3]{3}$ have the form $\\pm \\sqrt{2} + \\zeta_3^k \\sqrt[3]{3}$?", "Solution_6": "[quote=\"ZetaX\"]And when someone is really bored, he could take all the $6$ the conjugates $a_1,...,a_6$ of $\\sqrt{2}+\\sqrt[3]{3}$, the minimal polynomial is $(x-a_1)(x-a_2)...(x-a_6)$ then (the conjugates are $\\pm \\sqrt{2}+\\zeta_3^{k}\\sqrt[3]{3}$).[/quote]\r\n\r\nI guess this is the only way out.", "Solution_7": "[quote=\"-oo-\"]The result is $x^6 - 6 x^4 - 6 x^3 + 12 x^2 - 36 x + 1$. ;)\n\nIt is easy to see that the conjugates of $\\sqrt{2} + \\sqrt[3]{3}$ have the form $\\pm \\sqrt{2} + \\zeta_3^k \\sqrt[3]{3}$?[/quote]\r\n\r\nCan you please write in detail your solution? :?", "Solution_8": "He just did what I said above (at least I think so), so what is not clear\u00bf\r\n(that these are the conjugates can be seen by considering the extension of adjuncting $\\sqrt{2}$ to $\\mathbb{Q}[\\sqrt[3]{3}]$ and converse; in total, we don't even need to prove it formaly since what we get has the desired degree and the desired coefficients ;) )", "Solution_9": "hm.. as far as i can see...\r\nit's easy to see that $\\alpha$ has to be a root of this polynomial. moreover, $\\sqrt2 + \\sqrt[3]{3}\\in\\mathbb{Q}(\\sqrt2, \\sqrt[3]{3})$, hence $[\\mathbb{Q}(\\sqrt2 + \\sqrt[3]{3}): \\mathbb{Q}]|[\\mathbb{Q}(\\sqrt2, \\sqrt[3]{3}): \\mathbb{Q}] = 6$. but trivially $\\sqrt2 + \\sqrt[3]{3}$ can't have a minimal polynomial of degree less then $4$ (you can see it by \"brute force\").", "Solution_10": "Ok, we can [i]verify[/i] that the result is correct. But how can one [i]see[/i] that the conjugates have the desired form? I don't know how to investigate the galois group here ...", "Solution_11": "@-oo-: how exactly is your question to understand\u00bf (I find it very \"naturally\" that the conjugates come from conjugating each summand for itself)\r\n\r\nJust a proof that $-\\sqrt{2}+\\sqrt[3]{3}$ is a conjugate of $\\sqrt{2}+\\sqrt[3]{3}$ (the others are similar):\r\nOver $\\mathbb{Q}[\\sqrt[3]{3}]$ we have the irreducible polynomial $x^2-2$ (to see irreducibility ony could just calculate through it to see that there is no root; a more convenient but more theoretic way would be to use Eisenstein). Thus adjuncting $\\sqrt{2}$ is possible, giving us an automorphism that sends $\\sqrt{2}$ onto $-\\sqrt{2}$, but lets $\\mathbb{Q}[\\sqrt[3]{3}]$ fixed." } { "Tag": [ "quadratics" ], "Problem": "if [i]p[/i] and [i]k[/i] are the roots of [i]x[/i] in the equation $x^2 + px + k$ and $p \\ne k$ then find the values of [i]p[/i] and [i]k[/i].", "Solution_1": "[quote=\"akimatsu\"]if [i]p[/i] and [i]k[/i] are the roots of [i]x[/i] in the equation $x^2 + px + k$ and $p \\ne k$ then find the values of [i]p[/i] and [i]k[/i].[/quote]\r\n[hide]\n$p+k=-p$\n$pk=k$\n$2p=-k$\n$-2p^2=-2p$\n$2p^2-2p=0$\n$p_1=0$, $p_2=1$\nBut if $p=0$ then $k=0$ but $p\\neq k$ so $\\boxed{p=1}$, $\\boxed{k=-2}$[/hide]" } { "Tag": [], "Problem": "which do you think was better of the beijing olympics. write which one was better and why?", "Solution_1": "If I had to pick, I'd choose Opening Ceremony. There was slightly more pizazz in it, and you actually wanted to see what would happen next. In the Closing Ceremony, I was thinking like \"finish it up\", so there's my reason. I feel bad for London though, they're gonna have to match these 2 spectacular ceremonies.", "Solution_2": "both ceremonies were really good because they both illustrated Chinese history and culture, as well as modern themes and stuff. but if i had to choose i guess opening ceremony, since it had better effects and the theme was much clearer. plus london kinda had a not-as-good show in the closing ceremony compared to china, so it kind of effected the ending too." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "Prove that a square matrix of $n\\times n$ where $n=p\\cdot q$ can be written as a sum of $p$ matrixes of rank $q$. Prove that $I_n$ cannot be written as a sum of less than $p$ matrixes of rank $q$.", "Solution_1": "This seems to be a generalisation for a problem from the district round a couple of year ago.\r\n\r\n\r\nFor the second part:\r\n $rank(A_1+A_2+...+A_t) \\leq rank(A_1)+rank(A_2)+...+rank(A_t)$ which means that $p\\leq t$ contradiction beacause $t b$ to the ellipse ${\\frac{x^2}{a^2}} + {\\frac{y^2}{b^2}} = 1$ is\r\n\r\n (a) 3 (b) 4\r\n (c) 2 (d) 1\r\n \r\n\r\nI am getting (c) for both problems.", "Solution_1": "Q1) If by external point, the question means a particular point, then there is only one line that can be drawn through that point which will be normal to the ellipse.\r\nIf, however, the questions means [i]any[/i] point, then there is an inifinite number of such points and hence an inifinite number of normals. \r\nI suspect the answer should be (d)\r\n\r\nQ2) This ellipse is centred on the origin. The point (0,c) is a point on the y-axis somewhere above the top curve of the ellipse. The only line that can be drawn through this point [i]and[/i] normal to the ellipse is the y-axis itself. So the answer is (d), again.", "Solution_2": "[quote=\"Fermat2\"]Q1) If by external point, the question means a particular point, then there is only one line that can be drawn through that point which will be normal to the ellipse.\nIf, however, the questions means [i]any[/i] point, then there is an inifinite number of such points and hence an inifinite number of normals. \nI suspect the answer should be (d)\n\nQ2) This ellipse is centred on the origin. The point (0,c) is a point on the y-axis somewhere above the top curve of the ellipse. The only line that can be drawn through this point [i]and[/i] normal to the ellipse is the y-axis itself. So the answer is (d), again.[/quote]\r\n\r\nI did the following :-\r\nThe given equation is of the form $\\frac{x^2}{a^2} {+} \\frac{y^2}{b^2} =1$\r\nThen when you find the equation to a normal with slope $(m)$ \r\nthe equation wold be ${y} {=} {mx} {+} {\\frac{(b^2 - a^2)m}{\\sqrt{a^2 + {b^2}{m^2}}}}$i.e, in general four normals can be drawn from an external point. \r\nWhen the point is $(0,c)$ the equation reduces to ${{c^2}(a^2 +{b^2}{m^2})} {=} {{m^2}(b^2 - a^2)^2}$\r\nso we can have two values of m. Hence 2 such normals.\r\nPoint out the mistake if i have done.\r\nBye. :lol:", "Solution_3": "The direction of a normal to the curve (ellipse in this case) $F(x, y) = \\frac{x^2}{a^2} + \\frac{y^2}{b^2} - 1 = 0$ is $\\text{grad} F = (\\frac{\\partial F}{\\partial x},\\ \\frac{\\partial F}{\\partial y}) = (\\frac{2x}{a^2},\\ \\frac{2y}{b^2})$. This gradient cannot be zero on the ellipse, because if the x-component is zero, $\\frac{2x}{a^2} = 0,\\ x = 0$, then from the ellipse equation $\\frac{y^2}{b^2} = 1,\\ y = \\pm b,\\ \\frac{2y}{b^2} = \\pm \\frac{2}{b} \\neq 0$, the y-component is non-zero, and the other way around. Since $\\text{grad} F = (0,\\ \\frac{2}{b})$ is an external normal at ellipse point (x, y) = (0, b), $\\text{grad} F$ is an external normal at any ellipse point (x, y). Let P = (p, q) be an arbitrary point not on the ellipse main axes, i.e., $p \\neq 0, q \\neq 0$, and let U = (u, v) be the intersection of the normal to the ellipse from P with the ellipse. Since U is on the ellipse, $\\frac{u^2}{a^2} + \\frac{v^2}{b^2} = 1$. Since $p \\neq 0$, the ellipse normals at (0, b) and (0, -b) do not pass through the point P, hence $u \\neq 0$. Likewise, since $q \\neq 0$, the ellipse normals at (a, 0) and (-a, -) do not pass through the point P, hence $v \\neq 0$. The equation of the ellipse normal at U = (u, v) is then $y - v = m (x - u)$ where the slope $m = \\frac{\\partial F}{\\partial y} : \\frac{\\partial F}{\\partial x}|_{(x,y) = (u,v)} = \\frac{a^2v}{b^2u}$. If this line passes through the point P = (p, q), then $q - v = \\frac{a^2v}{b^2u} (p - u)$. Together with $\\frac{u^2}{a^2} + \\frac{v^2}{b^2} = 1$, this yields 2 equations for (u, v). For example, rearranging and squaring the 1st equation and then substituting the 2nd equation leads to a quartic equation for v:\r\n\r\n$\\frac{avp}{b^2q + (a^2 - b^2) v} = \\frac{u}{a},\\ \\ (\\frac{avp}{b^2q + (a^2 - b^2) v})^2 = \\frac{u^2}{a^2} = 1 - \\frac{v^2}{b^2}$\r\n\r\nThis indicates that up to 4 distinct normal lines to the ellipse from the point P may exist.\r\n\r\nWe can parametrize the ellipse equation as $x = a \\cos t,\\ y = b \\sin t,\\ t \\in (0, 2\\pi)$. The gradient is then $\\text{grad} F = ( \\frac{2 \\cos t}{a},\\ \\frac{2 \\sin t}{b})$ and $|\\text{grad} F| = \\frac{2}{ab} \\sqrt{b^2 \\cos^2 t + a^2 \\sin^2 t}$, so that the exterior unit normal to the ellipse is $\\vec n = \\frac{1}{\\sqrt{b^2 \\cos^2 t + a^2 \\sin^2 t}} (b \\cos t,\\ a \\sin t)$. The ellipe curvature $\\kappa$ is given by\r\n\r\n$\\kappa = \\frac{|x''y' - y''x'|}{(x'^2 + y'^2)^{3/2}} = \\frac{|(-a \\cos t)(-b \\cos t) - (-b \\sin t)(-a \\sin t)}{(a^2 \\sin^2 t + b^2 \\cos^2 t)^{3/2}} = \\frac{ab}{(a^2 \\sin^2 t + b^2 \\cos^2 t)^{3/2}}$\r\n\r\nThe normal line to an ellipse tangent from the tangency point passes through the center of a circle with radius $R = \\frac{1}{\\kappa}$ tangent to the ellipse. If an ellipse point has the radius vector $\\vec r = (x, y) = (a \\cos t, b \\sin t)$, the center C of this circle has the radius vector $\\vec r_C = \\vec r - R \\vec n$, \r\n\r\n$x_C = a \\cos t - \\frac{(a^2 \\sin^2 t + b^2 \\cos^2 t)^{3/2}}{ab} \\cdot \\frac{b \\cos t}{\\sqrt{b^2 \\cos^2 t + a^2 \\sin^2 t}} =$\r\n\r\n$= \\frac{\\cos t}{a}\\ [a^2(1 - \\sin^2 t) - b^2 \\cos^2 t] = \\frac{a^2 - b^2}{a}\\ \\cos^3 t$\r\n\r\nand similarly,\r\n\r\n$y_C = \\frac{a^2 - b^2}{b}\\ \\sin^3 t$\r\n\r\nThe locus of the centers of these circles is the envelope of the normals to the ellipse tangents at the tangency points (the ellipse evolute). Elliminating the parameter t from the parametric equations of the evolute and skipping the subscripts C from $x_C, y_C$, we get its equation as\r\n\r\n$(ax)^{2/3} + (by)^{2/3} = (a^2 - b^2)^{2/3} (\\cos^2 t + \\sin^2 t) = (a^2 - b^2)^{2/3}$\r\n\r\nThis is an astroid with the main axes $\\frac{|a^2 - b^2|}{a},\\ \\frac{|a^2 - b^2|}{b}$ (kind of astroid, usually, astroid is supposed to have equal main axes). Assume again, that the arbitrary point P does not lie on the ellipse main axes. If the point P lies inside the ellipse evolute, 4 distinct normal lines from P to the ellipse exist, if the point P lies on the ellipse evolute, 3 distinct normal lines from P to the ellipse exist, and if the point P lies outside the ellipse evolute, 2 distinct normal lines from P to the ellipse exist. If the point P lies on one of the ellipse main axes, 2 of the normal lines become identical.\r\n\r\nApplying these result to the question (Q1), the ellipse semimajor and semiminor axes are a = 3, b = 2. The evolute has 2 vertices on the x-axis at $(\\frac{a^2 - b^2}{a},\\ 0)$ and $(-\\frac{a^2 - b^2}{a},\\ 0)$ or $(\\frac 5 3,\\ 0)$ and $(-\\frac 5 3,\\ 0)$. These 2 vertices lie in the ellipse interior. On the other hand, the evolute has 2 vertices on the y-axis at $(0,\\ \\frac{a^2 - b^2}{b})$ and $(0,\\ -\\frac{a^2 - b^2}{b})$ or $(\\frac 5 2,\\ 0)$ and $(-\\frac 5 2,\\ 0)$. These 2 vertices lie in the ellipse exterior and so does a part of the evolute. Consequently, exterior points of the ellipse exist, which lie inside the evolute, i.e., with 4 distinct normal lines to the ellipse. Obviously, exterior points of the ellipse exist, which lie on or outside the evolute, i.e. with 3 or 2 distinct normal lines to the ellipse. Without an additional info about the exterior point P, it is not possible to say, whether the answer (c) is correct or whether the correct answer is not on the list. Therefore, the problem is wrong. Ditto the question (Q2). Without an additional info about the ellipse main axes a, b, it is not possible to decide, whether the evolute lies completely inside the ellipse or whether some part of it sticks out. But if it sticks out along the y-axis, i.e., if $\\frac{a^2 - b^2}{b} > b,\\ a^2 > 2b^2,\\ a > b \\sqrt 2$, then either 1 or 3 distinct normal lines to the ellipse from an exterior point on the y-axis may exist and again, it is not possible to choose among the answers (a), (d). Therefore, the problem is also wrong." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all primes of the form 4k+1.\r\n(I think there are finitly many numbers of this form.)", "Solution_1": "There are $\\infty$ many ;)\r\nProof:\r\nThe equation $x^2 \\equiv -1 \\mod p$ cannot have solutions for primes $p \\equiv 3 \\mod 4$ since that would give (using that $p-1=4k+2$) us $1 \\equiv x^{p-1} = x^{4k+2} = (x^4)^k \\cdot x^2 \\equiv -1 \\mod p$.\r\nNow assume there are only finetely many $\\equiv 1\\mod 4$, then let their product be $P$.\r\nNow look at $n=(2P)^2+1$:\r\nIt is odd, and coprime to all primes $\\equiv 1 \\mod 4$ contained in $P$. But any prime $q$ dividing $n$ gives $(2P)^2 \\equiv -1 \\mod q$, thus (remember $q \\not\\equiv 0,2 \\mod 4$) $q \\equiv 1\\mod 4$, which is clearly not possible, contradiction." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ F$ be a field with characteristic $ p$ and $ E$ a finite extension of $ F$. Prove that there is a non-negative integer $ n$ such that $ [E: F]\\equal{}p^n[E: F]_s$.", "Solution_1": "Consider the maximal separable intermediate extension." } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "In a convex pentagon $ABCDE$, $\\angle ABC = \\angle AED = 90 ^\\circ$. Let $F$ be the midpoint of $CD$. If $FB=FE$, prove that $\\angle BAC = \\angle DAE$.", "Solution_1": "Let $M$ and $N$ be the midpoints of $DA$ and $CA$. Then: \r\n$MF=\\frac12 AC=NB$ and $NF=\\frac12 DA=EM$ (because triangles $EAD$ and $ABC$ are rightangled and $DA$ and $AC$ are hypotenuses). $FE=FB$ from the text so $\\triangle EMF \\cong \\triangle FNB$. \r\n\r\nNow problem is done, because $\\angle EAD=\\frac12 \\angle EMD=\\frac12(\\angle EMF-\\angle DAC)=\\frac12(\\angle FNB-\\angle DAC)=\\frac12 \\angle CNB=\\angle BAC$.\r\n\r\nbye" } { "Tag": [], "Problem": "Let's say we are chatting computers. We all input something, and a computer outputs something. Please put it in some sort of computer code, or typeset. Once the game has gotten a few pages, we may be able to put in phrases. I'll make that rule starting in the 21st post in this topic. This is almost a conversation.\r\n\r\ninput: prompt x,y,z", "Solution_1": "i dont get it", "Solution_2": "ERROR: LAWL", "Solution_3": "is this like teh slot machine?", "Solution_4": "ALMOST. You can input calculator program threads, phrases, but only things you can type. Like this: \r\n\r\noutput:\r\nx=?\r\ny=?\r\nz=?\r\n\r\ninupt: (1,2,3)\r\n\r\nsometimes it can be funny.", "Solution_5": "input: $x\\star y = \\frac{(\\frac{97332}{5x})^{y}\\times \\frac{x-y}{y-x}}{xy}\\times \\sqrt{x}$\r\n$value= OPERATION(find): \\infty^{E}\\star\\sqrt{\\pi}$", "Solution_6": "output: error 1\r\ninput: $(a,b,c,d,e,...,x,y,z)$", "Solution_7": "output: error, omega does not compute\r\n\r\ninput: 4 store omega", "Solution_8": "Output: A can of coca cola\r\n\r\nInput: 2 dollars", "Solution_9": "output: Thank you for the moolah. Now I have enough money to buy a computer.\r\ninput: Output 2 dollars" } { "Tag": [ "calculus", "integration", "function", "limit", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Find all continuous functions $f : \\mathbb R \\to \\mathbb R$ such that:\r\n(a) $\\lim_{x \\to \\infty}f(x)$ exists;\r\n(b) $f(x) = \\int_{x+1}^{x+2}f(t) \\, dt$, for all $x \\in \\mathbb R$.", "Solution_1": "Let $\\lim_{x\\to \\infty}f(x) = L$. Then for every $\\epsilon>0$, there exists $C$ such that $|f(x)-L|<\\epsilon$ whenever $x>C$.\r\n\r\nNow let $x>C$ and $I_{k}=(x-k, \\infty)$. Then $L-\\epsilon < f < L+\\epsilon$ on $I_{0}$, so integrating both side of (1) on $I_{0}$ gives\r\n\r\n$L-\\epsilon < f(x-1) < L+\\epsilon$\r\n\r\nor\r\n\r\n$L-\\epsilon < f(x) < L+\\epsilon$ on $I_{1}$\r\n\r\nIf we recursively repeat this process, then for any $r \\in \\mathbb{R}$, we have $L-\\epsilon < f(r) < L+\\epsilon$. But this inequality holds for arbitrary $\\epsilon>0$, we conclude that $f(r) = L$. Therefore, $f$ is a constant function.\r\n\r\n\r\n\r\n-o- but I can't sure my proof is right...", "Solution_2": "It's right. I would probably use $\\le\\epsilon$ rather that $<\\epsilon$, just to avoid having to justify strict inequalities after integration.", "Solution_3": "Or we can start with an $x_{0}$ and construct an increasing to infinity sequence $x_{n}$ that is constant (we use Lagrange for example). Taking the limit of $f(x_{n})$ it is L=$\\lim_{x\\to\\infty}f(x)$ but also $x_{0}$ so the problem is solved. A quick computation shows that all constant functions verify the hyphotesis .", "Solution_4": "[quote=\"sos440\"]$I_{k}=(x-k, \\infty)$\n...\nThen $L-\\epsilon < f < L+\\epsilon$ on $I_{0}$, so integrating both side of (1) on $I_{0}$ gives\n\n$L-\\epsilon < f(x-1) < L+\\epsilon$\n[/quote]\r\nUnless I'm misunderstanding, don't you mean $I_{k}=(x-k, x-k+1)$?", "Solution_5": "[quote=ciprian]Or we can start with an $x_{0}$ and construct an increasing to infinity sequence $x_{n}$ that is constant (we use Lagrange for example). Taking the limit of $f(x_{n})$ it is L=$\\lim_{x\\to\\infty}f(x)$ but also $x_{0}$ so the problem is solved. A quick computation shows that all constant functions verify the hyphotesis .[/quote]\n\n\nCould you provide full solution" } { "Tag": [ "calculus", "geometry", "derivative", "integration" ], "Problem": "Question:\r\nAre there any books that anyone would recommend for someone just starting out in physics? (aka knows about nothing at all) I am looking to participate in physics competitions sometime in the future, but as of right now, I know next to nothing about physics.\r\n\r\nThanks!", "Solution_1": "What is your level of math? Whether or not you know calculus will impact what books I recommend.", "Solution_2": "Uh...I know most everything before calculus (Algebra I and II, Geometry, Precalculus) and a little bit of calculus (basic limits, some derivatives)\r\nI will be takin calculus next year.", "Solution_3": "If you're not already comfortable with calculus (by which I mean that taking derivatives, limits, and integrals is the easy part), then I will recommend a book by Giancoli. He has another one which uses calculus so try to avoid that one. I used this book to self-study and I really liked it. There are good explanations and lot's of good problems." } { "Tag": [], "Problem": "Suppose the operation # is defined on the set of real numbers as $a\\#b=a+ab$. What is the identity for this operation?", "Solution_1": "[quote=\"4everwise\"]Suppose the operation # is defined on the set of real numbers as $a$#$b=a+ab$. What is the identity for this operation?[/quote] There isn't one.\r\n\r\nSuppose an identity $e$ for $a$#$b$ exists.\r\n\r\nThen by definition of an identity element, $e$#$a = e +ea = a$ and likewise $a$#$e = a + ae = a$.\r\n\r\n$e$#$a = a \\Rightarrow e =\\frac{a}{1+a}$\r\n\r\n$a$#$e = a \\Rightarrow ae = 0$\r\n\r\nIf $a \\neq 0$ then $ e = 0 \\;\\&\\; e \\neq 0$ ; this follows from the supposition and therefore the supposition is false." } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Two circles touch each other externally at $ A$ and touch a common tangent at $ B$ and $ C$ $ (B \\not\\equal{} C)$. Let $ D$ and $ E$ be the points such that $ BD$ and $ CE$ are diameters of the circles. Prove that points $ D,A,C$ are collinear.", "Solution_1": "Construct a common tangent $ l$ which passes through $ A, l\\cap BC\\equal{}\\{M\\}$\r\n$ \\Rightarrow MB\\equal{}MC\\equal{}MA \\Rightarrow \\angle BAC\\equal{}90^o \\rightarrow$ QED" } { "Tag": [ "algorithm", "Gauss", "linear algebra", "matrix", "parameterization" ], "Problem": "I was thinking of writing a program to balance chemical equations for me, and since that involves converting the equation into a system of linear equations and then solving that, I was wondering how I should go about doing that? I don't know how to use Gaussian Elimination with matrices containing zero elements, so that seems out.\r\n\r\nEdit: Ok, scratch that, Gaussian Elimination does work with matrices with zero elements, but the number of linear equations have to be at least equal to the number of variables (i.e. the number of elements in the equation has to be at least equal to the number of chemicals, which often doesn't happen in chemical equations).", "Solution_1": "I would suggest setting one variable to 1, then figuring out all the other variables, then multiplying through to clear denominators. Of course you will need to make a rational class or something silly like that.", "Solution_2": "[quote=\"probability1.01\"]I would suggest setting one variable to 1, then figuring out all the other variables, then multiplying through to clear denominators. Of course you will need to make a rational class or something silly like that.[/quote]\r\nYeah, but doesn't Gaussian elimination require the same number of equations as variables? Every iteration you set one column to zero, so after (n-1) iterations you have 1 last variable = 1 constant, and then work backwards from that.", "Solution_3": "In a balancing equations problem, you will have more variables than equations.", "Solution_4": "eghm, when gauss-reducing, the program must be aware of the pivots chosen iteration by iteration, otherwise errors can be greatly amplyfied, or \"divide by 0\" problem can arise.\r\nhaving more variables than equations is not that big problem, there will be a set of solutions instead of a unique solution, anyway, by gaussian elimination in the later case you obtain a upper hessenberg matrix, in this form, once chosen the variables to be free parameters of the system, just solve via back-substitution as always.\r\nI would suggest to divide logically the program in two parts, first gaussian elimination, than once chosen the values of the free parameters, solve by backsubstitution.\r\nAnother way may be chosing the free parameters in the beginning, and then use an iterative method for solving the system..." } { "Tag": [ "geometry", "trigonometry", "LaTeX", "analytic geometry", "angle bisector", "area of a triangle", "Pythagorean Theorem" ], "Problem": "ABC is a triangle with angle B=90.\r\nAD is an angle bisector.\r\nE lies on the side AB with AE=3 and BE=9 and F lies on AC with AF=10 and FC=27.\r\nEF meets AD at G.\r\nFind the nearest integer to area GDCF.\r\n\r\nthank you.......", "Solution_1": "This would be an easy question if calculators were allowed. Somehow, because of the \"find the nearest integer\" I don't think they are.\r\n\r\nAre they allowed?", "Solution_2": "NO .....CALCULATOR IS NOT ALLOWED\r\nWE R ASKED TO GIVE A SOLUTION..... :)", "Solution_3": "reply someone.......please", "Solution_4": "[hide=\"obvious hint\"]use the angle bisector theorem[/hide]\n[hide=\"helpful hint \"]\nUse the fact that the area of a triangle is $ \\frac{1}{2}ab\\sin{C}$[/hide][/hide]", "Solution_5": "first, no need to shout. we are here to help\r\nyour problem:\r\n\r\nBC=35 by the pythagorean theorem\r\nBecause [AEF] = (1/2) * 3 * 10 * sin(EAF) and sin(EAF) = 35/37,\r\n[AEF] = 15*35/37\r\n\r\nIt is not difficult to compute tan(BAC/2) = 5/7\r\nthus BD = 60/7\r\n\r\n[ABD] = 360/7\r\n\r\nAll that needs to be computed is [AEG]\r\n\r\nLet [AEG] = a, [AFG] = b\r\na+b = 15*35/37\r\na = (1/2)*sin(BAC/2)*3*AG\r\nb = (1/2)*sin(BAC/2)*10*AG\r\na/b = 3/10\r\n10a=3b\r\n\r\na*13/3 = 15*35/37\r\na = 35*45/(13*37)\r\n\r\n[ABD] + [AEF] - [AEG] = 15*35/37 + 360/7 - 35*45/(13*37)\r\n= 15(35/37 + 24/7 - 105/(13*37))\r\n\r\nNow, [GFCD] = [ABC] - 15(35/37 + 24/7 - 105/(13*37))\r\n= 15(14 - 35/37 - 24/7 + 105*(13*37))\r\n=15(74/7 - 35/37 + 105/(13*37))\r\n=15(74/7 -350/(13*37))\r\n=15*33144/3367\r\n\r\nthis is where it ends with a calculator it is simple to find it.\r\nI have no clue how to approximate that", "Solution_6": "you could divide. And you should learn $ LaTeX$... there's a great guide on this site (look at the left taskbar).\r\n\r\nCoordinates would be interesting here :ninja:", "Solution_7": "I've used latex, just I was too lazy to use it for those geometric things lol" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z \\in R $ . Prove that : \r\n $ (x^2+3)(y^2+3)(z^2+3) \\geq \\frac{4}{27}(3xy+3yz+3xz+xyz)^2 $", "Solution_1": "Replace x by 3x and expand we have to prove\r\n$23x^2y^2z^2+5\\sum x^2y^2 + 3\\sum x^2 +1 \\ge 8\\sum x^2yz + 8\\sum x^2y^2z$\r\nIt can be resolved into three inequalities\r\n1. $\\sum x^2y^2 + 2x^2y^2z^2 + 1 \\ge 2\\sum x^2yz$ (APMO 04 no.5)\r\n2. $\\sum x^2 + 3x^2y^2z^2 \\ge 2\\sum x^2yz$ (amgm)\r\n3. $\\sum x^2y^2 + 3x^2y^2z^2 \\ge 2\\sum x^2y^2z$ (amgm)\r\n(1)+3*(2)+4*(3)\r\nThen it's done.", "Solution_2": "Really nice , Siuhochung ! \r\n You used $ \\sum x^2y^2 + 2x^2y^2z^2 + 1 \\ge 2\\sum x^2yz $ in my 2 posts :lol: \r\n The problem has a nicer solution", "Solution_3": "then can you post your solution? When I meet this type of problem I always expand, expand and expand. :D :P", "Solution_4": "Change the problem into : $ (a^2+3)(b^2+3)(c^2+3) \\geq 4(a+b+c+1)^2 $ , where $ a=\\frac{3}{x} $ . The problem becomes smarter ;)" } { "Tag": [ "LaTeX", "induction", "Euler" ], "Problem": "Prove that-\r\n\r\n1+(1/2)+(1/3).....1/n is not an integer(too lazy to type latex :D )", "Solution_1": "hint :\r\n[hide]\nI'm too lazy to answer in details. ;) \n\nBy induction, prove it is odd/even \n\nNo doubt it was answered previously on Mathlinks before[/hide]", "Solution_2": "[hide]Let $a$ be the largest integer such that $2^a\\leq n$. Since $2*2^a=2^{a+1}>n$, there is exactly 1 integer from 1 to n divisible by $2^a$.\n\nWhen writing $\\sum_{k=1}^n \\frac{1}{k}$ under a common denominator, the numerator of the term when $k=2^a$ is odd, but for all other $k$ it will be even. The sum is then\n(odd number)/(multiple of $2^a$).\n\nIf $n>1$, then $a\\in \\mathbb{N}$, so the denominator is not 1. Then the sum cannot be an integer.[/hide]", "Solution_3": "[hide]euler showed that the harmonic series is bounded and convergent to $\\frac{\\pi^2}{6} \\approx 1.645$, so the only integer it can be is $1$, (which occurs), and besides that, it goes in the range $(1,1.645)$, so it is never an integer\n\n(i am sure that you are not counting $n=1$)[/hide]", "Solution_4": "Wrong series, Altheman.", "Solution_5": "woops...i was thinking $\\sum_{k=1}^{\\infty} \\frac{1}{k^2}$" } { "Tag": [ "geometry" ], "Problem": "E and F are variable points on the sides AD and CD, respectively, of square ABCD such that DE=CF. BE and BF divide AC into three segments AG, GH, and HC which, as it turns out, can always be arranged to form a triangle. Prove that every such triangle has an angle of 60 degrees.", "Solution_1": "$AGE$ is similar with $CGB$ and\r\n$\\frac{y}{x+y}=\\frac{a}{b+c}$\r\n$FHC$ is similar with $BHA$ and\r\n$\\frac{x}{x+y}=\\frac{c}{a+b}$\r\n$\\frac{y}{x+y}+\\frac{x}{x+y}=\\frac{c}{a+b}+\\frac{a}{b+c}$\r\n$1=\\frac{c}{a+b}+\\frac{a}{b+c}$\r\n$a^2+c^2-ac=b^2$\r\n$a^2+c^2-2ac*cos60=b^2$\r\n\r\ntriangles with a,b,c has one 60 degrees at last.", "Solution_2": "I have one question: We've already know that the length of the three segments satisfy the equation, but do we have to prove that they can form a triangle togather before we say one angle is $60^o$?", "Solution_3": "no, the problem says that the segments can form a triangle\r\n\r\nbtw gunes nice job" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c,x,y,z\\geq 0$. Prove that\r\n\r\n$ x(b\\plus{}c)\\plus{}y(c\\plus{}a)\\plus{}z(a\\plus{}b)\\geq 2\\sqrt{(xy\\plus{}yz\\plus{}zx)(ab\\plus{}bc\\plus{}ca)}$", "Solution_1": "[quote=\"easternlatincup\"]Let $ a,b,c,x,y,z\\geq 0$. Prove that\n\n$ x(b \\plus{} c) \\plus{} y(c \\plus{} a) \\plus{} z(a \\plus{} b)\\geq 2\\sqrt {(xy \\plus{} yz \\plus{} zx)(ab \\plus{} bc \\plus{} ca)}$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=186636" } { "Tag": [], "Problem": "Each morning Boris walks to school. At one-fourth of the way he passes the machine and tractor station; at one-third of the way, the railroad station. At the machine and tractor station its clock shows 7:30, and at the railroad station its clock shows 7:35.\r\nWhen does Boris leave his house, when does he reach school ?\r\n\r\n[i]from book \"The Moscow puzzles\" by Boris A. Kordemsky[/i]\r\n[u]Please hide Your solution . Write inside Hide ... /Hide[/u]", "Solution_1": "[hide=\"Solution\"]\n\nIn $ 5$ minutes, Boris walks $ \\frac{1}{12}$ of the way, so the whole journey to school takes $ 60$ minutes. $ \\frac{1}{4}$ of the way takes $ 15$ minutes, so he left home at $ 7: 15$ and reaches school at $ 8: 15$.\n\n[/hide]", "Solution_2": "[hide]Time Left Home: 7:15[/hide]\n\n[hide]Time Arrive: 8:15[/hide]" } { "Tag": [ "vector", "trigonometry", "ratio", "analytic geometry", "graphing lines", "slope" ], "Problem": "Consider a river of width $ a$ located on the land to a latitude$ \\lambda$. In the river flows waters down with a speed$ v_0$ . Demonstrate that the water level to the right side of the river is superior to that of the left side in a quantity:\r\n\r\n$ \\Delta h$ $ \\equal{}$ $ \\dfrac{2a \\Omega v_0 Sin(\\lambda)}{g}$\r\n\r\n$ \\Omega$ $ \\equal{}$ Angular speed of the Earth", "Solution_1": "Solution.\r\n\r\nWe must assume that terms of order $ \\Omega^2$ are negligible. This means that the centripetal force is negligble in comparison with the coriolis force, and can be ignored.\r\n\r\nLet's define a set of orthogonal, unit basis vectors on every point on the Earth as follows:\r\n\r\n$ \\bf{e_1}$ directed west to east.\r\n$ \\bf{e_2}$ directed south to north.\r\n$ \\bf{e_3}$ directed in the outwards radial direction.\r\n\r\nThe river has velocity $ {\\bf v_0} = (\\, \\pm v_0 \\, , \\, 0 \\, , \\,0 \\,)$. The sign depends on whether the river is flowing west to east, or east to west. In the following argument, the top symbol (in the $ \\pm$ or $ \\mp$) always represents west to east.\r\n\r\nSo the angular velocity is $ {\\bf \\omega} = (\\, 0 \\, , \\, \\Omega \\cos{\\lambda} \\, , \\, \\Omega \\sin{\\lambda} \\,)$. This holds for all latitudes, positive and negative. \r\n\r\nThe only fictitious force is the coriolis. (The water is not accelerating as measured by an observor corotating with the Earth, thus the \"apparent force\" term dissapears). We introduce the mass term $ m$, which is the mass assosiated with some elementary volume. But this really isn't important, and serves more importantly as keeping dimensional consistancy.\r\n\r\n$ {\\bf F} = m(2 {\\bf \\omega} \\times {\\bf v_0}) = m(\\, 0 \\, , \\, \\pm 2 \\Omega v_0 \\sin{\\lambda} \\, , \\, \\mp 2 \\Omega v_0 \\cos{\\lambda} \\, )$\r\n\r\nThe two forces acting on the water at the surface is a sort of \"physical\" force, $ {\\bf T}$ (rather like tension in a string) from the water underneath, and the gravitational force $ m{\\bf g}$. Note that $ {\\bf g}=(\\, 0 \\, , \\, 0, \\, -g \\, )$. It is a well established fact that the equipotential surfaces of a constant (in time and space) vector field, are orthogonal, flat planes. Since the water surface forms and equipotential of $ {\\bf T}$, then what we ultimately require is the direction of $ {\\bf T}$.\r\n\r\n$ {\\bf F} = {\\bf T} + m{\\bf g}$\r\n\r\n$ \\Rightarrow {\\bf T} = m(\\, 0 \\, , \\, \\pm 2 \\Omega v_0 \\sin{\\lambda} \\, , \\, g \\mp 2 \\Omega v_0 \\cos{\\lambda} \\,)$\r\n\r\nWe now come to finding the gradient of $ {\\bf T}$ which is the ratio of it's components in the positive directions of $ {\\bf e_3}$ (up) and $ {\\bf e_2}$ (north).\r\n\r\n$ m_{T} = \\frac{g \\mp 2 \\Omega v_0 \\cos{\\lambda}}{ \\pm 2 \\Omega v_0 \\sin{\\lambda}}$\r\n\r\nSo the perpendicular gradient, in this $ ({\\bf e_2}, {\\bf e_3})$ plane gives the slope of the river surface as height over width. This is because the width is in the north-south direction.\r\n\r\n$ m_{surface} = -\\frac{1}{m_{T}} = \\frac{\\mp 2 \\Omega v_0 \\sin{\\lambda}}{g \\mp 2 \\Omega v_0 \\cos{\\lambda}}$\r\n\r\nThe term ${ 2 \\Omega v_0 \\cos{\\lambda}}$ is very small in comparison with $ g$, and the denominator is large in comparison to the numerator so this can be written as:\r\n\r\n$ m_{surface} = \\frac{\\mp 2 \\Omega v_0 \\sin{\\lambda}}{g}$\r\n\r\nWe take a minus if the river flows west to east, and a plus if the river flows east to west. (by the top sign, bottom sign consistancy used). \r\n\r\nAssume that the river flows on the northern hemisphere as to make $ \\lambda$ and hence $ \\sin{\\lambda}$ positive. \r\n\r\nIn the west to east case the left is the north bank and the right is the south bank. But then the surface gradient is negative, i.e the further north - to the left bank - the more shallow the water. The left bank undermines the right.\r\n\r\nIn the east to west case the left is the south bank and the right is the north bank. But then the surface gradient is positive, i.e the further north - to the right bank - the more deep the water. The left bank undermines the right.\r\n\r\nHowever, in the southern hemisphere this does NOT hold. This is because $ \\lambda < 0$ making $ \\sin{\\lambda} < 0$. So the water gradients have opposite signs, and thus slope in the opposite directions. Thus lateral rivers flowing in the southern hemisphere undermine the right bank more than the left. These claims also hold when the river flows with a northern or southern component, on each hemisphere. \r\n\r\nThe last special case is that corresponding to a river flowing on the equator, where $ \\lambda = 0$. The river is flat and neither bank is undermined.\r\n\r\nNow the difference in the height of the water is clearly given by:\r\n\r\n$ \\Delta h = a \\cdot |m_{surface}| = \\frac{2 a \\Omega v_0 \\sin{|\\lambda|}}{g}$\r\n\r\nThe case where the river does not flow in the east-west direction can be tackled with similar analysis, replacing the vector $ {\\bf v_0}$ appropriately. However, it can be extended from this case. Consider several sets of parallel lines drawn on an inclined plane. Draw lines perpendicular to, and intersecting each pair of parallel lines. Now take a second plane which intersects this inclined plane. How do the components of gradients of these perpendicular lines parallel to this intersecting plane change according to the angles that the parallel lines make with each other on the original plane?" } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra solved" ], "Problem": "$A,B\\in M_{4}(R), \\; AB=BA, \\; det(A^2+B^2)=0, \\; det(A-B)=0$\r\n\r\nProve that $det(A+B)=4det(A)+4det(B)$", "Solution_1": "I think that grobber won't offend if i use ideas, that he proposed for the same last problems by Moubinool\r\nDenote $P(x)=det(A-xB)$, so it follows(from the condition) that $P(i)P(-i)=0$ and $P(1)=0$, therefore, since $P(x)$-polynomial of degree 4 with real coefficients it must be $P(i)=P(-i)=P(1)$, so $P(x)$ is of the form $P(x)=(x^2+1)(x-1)(xdetB-detA)$ (because the domiant coefficient $detB$ and $P(0)=detA$), so we see that $det(A+B)=P(-1)=4(detB+detA)$\r\nso we are done" } { "Tag": [ "Why the bump" ], "Problem": "Nine copies of a certain pamphlet cost less than $ \\$10.00$ while ten copies of the same pamphlet (at the same price) cost more than $ \\$11.00$. How much does one copy of this pamphlet cost?\r\n\r\n\\[ \\textbf{(A)}\\ \\$1.07 \\qquad\r\n\\textbf{(B)}\\ \\$1.08 \\qquad\r\n\\textbf{(C)}\\ \\$1.09 \\qquad\r\n\\textbf{(D)}\\ \\$1.10 \\qquad\r\n\\textbf{(E)}\\ \\$1.11\r\n\\]", "Solution_1": "[hide]e, its the only one that is more than 11 dollars when multiplied by 10[/hide]", "Solution_2": "[hide=\"Click for solution\"]\nQuickly, we note that $ \\$1.11 \\times 9=\\$9.99$, which is close to the upper bound of what we want, so we try multiplying by $ 10$, which yields $ \\11.10$, so this works. We can test every other answer choice to see that just $ \\boxed{\\textbf{(E)}}$ works.\n[/hide]" } { "Tag": [ "Asymptote", "geometry", "incenter", "circumcircle" ], "Problem": "In Asymptote, how do you draw perpendicular bisectors and angle bisectors?", "Solution_1": "You can find the incenter and draw lines through it and the vertices for the angle bisectors, but I'm not sure how to draw $\\perp$ lines, so I can't help you with the $\\perp$ bisectors.", "Solution_2": "The perpendicular bisectors meet at the circumcenter. I assume you can find the midpoint of a line easily enough, so just draw a line that passes through both it and the circumcenter.", "Solution_3": "Tip: don't actually draw the circumcenter or the midpoint; just don't do draw O; or draw M; or whatever." } { "Tag": [ "trigonometry", "inequalities", "geometry unsolved", "geometry" ], "Problem": "Given ABCD on the sides AB, BC, CD, DA choosen points M, N, P, Q respectively.\r\nAM=BN=CP=DQ also given that MNPQ - quadralaterial. Prove that ABCD also quadralaterial", "Solution_1": "[quote=\"Alisher\"]MNPQ - quadralaterial[/quote]\r\nby \"quadralaterial\" do you mean cyclic? :maybe: \r\n\r\n(we call $ MNPQ$ cyclic iff $ M,N,P,Q$ lie on a circle)", "Solution_2": "sorry for my english I can't translate it. But \r\n I mean square. If given MNPQ so MN=NP=PQ=QM ,MN perpendicular NP, NP perpendicular PQ, PQ perpendicular QM, QM perpendicular MN", "Solution_3": "I'm sorry. I mean square", "Solution_4": "Isn't it obvious because triangle AMQ , MBN , NCP ,PQD are equilateral so all side of ABCD is equal and all angle's are 90' If I'm not mistaken", "Solution_5": "this problem was given in an Iranian olympiad like 12 years ago!\r\nanyway here's the solution:\r\n\r\nlet $ \\angle AQM \\equal{} \\alpha,\\angle NMB \\equal{} \\beta,\\angle PNC \\equal{} \\gamma,\\angle PDQ \\equal{} \\lambda$,we will show that $ \\alpha\\equal{}\\beta\\equal{}\\gamma\\equal{}\\lambda$.\r\n\r\nfirst of all let $ MN \\equal{} NP \\equal{} PQ \\equal{} QM \\equal{} a$ and let $ AM \\equal{} BN \\equal{} CP \\equal{} DQ \\equal{} r$.\r\n\r\nnow WLOG assume that $ \\alpha$ be the greatest among $ \\alpha,\\beta,\\gamma,\\lambda$.\r\n\r\nnow according to $ \\sin$ law in $ \\triangle AQM$ we have:\r\n\r\n$ \\frac {r}{\\sin (90 \\minus{} \\beta)} \\equal{} \\frac {a}{\\sin (90 \\minus{} \\alpha \\plus{} \\beta)}$\r\n\r\n$ \\Rightarrow\\frac {r}{\\cos\\beta} \\equal{} \\frac {a}{\\cos (\\alpha \\minus{} \\beta)}$\r\n\r\n$ \\Rightarrow\\frac {a}{r} \\equal{} \\frac {\\cos (\\alpha \\minus{} \\beta)}{\\cos\\beta}$\r\n\r\n$ \\Rightarrow 1 \\plus{} \\tan\\alpha\\cdot\\tan\\beta \\equal{} \\frac {a}{r\\cos\\alpha}$ (1)\r\n\r\nsimilary we conclude that:\r\n\r\n$ 1 \\plus{} \\tan\\beta\\cdot\\tan\\gamma \\equal{} \\frac {a}{r\\cos\\beta}$ (2)\r\n\r\n$ 1 \\plus{} \\tan\\gamma\\cdot\\tan\\lambda \\equal{} \\frac {a}{r\\cos\\gamma}$ (3)\r\n\r\n$ 1 \\plus{} \\tan\\lambda\\cdot\\tan\\alpha \\equal{} \\frac {a}{r\\cos\\lambda}$ (4)\r\n\r\nnow note that $ \\alpha\\geq \\lambda$ thus $ \\frac 1{\\cos\\alpha}\\geq\\frac 1{\\cos\\lambda}$ so according to (1),(4) we get that $ \\tan\\beta\\geq\\tan\\lambda$ hence $ \\boxed{\\beta\\geq\\lambda}$.\r\n\r\nthus we will have:\r\n\r\n$ 1 \\plus{} \\tan\\gamma\\cdot\\tan\\beta\\geq 1 \\plus{} \\tan\\gamma\\cdot\\tan\\lambda$\r\n\r\nand also by (2),(3) we get that $ \\cos\\beta\\geq\\cos\\gamma$ hence $ \\boxed{\\beta\\geq\\gamma}$.\r\n\r\nagain we have $ \\alpha\\geq\\gamma$ hence by a similar way we get that $ \\boxed{\\lambda\\geq\\gamma}$.\r\n\r\nnow by the inequalities in the boxes we get that:\r\n\r\n$ \\boxed{\\alpha\\geq\\beta\\geq\\lambda\\geq\\gamma}$ (*)\r\n\r\nthus we will have:\r\n\r\n$ 1 \\plus{} \\tan\\beta\\cdot\\tan\\gamma\\leq 1 \\plus{} \\tan\\alpha\\cdot\\tan\\lambda$ (5)\r\n\r\nnow using this inequality and (2),(4) we get that $ \\beta\\leq\\lambda$ so according to (*) we'll have:\r\n\r\n$ \\beta \\equal{} \\lambda$\r\n\r\nnow by (5) we get that $ \\alpha \\equal{} \\gamma$ thus:\r\n\r\n$ \\alpha \\equal{} \\beta \\equal{} \\gamma \\equal{} \\lambda$\r\n\r\nnow the rest is pretty easy." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Let P be a polynomial that belongs to $R[x]$, $P(x)=x^3+ax^2+bx+c$, a<>0, with the solutions $x_1, x_2, x_3$ which are distinct 2 by 2. \r\na) $\\frac{x_1}{P'(x_1)} + \\frac{x_2}{P'(x_2)} +\\frac{x_3}{P'(x_3)}$ is equal to ?\r\nb) Let $Q$ be a polynomial of $degre\\; 1$ that belongs to $R[x]$, what is the value of $\\frac{Q(x_1)}{P'(x_1)} + \\frac{Q(x_2)}{P'(x_2)} +\\frac{Q(x_3)}{P'(x_3)}$", "Solution_1": "The first sum is $\\frac{x_1} {(x_1-x_2)(x_1-x_3)} +\\frac{x_2} {(x_2-x_1)(x_2-x_3)} +\\frac{x_3} {(x_3-x_1)(x_3-x_2)} =0$ be clearing the denominators and straightforward checking.\r\n\r\nThe second is $0$ too. You only have to denote $Q(x) = ax+b$ and use the same way as above (note that the part $ax$ is done by the first sum).\r\n\r\nPierre.", "Solution_2": "thanks for solving it !" } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "calculus computations" ], "Problem": "$ I \\equal{} \\int_{_0}^{^{\\frac {\\pi}{4}}} x \\tan x dx$ ??\r\n\r\nAnybody help me ? Thanks !!", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1072333#1072333]this might help[/url] :)", "Solution_2": "$ \\textbf I \\equal{} \\int_0^{\\dfrac{\\pi}{4}}\\;x\\cdot \\tan\\,x\\;\\textbf dx\\; \\equal{} \\;\\int_0^1\\dfrac{y\\cdot\\tan^{ \\minus{} 1}\\,y}{1 \\plus{} y^2}\\;\\textbf dy\\qquad (\\text{where}\\quad y\\equal{}\\tan\\,x\\; )$\r\n\r\n$ \\equal{} \\dfrac{1}{2}\\int_0^1\\tan^{ \\minus{} 1}\\,y\\cdot\\textbf d\\left(\\ln\\,(1 \\plus{} y^2)\\right)$\r\n\r\n$ \\equal{} \\dfrac{1}{2}\\left(\\tan^{ \\minus{} 1}\\,y\\cdot \\ln\\,(1 \\plus{} y^2)\\bigg|_0^1 \\minus{} \\int_0^1\\dfrac{1}{1 \\plus{} y^2}\\;\\ln\\,(1 \\plus{} y^2)\\;\\textbf dy\\right)$\r\n\r\n$ \\equal{} \\dfrac{1}{2}\\left(\\dfrac{\\pi}{4}\\;\\ln\\,2 \\minus{} \\left( \\dfrac{\\pi}{2}\\;\\ln\\,2\\; \\minus{} \\;\\textbf K_{Catalan}\\right)\\right)$\r\n\r\n$ \\equal{} \\dfrac{1}{2}\\left(\\textbf K_{Catalan} \\minus{} \\dfrac{\\pi}{4}\\;\\ln\\,2\\right)$\r\n\r\nthe integral $ \\int_0^1\\dfrac{1}{1 \\plus{} y^2}\\;\\ln\\,(1 \\plus{} y^2)\\;\\textbf dy$ \r\nhas been shown many times on this forum.", "Solution_3": "I've got the answer. Thank all of you , Madness , Missan !! :lol:" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "if abc=1 prove that 1/a^3(b+c) +1/b^3(c+a) +1/c^3(a+b) is greater than or equal to 3/2", "Solution_1": "This is well known, it is number 10 on [url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url], four solutions are given. The third (Cauchy) is the easiest." } { "Tag": [ "AMC", "AIME", "MATHCOUNTS", "AMC 12", "USA(J)MO", "USAMO", "number theory", "\\/closed" ], "Problem": "Does the Intermediate Number Theory Class mostly help for aime, usamo.\r\nBecause the Introduction to Number theory class problems were much easier and did not cover aime level problems.", "Solution_1": "Yes, it covers all the important topics in AIME Number Theory, as well as some easy olympiad problems. \r\nIn my opinion, it's a lot harder than the Introduction class, because the Intro only has MAHTCOUNTS, AMC 10, and easy AMC 12 NT problems.", "Solution_2": "How come there isn't an Intermediate NT textbook?", "Solution_3": "Remy i have noticed that some problems in the AMC 12 are extremely hard (20- 25) yet they are not covered in intro to nt.", "Solution_4": "Oh yeah, Intermediate NT also covers hard AMC 12 problems too. I think AMC12 has a big range of difficulty, and the Intermediate class will include the last few problems as well (while the Intro covers the easy one).\r\n\r\n@maybach: The staff doesn't have any plans to publish the textbook yet. Writing a book is not an easy task, and it takes a long time to do so. I heard that they will publish the Interm NT textbook [i]someday[/i], but not in the close future.", "Solution_5": "i notice the Intermediate NT class only have 8 class, and no book, the class just introduce some concepts? does it offer enough problems? when will the class start?", "Solution_6": "If you want a supplementary book try [url=http://www.amazon.com/exec/obidos/ASIN/081763245X/artofproblems-20]Number Theory: Structure, Examples, and Problems[/url] by Titu Andreescu and Dorin Andrica.\r\n\r\nThe book is a bit more difficult than the class.", "Solution_7": "[quote=\"ic2149\"]i notice the Intermediate NT class only have 8 class, and no book, the class just introduce some concepts? does it offer enough problems? when will the class start?[/quote]\r\nFuture classes are not scheduled yet, so you will have to wait. As far as I remember, it introduced concepts, proved them, and had 7-10 problems every week, along with around 14 message board problems each week for homework. However, there's no Challenge Sets like most subject courses do.", "Solution_8": "So, remy did you feel a difference in your number theory skills?\r\nDid it raise your'e amc 10/ 12/ aime/ usamo :| ?", "Solution_9": "Is there going to be a Intermediate NT book in a few years??", "Solution_10": "[quote=\"ic2149\"]i notice the Intermediate NT class only have 8 class, and no book, the class just introduce some concepts? does it offer enough problems? when will the class start?[/quote]\r\n\r\nThere are plenty of problems. We haven't set a date for the course yet.\r\n\r\nWe are not sure when there will be a corresponding book.", "Solution_11": "Are there other books that are used as follow up to Intro to NT?", "Solution_12": "[quote=\"AlphaBetaTheta\"]So, remy did you feel a difference in your number theory skills?\nDid it raise your'e amc 10/ 12/ aime/ usamo :| ?[/quote]\r\n\r\nEh I don't know. I took it this summer, so I didn't have a chance to apply them yet. XD But I think I learned important things from the course." } { "Tag": [ "algebra", "system of equations", "algebra solved" ], "Problem": "Solve the following system of equations:\r\n\r\nx + [y] + {z} = 3.9 (1)\r\ny + [z] + {x} = 3.5 (2)\r\nz + [x] + {y} = 2 (3)\r\n\r\n[a] denotes integer part and {a} denotes its fractional part, a = [a] + {a}", "Solution_1": "You add them all and get :\r\nx+y+z=4,7 (4)\r\nthen you do (4)-(1);(4)-(2);(4)-(3)\r\nand you get x,y,z! :D \r\n\r\nx=1,7\r\ny=2,8\r\nz=0,2\r\n\r\nsimple! cheers! :D :D" } { "Tag": [ "geometry", "geometric transformation", "reflection" ], "Problem": "i was wondering what are some good jobs that are available to someone who intends to major in mathematics.\r\ni know that i could be an actuary but what else is there.\r\nif you respond, could you please also provide a short description of the job. \r\n\r\n thanks ahead of time. :)", "Solution_1": "Here's a page which includes short descriptions:\r\n\r\nhttp://www.cs.xu.edu/math/mathcareers.html\r\n\r\nAnd, here's one with links to other math-related-careers pages (it's not as good as the earlier one, but does have a little bit of additional info):\r\n\r\nhttp://cerebro.cs.xu.edu/~smbelcas/mathcar.html", "Solution_2": "Your can do quantitative finance job or financial engineering,\r\nhere is the introduction:\r\n[url=http://www.quantfinancejob.com]QuantFinanceJob.com[/url]", "Solution_3": "This is by no means a common career path (and probably doesn't even count as a job per se), but here's what one math major is doing: http://www.ayria.com", "Solution_4": "Maybe not completely unique. Phil Alvin, whose rock bands over the years have included the Blasters, has an M.S. in math from my university.", "Solution_5": "What kind of career options are available to me if I major in pure mathematics for a BSc.?\r\n\r\nThat is exclusive of teaching and statistics related occupations.", "Solution_6": "We have a poster here in our office from the American Mathematical Society. It says, in big letters near the top, \"What can I do with a math degree?\"\r\n\r\nThe listed options are:\r\nActuary\r\nAnimator\r\nAppraiser\r\nBenefits Administrator\r\nClimate Analyst\r\nCollege Professor\r\nCommodities Trader\r\nCryptanalyst\r\nEpidemiologist\r\nForeign Exchange Trader\r\nForensic Analyst\r\nMarket Research Analyst\r\nPollster\r\nPopulation Ecologist\r\nPublic Utilities Analyst\r\nQuantitative Analyst\r\nResearch Scientist\r\nStatistician\r\nTeacher\r\nTechnical Writer\r\nUrban Designer\r\n\r\nSee also [url]http://www.ams.org/employment[/url]", "Solution_7": "I think you have to be careful though - these are jobs that some math major would be able to get, but a BSc in pure math will not automatically qualify you for very much at all. You should probably reflect a little on what you want to do now and take some electives in a relevant subject or maybe do some volunteer work. Your math skills will be invaluable, but you need to have displayed an interest in other career paths. Otherwise people will assume that you are only interested in math and are just settling for something else, and they won't want to hire you.\r\n \r\nI should know - I graduated with a BSc in pure math a few years ago and couldn't find any job, so ended up continuing on with my part-time job as a photocopying assistant and doing other odd unskilled jobs for nearly a year before coming back to university and starting my phd. I live in fear of the day I have to graduate - don't turn out like me!", "Solution_8": "Thanks for the advice. yeah i planned to take electives in areas such as physics and programming or some economics.\r\nbut thanks.", "Solution_9": "Here's a recently-updated mathematics careers webpage:\r\n\r\nhttp://www.toroidalsnark.net/mathcareers.html\r\n\r\n(Okay, I'll admit it, I made the page)", "Solution_10": "see http://www.ams.org/employment\r\n\r\nps: I like your malmsteen avatar ;)", "Solution_11": "haha thanks :D", "Solution_12": "Whenever this question gets asked, it seems like \"Consultant\" or \"Investment Banker\" are never listed...is that just because the list is geared towards possible interests of math majors, or is there really a prejudice against them in those two fields (since both fields require a high degree of socialization and schmoozing, which math majors stereotypically do not possess)/?", "Solution_13": "[quote]Whenever this question gets asked, it seems like \"Consultant\" or \"Investment Banker\" are never listed...is that just because the list is geared towards possible interests of math majors, or is there really a prejudice against them in those two fields (since both fields require a high degree of socialization and schmoozing, which math majors stereotypically do not possess)/?[/quote]\r\n\r\nSurprisingly, many, if not most, applied math and statistics folks end up doing \"consulting\" work. Their clients might be internal to an organization but virtually all are working with scientists, engineers, and other specialists to solve problems critical to the missions of their organizations. \r\n\r\nInterpersonal skills are critical in just about all careers. Applied math/stat folks are often surprisingly good when communicating with other scientists and professionals in a work environment. The sterotypes probably originate in the often failed attempts to educate students.\r\n\r\nThere's a difference between communicating effectively about how your expertise can be utilized to address a problem and \"socializing/schmoozing.\" You don't necessarily have to have a wide circle of friends or be the life of the party to succeed in careers that require personal interaction." } { "Tag": [], "Problem": "find all the answer of:\r\n$2a^2+3b^2-5c^2=1997$", "Solution_1": "Are a, b and c integers?", "Solution_2": "Are you sure that the question is to find [b]all[/b] integer solutions?" } { "Tag": [], "Problem": "1. Let x= a+b+c. Show that $ (a\\plus{}b\\plus{}c)^2\\plus{}(3\\minus{}a\\minus{}b\\minus{}c)^2\\equal{} 2(x\\minus{}3/2)^2\\plus{}9/2$\r\n\r\n2. Let $ t\\equal{}xy\\plus{}yz\\plus{}xz$ Find $ (x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz)^2$ in terms of t.", "Solution_1": "[quote=\"bigman\"]1. Let x= a+b+c. Show that $ (a \\plus{} b \\plus{} c)^2 \\plus{} (3 \\minus{} a \\minus{} b \\minus{} c)^2 \\equal{} 2(x \\minus{} 3/2)^2 \\plus{} 9/2$\n\n2. Let $ t \\equal{} xy \\plus{} yz \\plus{} xz$ Find $ (x^3 \\plus{} y^3 \\plus{} z^3 \\minus{} 3xyz)^2$ in terms of t.[/quote]\r\n\r\n1. LHS= (a+b+c)^2+(3-(a+b+c))^2\r\n\r\nSub in x, x^2+(3-x)^2\r\n\r\nx^2+(9-6x+x^2)\r\n\r\n2x^2-6x+9\r\n\r\nComplete square:\r\n\r\n=2(x - 3/2)^2 + 9/2\r\n\r\n2. Similar method but consider the expansion of a^3+b^3+c^3", "Solution_2": "[quote=\"bigman\"]1. Let x= a+b+c. Show that $ (a + b + c)^2 + (3 - a - b - c)^2 = 2(x - 3/2)^2 + 9/2$\n\n2. Let $ t = xy + yz + xz$ Find $ (x^3 + y^3 + z^3 - 3xyz)^2$ in terms of t.[/quote]\r\n\r\n1. Since $ x=a+b+c$, using that substitution property of inequality. \r\nThe $ LHS$ becomes:\r\n$ x^2+(3-x)^2$\r\nExpanding, we get:\r\n$ x^2+x^2-6x+9 \\implies \\boxed{2x^2-6x+9}$\r\n\r\nExpanding the RHS, we get:\r\n$ 2(x^2-3x+9/4)+9/2 \\implies \\boxed{2x^2-6x+9}$\r\n\r\nTherefore, we have proven that $ (a + b + c)^2 + (3 - a - b - c)^2 = 2(x - 3/2)^2 + 9/2.$\r\n\r\n2. Expansion of the expression. (Not very much progress.)\r\n\r\n$ (x^3 + y^3 + z^3 - 3xyz)^2$ \r\n$ ([x^3 + y^3] + [z^3 - 3xyz])^2$\r\n\r\n$ (x^6+2x^3y^3+y^6)+2(x^3z^3-3x^4yz+y^3z^3-3xy^4z)+(z^6-6xyz^4-9x^2y^2z^2)$\r\n\r\n$ x^6+2x^3y^3+y^6+2x^3z^3-6x^4yz+2y^3z^3-6xy^4z+z^6-6xyz^4-9x^2y^2z^2$\r\n\r\nI'm stuck.", "Solution_3": "wow. complex math. WHAT GRADE IS THIS STUFF FOR?!?!?!" } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Where are the points in a regular triangle which has the following special property: from the distances to the sides (from the points) a triangle can be constructed.\r\n\r\nThe solution can be seen in the figure attached but I don't know how to prove it. [geogebra]5727740c53fa39d13ead50e74152a9033d0b9418[/geogebra]", "Solution_1": "Let the distance from the point to AB be p, to AC be q, to BC be r, side length of regular triangle be l.\r\n\r\nObviously, ABp+ACq+BCr=lr=2S, where S is the area of ABC.\r\n\r\nsince p,q,r need to be able to form a triangle, we need p+q>r, which is equivalent to rlq and q+r>p give the point must lie in ACFG and ABFE. Combine all, the point must lie within EFG." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) = 8. Prove that\r\n\r\n ab + bc + ca <= a + b + c.", "Solution_1": "Well by using the well known inequality \\[9(a+b)(b+c)(c+a)\\geq 8(a+b+c)(ab+bc+ca),\\] and $(a+b+c)^{2}\\geq 3(ab+bc+ca),$we obtain $(ab+bc+ca)\\leq 3$.\r\nAnd now, we obtain \\[(a+b+c)^{2}\\geq 3(ab+bc+ca)\\geq (ab+bc+ca)^{2},\\] thus $a+b+c\\geq ab+bc+ca$.", "Solution_2": "[quote=\"X-men\"]Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) = 8. Prove that\n\n ab + bc + ca <= a + b + c.[/quote]\nLet $a, b, c $ be positive real numbers such that $ (a + b)(b + c)(c + a) \\leq8.$ Prove that\n\n $ ab + bc + ca \\leq a + b + c.$" } { "Tag": [ "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "Let $P(x)$ be a polynomial with integer coeffiences . Let $k$ be a possitive integer . Prove that : There exists $n \\in N$ such that : $k \\mid P(1)+P(2)+...+P(n)$", "Solution_1": "$P(1)+...+P(k^2)\\equiv k(P(1)+...+P(k)) \\equiv 0 \\mod k$", "Solution_2": "[quote=\"Megus\"]$P(1)+...+P(k^2)\\equiv k(P(1)+...+P(k)) \\equiv 0 \\mod k$[/quote] \r\n Megus , Can you explain clearly ? I think it's wrong .", "Solution_3": "I think it's right.\r\n$P(m) \\mod k$ does only depend on $m \\mod k$, thus $P(m) \\equiv P(m+k) \\equiv P(m+2k) \\equiv ... \\mod k$." } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ m$ be a positive integer.Call a ring $ R$ a $ m$-ring,if $ x^m\\equal{}x$ for all $ x\\in R$.\r\n\r\nprove that a necessary and sufficient condition that a $ m$-ring be a Boolean ring is that $ m$\r\n\r\nis even and $ m\\minus{}1$ has no prime factor of the form $ 2^p\\minus{}1$.\r\n\r\nImam Ali (as):\r\n\r\nThe first fruit of forbearance is that people will sympathize with you and they will go against the man who offended you arrogantly.", "Solution_1": "this was discussed here:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=24885", "Solution_2": "Thanks for attention" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Assuming that $ f: \\mathbb{N}^* \\rightarrow\\mathbb{N}^*$ satisfying that:\r\nfor every $ n\\in \\mathbb{N}^*$, $ f(n\\plus{}1)>f(f(n))$\r\nProve that:\r\n$ f(n)\\equal{}n (n\\in\\mathbb{N}^*)$", "Solution_1": "This is an old problem. It's originate in IMO 1977. The solution can see in document below(exercise 20):" } { "Tag": [ "geometry", "trigonometry", "inequalities", "geometry proposed" ], "Problem": "Given a triangle with the area $S$, and let $a$, $b$, $c$ be the sidelengths of the triangle.\r\n\r\nProve that $a^{2}+4b^{2}+12c^{2}\\geq 32\\cdot S$.", "Solution_1": "\\[ \\boxed {\\ \\triangle ABC\\Longrightarrow a^2+4b^2+12c^2\\ge 32S\\ } \\]Hello, Flavian ! How are you ? The your proposed problem is nicely and easily.[hide=\"Here is a short metrical solution.\"]$D\\in (BC)$, $DC=2BD$ and [u]Stewart's relation[/u] $16AD^2+3a^2=4b^2+12c^2\\Longrightarrow$\n$a^2+4b^2+12c^2=4\\left(4AD^2+a^2\\right)\\ge 16aAD\\ge 16aAD\\sin \\widehat {ADB}=32S$, i.e.\n$a^2+4b^2+c^2\\ge 32S$, with equality if and only if $AD=\\frac a2$ and $AD\\perp BC\\ .$\n[b]Generalization.[/b] $\\boxed {\\ k>1\\Longrightarrow a^2+k^2b^2+k^2\\left(k^2-1\\right)c^2\\ge 4k^3S\\ }\\ .$\nFor $k: =2$ obtain the [b]Flavian[/b]'s proposed inequality.\nFor $k: =\\sqrt 2$ obtain the well-known inequality $a^2+2\\left(b^2+c^2\\right)\\ge 8\\sqrt 2S\\ .$[/hide] I offer you an another inequality: $\\triangle ABC,\\ k\\ge 1\\Longrightarrow$ $\\boxed {\\ 15a^2+5(k^2+k+1)b^2+(k-1)(3k^3+6k^2+4k+2)c^2\\ge 15(k+1)(k^2+1)S\\ }\\ .$", "Solution_2": "Thank you for the nice solution. I saw another generalisation for this inequality, but I think that is same with your. I am verry good, preparing for olympiad. But how are u? :)", "Solution_3": "Another generalization:\r\n \r\n[b]Generalization[/b] Let $ABC$ be a triangle with side lengths $a, b, c$ and with area $S$. If $m, n, p$ are three non-negative real numbers then\r\n $ma^2+nb^2+pc^2 \\geq 4\\sqrt{mn+np+pm}.S$\r\n\r\nFrom this you can lead to the following inequality: \r\n\r\nFor any positive $m, n, p$ such that $k = 2mn+2np+2pm-m^2+n^2+p^2 \\geq 0$ then $\\sum ma^2 \\geq 4\\sqrt{k}S + \\sum m(b-c)^2$", "Solution_4": "Very nice, Treegoner ! Thanks !" } { "Tag": [ "search", "function", "\\/closed" ], "Problem": "I have one example right away and I've seen another one but I cannot recall it now. If you go to the profile of [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=viewprofile&u=20118]you-know-who[/url], the post count shows 45 posts, but when you click on the \"Find all posts by ...\" link, you see none. The same happens if you try to use the search function and search by name. To be honest, I haven't seen a single post by this user, so I don't really know whether it is the post counter or the search engine that is malfunctioning. Or, perhaps, there is some other reason behind this. Anyway, I just thought it may be worth reporting.", "Solution_1": "Those posts the student made are posted in some forums you do not have permission to access (classes forums mainly). There is nothing wrong with the search system. \r\n\r\nThanks for reporting then." } { "Tag": [], "Problem": "How many common fractions with a value between 0 and 1 have 6\nas their denominator?", "Solution_1": "1/6, 2/6, 3/6, 4/6, and 5/6 are the five fractions, but the middle three all simplify, leaving only [b][u]2[/u][/b], 1/6 and 5/6" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Prove that for any positive integer n,\r\n\r\n1/4 ((2 + \\sqrt 3)2n + 1 + (2 - \\sqrt 3)2n + 1)\r\n\r\ncan be written as the sum of two squares of consecutive integers.", "Solution_1": "The problem is straightforward:\r\n Trying to find a such that a^2+(a+1)^2=the number given in the problem, we solve a second degree equation and find that:\r\n a=1/4{ (1+sqrt(3)*(2+sqrt(3))^n-(sqrt(3)-1)*(2-sqrt(3))^n-2). So, it is enough to prove that the number in the RHS is a positive integer. Let \r\n (2+sqrt(3))^n=a_n+sqrt(3)*b_n. Then (2-sqrt(3))^n=a_n-sqrt(3)*b_n and making the computations the RHS becomes 1/2(3b_n+a_n-1). So, it remains to prove that a_n+b_n is odd, which is trivial using Newton formula and identifying a_n and b_n from (2+sqrt(3))^n=a_n+sqrt(3)*b_n." } { "Tag": [], "Problem": "I can't figure out how to set this one up. Can anyone help me out? It makes my head hurt.\r\nThe larger of two numbers is four less than three times the smaller. Also, three times the larger number is six more than seven times the smaller. Find the two numbers.", "Solution_1": "[quote=\"Scrollphaser\"]I can't figure out how to set this one up. Can anyone help me out? It makes my head hurt.\nThe larger of two numbers is four less than three times the smaller. Also, three times the larger number is six more than seven times the smaller. Find the two numbers.[/quote]\r\nthe larger of two numbers : $ x>y$\r\nis four less than three times the smaller: $ x\\equal{}3y\\minus{}4$\r\nthree times the larger number is more than seven times the smaller: $ 3x\\equal{}7y\\plus{}6$\r\nthere ya go :D", "Solution_2": "Thanks. It didn't realize I needed two variables. \r\nYou lost me though. Are these the \"two numbers\" then?\r\nSmaller is 3x=7y+6\r\nLarger is x=3y-4\r\n?\r\nIf so, why wouldn't the smaller be\r\n3(3y-4)=7y+6\r\n?\r\nWhy did the problem ask for the two numbers if the answers would be expressions?\r\nThe instructions at the begining of this problem set say, \"Write an equation and use it to solve the problem. Make sure to answer the question asked by the problem.\"", "Solution_3": "[quote=\"Scrollphaser\"]Smaller is 3x=7y+6 \nLarger is x=3y-4 \n\nIf so, why wouldn't the smaller be \n3(3y-4)=7y+6 [/quote]\n\nIt is. From there, you need to solve for y and plug that into either of your original equations to find x. \n\n[quote=\"Scrollphaser\"]Why did the problem ask for the two numbers if the answers would be expressions?[/quote]\r\n\r\nThe answers are numbers. However, you have to use the expressions to find the numbers.", "Solution_4": "[quote=\"Scrollphaser\"][hide]Thanks. It didn't realize I needed two variables. \nYou lost me though. Are these the \"two numbers\" then?\nSmaller is 3x=7y+6\nLarger is x=3y-4\n?\nIf so, why wouldn't the smaller be\n3(3y-4)=7y+6\n?\nWhy did the problem ask for the two numbers if the answers would be expressions?\nThe instructions at the begining of this problem set say, \"Write an equation and use it to solve the problem. Make sure to answer the question asked by the problem.\"[/hide][/quote]\r\n\r\nFinding numbers from expressions: Normally you need $ a\\plus{}1$ equations in $ a$ variables to find the numerical solutions, except in the case of Diophantine equations, or when certain restrictions hold.\r\n\r\nFor simple questions, simply substitute. We know that $ x\\equal{}3y\\minus{}4$, so $ 3x\\equal{}9y\\minus{}12$. Then $ 9y\\minus{}12\\equal{}7y\\plus{}6$. $ 2y\\minus{}12\\equal{}6$, $ 2y\\equal{}18$, $ y\\equal{}9$.", "Solution_5": "[quote=\"ThetaPi\"]\nFinding numbers from expressions: Normally you need $ a \\plus{} 1$ equations in $ a$ variables to find the numerical solutions,...[/quote]\r\n\r\n :huh: $ a\\plus{}1$??", "Solution_6": "[quote=\"leoxnlin\"][hide][quote=\"ThetaPi\"]\nFinding numbers from expressions: Normally you need $ a \\plus{} 1$ equations in $ a$ variables to find the numerical solutions,...[/quote]\n\n :huh: $ a \\plus{} 1$??[/hide][/quote]\r\n\r\nOh sorry. It should be $ a$ equations for $ a$ variables." } { "Tag": [ "function", "limit", "calculus", "calculus computations" ], "Problem": "Denote with $ A$ the set of all functions $ f: [\\minus{}1,1]\\rightarrow\\mathbb{R}$, for which the graph is a part of the unit circle $ T\\equal{}\\{(x,y)\\in \\mathbb{R}^{2}|x^{2}\\plus{}y^{2}\\equal{}1\\}.$ Find all continuous functions in $ A$.\r\n\r\n[hide]One possible solution: Let us study the continuity in the point $ x\\equal{}0$. We know that $ f(0)\\equal{}\\pm 1$. We choose $ \\epsilon\\equal{}1$. We find the greatest $ \\sigma$ $ \\left( \\sigma \\in (0,1] \\right)$, so that for all $ x$ satisfying $ |x|<\\sigma$, we have $ |f(x)\\minus{}f(0)|<1$. If $ \\sigma\\equal{}1$, then all the points are on the same side of $ x$-axis. So we have $ 2$ continuous solutions. But if $ \\sigma<1$, then we study the points $ x\\equal{}\\pm \\sigma$. If $ f(\\pm \\sigma)$ or on the other side of the $ x$-axis than $ f(0)$, then the function is not continuous in $ x\\equal{}\\pm \\sigma.$ So they are on the same side as $ f(0)$. We again study continuity of the function in the points $ x\\equal{}\\pm \\sigma$ for $ \\epsilon\\equal{}|f(\\sigma)|$. If there do not exist corresponding $ \\sigma$-s, then the function is not continuous in points $ x\\equal{}\\pm \\sigma$. But if they exist, then the initial $ \\sigma$ (for continuity in $ x\\equal{}0$) was not the greatest. Contradiction. [/hide]", "Solution_1": "Intuitively it is obvious, that the only possible solutions are $ y=\\sqrt{x^{2}+1}$ and $ y=-\\sqrt{x^{2}+1}$, but sometimes it is tough to write a formal proof. I thought of some possibilities:\r\n\r\n[list]2. possibility: \nLet us suppose that $ f$ is continuous and that there are two points, which lie in different sides of $ x$-axis. WLOG we can say $ f(x_{0})=-\\sqrt{x_{0}^{2}+1}, f(x_{1})=\\sqrt{x_{1}^{2}+1}$ and $ x_{0} m$\r\n\r\nThen, $ S\\equal{}\\{1,2,\\dots ,r\\plus{}n\\minus{}1\\}$ is the maximal subset." } { "Tag": [ "calculus", "integration", "calculus computations" ], "Problem": "Evaluate:\r\n\r\n$\\int_{0}^\\infty \\frac{dx}{\\sqrt{x}}$\r\n\r\nI'm not exactly sure how to convert this from an integral to a limit problem. Integration is fine for me, but I just need help setting up the problem. Any help?", "Solution_1": "This integral doesn't converge anywhere. So, I am not sure you would get anything even if you converted the problem to computing limits.\r\n\r\nIn fact, $\\int_{0}^{\\infty}\\frac{1}{x^{\\alpha}}\\,dx$ converges only if $\\alpha > 1$. In your problem, $\\alpha = \\frac12 < 1.$", "Solution_2": "I don't think $\\int_{0}^\\infty x^{t}$ ever converges. If $t <-1$, it blows up too much near 0. If $t >-1$, it doesn't decay enough $\\to \\infty$. If $t =-1$, it has both problems. \r\n\r\nIf you write it as $\\int_{a}^{b}x^{t}$ and evaluate the integral, and then take the limits as $a\\to 0$ and $b\\to \\infty$ you can prove the divergence." } { "Tag": [ "trigonometry", "ARML", "calculus", "derivative", "integration", "articles", "search" ], "Problem": "Hi all,\r\n\r\nI noticed something strange as I went from grade-school mathematics to college mathematics: I used to be horrible at manipulating symbols and equations, esp trig equations, and would frequently have trouble seeing the \"tricks\" used to manipulate an equation into solvability. Yet, when I started doing higher math proofs came much quicker to me.\r\n\r\nDo you guys think that there is an actual difference in \"elementary\" vs. \"higher\" mathematics? Could a good grade school mathematician not be very good with proofs and vice versa? Or is it a matter of training?", "Solution_1": "computational and rigorous mathematics may seem to have a different type of ora about them, but they do both belong to the same beast. in grade school mathematics you basically get a survey of various categories of mathematics but you don't get to justify much. this justification is the actual mathematics and in the higher level math courses i believe you begin to see how everything ticks and tocks and start to answer more of the \"why?\" and \"how?\" questions.", "Solution_2": "[quote=\"joshuatan\"]Do you guys think that there is an actual difference in \"elementary\" vs. \"higher\" mathematics? Could a good grade school mathematician not be very good with proofs and vice versa? Or is it a matter of training?[/quote]\r\n\r\nInteresting question. I think a lot of it is training...if you're not taught to ask \"why\" and \"how\", I've seen good grade school mathematicians be TERRIBLE at proofs. \r\n\r\nMore commonly these days, I run into students and most of my professors who only do rigorous proofs, and can't add or subtract. Like in my last linear algebra class, everytime the prof started writing numbers on the board, we'd all groan and wince (including the professor) because we're so bad with numbers. We didn't do a single correct numeric example all term, it always took 20 minutes to figure out where the mistakes were hahaha. I suspect this is from lack of practice though!", "Solution_3": "Well, I'm now a graduate student, but not so long ago I was a high school student at Stuyvesant, in New York. I still go back and visit from time to time, and I've found that the math team kids there are often quicker at a variety of tasks now than I am. Actually, at ARML some of the coaches were joking that we should have a coach vs. team contest, and I was rather pessimistic about the coaches' chances in such a hypothetical matchup. On the other hand, I find that in writing contests, there are some ideas, some modes of thought, some types of attack on certain problems that are \"trivial\" to me now, but that I wouldn't have thought of immediately even at my math team prime.\r\n\r\nSorry if this is a bunch of rambling anecdotes, rather than a coherent comment. Let me add one further: I think that many competent students first run into trouble in mathematics at the level of calculus. (Some, of course, have run into trouble much earlier; but let's focus on the ones who make it through algebra untroubled.) I think there are two reasons that calculus is hard: it requires certain technical skills (a mastery of algebraic and trigonometric manipulations, for example) that not all students have, but it also has certain conceptual barriers (the notions of limits, derivatives, antiderivatives, integrals, etc., and what these things all [i]mean[/i] -- how their abstract definitions are related to their content). To me, at least, these have always seemed to be distinct barriers that must be overcome separately, making me think the answer to the question is that they really are different.", "Solution_4": "A little bit ago I posted a link to a famous article called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1296484634&t=192363]Lockhart's Lament[/url]. His basic argument is that the math of grade school is quite different from higher math and that we are losing a bunch of potential students who don't realize how fun math can be. Grade school students think of math as numbers, most people who have been through higher stuff hear the the word math think of tons of things almost none of which involve numbers.\r\n\r\nI agree. There are some good comments on the thread about it too. :)", "Solution_5": "Of course it's different, but it's silly to think that one can understand higher mathematics without an understanding of elementary mathematics. To put it plainly, higher mathematics is built on elementary mathematics. There is no such thing as a competent mathematician who \"can't\" do elementary mathematics. They might mess up mechanical manipulations, but that's just mental laziness. I actually hate it when profs say that they can't get simple computations right at the blackboard. What they are really saying is, \"I don't take my job as a teacher seriously enough to make the necessary effort to do this right.\" This is especially true since the prof should have worked out the problem head of time anyway.", "Solution_6": "Well, when helping people or tutoring I almost always appeal to the stupid computations argument, and I don't believe that it is lack of taking teaching seriously. On the contrary, I think it is of the utmost importance for students to realize that the answer doesn't matter and one shouldn't waste their time fighting for the proper answer if it isn't coming quickly. \r\n\r\nI still recall a calc test with four problems on it. I got three answers wrong and one correct but I didn't work out the answer, I just wrote, the answer is the two solutions to this quadratic. I got a 99%. My prof wrote -1 for incorrect answers at the top.\r\n\r\nThere are always computers and calculators in the real world.\r\n\r\nOf course, you shouldn't always get every numerical answer wrong. I'm not advocating that, but I have definitely seen students go into a test unprepared, because they got caught up in trying to make an incorrect answer match the back of the book. If someone was there to tell them that they had the right concept and the error was meaningless, they would have learned more. Showing that that attitude is OK to adopt in those situations is a good teaching strategy. At least it helped me in lots of situations.", "Solution_7": "To all who took the time to reply: thanks for your thoughts!\r\n\r\nI kind of feel that Lockhart is too much of an idealist for my taste, but then again I'm not American so I don't really know whether I would have had the same reaction had I actually been in the K-12 system. I didn't really like how he seemed to come across as (over)emphasising intuition at the expense of rigour, or implying that formalism somehow makes the subject dry (because as you get more and more abstract, I think you need to have formalism to keep your intuition in line). However perhaps he is rebelling against a system that has veered too far to one end, and therefore feels that an extremist(?) seeming argument is necessary to wake people up.", "Solution_8": "HilbertThm90, you certainly have a point. In particular, every reasonable grader is lenient with computational mistakes made under timed conditions. But that argument fades away when it comes to homework or pretty much anything else. In the real world, students will become doctors or engineers or whatever, and just like in elementary school, the right answers do matter.\r\n\r\nMore rebuttal: The techniques that one uses to avoid \"stupid\" mistakes are good habits that encourage clear thought anyway---writing out your work clearly, not taking huge steps, taking your time, and (my favorite) trying to work the problem in a way that minimizes computation. It's also important to realize that in the scenario that was being contemplated (math professor teaching first-year college students), there is a huge chasm between what the professor considers to be mindless manipulation and what the students consider to be mindless manipulation. It's all too easy to start going down a slippery slope from \"this is just arithmetic\" to \"this is just some algebraic manipulation\" to \"oh, you've seen this a bunch of times and I'm sure you can do it yourself, so it's no biggie that I just royally F'ed this computation.\" But I guess when it comes down to it, I concede that it's partly an issue of personal style. \r\n\r\nAs for the \"back of the book\" issue, I have to disagree. Students rightfully want to know if what they are doing is correct. (The ones that already know that they're essentially right don't need help.) Checking against the back of the book is a reasonable method of verifying that you are doing things right. Of course, it doesn't tell you if you are doing things wrong for the reason you mention---maybe it was just a computational mistake. But that means that if you are careful to avoid making computational errors, you can be more confident that there is a substantive error. (But I do agree that sometimes students get a little obsessed with the back of the book thing.)\r\n\r\nRandom remark: Most high school students take multiple choices tests in school, and even if they don't, then their most important tests (SAT, AP) still have a lot of multiple choice. So from a practical point of view (at least for high school tutoring), it's important for your students to avoid computation errors.\r\n\r\nOkay, I've strayed far from the topic of the thread. Sorry." } { "Tag": [ "function", "calculus", "inequalities unsolved", "inequalities" ], "Problem": "Let $ x,y,z $ satisfying : $ x^2 +y^2 =2 $ and $ z^2+2z(x+y)=8 $ \r\n Find max of $ A=z(y-x) $", "Solution_1": "By a little computation we have $ (y-x)^2=8-\\frac{16}{z^2}-\\frac{z^2}{4}$\r\nFrom there, we can go by 2 way:\r\n 1> $ (z(y-x))^2=8z^2-16-\\frac{z^4}{4}$ leads exactly to $ -4\\sqrt3 \\leq z(y-x) \\leq 4\\sqrt3.$ ( because RHS is a one variable function)\r\n or 2> $ 8= \\frac{16}{z^2}+4\\frac{z^2}{16}+3\\frac{(y-x)^2}{3} \\geq 8(\\frac{z^6}{16^3}\\frac{(y-z)^6}{3^3})^{1/8}$ then get the answer is $ 4\\sqrt3$\r\n\"TT loves TT\" :D :D :D :D :D :D :D :D :D :D :D :D :D", "Solution_2": "tot ,tot . May be you are very lucky :P :P :D :D", "Solution_3": "These seem like they will be simple using Lagrange Multipliers", "Solution_4": "Can you tell mw what is this Lagrange method,can we use it to find the min,max point. I have heard about it before but I don't know it.", "Solution_5": "Using multivariable calculus we find that a necessary condition for an extremum is that, given $f(x,y,z)=0$, with constraints $c_1(x,y,z)=k_1$, $c_2(x,y,z)=k_2$, ..., $c_j(x,y,z) = k_j$ is that:\r\n\r\n$\\nabla f(x,y,z) = a_1\\nabla c_1(x,y,z) + a_2\\nabla c_2(x,y,z) + \\dotsc + a_j \\nabla c_j(x,y,z)$.\r\n\r\nwhere $a_1,a_2,\\dotsc a_j$ are constants and where $\\nabla$ is the gradient operator.\r\n\r\nThis leads to a system of $3+j$ equations in $3+j$ variables, if we expand it in scalar form.", "Solution_6": "Nice solution, darkmaster. My calculus approach just got me bogged down.", "Solution_7": "Let $ x,y,z $ be reals such that $ x^2 +y^2 =2 $ and $ z^2+z(x+y)=8 .$ Prove that\n $$ -6\\leq z(y-x) \\leq 6$$ \n\n" } { "Tag": [ "algebra", "polynomial", "induction", "search", "algebra unsolved" ], "Problem": "P(x) is the 100. degree polynomial and P(k)=2^k where k= 1,2,3,...,100 then fin P(101 )? \r\n\r\nThe solution is not 2^101 :)", "Solution_1": "yes, it is $ 2^{101}\\minus{}2$(by induction or Lagrange's Theorem)", "Solution_2": "[quote=\"shoki\"]yes, it is $ 2^{101} \\minus{} 2$(by induction or Lagrange's Theorem)[/quote]\r\n\r\ncould you send the solution by Lagrange ?", "Solution_3": "Sorry, that involves a lot of computations but if u know the theorem that's easy(it needs only computations) if not,search it, u will certainly find it :wink:", "Solution_4": "I think the answer is $ 2^{101}\\minus{}1$, not $ 2^{101}\\minus{}2$...\r\n\r\nJust construct the finite difference table and note that the 100th differences are constant.", "Solution_5": "there is a problem... epsilon should give the value of $ P(101)$ (because its degree is $ 100$),if not, we can't solve it.\r\ni was wrong by saying that $ P(101) \\equal{} 2^{101} \\minus{} 2$.in fact there was a problem like this(and i think that epsilon's problem is this in fact):\r\nIf $ P(x)$ is a polynomial with $ deg(P(x)) \\equal{} n$ such that for $ k \\equal{} 1,2,...,n \\plus{} 1$ we have $ P(k) \\equal{} 2^{k}$. then prove that \r\n$ P(n \\plus{} 2) \\equal{} 2^{n \\plus{} 2} \\minus{} 2$.", "Solution_6": "This is a very well-known problem. The polynomial is\r\n\r\n$ P(x)\\equal{}\\binom{x}{0}\\plus{}\\binom{x}{1}\\plus{}\\cdots\\plus{}\\binom{x}{100}$" } { "Tag": [ "calculus" ], "Problem": "Has anyone here taken AP Statistics? If so, did you take a class, or did you study on your own? I'm taking Calculus right now and I want to get all the math AP tests done this year...any thoughts would be appreciated.", "Solution_1": "I took a class through my high school. It was quite ridiculous since statistics is heavily dependent upon calculus (all those areas...), but yet we didn't use any calculus. You would probably be much better off getting a good book and learning it on your own.", "Solution_2": "[quote]You would probably be much better off getting a good book and learning it on your own.[/quote]\r\n\r\nMany things are like that...", "Solution_3": "Yeah, now I regret skipping math...if I'd know AoPS two years ago, then I would've slept through math and studied with AoPS...yet since I learned about AoPS through high school, there's a paradox...", "Solution_4": "Tare, r u a middle or high skool student?", "Solution_5": "both, weird, ain't it? :) \r\n\r\nHere, you could say I'm in high school since I'm in sophomore math, but usually (well, for all the other classes) I go to middle.", "Solution_6": "i did that for 2 yrs in middle school too...but ur still considered a middle skooler", "Solution_7": "Do you know of any good books that would teach me all the skills required?", "Solution_8": "Me too, this is my second year.\r\n...though this is a [b]math [/b]forum, so we might as well as go with what year math we're in...\r\n\r\nI'm not sure about any good books, though." } { "Tag": [ "videos", "function", "search", "complex numbers", "complex analysis", "AMC" ], "Problem": "What exactly is Birch and Swinnerton Dyer?", "Solution_1": "I won't tell you exactly what it is, but it is a statement about the Taylor expansion of a function. The only two of the Millenium Problems I think I could explain to anyone with any coherence whatsoever are the Poincar Conjecture and the Riemann Hypothesis. You can read the CMI official problem statements and/or watch the video lectures. I watched the video for the Riemann Hypothesis, but it doesn't have any particularly deep mathematical content. Still, if you are interested, you should watch the one on Birch Swinnerson-Dyer.", "Solution_2": "Short statement of the problem:\r\nhttp://www.claymath.org/Millennium_Prize_Problems/Birch_and_Swinnerton-Dyer_Conjecture/\r\n\r\nLink to PDF of the fuller problem statement:\r\nhttp://www.claymath.org/Millennium_Prize_Problems/Birch_and_Swinnerton-Dyer_Conjecture/_objects/Official_Problem_Description.pdf\r\n\r\nLink to a lecture about it:\r\nhttp://www.claymath.org/Millennium_Prize_Problems/Birch_and_Swinnerton-Dyer_Conjecture/_objects/Lecture_by_Fernando_Rodriguez-Villegas_at_UT.ram", "Solution_3": "At first I thought this was about Tarepandas :shock: :lol: \r\n\r\nThe only thing to add would be that \"Birch and Swinnerton Dyer\" is just my rank, not my avatar. My avatar is that oddly colored squiglly thing that I made (see if you can figure out the mathematical meaning behind it). For mathfanatic it's the pear, for Syntax Error it's that surfur dude, etc. It's usually a picture.\r\nBirch and Swinnerton Dyer is the longest of the 7 millenium prize problems, thus it was selected to be the highest rank. From experience I assume that the lowest rank (P vs. NP) features the problem with the shortest title and the last (Birch and Swinnerton Dyer) is the longest since the difficulty level of each is not in order, a la ComplexZeta I believe. If it was ordered from easiest to hardest it should start with the Reimann Hypothesis...or something like that.", "Solution_4": "Actually it would end with the Riemann Hypothesis since that's the hardest. It's also the most interesting, but that's just personal bias speaking. (Note that my user name is ComplexZeta, for example...)", "Solution_5": "Oh whoops, sorry.\r\nWhat does \"ComplexZeta\" mean anyways?", "Solution_6": "I thought it was either the complex (a+bi) part of the Riemann Zeta function or somat, or else it's just two math words spliced, diced, and spiced together...but now watch Simon prove me wrong.", "Solution_7": "mathfanatic is basically right. Complex refers to complex numbers and complex analysis (which is also known by other names, such as complex variables and complex function theory), which is a field I really enjoy. Zeta refers to the Riemann zeta function. Then I just put them together. If you search through the forum memberlist, you'll see a member named RealZeta. He's a friend of mine who made that username because he was inspired by mine.", "Solution_8": "[quote=\"ComplexZeta\"]...inspired by mine.[/quote]\r\n\r\nOh. I thought he was trying to be funny.", "Solution_9": "So now we know over half the namings of the most active members in AoPS (12 I believe)", "Solution_10": "??? you mean the meaning behind the names??? well.. you can figure mine out really easily. (It wasn't really supposed to have a meaning... lol)", "Solution_11": "Yup, like mathfanatic's name coming from being a math fanatic and such", "Solution_12": "mmm i was pondering \"mathamaniac\" for my username but i went with syntax error because i was thinking of calculators. ah jes", "Solution_13": "Personally I hate Syntax Errors, they are [b]so[/b] annoying, especially when you're rushing in the competition and it keeps on saying \"Error: Syntax\" over and over and over...\r\nDoesn't mean you Neal :) \r\n\r\nIs there a relationship between the length of the title of the 7 Millenium Prize Problems and how hard they are to understand? (Not necessarily their difficulty level) :lol:" } { "Tag": [ "vector", "algebra", "polynomial", "linear algebra", "linear algebra unsolved" ], "Problem": "The following is a Theorem from Friedberg, Insel and Spence.\r\n\r\nTheorem 7.1. Let $ T$ be a linear operator on a vector space $ V$, and let $ \\lambda$ be an eigenvalue of $ T$. If $ K_\\lambda$ represents the generalized eigenspace of $ T$ corresponding to $ \\lambda$, then for any scalar $ \\mu \\neq \\lambda$, the restriction of $ T-{\\lambda}I$ to $ K_\\lambda$ is one-to-one.\r\n\r\nLet $ x \\in K_\\lambda$ and $ (T-{\\lambda}I)(x) = 0$. By way of contradiction, suppose that $ x\\neq 0$. Let p be the smallest integer for which $ {(T-{\\lambda}I)}^p(x)=0$, and let $ y={(T-{\\lambda}I)}^{p-1}(x)$. Then $ (T-{\\lambda}I)(y)={(T-{\\lambda}I)}^p(x)=0$ and hence $ y \\in E_\\lambda$. \r\n\r\nFurthermore, ${{ (T-{\\mu}I)(y)=(T-{\\mu}I)(T-{\\lambda}I})^{p-1}(x)=(T-{\\lambda}I})^{p-1}(T-{\\mu}I)(x)=0$\r\n\r\nAt this point, what's the justification for commuting $ (T-{\\mu}I)$ and ${ {(T-{\\lambda}I})}^{p-1}$? Thanks, I'm having a hard time following that step, but after that, everything is clear.", "Solution_1": "$ A$ always commutes with any power of $ A.$\r\n\r\nFor instance: $ A\\cdot A^m\\equal{}A^{m\\plus{}1}\\equal{}A^m\\cdot A,$ and all I've done there is rearrange some parentheses.\r\n\r\nMore generally, $ A$ commutes with any polynomial in $ A.$", "Solution_2": "Ah, that makes sense. I got tripped up on the $ \\mu$ and $ \\lambda$ constants, but I guess they don't interfere since they are coefficients of $ I$. But if it were some other linear operator $ U\\neq I$, the commutativity wouldn't hold. \r\n\r\nThanks, KM." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given $ a, b, c > 0$. Prove that:\r\n\r\n$ \\sqrt[3]{\\frac{(a\\plus{}b)(b\\plus{}c)}{ab}} \\plus{} \\sqrt[3]{\\frac{(b\\plus{}c)(c\\plus{}a)}{bc}} \\plus{} \\sqrt[3]{\\frac{(c\\plus{}a)(a\\plus{}b)}{ca}} \\geq$$ \\frac{6}{\\sqrt[9]{2}}. \\sqrt[9]{\\frac{ab^2\\plus{}bc^2\\plus{}ca^2\\minus{}abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}}$\r\n :)", "Solution_1": "The hint: by Am-Gm, we can prove the stronger:\r\n\r\n$ \\frac{3}{4\\sqrt[3]{4}}( \\sqrt[3]{\\frac{(a\\plus{}b)(b\\plus{}c)}{ab}}\\plus{}\\sqrt[3]{\\frac{(b\\plus{}c)(c\\plus{}a)}{bc}}\\plus{}$$ \\sqrt[3]{\\frac{(c\\plus{}a)(a\\plus{}b)}{ca}}) \\geq\\ 2\\plus{} \\frac{ab^2\\plus{}bc^2\\plus{}ca^2\\minus{}abc}{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}$\r\n :)" } { "Tag": [ "complex numbers", "geometry proposed", "geometry" ], "Problem": "Let $ABC$ be a triangle and $A_1$, $B_1$, $C_1$ the midpoints of the sides $BC$, $CA$ and $AB$ respectively. Prove that if $M$ is a point in the plane of the triangle such that \\[ \\frac{MA}{MA_1} = \\frac{MB}{MB_1} = \\frac{MC}{MC_1} = 2 , \\] then $M$ is the centroid of the triangle.", "Solution_1": "Isn't it a bit too trivial? Simply consider the Apollonian Circles...", "Solution_2": "[quote=\"mecrazywong\"]Isn't it a bit too trivial? Simply consider the Apollonian Circles...[/quote]\r\n\r\nJust love the little things one finds are being claimed trivial from time to time\r\n\r\nDaniel", "Solution_3": "[quote=\"mecrazywong\"]Isn't it a bit too trivial? Simply consider the Apollonian Circles...[/quote]Remember this is for junior team selection, and it is supposed to be an easy problem on the test. It is not so trivial for a 14-year old :)", "Solution_4": "it's fairly easy with complex numbers also.\r\n\r\nlet afix of point $X$ be $x$, and $m=0$ . From conditions above we have $|a|=|b+c|, |b|=|a+c|, |c|=|a+b|$, and if we rewrite $a$ as $x+iy$, $b$ as $m+ni$ and $c$ as $p+qi$ we obtain $x^2+y^2=(m+p)^2+(n+q)^2$, $m^2+n^2=(x+p)^2+(y+q)^2$ and $p^2+q^2=(m+x)^2+(y+n)^2$ . After suming this $(x+m+p)^2+(y+n+q)^2=0$, hence $a+b+c=0$ and $M$ is it's centroid.", "Solution_5": "Here is my solution:$BM=2MB_{1},CM=2MC_{1}$ and $C_{1}B_{1}=\\frac{BC}2$ therefore $\\frac{C_{1}M}{CM}=\\frac{C_{1}B_{1}}{CB}=\\frac{B_{1}M}{BC}=2 \\iff \\triangle C_{1}B_{1}M$ ~ $\\triangle CBM$,thus $\\angle{CBM}=\\angle{C_{1}B_{1}M}$.But $C_{1}B_{1}$ is paralel to $BC$ therefore $B_{1},M,B$ are colinear.By the same manner we prove that $C_{1},M,C$ are colinear, hence M is the centroid of the triangle and the problem is solved!", "Solution_6": "We'll find after some computations that MG=0 (we'll use Leibinz theorem)", "Solution_7": "[quote=Valentin Vornicu]Let $ABC$ be a triangle and $A_1$, $B_1$, $C_1$ the midpoints of the sides $BC$, $CA$ and $AB$ respectively. Prove that if $M$ is a point in the plane of the triangle such that \\[ \\frac{MA}{MA_1} = \\frac{MB}{MB_1} = \\frac{MC}{MC_1} = 2 , \\] then $M$ is the centroid of the triangle.[/quote]\n\n\nLet $G$ be the centroid of triangle $ABC$. Let $A'$ be the reflection of $A$ about $A_1$. See that $(A,A_1;G,A') = -1$ so $M \\in \\odot (GA')$ (Appolonius, as $\\frac{MA}{MA_1} = 2$). Similarly. define $B',C'$. We see that $M$ is the point of concurrence of $\\odot (GA'),\\odot (GB'),\\odot (GC')$. It is easy to see that these three circles cannot be coaxial. So, $M=G$. $\\blacksquare$ " } { "Tag": [ "puzzles" ], "Problem": "Happened already? Has this that states which another, is there inexplicable and bizarre? More. Even something by replaced, be (and disappear) -- Instantly will it here. It is why, and for is, universe the what: discovers anyone, ever. If. That states which theory? A is there.", "Solution_1": "Hitchhiker's Guide to the Galaxy? Answer and Question to Life, the Universe, and Everything.", "Solution_2": "[hide=\"Solution\"]\nReverse the wording, and reorder the punctuation (it was pretty awkward to read it right?)\n\nThere is a theory which states that if ever anyone discovers what the universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another which states that this has already happened.\n\n(This is what?)\n\nAs brightknight says, it's a quote from HG2G (the radio show actually)\n\n[/hide]" } { "Tag": [ "blogs", "AMC", "AIME", "AMC 10" ], "Problem": "I plan to write a bunch of essays about AMC. The first one - Should You Check Your Answers During Tests? - discusses a mathematicians approach to the question. In my essay I use AMC 10/12 for my sample calculations. The essay is on my blog:\r\nhttp://blog.tanyakhovanova.com/?p=201", "Solution_1": "YES You should check your answers during AMCs\r\nI didn't last year and i still feel bad about it\r\n\r\nOn AMC 10A i got a 114 (cutoff for aime was 120)\r\nwhen I checked my answers with the online answers i saw that i got 4 problems in the first 20 wrong\r\n\r\nThen, on AMC 10B, i got another 114, and when I checked my answers, i saw that i got number 1, 3, 9 and 16 wrong\r\n\r\nI think ppl like me tend to check their answrs to only that last 5 problems, assuming that they make no stupid mistakes on the first 20", "Solution_2": "On the other hand, a lot of people like me need the full 75 minutes to finish all 25 questions. So I would say that you should finish as many questions as possible, THEN check your work.", "Solution_3": "in fact, I can't finish the Amc 12 with the full 75 min...though my accuracy in the first 17 problems is usually very good...", "Solution_4": "There can be different forms of checking. For example, you can check your bubbling (it would be frustrating to get one wrong like this), you can re-read a question to make sure you interpreted the question correctly, you can check your arithmetic, you can check your method, ... You have to be able to balance this as well as time.", "Solution_5": "It also depends on your own ability. If you're a neat person who rarely makes mistakes, then you don't need to check. However, if you make lots of computational errors, you should check.", "Solution_6": "do not check ur answers \r\n\r\nthe AMCs are designed NOT to trick you on the multiple choice\r\nso if ur answer matches one on the multiple choice, then its probably right and if it doesnt match, it means you made a careless mistake\r\n\r\ndont guess either, unless you have narrowed it down to 2-3 but i dont really see how thats possible , well actually it might be but its not too often.\r\n\r\ni think you should first try to solve all the problems and then think about checking", "Solution_7": "iYOA, I would say that isn't entirely true. That sort of approach also would do no good on the AIME. \r\n\r\nAMC 10 tries to trick you, just not as much as they could. (They would give you a break on some difficult questions, but they still try to trick you a bit.)\r\n\r\nI would say check them.", "Solution_8": "iYOA, I would say that isn't entirely true. That sort of approach also would do no good on the AIME. \r\n\r\nAMC 10 tries to trick you, just not as much as they could. (They would give you a break on some difficult questions, but they still try to trick you a bit.)\r\n\r\nI would say check them.", "Solution_9": "well with AIME its very important to check your work especially if you make lots of careless mistakes\r\nbut not so much on AMC. i think in the time it takes for you to find a mistake, you could be solving more problems.\r\n\r\nand i did read somewhere that the AMC is not designed to trick a person even though they knew how to solve the certain problem.\r\n\r\nso i think checking should come last, only when you think you have solved all the problems you possibly can. (this only pertains to AMC)\r\n\r\n\r\non AIME it all depends on the person i guess.", "Solution_10": "That I agree with. I think you should solve them all before checking.", "Solution_11": "While doing the test, you should read each question carefully. Read it more than once to make sure that you haven't misread it. Then make sure that your answer makes sense in the context of the question. (This probably doesn't count as checking, but its still very important.) Also, even if you don't have time to do anything else, always check your bubbling at the end. It only takes half a minute and could save you from missing a question." } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Prove the following inequality.\r\n\r\n$ \\left(\\int_0^1 (x^2\\plus{}px\\plus{}q)\\ dx\\right)^2\\leq \\int_0^1 (x^2\\plus{}px\\plus{}q)^2\\ dx\\minus{}\\frac{1}{180}$", "Solution_1": "iff\r\n\r\n$ 0 \\le \\frac{(p\\plus{}1)^2}{12}$ \r\n\r\ncool one.", "Solution_2": "O.K. :)", "Solution_3": "[quote=\"kenn4000\"]iff\n\n$ 0 \\le \\frac{(p\\plus{}1)^2}{12}$ \n\ncool one.[/quote]\n\nI also got same by brute force technique\n\n[b][size=150]Is there any elegant way ?[/size]\n[/b]", "Solution_4": "Interesing.\nA Generalization\n\n$\\boxed{\\boxed{\n\\left(\\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right) \\, dx\\right)^2\\leq \\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right)^2 \\,\n dx-\\frac{2^{-1-4 n} \\pi (n!)^2}{\\Gamma \\left(\\frac{1}{2}+n\\right) \\Gamma \\left(\\frac{3}{2}+n\\right)}\n}}$\n\nWith $\\alpha_n = 1, n \\ge 1$, and also, the constant is optimal.", "Solution_5": "[quote=\"EulerIntegral\"]Interesing.\nA Generalization\n\n$\\boxed{\\boxed{\n\\left(\\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right) \\, dx\\right)^2\\leq \\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right)^2 \\,\n dx-\\frac{2^{-1-4 n} \\pi (n!)^2}{\\Gamma \\left(\\frac{1}{2}+n\\right) \\Gamma \\left(\\frac{3}{2}+n\\right)}\n}}$\n\nWith $\\alpha_n = 1, n \\ge 1$, and also, the constant is optimal.[/quote]\n\ncan u provide some explanation", "Solution_6": "[quote=\"EulerIntegral\"]Interesing.\nA Generalization\n\n$\\boxed{\\boxed{\n\\left(\\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right) \\, dx\\right)^2\\leq \\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right)^2 \\,\n dx-\\frac{2^{-1-4 n} \\pi (n!)^2}{\\Gamma \\left(\\frac{1}{2}+n\\right) \\Gamma \\left(\\frac{3}{2}+n\\right)}\n}}$\n\nWith $\\alpha_n = 1, n \\ge 1$, and also, the constant is optimal.[/quote]\n\nWhere the hell did this come from?", "Solution_7": "I made this and proposed as a problem.", "Solution_8": "$f(\\vec{\\alpha})=\\int_{0}^{1}(\\sum_{i=0}^{n}\\alpha_{i}x^{i})^{2}\\, dx-(\\int_{0}^{1}(\\sum_{i=0}^{n}\\alpha_{i}x^{i})\\, dx)^{2} $\n\n$=\\sum_{i=0}^{n}\\alpha_{i} \\int_0^1\\sum_{j=0}^{n}\\alpha_{j}x^{i+j}) dx-( \\sum_{i=0}^n \\frac{\\alpha_i}{i+1} )^2$\n\n$=\\sum_{i=0}^{n}\\alpha_{i} \\sum_{j=0}^{n}\\frac{\\alpha_{j}}{1+i+j} - \\sum_{i=0}^{n} \\frac{\\alpha_i}{i+1} \\sum_{j=0}^n \\frac{\\alpha_j}{j+1}$\n\n$=\\sum_{i=0}^{n}\\alpha_{i} \\sum_{j=0}^{n} \\alpha_j ( \\frac{1}{i+j+1}-\\frac{1}{(i+1)(j+1)})$\n\n$=\\vec{\\alpha}^T A \\vec{\\alpha}$ for the symmetric $A_{ij}= \\frac{1}{1+i+j}-\\frac{1}{(1+i)(1+j)}$\n\nso if we find the min of this function, hopefully its that crazy constant", "Solution_9": "[quote=\"EulerIntegral\"]Interesing.\nA Generalization\n\n$\\boxed{\\boxed{\n\\left(\\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right) \\, dx\\right)^2\\leq \\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right)^2 \\,\n dx-\\frac{2^{-1-4 n} \\pi (n!)^2}{\\Gamma \\left(\\frac{1}{2}+n\\right) \\Gamma \\left(\\frac{3}{2}+n\\right)}\n}}$\n\nWith $\\alpha_n = 1, n \\ge 1$, and also, the constant is optimal.[/quote]\n\n\nI just simplified the constant\n\n$\\boxed{\\boxed{\n\\left(\\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right) \\, dx\\right)^2\\leq \\int_0^1 \\left(\\sum _{i=0}^n \\alpha _i x^i\\right)^2 \\,\n dx-\\frac{ (n!)^4}{(2n)! (2n+1)!}\n}}$\n\nWith $\\alpha_n = 1, n\\in\\mathbb{N}$", "Solution_10": "[quote=\"EulerIntegral\"]I made this and proposed as a problem.[/quote]\n\ncan u give your solution" } { "Tag": [ "AMC", "AMC 12", "AMC 12 A" ], "Problem": "I took it in BEIJING,i want to konw my score", "Solution_1": "Look further down in the forums for the thread.\r\n\r\nEdit: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132366[/url]", "Solution_2": "luckily i got 150" } { "Tag": [ "inequalities", "geometry", "circumcircle", "inradius", "function", "trigonometry", "inequalities proposed" ], "Problem": "Let $ ABC$ be an acute triangle with three sides $ a,b,c$ and $ R , r$ is the circumradius and the inradius of the triangle.Prove that:\r\n$ \\sqrt{\\frac{b+c-a}{a}}+\\sqrt{\\frac{c+a-b}{b}}+\\sqrt{\\frac{a+b-c}{c}} \\ge \\sqrt{9+\\frac{1}{2}(1-\\frac{2r}{R})^{2}}$", "Solution_1": "[quote=\"quykhtn-qa1\"]Let $ ABC$ be an acute triangle with three sides $ a,b,c$ and $ R , r$ is the circumradius and the inradius of the triangle.Prove that:\n$ \\sqrt {\\frac {b + c - a}{a}} + \\sqrt {\\frac {c + a - b}{b}} + \\sqrt {\\frac {a + b - c}{c}} \\ge \\sqrt {9 + \\frac {1}{2}(1 - \\frac {2r}{R})^{2}}$[/quote]\r\nSince this inequality,we have:\r\n$ \\sqrt{\\frac{b+c-a}{a}}+\\sqrt{\\frac{c+a-b}{b}}+\\sqrt{\\frac{a+b-c}{c}} \\ge 3$\r\n\r\n :)", "Solution_2": "[quote=\"quykhtn-qa1\"]Let $ ABC$ be an acute triangle with three sides $ a,b,c$ and $ R , r$ is the circumradius and the inradius of the triangle.Prove that:\n$ \\sqrt {\\frac {b + c - a}{a}} + \\sqrt {\\frac {c + a - b}{b}} + \\sqrt {\\frac {a + b - c}{c}} \\ge \\sqrt {9 + \\frac {1}{2}(1 - \\frac {2r}{R})^{2}}$[/quote]\r\nThis inequality very nice and not easy.No one like it? :)", "Solution_3": "[quote=\"quykhtn-qa1\"]Let $ ABC$ be an acute triangle with circumradius $ R$ and inradius $ r$ .Prove that\n\n$ \\sqrt {\\frac {b + c - a}{a}} + \\sqrt {\\frac {c + a - b}{b}} + \\sqrt {\\frac {a + b - c}{c}} \\ge \\sqrt {9 + \\boxed {\\frac {1}{2}}\\left(1 - \\frac {2r}{R}\\right)^{2}}$ .[/quote]\n[color=darkblue][b]I think that is better $ 2$ than $ \\frac 12$ in above proposed inequality.\n\nI dedicate my proof to my friend, the student[/b][/color] [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=61896][color=darkred][b][u]Mateescu Constantin, Pitesti, RO[/u].[/b][/color][/url]\n\n\n[quote=\"Virgil Nicula\"][color=darkred][b][u]Lemma 1[/u].[/b] In any [u]acute triangle[/u] $ ABC$ there is the inequality $ \\boxed {\\ \\frac {a^3 + b^3 + c^3}{abc}\\ \\ge\\ 1 + \\frac Rr\\ }\\ \\ (1)$ .[/color][/quote]\n[color=darkblue][b]Proof.[/b] Apply the [u][b]Walker[/b]'s inequality[/u], i.e. $ s^2\\ge 2R^2 + 8Rr + 3r^2\\ (*)$ . Indeed, \n\n$ \\frac {a^3 + b^3 + c^3}{abc} = 3 + \\frac {a^3 + b^3 + c^3 - 3abc}{abc} =$ $ 3 + \\frac {(a + b + c)\\left[\\left(a^2 + b^2 + c^2\\right) - (ab + bc + ca)\\right]}{abc} =$\n\n$ 3 + \\frac {s^2 - 3r^2 - 12Rr}{2Rr} =$ $ \\frac {s^2 - 3r^2 - 6Rr}{2Rr}\\ \\stackrel{(*)}{\\ge}\\ \\frac {\\left(2R^2 + 8Rr + 3r^2\\right) - 3r^2 - 6Rr}{2Rr} =$ $ 1 + \\frac Rr$ .[/color]\n\n[quote][color=darkred][b][u]Lemma 2[/u].[/b] In any [u]acute triangle[/u] $ ABC$ there is the inequality $ \\boxed {\\ \\sin\\frac A2 + \\sin\\frac B2 + \\sin\\frac C2\\ \\ge\\ \\frac 54 + \\frac {r}{2R}\\ }\\ \\ (2)$ .[/color][/quote]\n[color=darkblue][b][u]Proof[/u].[/b] Apply the [u][b]Popoviciu[/b]'s inequality[/u] to the concave function $ \\mathrm{COS}$ on the interval $ \\left (0\\ ,\\ \\frac {\\pi}{2}\\right )$ :\n\n$ \\cos A + \\cos B + \\cos C +$ $ 3\\cos\\frac {A + B + C}{3}\\ \\le\\ 2\\cdot\\left(\\cos\\frac {A + B}{2} + \\cos\\frac {B + C}{2} + \\cos\\frac {C + A}{2}\\right)\\ \\Longleftrightarrow$\n\n$ 1 + \\frac rR + \\frac 32\\ \\le\\ 2\\cdot\\left(\\sin\\frac A2 + \\sin\\frac B2 + \\sin\\frac C2\\right)\\ \\Longleftrightarrow\\ \\sin\\frac A2\\ +$ $ \\ \\sin\\frac B2\\ + \\ \\sin\\frac C2\\ \\ge\\ \\frac 54\\ + \\ \\frac r{2R}$ .[/color]\n\n[quote][color=darkred]Prove that in any [u]acute triangle[/u] $ ABC$ there is the inequality $ \\boxed {\\ \\sum\\sqrt {\\frac {b + c - a}{a}}\\ \\ge\\ \\sqrt {5 + \\frac Rr + \\frac {4r}{R}}\\ \\ge\\ \\sqrt {9 + 2\\cdot\\left(1 - \\frac {2r}{R}\\right)^2}\\ \\ge\\ 3\\ }$ .[/color][/quote] \r\n[color=darkblue][b][u]Proof[/u].[/b] $ \\left(\\sum\\sqrt {\\frac {b + c - a}{a}}\\right)^2 =$ $ \\sum\\frac {b + c - a}{a} + 2\\cdot\\sum\\sqrt {\\frac {(a + b - c)(a + c - b)}{bc}} =$ $ \\sum\\left(\\frac ab + \\frac ba\\right) - 3 + 4\\cdot\\sum\\sqrt {\\frac {(s - b)(s - c)}{bc}} =$\n\n$ \\sum\\frac {c^2 + 2ab\\cdot\\cos C}{ab} - 3 + 4\\cdot\\sum\\sin\\frac A2 =$ $ \\sum\\frac {c^2}{ab} + 2\\sum\\cos C - 3 + 4\\cdot\\sum\\sin \\frac A2 =$ $ \\frac {a^3 + b^3 + c^3}{abc} + 2\\cdot\\left(1 + \\frac rR\\right) - 3 + 4\\cdot\\sum\\sin\\frac A2\\ \\stackrel{(1)\\wedge (2)}{\\ge}$ \n\n$ 1 + \\frac Rr + 2\\left(1 + \\frac rR\\right) - 3 + 5 + \\frac {2r}{R} =$ $ 5 + \\frac Rr + \\frac {4r}{R}\\ \\Longrightarrow\\ \\sum\\sqrt {\\frac {b + c - a}{a}}\\ \\ge\\ \\sqrt {5 + \\frac Rr + \\frac {4r}{R}}$ . Prove easily that and [u]right inequality[/u] is truly.\n\nIndeed, $ \\sqrt {5 + \\frac Rr + \\frac {4r}{R}}\\ge\\sqrt {9 + 2\\cdot\\left(1 - \\frac {2r}{R}\\right)^2}$ $ \\Longleftrightarrow 5 + \\frac Rr + \\frac {4r}{R}\\ge 9 + 2\\cdot\\left(1 - \\frac {2r}{R}\\right)^2\\Longleftrightarrow$ $ \\frac {(R - 2r)^2}{Rr}\\ge 2\\cdot\\left(1 - \\frac {2r}{R}\\right)^2\\Longleftrightarrow R\\ge 2r$ .[/color]", "Solution_4": "We can also prove that:\n$ \\sqrt{\\frac{b+c-a}{a}}+\\sqrt{\\frac{c+a-b}{b}}+\\sqrt{\\frac{a+b-c}{c}} \\ge \\sqrt{9+3(1-\\frac{2r}{R})^2} $\nFor any acute triangle $ ABC $ ,we have:\n$ sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2} \\ge m+\\frac{(3-2m)r}{R} $ ,where $ m\\le k=\\frac{\\sqrt{4-2\\sqrt{2}}+3\\sqrt{2}-5}{3\\sqrt{2}-4} $\nWe can check that: $ k>\\frac{4}{3}>\\frac{5}{4} $.\nWith $ m=\\frac{4}{3} $, we have Jackgarfunkel inequality.", "Solution_5": "$ \\sin\\frac{A}{2}+\\sin\\frac{B}{2}+\\sin\\frac{C}{2} \\ge m+\\frac{(3-2m)r}{R} $ ,where $ m\\le k=\\frac{\\sqrt{4-2\\sqrt{2}}+3\\sqrt{2}-5}{3\\sqrt{2}-4}. $\n\na hard result.\n\nyou why \n\\ssin\\frac{A}{2},ssin\\frac{A}{2}.", "Solution_6": "[quote=\"quykhtn-qa1\"]\nFor any acute triangle $ ABC $ ,we have:\n$ sin\\frac{A}{2}+sin\\frac{B}{2}+sin\\frac{C}{2} \\ge m+\\frac{(3-2m)r}{R} $ ,where $ m\\le k=\\frac{\\sqrt{4-2\\sqrt{2}}+3\\sqrt{2}-5}{3\\sqrt{2}-4} $\nWe can check that: $ k>\\frac{4}{3}>\\frac{5}{4} $.\nWith $ m=\\frac{4}{3} $, we have Jackgarfunkel inequality.[/quote]\n\nThis is indeed quite a strong inequality and the main difficulty in proving it by the use of some classic methods is due to the special equality case which occurs when $\\triangle\\, ABC$ is either [b]equilateral[/b] or [b]right-isosceles[/b]. However, we can overcome this \"inconvenience\" by making use of the following [b]VERY STRONG[/b] inequality, the [u][b]equality case[/b] of which is attained for any [b]isosceles triangle[/b][/u]:\n\\[\\boxed{\\ \\sum\\, \\sin\\frac A2\\, \\ge\\, \\sqrt {\\frac {9R-2r}{4R}+\\sqrt {\\frac {s^2+r^2+2Rr}{2R^2}}}-\\frac 12\\ }\\ \\ \\ \\ .\\]\nI derived this amazing (although not aesthetically pleasing) inequality by using the identity \n\\[\\boxed{\\ \\sum\\, \\cos\\frac {B-C}2=\\left(\\frac 12+\\sum\\, \\sin\\frac A2\\right)^2+\\frac r{2R}-\\frac 54\\ }\\]\nand the following beautiful inequality, which I proved it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2315537#p2315537][b]here[/b][/url], that is:\n\\[\\boxed{\\ \\sum\\, \\cos\\frac {B-C}2\\, \\ge\\, 1+\\sqrt {\\frac {s^2+r^2+2Rr}{2R^2}}\\ }\\ \\ \\ \\ .\\]\nThus, it remains to show that:\n\\[\\sqrt {\\frac {9R-2r}{4R}+\\sqrt {\\frac {s^2+r^2+2Rr}{2R^2}}}-\\frac 12\\, \\ge\\, k+\\frac {\\left(3-2k\\right)r}R\\]\nwhich further simplifies to:\n\\[\\boxed{\\ s^2\\, \\ge\\, \\frac {\\left[\\left(4k^2+4k-8\\right)R^2+\\left(-16k^2+16k+14\\right)Rr+\\left(6-4k\\right)^2r^2\\right]^2-8R^2\\left(2Rr+r^2\\right)}{8R^2}\\ }\\ \\ \\ \\ ,\\]\nwhere $k=\\tfrac{\\sqrt{4-2\\sqrt 2}+3\\sqrt 2-5}{3\\sqrt 2-4}$ is the best possible constant for the type of inequality in question. Finally, the latter inequality can be proven by using the well known results which are valid in [b]non-obtuse[/b] triangles, viz. $\\boxed{\\, s^2\\, \\ge\\, 2R^2+8Rr+3r^2\\, }$ ([u][b]Walker[/b][/u]) and $\\boxed{\\, s\\, \\ge\\, 2R+r\\, }$ . Specifically, it can be shown - at least by the use of a computer programme - that:\n\\[\\boxed{\\ \\begin{array}{rrrr}\n2R^2+8Rr+3r^2\\ge\\tfrac {\\left[\\left(4k^2+4k-8\\right)R^2+\\left(-16k^2+16k+14\\right)Rr+\\left(6-4k\\right)^2r^2\\right]^2-8R^2\\left(2Rr+r^2\\right)}{8R^2} & \\text{holds for} & 2\\le\\tfrac Rr\\le 1+\\sqrt 2 \\\\ \\\\ \n\\left(2R+r\\right)^2\\ge\\tfrac {\\left[\\left(4k^2+4k-8\\right)R^2+\\left(-16k^2+16k+14\\right)Rr+\\left(6-4k\\right)^2r^2\\right]^2-8R^2\\left(2Rr+r^2\\right)}{8R^2} & \\text{holds for} & \\frac Rr\\ge 1+\\sqrt 2\\end{array}\\ } \\ \\ \\ \\ \\ .\\] \nWe are done. :)", "Solution_7": "[quote=quykhtn-qa1]\n\\[\\left\\{\\begin{array}{c}\\triangle ABC \\\\\\\\ \\max\\{A, B, C\\} \\le 90^{\\circ} \\\\\\\\ k=\\frac{\\sqrt{4-2\\sqrt{2}}+3\\sqrt{2}-5}{3\\sqrt{2}-4} \\end{array}\\right\\|\\ \\ \\implies\\ \\ \\boxed{\\boxed{\\sin\\frac A2 + \\sin\\frac B2 + \\sin\\frac C2 \\ge k + \\left(3 - 2k\\right)\\cdot\\frac rR}}\\]\n[/quote]\n\nThis slightly sharper version of Jack Garfunkel's inequality can also be proven using the following STRONG results:\n\n\\[\n\\boxed{\n\\begin{array}{lll}\n\\sin\\frac A2 + \\sin\\frac B2 + \\sin\\frac C2 & \\ge & \\frac 12 + \\sqrt{1 - \\sqrt{1 - \\frac {2r}R}} + \\frac 12 \\sqrt{1 - \\frac {2r}R} \\\\\\\\\n\\sin\\frac A2 + \\sin\\frac B2 + \\sin\\frac C2 & \\ge & \\frac{\\sqrt{2}}2 + \\sqrt{\\frac 12 + \\frac {\\sqrt{2} - 1}2\\cdot\\frac rR}\n\\end{array}\n}\\ .\n\\]\n\n[url=http://www.artofproblemsolving.com/community/c6h1221278p6108147][b]The first inequality[/b][/url] holds in any triangle and [u]attains equality for [b]most isosceles triangles[/b][/u]; the second one holds in non-obtuse-angled triangles, [u]attains equality for any [b]right-angled triangle[/b][/u] and it is a simple consequence of identity [b]141.2[/b] from [url=http://www.artofproblemsolving.com/community/c6h412623p2481722][b]here[/b][/url] (just replace $\\phi$ by $\\pi / 2$). Now we can prove the conclusion in two steps as in my previous post, namely:\n\n\\[\n\\boxed{\n\\begin{array}{lllcr}\n\\frac 12 + \\sqrt{1 - \\sqrt{1 - \\frac {2r}R}} + \\frac 12 \\sqrt{1 - \\frac {2r}R} & \\ge & k + \\left(3 - 2k\\right)\\cdot\\frac rR & \\text{holds for} & 2\\le\\frac Rr\\le 1 + \\sqrt{2} \\\\\\\\\n\\frac{\\sqrt{2}}2 + \\sqrt{\\frac 12 + \\frac {\\sqrt{2} - 1}2\\cdot\\frac rR} & \\ge & k + \\left(3 - 2k\\right)\\cdot\\frac rR & \\text{holds for} & \\frac Rr \\ge 1 + \\sqrt{2}\n\\end{array}\n}\\ .\n\\]" } { "Tag": [ "function", "trigonometry" ], "Problem": "I was just wondering if there is any formula or quick method to calculate the sum of the factors of a number. I have seen problems involving this calculations a number of times. :)", "Solution_1": "first prime factorize the number\r\nyou will get\r\n$ p_1^{e_1}p_2^{e_2}p_3^{e_3} \\ldots p_n^{e_n}$\r\nthe sum of the number of prime factors is\r\n$ \\prod^n_{k\\equal{}1} \\sum^{e_k}_{l\\equal{}0}p^l$\r\nif that is kinda complicated, i will demonstrate this technique for the number 45360\r\n$ 45360\\equal{}2^4 \\cdot 3^4 \\cdot 5^1 \\cdot 7^1$\r\nsum of factors= $ (1\\plus{}2\\plus{}4\\plus{}8\\plus{}16)(1\\plus{}3\\plus{}9\\plus{}27\\plus{}81)(1\\plus{}5)(1\\plus{}7)$\r\nequals some big number\r\nthe reasoning behind this formula is quite complicated but also quite elegant", "Solution_2": "[quote=\"stevenmeow\"]first prime factorize the number\nyou will get\n$ p_1^{e_1}p_2^{e_2}p_3^{e_3} \\ldots p_n^{e_n}$\nthe sum of the number of prime factors is\n$ \\prod^n_{k \\equal{} 1} \\sum^{e_k}_{l \\equal{} 0}p^l$\nif that is kinda complicated, i will demonstrate this technique for the number 45360\n$ 45360 \\equal{} 2^4 \\cdot 3^4 \\cdot 5^1 \\cdot 7^1$\n[/quote]\r\nFrom this step let me u tell the method in which I solved this question.\r\ndefine a func f(n,k)=(kn-1)/(k-1). where k is the prime factor and n is the pth value of k.\r\nif we take this example only.\r\nf(4,2)=(8-1)(2-1)=7\r\nsimilarly find for others and multiply them.\r\n If you want to find the sum of factors of 720\r\nf(16)=(2*16-1)/(2-1) = 31\r\nf(9) = (3*9-1)/(3-1) = 13\r\nf(5) = (5*5-1)/(5-1) = 6\r\nur answer is 31*13*6 = 2418", "Solution_3": "Note, this is a function called sum-of-divisors function (Denoted $ \\sigma(x)$).", "Solution_4": "This method intuitively makes sense. To take a small example, $ 24\\equal{}2^3\\cdot3$ we get $ (1\\plus{}2^1\\plus{}2^2\\plus{}2^3)(1\\plus{}3)$ Expanding this we get $ 1\\plus{}2^1\\plus{}2^2\\plus{}2^3\\plus{}3\\plus{}2^13^1\\plus{}2^23^1\\plus{}2^33^1$ What we just listed out are all of the factors of 24.", "Solution_5": "Wikipedia also gives a nice trigonometric series:\r\n$ \\sigma(n)\\equal{}\\sum_{k\\equal{}1}^{n}\\sum_{j\\equal{}1}^{k}\\cos\\left(\\frac{2\\pi jn}{k}\\right)$.\r\n\r\nPerhaps not the easiest thing to evaluate on paper, but it's interesting nonetheless. \r\n\r\nSee http://en.wikipedia.org/wiki/Divisor_function" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c > 0$\r\n\r\nProve that:\r\n$ \\sqrt \\frac {a^3}{a^3 \\plus{} (b \\plus{} c)^3} \\plus{} \\sqrt \\frac {b^3}{b^3 \\plus{} (b \\plus{} c)^3} \\plus{} \\sqrt \\frac {c^3}{c^3 \\plus{} (a \\plus{} b)^3} \\ge 1$\r\n\r\n\r\nI didn't prove that, so I'll get a bad mark :(\r\n\r\nP/s: Sorry, I'm a newbie, so if I posted it in wrong box, please del it. :blush:", "Solution_1": "Do a normalization: $ a\\plus{}b\\plus{}c\\equal{}1$, the inequality becomes:\r\n\r\n$ \\sum_{cyc}{\\sqrt{\\frac{a^3}{1\\minus{}3a\\plus{}3a^2}}} \\geq 1$\r\n\r\nWe choose appropriate constants $ \\alpha, \\beta$ such that\r\n\r\n$ \\sqrt{\\frac{x^3}{1\\minus{}3x\\plus{}3x^2}} \\geq \\alpha x \\plus{} \\beta$ for all $ x\\in [0,1]$\r\n\r\nand that $ \\alpha \\plus{} 3 \\beta \\equal{} 1$.\r\n\r\nDone!", "Solution_2": "$ \\sqrt {\\frac {a^3}{a^3 \\plus{} (b \\plus{} c)^3}}\\geq\\frac {a^2}{\\sum a^2}$\r\n$ \\Leftrightarrow a^3(\\sum a^2)^2\\geq a^4[a^3 \\plus{} (b \\plus{} c)^3]$\r\n$ \\Leftrightarrow a^3(b^4 \\plus{} c^4) \\plus{} 2a^3\\sum a^2b^2 \\minus{} a^4(b^3 \\plus{} c^3) \\minus{} 3a^4(b^2c \\plus{} bc^2)\\geq 0$\r\n$ \\Leftrightarrow (b^4 \\plus{} c^4) \\plus{} 2\\sum a^2b^2 \\minus{} a(b^3 \\plus{} c^3) \\minus{} 3a(b^2c \\plus{} bc^2)\\geq 0$\r\n$ \\Leftrightarrow (b^2 \\plus{} c^2)^2 \\plus{} 2a^2(b^2 \\plus{} c^2)\\geq a(b \\plus{} c)^3$\r\nthat $ LHS\\geq \\frac {(b \\plus{} c)^4}{4} \\plus{} a^2(b \\plus{} c)^2 \\equal{} (b \\plus{} c)^2[\\frac {(b \\plus{} c)^2}{4} \\plus{} a^2]\\geq a(b \\plus{} c)^3$\r\n$ \\Rightarrow \\sum\\sqrt {\\frac {a^3}{a^3 \\plus{} (b \\plus{} c)^3}}\\geq\\sum\\frac {a^2}{\\sum a^2} \\equal{} 1$", "Solution_3": "this inequality is very easy :lol: . THANKS VERY MUCH", "Solution_4": "[quote=\"lia_h1\"]Let $ a,b,c > 0$\n\nProve that:\n$ \\sqrt \\frac {a^3}{a^3 \\plus{} (b \\plus{} c)^3} \\plus{} \\sqrt \\frac {b^3}{b^3 \\plus{} (b \\plus{} c)^3} \\plus{} \\sqrt \\frac {c^3}{c^3 \\plus{} (a \\plus{} b)^3} \\ge 1$\n\n\nI didn't prove that, so I'll get a bad mark :(\n\nP/s: Sorry, I'm a newbie, so if I posted it in wrong box, please del it. :blush:[/quote]\r\n$ a\\plus{}b\\plus{}c\\equal{}1$ then Jensen :wink:", "Solution_5": "[quote=\"lia_h1\"]Let $ a,b,c > 0$\n\nProve that:\n$ \\sqrt \\frac {a^3}{a^3 \\plus{} (b \\plus{} c)^3} \\plus{} \\sqrt \\frac {b^3}{b^3 \\plus{} (b \\plus{} c)^3} \\plus{} \\sqrt \\frac {c^3}{c^3 \\plus{} (a \\plus{} b)^3} \\ge 1$\n\n\nI didn't prove that, so I'll get a bad mark :(\n\nP/s: Sorry, I'm a newbie, so if I posted it in wrong box, please del it. :blush:[/quote]\r\n\r\nSee http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62256" } { "Tag": [ "vector", "inequalities", "topology", "triangle inequality", "real analysis", "real analysis unsolved" ], "Problem": "a) Let $ \\parallel{} \\parallel{}$ be any norm on $ \\mathbb{R}^{m}$ and let $ B \\equal{} \\{x \\in \\mathbb{R}^{m} | \\parallel{}x\\parallel{} \\le 1\\}$. Prove that $ B$ is compact.\r\n\r\nb) Show that any 2 norms on $ \\mathbb{R}^{m}$ are equivalent (i.e. for $ \\parallel{} \\parallel{}_{1}$ and $ \\parallel{} \\parallel{}_{2} \\exists c > 0$ s.t. $ \\parallel{}x\\parallel{}_{1} \\le c\\parallel{}x_{2}\\parallel{}$ and $ \\parallel{}x\\parallel{}_{2} \\le c\\parallel{}x_{1}\\parallel{}, \\forall x \\in \\mathbb{R}^{m}$)\r\n\r\nMy biggest challenge is with the second part of this question, because how is it that for a norm which is a vector with distance and magnitude be equal to some other vector due to a specified $ c$?", "Solution_1": "If you have trouble in the abstract setting, try something more concrete first.\r\nFor example, in $ \\mathbb{R}^2$ consider the two norms $ \\|(x,y)\\|_1 \\equal{} \\sqrt{x^2\\plus{}y^2}$ and $ \\|(x,y)\\|_2 \\equal{} |x|\\plus{}|y|$.", "Solution_2": "For a), can I use the fact that $ \\mathbb{R}^{m}$ is a metric space and try to show that it contains the limits of all its Cauchy sequences to prove completeness. And then from there....I could use some help or at least how I could use the inequality $ \\parallel{} x \\parallel{} \\le 1$", "Solution_3": "a) is a straightforward consequence of the Heine-Borel theorem. For b) try picking a basis and using the triangle inequality along with the list of norms of each basis vector.", "Solution_4": "could you explain what you mean by list of norms for each basis vector is? Thanks", "Solution_5": "I mean pick a basis $ e_1, e_2, ... e_m$, look at the list of norms $ \\parallel{}e_1\\parallel{}, \\parallel{}e_2\\parallel{}, ... \\parallel{}e_m\\parallel{}$, and use it to bound the norm of an arbitrary vector.", "Solution_6": "So was the idea to use the linear independence of a basis and triangle inequality , as suggested, to come up with the constant $ c$?", "Solution_7": "for a) is there any way to form a closed ball in $ B$ so that I can show that it is closed? I can't think of any other way to show that it might be closed unless there are other ideas?", "Solution_8": "The question is slightly ambiguous; until you know that two norms on $ \\mathbb{R}^m$ generate the same topology it's not clear whether \"compact\" means \"compact with respect to the given norm\" or \"compact with respect to the usual norm.\" In the former case, the ball is automatically closed with respect to the given norm, since the norm is continuous with respect to itself. In the latter case, part b) implies that any two topologies generated by a norm are equivalent." } { "Tag": [], "Problem": "Is there such a number that if you increase and decrease it by 15, the digits are still the same as the original number (just arranged)?", "Solution_1": "probably, but I haven't figured out what it is ;) .", "Solution_2": "Well if you increase it and decrease it by 15 you still have the same number ... right?", "Solution_3": "To reverse the digits of a two-digit number, you have to increase/decrease it by a multiple of $9$.\r\nTo reverse the digits of a three-digit number, you have to increase/decrease it by a multiple of $99$.\r\nTo reverse the digits of a four-digit number, you have to increase/decrease it by a multiple of $999$.\r\nAND SO ON...\r\n\r\nThus the problem has no solution.", "Solution_4": "You do not have to reverse the digits, just rearrange them.", "Solution_5": "Oh yeah, but either way my point is that no matter how large the number is, to rearrange the digits in any way you have to increase/decrease the number by a multiple of $9$.", "Solution_6": "When I first did the problem, I didn't think that there was a solution either.", "Solution_7": "[quote=\"tarquin\"]Well if you increase it and decrease it by 15 you still have the same number ... right?[/quote]\r\n\r\nI don't get it. I think tarquin's right.", "Solution_8": "I give up. Forget it everyone", "Solution_9": "I can prove that there is no possible solution. \r\n\r\nFirst, no number under 100 will work, because to rearrange the digits of one of these numbers, you must reverse them. Second, any number between 15 and 85 mod 100 will not work, because the digits beyond the hundreds digits will not change, and it will follow the same rules as numbers under one hundred. Finally, numbers between 85 and 99 and 0 to 15 mod 100 will also not work, because, if there is a number, it will work for both the positive and negative parts. Therefore, you will have at least one number of the resulting 3, that is between 15 and 85 mod 100, that must work for either the positive or the negative. Since none of these numbers works for either, we can conclude that no numbers at all will work." } { "Tag": [ "ratio", "trigonometry", "geometry proposed", "geometry" ], "Problem": "Let $ ABC$ be a triangle and $ P$ is a point lie on bisector of $ A$, let $ A'B'C'$ be pedal triangle of $ P,PA\\cap B'C' \\equal{} I,PA'\\cap B'C' \\equal{} M$ prove that $ \\frac {B'C'}{IM}\\sin\\frac {|B \\minus{} C|}{2} \\equal{} 2\\frac {b \\plus{} c}{a}\\cos\\frac {A}{2}$.", "Solution_1": "Hi is it nice :blush: ?" } { "Tag": [ "symmetry" ], "Problem": "Find the positive reals $ x_{1},...,x_{10}$ satisfying the sytem of equations $ (x_{1},...,x_{k})(x_{k}+...,x_{10})=1$ $ k=1,...,10$", "Solution_1": "[quote=\"mdk\"]Find the positive reals $ x_{1},...,x_{10}$ satisfying the sytem of equations $ (x_{1},...,x_{k})(x_{k}+...,x_{10})=1$ $ k=1,...,10$[/quote]\r\nPlease edit your problem.", "Solution_2": "The problem looks correct to me", "Solution_3": "Maybe he means the shortage of $ +$ signs.", "Solution_4": "Let $ s_{k}=x_{1}+x_{2}+...+x_{k}$; $ s_{10}=t$\r\n\r\n$ x_{1}=\\frac{1}{t}=x_{10}$\r\n\r\n$ \\left(\\frac{1}{t}+x_{2}\\right)\\left(t-\\frac{1}{t}\\right)=1$\r\n\r\n$ x_{2}=\\frac{1}{(t)(t^{2}-1)}$\r\n\r\n$ \\left(\\frac{1}{t}+\\frac{1}{(t)(t^{2}-1)}+x_{3}\\right)\\left(t-\\frac{1}{t}-\\frac{1}{(t)(t^{2}-1)}\\right)=1$\r\n\r\n$ x_{3}=\\frac{1}{(t)(t^{2}-1)(t^{2}-2)}$\r\n\r\n$ \\left(\\frac{1}{t}+\\frac{1}{(t)(t^{2}-1)}+x_{3}\\right)\\left(t-\\frac{1}{t}-\\frac{1}{(t)(t^{2}-1)}\\right)=1$\r\n\r\n$ x_{4}=\\frac{1}{(t)(t^{2}-2)(t^{4}-3t^{2}+1)}$\r\n\r\n$ x_{5}=\\frac{1}{(t)(t^{4}-4t^{2}+3)(t^{4}-3t^{2}+1)}$\r\n\r\n\r\nIt follows through symmetry that $ x_{1}=x_{10}$, $ x_{2}=x_{9}$,...,$ x_{j}=x_{11-j}$.\r\n\r\nAdding all those up, we get $ t^{6}-6t^{4}+9t^{2}-2=0$\r\n\r\n$ t^{2}=2\\pm\\sqrt{3},2$\r\n\r\nWe throw out some solutions to get:\r\n$ t=\\frac{\\sqrt{6}-\\sqrt{2}}{2}, \\frac{\\sqrt{6}+\\sqrt{2}}{2}$\r\n\r\nThis gives the solutions:\r\n\r\n$ (x_{1},x_{2},x_{3},x_{4},x_{5})=\\left( \\frac{\\sqrt{6}-\\sqrt{2}}{2}, \\frac{2\\sqrt{2}-\\sqrt{6}}{2}, \\frac{2\\sqrt{3}-3\\sqrt{2}}{6}, \\frac{9\\sqrt{2}-5\\sqrt{3}}{6}, \\frac{3\\sqrt{3}-5\\sqrt{2}}{4}\\right)$\r\n\r\nEDIT: woops; only positive reals.", "Solution_5": "wow, I really didn't see that, my fault. Since I couldn't edit my post, here it goes again\r\n\r\nFind the positive reals $ x_{1},...,x_{10}$ satisfying the sytem of equations $ (x_{1},...,x_{k})(x_{k}+...+x_{10})=1$ $ k=1,...,10$" } { "Tag": [ "function", "calculus", "integration", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ J_{n}(x)$ be a Bessel function.\r\n\r\nShow that:\r\n\r\n1) $ \\int_{0}^{\\infty}J_{n}(x) dx = 1$\r\n\r\n2) $ \\int_{0}^{\\infty}\\frac{J_{n}(x)}{x}dx = \\frac{1}{n}$", "Solution_1": "Perhaps use the integral form definition for the Bessel functions to get a double integral that may be easier to work with?", "Solution_2": "I was able to show this only for odd positive integer $ n$.\r\n\r\nStep 1)\r\n\r\n\\begin{eqnarray*}\\int_{0}^{\\infty}J_{1}(x) \\; dx & = & \\lim_{s\\to0^{+}}\\int_{0}^{\\infty}J_{n}(x) e^{-sx^{2}}\\; dx\\\\ & = & \\lim_{s\\to0^{+}}\\int_{0}^{\\infty}\\sum_{l=0}^{\\infty}\\frac{(-1)^{l}}{l!(1+l)!}\\frac{x^{2l+1}}{2^{2l+1}}e^{-sx^{2}}\\; dx\\\\ & = & \\lim_{s\\to0^{+}}\\sum_{l=0}^{\\infty}\\frac{(-1)^{l}}{l!(1+l)!}\\frac{1}{2^{2l+1}}\\int_{0}^{\\infty}x^{2l+1}e^{-sx^{2}}\\; dx\\\\ & = & \\lim_{s\\to0^{+}}\\sum_{l=0}^{\\infty}\\frac{(-1)^{l}}{l!(1+l)!}\\frac{1}{2^{2l+1}}\\int_{0}^{\\infty}\\left( \\frac{t}{s}\\right)^{l}e^{-t}\\; \\frac{dt}{2s}\\quad \\quad (\\text{where }t = sx^{2})\\\\ & = & \\lim_{s\\to0^{+}}\\sum_{l=0}^{\\infty}\\frac{(-1)^{l}}{(1+l)!}\\frac{1}{(4s)^{l+1}}\\\\ & = & \\lim_{s\\to0^{+}}\\int_{0}^{1/(4s)}e^{-x}\\; dx \\\\ & = & 1 \\end{eqnarray*}\r\n\r\n\r\n\r\nStep 2) For $ n \\geq 2$, the differential equation $ x^{2}J_{n}''(x)+x J_{n}'(x)+(x^{2}-n^{2}) J_{n}(x) = 0$ implies that\r\n\r\n$ \\int_{0}^{\\infty}J_{n}''(x) \\; dx+\\int_{0}^{\\infty}\\frac{J_{n}'(x)}{x}\\; dx+\\int_{0}^{\\infty}\\left( 1-\\frac{n^{2}}{x^{2}}\\right) J_{n}(x) \\; dx = 0$\r\n\r\nor\r\n\r\n$ \\int_{0}^{\\infty}J_{n}(x) \\; dx = (n^{2}-1)\\int_{0}^{\\infty}\\frac{J_{n}(x)}{x^{2}}\\; dx \\quad \\quad \\quad \\cdots \\; (*)$\r\n\r\nUsing the relation $ \\frac{d}{dx}[x^{n}J_{n}(x)] = x^{n}J_{n-1}(x)$ and integration by parts,\r\n\r\n$ \\int_{0}^{\\infty}\\frac{J_{n}(x)}{x^{k}}\\; dx \\; = \\; (n+k+1)\\int_{0}^{\\infty}\\frac{J_{n+1}(x)}{x^{k+1}}\\; dx$\r\n\r\nfor $ 0 \\geq k \\geq n$. Plugging it to $ (*)$ gives $ \\int_{0}^{\\infty}J_{n-2}(x) \\; dx = \\int_{0}^{\\infty}J_{n}(x) \\; dx.$", "Solution_3": "Number 2) can be solved as follows:\r\n\r\n$ \\int_{0}^{\\infty}\\frac{f(t)}{t}dt=\\int_{0}^{\\infty}F(s)ds$, where \r\n\r\n$ F$ is the Laplace transform of $ f$.\r\n\r\nNow, we know that the Laplace transform of the Bessel function $ J_{n}$ is given by:\r\n\r\n$ F(J_{n})(p)=\\frac{(-p+\\sqrt{p^{2}+1})^{n}}{\\sqrt{p^{2}+1}}$, and hence\r\n\r\n$ \\int_{0}^{\\infty}\\frac{J_{n}(x)}{x}dx=\\int_{0}^{\\infty}\\frac{1}{(p+\\sqrt{p^{2}+1})^{n}\\sqrt{p^{2}+1}}dp$\r\n\r\n$ =\\frac{-1}{n(p+\\sqrt{p^{2}+1})^{n}}\\biggm|_{0}^{\\infty}=\\frac{1}{n}$." } { "Tag": [ "linear algebra", "matrix", "Gauss", "algebra", "system of equations", "linear algebra unsolved" ], "Problem": "I have to solve system of equations using Gauss-Zeidel method, this method is only appliable to systems whose matrix are diagonally dominant (http://en.wikipedia.org/wiki/Diagonally_dominant_matrix) . My system of equations is:\r\n\r\n2.7*x1+3.3*x2+1.3*x3 = 2.1;\r\n3.5*x1-1.7*x2+2.8*x3 = 1.7;\r\n4.1*x1+5.8*x2-1.7*x3 = 0.8;\r\n\r\nAs you see it's matrix is not diagonally dominant, how can i transform it, that i could solve using Gauss-Zeidel method wit it. Thank you.", "Solution_1": "Your mean?" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Does there exist a continuos $f: R \\rightarrow R$ such that:\r\n$f(x)$ is rational iff $x$ is rational?", "Solution_1": "Yup! $f(x)=x$ :D. Maybe you meant $f(x)$ is [b]irrational[/b] iff $x$ is rational? In that case the answer is no:\r\n\r\n$f$ must take two different irrational values, so it takes all the irrational values in an interval because it's continuous so it has Darboux property. This means that it takes $\\mathbb Q$ to $\\mathbb {(R\\setminus Q)}\\cap (a.b)$, and this can't be done because the second set has continuum cardinality and the first set is countable.", "Solution_2": "You see f(x)+x is continious and also it can not get rational numbers.Contradiction", "Solution_3": "Sorry for revising this topic after an year or so, but I think there is an easier solution to this problem (please let me know if I am repeating what someone else has already written in another topic):\r\n\r\nDoes there exist a continuous function $f: \\mathbb{R}\\to\\mathbb{R}$ such that $f(x) \\in \\mathbb{Q}$ if and only if $x \\in \\mathbb{R}\\setminus \\mathbb{Q}$?\r\n\r\nSuppose such a function $f(x)$ exists. Consider the function $f(f(x))$. We have that $f(f(x))\\in \\mathbb{Q}$ if and only if $x \\in \\mathbb{Q}$. Therefore the functions $g(x): = f(x) - f(f(x))$ and $h(x) = f(x) + f(f(x))$ enjoy the property $g(x), h(x) \\in \\mathbb{R}\\setminus\\mathbb{Q}$ for all $x\\in\\mathbb{R}$. And since $g(x)$ and $h(x)$ are continuous functions, they must be identically equal to (irrational) constants, which means that $f(x) = \\frac{1}{2}(g(x) + h(x))$ must be identically equal to a constant, which is the desired contradiction.\r\n\r\n\r\nCheers," } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$a,b,c,d,e$ are positive real numbers, such that $a^2+b^2+c^2+d^2+e^2\\le 1$. Prove that $\\frac{1}{1+ab}+\\frac{1}{1+bc}+\\frac{1}{1+cd}+\\frac{1}{1+de}+\\frac{1}{1+ea}\\ge \\frac{25}{6}$", "Solution_1": "[quote=\"Beat\"]$a,b,c,d,e$ are positive real numbers, such that $a^2+b^2+c^2+d^2+e^2\\le 1$. Prove that $\\frac{1}{1+ab}+\\frac{1}{1+bc}+\\frac{1}{1+cd}+\\frac{1}{1+de}+\\frac{1}{1+ea}\\ge \\frac{25}{6}$[/quote]\r\nVery easy.\r\nWe have $1\\ge a^2+b^2+c^2+d^2+e^2\\ge ab+bc+cd+de+ea$\r\nAnd $\\frac{1}{1+ab}+\\frac{1}{1+bc}+\\frac{1}{1+cd}+\\frac{1}{1+de}+\\frac{1}{1+ea}\\ge\\frac{25}{5+ab+bc+cd+de+ea}\\ge\\frac{25}{6}$ :lol:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $ S(n)$ denote the sum of the digits of a natural number $ n$ (in base $ 10$). Prove that for every $ n$, $ S(2n) \\le 2S(n) \\le 10S(2n)$. Prove also that there is a positive integer $ n$ with $ S(n)\\equal{}1996S(3n)$.", "Solution_1": "Already posted. It is trivial that S(ax)\u2264aS(x) for all a in {1,2,...,9} so S(2n)\u22642S(n)=2S(10n)=2S(5*2n)\u2264(2*5)S(2n)=10S(2n).\r\nChoose n:=[100*(10^k+2)/3]+6 so 1996S(3n)=S(n).", "Solution_2": "[quote=bboypa]Already posted.[/quote]\n\nWhere\n", "Solution_3": "[quote=bboypa]2S(n)=2S(10n)[/quote] \nwrong)\n\n", "Solution_4": "bump this", "Solution_5": "[quote=mistakesinsolutions][quote=bboypa]2S(n)=2S(10n)[/quote] \nwrong)[/quote]\n\nWhy is it wrong?\nThis is correct because $S(n) = S(10n)$ obviously" } { "Tag": [ "quadratics" ], "Problem": "12Xcubed - 61Xsquared+2x + 30 = 0 \r\n\r\nRoot = -2/3\r\n\r\nPlease Help!", "Solution_1": "hello, do you want to solve the equation $ 12x^3\\minus{}61x^2\\plus{}2x\\plus{}60\\equal{}0$?\r\nSonnhard." } { "Tag": [ "modular arithmetic", "LaTeX" ], "Problem": "Find all positive integer solutions to $ 7^x\\minus{}3^y\\equal{}4$.", "Solution_1": "x=y=1 is a solution, I claim there are no others.\r\n\r\nConsider the equation mod 8. $ ( \\minus{} 1)^x \\minus{} 3^y\\equiv 4$ gives x,y odd.\r\n\r\nNow, if $ y\\ge 2$ consider the equation mod 9. $ 7^x\\equiv 4$ so a quick check shows that 7^x cycles 7,4,1,7,.. so x is congruent to 0 mod 3. Therefore x is congruent to 3 mod 6. Let x=3+6k with $ k\\ge 0$.\r\n\r\nMod 13, we have $ 7^{3 \\plus{} 6k} \\equiv 5,8$, and $ 3^y \\equiv 1,3,9$ The only pair that subtracts to 4 mod 13 is (5,1), so $ 3^y \\equiv 1$ mod 13. A quick check shows that 3^y cycles 1,3,9,1,3,9,... so x is congruent to 0 mod 3.\r\n\r\nLet x=3m with $ m\\ge 1$.\r\n\r\n$ (7^{1 \\plus{} 2k})^3 \\minus{} (3^m)^3 \\equal{} p^3 \\minus{} q^3 \\equal{} (p \\minus{} q)(p^2 \\plus{} pq \\plus{} q^2) \\equal{} 4$ so $ p^2 \\plus{} pq \\plus{} q^2|4$ but\r\n$ p^2 \\plus{} pq \\plus{} q^2 \\ge q^2 \\ge 9$ so we have a contradiction!", "Solution_2": "You can use \\equiv $ \\equiv$ instead of \\cong $ \\cong$ for congruences.", "Solution_3": "Thx, post edited.", "Solution_4": "Another tip: $ 7 \\equiv 3 \\pmod{4}$ is a good way to include the modulus in the $ \\text{\\LaTeX}$ code." } { "Tag": [ "videos" ], "Problem": "Does anybody here play DotA (Defense of the Ancients)? Discuss.", "Solution_1": "ya i play a wicked shadow shaman, witch doctor, and axe in league play", "Solution_2": "I've heard the song... I may sound like an idiot but their is a song called DotA by Basshunter and the music video is all of them playing DotA.", "Solution_3": "Oh... so thats what Dota is. I thought it was Dork_of_the_ages or something", "Solution_4": "I play DotA. My favorite hero is Rikimaru.", "Solution_5": "[quote=\"echen\"]I've heard the song... I may sound like an idiot but their is a song called DotA by Basshunter and the music video is all of them playing DotA.[/quote]\r\n\r\nDotA is the most popular song in FFR. (I wonder if the reason for that is the game... the song isn't that good)", "Solution_6": "Hm, I really suck and I'm not really experienced, but I kinda like silencer. Glaives is awesome for harassing and for dps, and global silence and pretty awesome. Although I have to say, the other two skills are incredibly useful (for last word, a lot of spells are out of 700 range, and even if they aren't they can just animation cancel it).", "Solution_7": "I play a wide variety of heroes. My favorite and the best has to be anti-mage though. I also like broodmother for mad pushing, rooftrellen and pandaren brewmaster for tanking, mortred for late-game carrying. (Mortred got really better with stifling dagger whereas the old shadow strike was virtually useless)", "Solution_8": "I'm playing regulary with some friends since half a year, but about 2 till 6 human players only. I dislike some stressful heros (especially those that have lot of clones/minions one needs to take too much care of, worst one being Geomancer), but there is no hero I like most (Ursa Warrior was once my favorite, but others get better at time).", "Solution_9": "DotA is extremely popular with the Asians primarily at my high school; I believe one crafty one placed DotA on the school network (hidden among teacher files) when he hacked some of the higher level accounts. (Along with Counter Strike :P ). \r\n\r\nI've never played Warcraft, actually - because my dad never buys computer games, but mostly it's because FPS's are better? :D", "Solution_10": "glaives suck now in the game i play :P", "Solution_11": "I play a different hero every time.\r\nbut my favorites have been\r\n-Traxex(Drow Ranger)\r\n-Clinks(Bone Fletcher)\r\n-Puck(idk.)\r\n-Ursalis(Ursa)\r\n-Axe(idk)\r\n\r\nI also have seperate builds for each kind of hero\r\n\r\n^*-Agility-*^\r\n=-Ursa\r\n=-Troll\r\n=-Drow\r\n=-B. Fletcher\r\n-Bracer.\r\n-Wraith Band.\r\n-Healing Salve\r\n-Perseverance\r\n-sell wraith band\r\n-power treads\r\n-Helm of Dominator(or Vladmir's Offering)\r\n-Yasha\r\n-Sell Bracer\r\n-Sange\r\n-SnY\r\n-Butterfly\r\n-Radiance.\r\ni pretty much destroy everyone after that :D\r\n\r\n^*-Strength-*^\r\n=-Axe\r\n=-Crixalis\r\n-Vanguard\r\n-Boots of Travel\r\n-Vladmir's Offering(or Helm of Dominator.)\r\n-BattleFury\r\n-Monkey King Bar\r\n-SnY\r\nthis is a very professional build,\r\nyou're going to need early kills\r\nand first bloods to be able to afford.\r\nbut if you manage to get all of these\r\nyou will dominate the game.\r\n\r\n^*-Intelligence-*^\r\n-Pugna\r\n-Invoker(SO HARD TO USE.)\r\nI don't use intelligence that much.\r\nbut i know a little bit about the\r\nearly build.\r\n\r\n-Helm of Rejuvination\r\n-Bracer\r\n-Perseverance\r\n-Oblivion Staff\r\n-Mystic Wand\r\n-i don't really know.\r\nhaven't got far enough as a int hero.\r\n\r\ni'll post up whatever i can once i get even MORE experienced.\r\n\r\n\r\nRATE ME HIGH IF THIS HELPED YOU IN ANY WAY! :]\r\n\r\nthanks.", "Solution_12": "[quote=\"AznLeeBoy\"]I play a different hero every time.\nbut my favorites have been\n-Traxex(Drow Ranger)\n-Clinks(Bone Fletcher)\n-Puck(idk.)\n-Ursalis(Ursa)\n-Axe(idk)\n\nI also have seperate builds for each kind of hero\n\n^*-Agility-*^\n=-Ursa\n=-Troll\n=-Drow\n=-B. Fletcher\n-Bracer.\n-Wraith Band.\n-Healing Salve\n-Perseverance\n-sell wraith band\n-power treads\n-Helm of Dominator(or Vladmir's Offering)\n-Yasha\n-Sell Bracer\n-Sange\n-SnY\n-Butterfly\n-Radiance.\ni pretty much destroy everyone after that :D\n\n^*-Strength-*^\n=-Axe\n=-Crixalis\n-Vanguard\n-Boots of Travel\n-Vladmir's Offering(or Helm of Dominator.)\n-BattleFury\n-Monkey King Bar\n-SnY\nthis is a very professional build,\nyou're going to need early kills\nand first bloods to be able to afford.\nbut if you manage to get all of these\nyou will dominate the game.\n\n^*-Intelligence-*^\n-Pugna\n-Invoker(SO HARD TO USE.)\nI don't use intelligence that much.\nbut i know a little bit about the\nearly build.\n\n-Helm of Rejuvination\n-Bracer\n-Perseverance\n-Oblivion Staff\n-Mystic Wand\n-i don't really know.\nhaven't got far enough as a int hero.\n\ni'll post up whatever i can once i get even MORE experienced.\n\n\nRATE ME HIGH IF THIS HELPED YOU IN ANY WAY! :]\n\nthanks.[/quote]\r\n\r\nI don't recommend Sange n Yasha for any heroes except a select few (like the Juggernaut). Also, you should never build a perserverance unless you are planning to use it for some higher level item.\r\n\r\nPugna can be built in varieties of ways. One of my friend loved to stick a Dagon on him.\r\n\r\nAs for Invoker, here is three items I always use with him:\r\n\r\n-Guinsoo: A must for Kael\r\n-Linken: For defense\r\n-Boots of Travel: mobility" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let\n\\[ \\left(a_nx^n\\plus{}\\dots \\plus{}a_1x\\plus{}a_0\\right)(b_kx^k\\plus{}\\dots \\plus{}b_1x\\plus{}b_0)\n\\equal{} 1\\plus{}x\\plus{}x^2\\plus{}\\dots\\plus{}x^{n\\plus{}k}\\]\nwhere the coefficients $a_i, b_i$ are non-negative reals and $ a_0\\equal{}b_0\\equal{}1$.\n\nProve every $ a_i$ and every $ b_i$ is either $0$ or $1$.", "Solution_1": "any idea? :|", "Solution_2": "Try some complex number substitution perhaps.", "Solution_3": "It is sufficient to show that the $a_i$ and $b_i$ are integers. :D (Easy to see why, I'll leave that to the reader.)\n\nNow, all the roots of both LHS polynomials $a(x)$ and $b(x)$ are $(n+k+1)$th primitive roots. Those roots have certain [url=http://en.wikipedia.org/wiki/Cyclotomic_polynomial]cyclotomic polynomials[/url] as minimal polynomials. As those divide either $a(x)$ or $b(x)$, we can write $a(x)$ and $b(x)$ as products of cyclotomic polynomials, possibly with a constant factor. But as all absolute terms are $1$ and $a_0=b_0=1$, the factor is $1$. And so $a_i,b_i\\in\\mathbb Z$ qed." } { "Tag": [ "group theory", "superior algebra", "superior algebra solved" ], "Problem": "Let $ G$ be a group with this property that for every two elements $ a,b\\in G$, $ (ab)^{n}=a^{n}b^{n}$, which $ n$ is a positive integer. Prove that for every elements like $ a,b\\in G$ the following holds : \r\n\r\n \\[ a^{n}b^{n-1}=b^{n-1}a^{n}\\]", "Solution_1": "$ (ab)^{n}\\equal{} a^{n}b^{n}\\Rightarrow a(ba)^{n\\minus{}1}b \\equal{} a^{n}b^{n}\\Rightarrow (ba)^{n\\minus{}1}\\equal{} a^{n\\minus{}1}b^{n\\minus{}1}\\Rightarrow$\r\n$ (ba)^{n}\\equal{} (ba)\\cdot(a^{n\\minus{}1}b^{n\\minus{}1})$. But $ (ba)^{n}\\equal{} b^{n}a^{n}\\Rightarrow b^{n}a^{n}\\equal{} ba^{n}b^{n\\minus{}1}\\Rightarrow b^{n\\minus{}1}a^{n}\\equal{} a^{n}b^{n\\minus{}1}$.", "Solution_2": "A more interesting problem with $ (ab)^{n}\\equal{} a^{n}b^{n}$ is to show that if $ n$ does not divide $ |G|$ (finite) then all $ n\\minus{}1$ degree elements commute with everything else.", "Solution_3": "This one follows from the fact that $ a\\to a^n$ is group homomorphism and since $ n$ does not divide $ |G|$ it follows that it is isomorphism and hence for any $ a\\in G$ there is some $ u\\in G$ so that $ u^n\\equal{}a$ but $ \\alpha^{n\\minus{}1}$ commutes with $ u^n$ an so commutes with $ a$. But $ a$ was arbitrary chosen so $ \\{\\alpha^{n\\minus{}1}|\\alpha\\in G\\}\\subset Z(G)$" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let $a,b$ integer positive numbers...\r\n\r\nProve that: \r\n\r\n$a^3 b - b^3a = 0 \\mod (6).$\r\nfor all $a,b$\r\n\r\n\r\n\r\nCarlos Bravo :)\r\nLma -PERU", "Solution_1": "$a^3b-b^3a=b(a-1)a(a+1)-a(b-1)b(b+1)$\r\nThe product of the three consecutive numbers is the multiple of $6$, we are done.", "Solution_2": "$a^3b-b^3a=ab(a-b)(a+b)$ first we prove 2 divides it. if one of the numers is even, we are done, otherwise $a-b$ is even so we are equally done. Now we prove 3 divides it. If 3 divides one of the numbers we are done, otherwise one between $a+b$ or $a-b$ is a multiple of $3$ since they are different and they cannot be 1 and 2 mod 3 because it would imply one of $a$ or $b$ to be $0$ mod 3" } { "Tag": [ "topology", "abstract algebra", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $G$ be a path-connected and locally path-connected topological group. $\\mu: G \\times G \\to G$ is its composition law.\r\n1/$\\phi: \\pi_{1}(G,1) \\times \\pi_{1}(G,1)$ sends a couple of (class of) paths $(\\gamma_{1}, \\gamma_{2})$ to a path in $\\pi_{1}(G,1)$ defined by\r\n$\\phi(\\gamma_{1}, \\gamma_{2})(t)=\\mu(\\gamma_{1}(t), \\gamma_{2}(t))$. Show that $\\phi$ is a well-defined morphism from $\\pi_{1}(G,1) \\times \\pi_{1}(G,1)$ towards $\\pi_{1}(G,1)$ (is $\\phi$ sends $(\\gamma_{1}, \\gamma_{2})$ towards $\\gamma_{1}. \\gamma_{2}$ where $.$ is the usual composition law of $\\pi_{1}(G,1)$).\r\n\r\n2/ show that $\\pi_{1}(G,1)$ is abelian.", "Solution_1": "is it [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=abelian+group+topological&t=120436]this[/url] one?", "Solution_2": "I admit I don't really understand what you have tried to show in this post cited above ?", "Solution_3": "well, i admit i didn't spend much time giving explanations, in that topic..\r\n\r\nthen: i think the first point is pretty straightforward (i didn't even try it).\r\nas regarding the second one.. just consider an homotopy between two loops $\\alpha, \\beta$ as a map from $S(I)$* to your space.\r\nnow $S(I)$ is homeomorphic to the square $I\\times I$, and you can choose this homeomorphism to send $[(0,0)]$ (the equivalence class of the point $(0,0)$, which is a point in $S(I)$) to $[0,0]$, and $[(1,1)]$ to $(1,1)$.\r\nfinally, consider the map $H: I\\times I \\to G$, $H: (s,t) \\mapsto \\alpha(s)\\beta(t)$.\r\nusing the homeomorphism defined before, what you get is an omotopy from $\\alpha * \\beta$ (which is the low-right border of your square) and $\\beta * \\alpha$ (which is its left-high border).\r\nby the way, the diagonal is your loop $\\alpha\\cdot \\beta$, where $\\cdot$ is the group operation.\r\n\r\n* $I = [0,1]$, and $S(X)$ is the suspension of $X$, which is $X\\times I$ where points of $X\\times\\{0\\}$ and points of $X\\times\\{1\\}$ are identified (and are two distinct points, not to one single point)." } { "Tag": [ "ratio", "number theory unsolved", "number theory" ], "Problem": "Find all [tex]a[/tex], [tex]b[/tex] integers, if:\r\n[tex](a^2+ab+b^2)/(ab-1)[/tex]\r\nis also an integer.", "Solution_1": "If I am not wrong then the equation: \r\n[tex] \\frac{a^2+b^2+1}{ab-1}=p [/tex]\r\nhave infinitely number solutions.And the least solutions are:\r\nFor [tex] p=3;a=4,b=2 [/tex].\r\nFor [tex] p=6;a=2,b=1 [/tex]. :)", "Solution_2": "edit: silly reply removed :blush:", "Solution_3": "I think it has already been posted. The topic was started by Valentin, if I remember correctly.", "Solution_4": "If you will look more carefully you will see that:\r\n- [tex]a,b[/tex] any integers\r\n(case when they have opposite signs is lost: [tex]\\frac{a^2-ab+b^2}{ab+1}=p[/tex] )\r\n- i asked for a general explicit solution (not only the possible values of that [tex]p[/tex]), but i doubt that it is possible :maybe:" } { "Tag": [ "function", "analytic geometry", "algebra", "polynomial" ], "Problem": "A function $f$ which takes real values, is symmetric about the y-axis satisfies and satisfies for $x,y,x+y \\neq 0:$ \\[f \\left( \\dfrac{1}{x+y} \\right) = f \\left( \\dfrac{1}{x} \\right) + f \\left( \\dfrac{1}{y} \\right) + 2xy - 1.\\] Find all functions $f.$", "Solution_1": "If a function is symmetric about the y-axis, it satisfies [tex]f(x) = f(-x)[/tex] for all x in its domain. Then we have:\r\n[tex]f(\\frac{1}{x+y}) = f(\\frac{1}{x}) +f(\\frac{1}{y}) +2xy-1[/tex]\r\n[tex]f(-\\frac{1}{x+y}) = f(-\\frac{1}{x}) +f(-\\frac{1}{y}) +2xy-1[/tex]\r\nBut we have [tex]f(\\frac{1}{x+y}) = f(-\\frac{1}{x+y}) [/tex], so we get\r\n[tex] f(\\frac{1}{x}) +f(\\frac{1}{-x}) = f(-\\frac{1}{y}) +f(-\\frac{1}{y})[/tex].\r\nThen we get that [tex]f(\\frac{1}{x}) +f(\\frac{1}{-x}) = c[/tex] where c is a constant. Again, we use that [tex]f(x) = f(-x)[/tex] , and then put in [tex]x^{-1} \\longrightarrow x[/tex]. Thus, [tex] f(x) = c/2 [/tex]. But plugging this into the original equations shows there are no such functions.\r\n\r\nMy function equation skills are quite bad, so please correct any errors :D", "Solution_2": "There is a function which satisfies the conditions. Please have another try. :)", "Solution_3": "By inspection, $f(x) = 1/x^2 + 1$ is a solution.", "Solution_4": "Can anyone tell me what I did wrong?", "Solution_5": "[quote=\"tetrahedr0n\"]If a function is symmetric about the y-axis, it satisfies [tex]f(x) = f(-x)[/tex] for all x in its domain. Then we have:\n[tex]f(\\frac{1}{x+y}) = f(\\frac{1}{x}) +f(\\frac{1}{y}) +2xy-1[/tex]\n[tex]f(-\\frac{1}{x+y}) = f(-\\frac{1}{x}) +f(-\\frac{1}{y}) +2xy-1[/tex]\nBut we have [tex]f(\\frac{1}{x+y}) = f(-\\frac{1}{x+y}) [/tex], so we get\n[tex] f(\\frac{1}{x}) +f(\\frac{1}{-x}) = f(-\\frac{1}{y}) +f(-\\frac{1}{y})[/tex].\n[/quote]\r\n\r\nInstead of this, you should have gotten\r\n\\[f \\left( \\frac{1}{x} \\right) - f \\left( -\\frac{1}{x} \\right) =\r\nf \\left( -\\frac{1}{y} \\right) - f \\left( \\frac{1}{y} \\right).\\]\r\nBut you know $f$ is symmetric, so both sides are 0. So this approach doesn't tell you anything.", "Solution_6": "Ah, my solution completely fails. Thank for pointing that out.", "Solution_7": "it's a little confusing using x and y where you state the y-axis. Is this the same y? Can you rewrite it in function of a,b if possible, :)", "Solution_8": "Peter, let me try to clear it up.\r\n\r\nThe cartesian coordinate plane has two lines that define it; the x-axis and the y-axis. They intersect at the origin. The equation of the x-axis is y=0 and the y-axis is x=0. A point in this plane is denoted by (x,y). The x part of (x,y) tells you how far away the point is from the y-axis and the y tells you how far away from the x-axis the point is. \r\n\r\nA function takes x-values and gives back y values. This means that f(x)=y. So if you want to graph the function you would plug in an x value and you would then get a y value. So then you plot the point (x,y). You could keep doing this until you have a suitable graph. \r\n\r\nSo there is a connection between the x's and y's and the y-axis. \r\n\r\nI hope that this helps.\r\n\r\nAsk again if you still don't understand.", "Solution_9": "No need to teach me about the carthesian plane Jolm! :D\r\n\r\nWhat I mean is: he's speaking about f(x) AND about f(y), and that is confusing since y=f(y) for this point, so he uses a dubious notation. Besides, there is [b]no reason at all[/b] even to mention the plane here :)\r\n\r\nJust watch as it goes:\r\nFind all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ which satisfy:\r\n\\[\\forall a \\in \\mathbb{R}: f(a)=f(-a)\\]\r\n\\[\\forall a,b \\in \\mathbb{R}_0, a\\not=-b : f\\left(\\frac{1}{a+b}\\right) = f\\left(\\frac{1}{a} \\right) + f\\left( \\frac{1}{b}\\right) + 2ab - 1\\]", "Solution_10": "I hope you don't mind adding: $a,b \\neq 0.$", "Solution_11": "I was pretty sure that you knew about the cartesian plane :D but I thought that maybe you had a different convention in your country. I am not too familiar with other countries' curricula. I just explained it in the case that that was what you had trouble with. I see what you mean now.", "Solution_12": "[quote=\"Peter VDD\"]\\[\\forall a,b \\in \\mathbb{R}_0, ...\\][/quote][quote=\"orl\"]I hope you don't mind adding: $a,b \\neq 0.$[/quote]\r\n\r\n:roll:\r\n\r\nJolm: yeah I understand, still I though you'd have figured out that when saying f(x)=y then f(y) would get a bit problematic :)", "Solution_13": "From:\r\n$f\\left(\\frac{1}{a+b}\\right) = f\\left(\\frac{1}{a} \\right) + f\\left( \\frac{1}{b}\\right) + 2ab - 1$\r\nSince the R.H.S. has a $2ab$ term and a constant only, can we assume that the function f is always of the form $f(x) = m/x^2 + n$, where $m$ and $n$ are constants?", "Solution_14": "Could you explain that in more detail BH?", "Solution_15": "[quote=\"anipoh\"]From:\n$f\\left(\\frac{1}{a+b}\\right) = f\\left(\\frac{1}{a} \\right) + f\\left( \\frac{1}{b}\\right) + 2ab - 1$\nSince the R.H.S. has a $2ab$ term and a constant only, can we assume that the function f is always of the form $f(x) = m/x^2 + n$, where $m$ and $n$ are constants?[/quote]\r\n\r\nNo, you can never assume that. Just like you cannot assume that a function from N to N will be a polynomial, or anything else. Just assume what is given :)" } { "Tag": [ "search" ], "Problem": "[u][b]Formal Instructions:[/u][/b]\r\nThe aim of the game is to be the last player remaining.\r\nEvery round, there is a time limit.\r\nBefore the time limit ends, all players must submit via PM a number between 0 and 100 to me. There is a specific submission format written below and you must abide by this format or else your submission WILL NOT BE COUNTED.\r\nAnd the end of the time limit (called a round), I tally up the submissions and scores in this fasion:\r\n- I post all the submissions (a fail to submit counts as 100).\r\n- I post what half the average submission was.\r\n- I post all players who submitted under half the average. These players are OUT.\r\n- Everybody gains points equal to their distance from half the average. Note: you do not want points!\r\n- If anybody has over 100 points total, they are declared OUT.\r\n- Submissions for the next round must be in [b]within 24 hours of the previous round[/b]. Yes, that is a new rule.\r\nWhen there is one player remaining, they win.\r\n\r\n[u][b]Submission format (has changed from Game 5)[/u][/b]\r\nSubmission Title: \"[u]Number[/u], [u]Username[/u] (Round [u]#[/u])\r\nSubmission Body: Any comments. If you would like me to read the body, out a * after the Submission Title.\r\n\r\n[u][b]Previous games[/u][/b]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1734033730&t=149694]Game 1[/url]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=151415]Game 2[/url]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=154425]Game 3[/url]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=326399784&t=168049]Game 4[/url]\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170873]Game 5[/url]\r\n\r\n[u][b]Sign-up list (12)[/u][/b]\r\nAstroPhys \r\nHamster1800 \r\nxpmath \r\nperfect628 \r\nshaggy75 \r\nNerd_of_the_Ages \r\nDonkeyKong \r\nRandomdragoon \r\nabacadaea\r\nKlebian\r\nhunter\r\nmz94\r\n\r\nAbove is the signup list from last game. If you would like to be moved off, post. Otherwise, the game starts in 3 days.\r\n\r\n[b][u]Note to returning players[/u][/b]\r\nThe rules have changed. Please read them.", "Solution_1": "ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!", "Solution_2": "Me too.!!!lette1!", "Solution_3": "/in", "Solution_4": "/out.\r\n\r\nThe message is too small. Please make the message longer before submitting.", "Solution_5": "Can I join? If so, SIGN ME UP!!!", "Solution_6": "[u][b]Sign-up list (13)[/u][/b]\r\nAstroPhys \r\nHamster1800 \r\nxpmath \r\nperfect628 \r\nshaggy75 \r\nNerd_of_the_Ages \r\nDonkeyKong \r\nRandomdragoon \r\nabacadaea\r\nKlebian\r\nhunter\r\nmz94\r\nerabel\r\n\r\nWow, big game.... Sign-ups still allowed. The game will start on Saturday.", "Solution_7": "Can I play?", "Solution_8": "[b]Sign-up list (14)[/u][/b]\r\nAstroPhys \r\nHamster1800 \r\nxpmath \r\nperfect628 \r\nshaggy75 \r\nNerd_of_the_Ages \r\nDonkeyKong \r\nRandomdragoon \r\nabacadaea\r\nKlebian\r\nhunter\r\nmz94\r\nerabel\r\nmnmath\r\n\r\nThat's enough. [b]Round 1 Due sometime tomorrow (Saturday)[/b].", "Solution_9": "There are quite possibly some people on that list (from last game) who do not know they are on this list\r\n\r\nCan you just remove them from the game instead of giving them 100s if they don't submit this round?\r\nOtherwise it's pretty cheap...", "Solution_10": "Okay, if anybody does not submit the first round, they are [i]removed from the game[/i].", "Solution_11": "Only 6 submissions....\r\n[b][size=200]ROUND 1 RESULTS[/size][/b]\r\n[hide=\"Submissions\"]Hamster: 37\nKlebian: 30\nxpmath: 30\nperfect: 25\nabacadaea: 25\nshaggy: 20[/hide][hide=\"Half The Average\"]Half the average was a stunning [b]14[/b][/hide][hide=\"Below Half the Average\"]Nobody[/hide][hide=\"Scores\"]Hamster: 24\nKlebian: 17\nxpmath: 17\nperfect: 12\nabacadaea: 12\nshaggy: 7[/hide][hide=\"Over 100\"]Nobody[/hide][hide=\"Players Remaining\"][u](6/14)[/u]\nHamster\nKlebian\nxpmath\nperfect\nabacadaea\nshaggy[/hide]", "Solution_12": "Darn, I was under the impression that the 100s would still count towards the average..(in which case I would have won first round :))", "Solution_13": "[quote=\"miyomiyo\"]Only 6 submissions....\n[b][size=200]ROUND 1 RESULTS[/size][/b]\n[hide=\"Submissions\"]Hamster: 37\nKlebian: 30\nxpmath: 30\nperfect: 25\nabacadaea: 25\nshaggy: 20[/hide][hide=\"Half The Average\"]Half the average was a stunning [b]14[/b][/hide][hide=\"Below Half the Average\"]Nobody[/hide][hide=\"Scores\"]Hamster: 24\nKlebian: 17\nxpmath: 17\nperfect: 12\nabacadaea: 12\nshaggy: 7[/hide][hide=\"Over 100\"]Nobody[/hide][hide=\"Players Remaining\"][u](6/14)[/u]\nHamster\nKlebian\nxpmath\nperfect\nabacadaea\nshaggy[/hide][/quote]\r\n\r\nWait a sec!!! No submissions are counted as 100 no more?? \r\n\r\nI had NOADSO(Northern Oklahoma A...something District String Orchestra) on Sat. , so I couldn't get online and thought you would count me as 100...\r\n\r\nAm I still in the game?\r\nP:S I am submitting this time..", "Solution_14": "[quote=\"DonkeyKong\"]Wait a sec!!! No submissions are counted as 100 no more??[/quote][quote=\"miyomiyo\"]Okay, if anybody does not submit the first round, they are removed from the game.[/quote]\r\n", "Solution_15": "[quote=\"miyomiyo\"][quote=\"DonkeyKong\"]Wait a sec!!! No submissions are counted as 100 no more??[/quote][quote=\"miyomiyo\"]Okay, if anybody does not submit the first round, they are removed from the game.[/quote]\n[/quote]\r\n\r\nBut I really couldn't get on, though...\r\n\r\nIt lasted through 8 AM in the morning and I came back home at 9 PM and had to sleep then....", "Solution_16": "So, like are we playing the game or not?", "Solution_17": "Sorry, I'm too busy to mod this game. :( \r\n\r\nFeel free to make your own HTA games..." } { "Tag": [ "geometry" ], "Problem": "Find:\r\n$\\left[\\frac{a-c}{(b-c)(a-b)}+\\frac{c-b}{(a-b)(c-a)}-\\frac{b-a}{(c-a)(b-c)}\\right]\\cdot\\left(\\frac{a-b}{2}\\right)$\r\n\r\nIn what contest would you see that kind of problem? The solution's quite simple and clever though. Maybe good enough to memorize.", "Solution_1": "[hide]The expression on the left can be somewhat simplified as\n$\\frac{-(c-a)^2-(b-c)^2+(a-b)^2}{(b-c)(a-b)(c-a)}$\nWhen multiplied by (a-b)/2, the a-b in the denominator is canceled out by the one in the numerator, so\n$\\frac{-(c-a)^2-(b-c)^2+(a-b)^2}{2(b-c)(c-a)}$\nExpand the numerator.\n$\\frac{-c^2+2ac-a^2-b^2+2bc-c^2+a^2-2ab+b^2}{2(b-c)(c-a)}$\nAll the squared terms cancel out, leaving\n$\\frac{2ac+2bc-2ab}{2(b-c)(c-a)}$\nWe can divide out a 2 from the numerator and denominator (and expand the denominator to see if anything pops out):\n$\\frac{ac+bc-ab}{(b-c)(c-a)}$\nThe numerator doesn't factor as far as I can tell. So, the above is the final answer.\n[/hide]", "Solution_2": "[quote=\"LynnelleYe\"][hide]\n$\\frac{-c^2+2ac-a^2-b^2+2bc-c^2+a^2-2ab+b^2}{2(b-c)(c-a)}$\nAll the squared terms cancel out, leaving\n$\\frac{2ac+2bc-2ab}{2(b-c)(c-a)}$\n[/hide][/quote]\n\n[hide]You might have missed the fact that the -c^2 terms don't cancel out, they add to -2c^2. Good work, but paraphrasing rusczyk, can you find a slicker way?[/hide]", "Solution_3": "Theres a small mistake there, when its fixed you'll see that it does factorise in the end :)", "Solution_4": "Originally I just expanded and condensed the stuff until I eventually got a solution of:[hide]-1[/hide]\r\n\r\n[quote=\"EFuzzy\"]\nGood work, but paraphrasing rusczyk, can you find a slicker way?[/quote]\r\n\r\nThen I read that. You can let a-c=x, b-c=y, and a-b=x-y and it becomes easier (it is still just manipulating algebra).\r\n\r\nIs there an even slicker way?", "Solution_5": "I would like to know the title of the Chinese book that is the source of the problem. That kind of problem is rather typical of junior-high math problems in China (or Taiwan).", "Solution_6": "I believe no problem like that appear on the real math competitions rarely. But, it's true that many geometry and algebra problems involve clever manipulation of x's and y's or a's,b's, and c's, it's good to play with them.\r\n\r\n :)", "Solution_7": "[hide]No, I don't think the answer is -1. :) [/hide]\n\nHint:\n[hide]In the first term, where will adding a \"-b+b\" term help? Is it legal?[/hide]\r\n\r\nThe title of the book was \u4e2d\u5b66\u6570\u5b66\u6f14\u7b97 \u4e00\u9898\u591a\u89e3", "Solution_8": "[quote=\"EFuzzy\"]The title of the book was \u4e2d\u5b66\u6570\u5b66\u6f14\u7b97 \u4e00\u9898\u591a\u89e3[/quote]\r\n\r\nThank you. I take it that the book regards a problem of this level as a JUNIOR-high math problem, right? It is at a very good level for this Getting Started forum, as it only takes relentless application of simple algebra. (Chinese-language textbooks characteristically apply simple ideas to knotty, multistep problems, so that the learners really get to know the basic ideas thoroughly.) \r\n\r\nI'd be very happy to see you and other participants here post more problems like that as challenge problems to this forum. Thanks for sharing.", "Solution_9": "[quote=\"EFuzzy\"]Find:\n$\\left[\\frac{a-c}{(b-c)(a-b)}+\\frac{c-b}{(a-b)(c-a)}-\\frac{b-a}{(c-a)(b-c)}\\right]\\cdot\\left(\\frac{a-b}{2}\\right)$[/quote]\r\n\r\nHere is my solution.\r\n\r\n[i]Note on moderator edit: I used my moderator powers to put the solution in hidden text, as is preferred in challenge problem threads.--tokenadult[/i]\r\n\r\n[hide]Paying attention to cyclic order of $a,b,c$ ,that is to say $a\\rightarrow b\\rightarrow c\\rightarrow a$ and $(a-b)\\rightarrow (b-c)\\rightarrow (c-a)\\rightarrow (a-b)$\n\n\n$\\left[\\frac{a-c}{(b-c)(a-b)}+\\frac{c-b}{(a-b)(c-a)}-\\frac{b-a}{(c-a)(b-c)}\\right]$\n\n$=-\\frac{c-a}{(a-b)(b-c)}-\\frac{b-c}{(c-a)(a-b)}+\\frac{a-b}{(b-c)(c-a)}$\n\n$=\\frac{-(c-a)^2-(b-c)^2+(a-b)^2}{(a-b)(b-c)(c-a)}$\n\n$=-\\frac{(b-c)^2+(c-a)^2-[{(b-c)+(c-a)}]^2}{(a-b)(b-c)(c-a)}$\n\n$=\\frac{2(b-c)(c-a)}{(a-b)(b-c)(c-a)}$\n\nTherefore the desired answer is 1.[/hide]\r\n\r\nYou can practice more at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=19463[/url]\r\n\r\nkunny", "Solution_10": "A very slick way we can simplify the terms is as follows:\r\n\r\n[hide]Take a look at the first term:\n$\\frac{a-c}{(b-c)(a-b)}$\n\nWe can add a $\\\"-b+b\\\"$ to the numerator without any harm:\n$\\frac{a-b+b-c}{(b-c)(a-b)}$\n\nNow we can seperate this into seperate fractions:\n$\\frac{a-b}{(b-c)(a-b)}+\\frac{b-c}{(b-c)(a-b)}$\n\n$\\frac{1}{b-c}+\\frac{1}{a-b}$\n\nAnd we can do this for all of the first 3 terms![/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Hi. I\u2019m new here. I need help with the following inequality please.\r\n\r\nLet $ a,b,c$ be positive real numbers with $ ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that $ (a \\plus{} b \\plus{} c)^2[1\\plus{}abc(a \\plus{} b \\plus{} c) ]\\geq4$.\r\n\r\nThis is part of another problem that I\u2019m trying to solve. I\u2019ve got the problem down to the above inequality, and if I can solve it I\u2019ll have solved my original problem. :wink:", "Solution_1": "[quote=\"Jane_Bennet\"]Hi. I\u2019m new here. I need help with the following inequality please.\n\nLet $ a,b,c$ be positive real numbers with $ ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that $ (a \\plus{} b \\plus{} c)^2[1 \\plus{} abc(a \\plus{} b \\plus{} c) ]\\geq4$.\n[/quote]\r\n$ (a \\plus{} b \\plus{} c)^2[1 \\plus{} abc(a \\plus{} b \\plus{} c) ]\\geq4\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a^2\\plus{}2ab)\\cdot\\sum_{cyc}(a^2b^2\\plus{}3a^2bc)\\geq4(ab\\plus{}ac\\plus{}bc)^3\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a^4b^2\\plus{}a^4c^2\\minus{}2a^3b^3\\plus{}3a^4bc\\minus{}a^3b^2c\\minus{}a^3c^2b\\minus{}a^2b^2c^2)\\geq0,$ which true by Muirhead.", "Solution_2": "Sorry, what is Muirhead? :(", "Solution_3": "[quote=\"Jane_Bennet\"]Sorry, what is Muirhead? :([/quote]\r\nMuirhead Inequality\r\nif $ a_1, a_2, ..., a_n, b_1, b_2, ....b_n$ are integer number such that:\r\n1) $ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_k \\ge b_1 \\plus{} b_2 \\plus{} ... \\plus{} b_k$ for $ n \\minus{} 1 \\ge k \\ge 1$ and\r\n2) $ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n \\equal{} b_1 \\plus{} b_2 \\plus{} ... \\plus{} b_n$ so:\r\n\r\n$ \\sum_{sym}x_1^{a_1}x_2^{a_2}...x_n^{a_n} \\ge \\sum_{sym}x_1^{b_1}x_2^{b_2}...x_n^{b_n}$ for all $ x_1, x_2, ..., x_n \\in R^{ \\plus{} }$", "Solution_4": "[quote=\"Jane_Bennet\"]Sorry, what is Muirhead? :([/quote]\r\nWithout Muirhead:\r\n$ \\sum_{cyc}(a^4b^2\\plus{}a^4c^2\\minus{}2a^3b^3\\plus{}3a^4bc\\minus{}a^3b^2c\\minus{}a^3c^2b\\minus{}a^2b^2c^2)\\equal{}$\r\n$ \\equal{}\\sum_{cyc}(a^4b^2\\minus{}2a^3b^3\\plus{}a^2b^4)\\plus{}\\sum_{cyc}(a^3\\minus{}abc)abc\\plus{}\\sum_{cyc}(2a^4bc\\minus{}a^3b^2c\\minus{}a^3c^2b)\\equal{}$\r\n$ \\equal{}\\sum_{cyc}(a\\minus{}b)^2a^2b^2\\plus{}\\frac{(a\\plus{}b\\plus{}c)abc}{2}\\cdot\\sum_{cyc}(a\\minus{}b)^2\\plus{}\\sum_{cyc}(a^3\\minus{}a^2b\\minus{}ab^2\\plus{}b^3)abc\\equal{}$\r\n$ \\equal{}\\sum_{cyc}(a\\minus{}b)^2\\left(a^2b^2\\plus{}\\frac{(a\\plus{}b\\plus{}c)abc}{2}\\plus{}(a\\plus{}b)abc\\right)\\geq0.$", "Solution_5": "No! It's very easy. I tell the truth!\r\nI put $ p\\equal{}a\\plus{}b\\plus{}c;q\\equal{}ab\\plus{}bc\\plus{}ca(\\equal{}1);r\\equal{}abc$ \r\nRewrite as follows:\r\n$ p^2(1\\plus{}pr)\\geq{4}$ very simple." } { "Tag": [], "Problem": "Show that no integer of the form 4a(8k + 7) is a sum of three squares.\r\n\r\nIs any integer not of the form 4a(8k + 7) a sum of three squares?", "Solution_1": "I think the last part should be \"... no sum of 3 ...\" ?", "Solution_2": "anyway, a)\r\nno number of form 8k+7 is writable as sum of 3 squares, so the 4^n factor will not chance anthing to this :?\r\n\r\nb) I suppose a is an integer? elseway it's pretty easy\r\nanyway, I think 22 is a good counterexample" } { "Tag": [ "geometry", "rectangle", "induction", "combinatorics proposed", "combinatorics" ], "Problem": "In a $2n\\times 2n$ table we have put some $n\\times1$ rectangles which they don't overlap and we can't put an $n\\times 1$ rectangle anymore. Find the maximum number of uncovered rectangles.", "Solution_1": "Any answer ????", "Solution_2": "I suppose $2n^2-n$ is the answer - right?", "Solution_3": "The answer is n+2.Prove that by induction on general statement", "Solution_4": "[quote=\"Megus\"]I suppose $2n^2-n$ is the answer - right?[/quote]\r\nYes, I think it's the right answer,too :) and we need $2n+1$ rectangles$n\\times 1$to cover it :roll:" } { "Tag": [], "Problem": "Prove that: \\[ \\binom{n}{0}\\plus{}\\binom{n}{1}\\plus{}\\binom{n}{2}\\plus{}\\binom{n}{3}\\plus{}\\ldots\\plus{}\\binom{n}{k}\\plus{}\\ldots\\plus{}\\binom{n}{n}\\equal{}2^n\\] \\[ k \\le n\\]", "Solution_1": "$ (1+x)^n=\\sum_{r=0}^n {}_nC{}_r x^r$.\r\n\r\nThe number of the sub-set which is made from distinct $ n$ elements.", "Solution_2": "[hide=\"Proof 1\"]The LHS is the sum over all $ i$ the number of subsets of $ \\{1, 2, \\ldots, n\\}$ with $ i$ elements, i.e., simply the number of subsets of $ \\{1, 2, \\ldots, n\\}$. However, $ 2^n$ gives this number as well (associate with each subset an n-bit long binary number, with the $ i$th bit corresponding to whether $ i$ is in that subset or not. [/hide]\n\n[hide=\"Proof 2 (slightly overkill. just slightly.)\"]$ e^{2x} \\equal{} 1 \\plus{} \\sum_{k\\equal{}1}^{\\infty} \\frac{2^k x^k}{k!} \\equal{} (e^x)(e^x) \\equal{} \\sum_{k\\equal{}1}^{\\infty} \\sum_{j\\equal{}0}^k \\frac{x^k}{j! (k \\minus{} j)!}$ (by simply multiplying their Taylor series together.) Equating the $ n$th coefficient gives us $ \\frac{2^n}{n!} \\equal{} \\sum_{k\\equal{}0}^n \\frac{1}{k! (n\\minus{}k)!}$, i.e., $ 2^n \\equal{} \\sum_{k\\equal{}0}^n {n \\choose k}$. [/hide]", "Solution_3": "In the #2 post put x=1 and the answer will be OK", "Solution_4": "Yes. :) \r\n\r\nIn short, Binominal Theorem : $ \\boxed{(a+b)^n=\\sum_{k=1}^n {}_nC{}_k a^{n-k}b^k}$", "Solution_5": "[quote=\"hasan4444\"]Prove that:\n\\[ \\binom{n}{0} \\plus{} \\binom{n}{1} \\plus{} \\binom{n}{2} \\plus{} \\binom{n}{3} \\plus{} \\ldots \\plus{} \\binom{n}{k} \\plus{} \\ldots \\plus{} \\binom{n}{n} \\equal{} 2^n\\]\n\n\\[ k \\le n\\]\n[/quote]\r\n\r\nI'm confused, isn't it implied by the ellipsis that $ k \\le n$ ? Also, how could $ \\binom {n}{k}$ ever have $ k > n$? How can you do n choose k if k is greater than n?", "Solution_6": "Yes \"zserf\" I'm also confused by what you are saying I mean logically $ k1, let p be a common prime divisor\r\n\r\non one hand y==1(mod p)\r\n==> y^n+...+y+1==n+1==0(mod p)\r\n\r\non the other hand p|x\r\n\r\ncontradiction because gcd(x,n+1)=1\r\n\r\n(2) from (1) it follows that y^n+...+y+1=z^n for some positive integer z, which is impossible because y^n < y^n+...+y+1 < (y+1)^n\r\n\r\nTherefore there are no positive integer solutions.", "Solution_2": "See also here \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=45697", "Solution_3": "It can be also solved by using Euler's totient theorem very easily. First assume x to be prime or prime to the power of something. Then applying inequality.", "Solution_4": "Suppose $n=p-1$ for any odd prime $p$. Then we have,\n$$x^{p-1} \\equiv 1 \\pmod {p}\\implies y^p-2 \\equiv 0\\pmod{p}$$\nThen we have,\n$$y \\equiv 3 \\pmod {p}$$\nIf $p$ is odd we must have $x$ even implying $y$ even. Contradiction.\nIf $p=2$, we have \n$n<2$. Contradiction.\n\nIf $n=p$, for some odd integer $p$.\n$$x^{p}+1 \\equiv y^{p+1} \\pmod{p}$$\nWe must have $x$ odd.Then, we see that $y$ is even.\n$$x^{\\phi{p}} \\equiv 1 \\pmod{p}$$\nI can't see motivation to proceed. Can anyone help me?", "Solution_5": "Note that this is equivalent to $x^n=y^{n+1}-1=(y-1)(y^n+y^{n-1}+...+y+1)$.\n\n[b]Lemma: [/b] $\\gcd{(y^n+y^{n-1}+...+y+1,y-1)}=\\gcd{(y-1,n+1)}$.\n[b]Proof of Lemma: [/b] Using Euclidian Algorithm, \n\\begin{align*}\\gcd{(y^n+y^{n-1}+...+y+1,y-1)}&=\\gcd{((-y^n+y^{n-1})+y^n+y^{n-1}+...+y+1,y-1)}\\\\&=\\gcd{(y^{n-1}+(y^{n-1}+...+y+1,y-1))}.\\end{align*}\nIt's easy to see that we can continuously replace $y^k$ with $y^{k-1}$ while keeping the $\\gcd$, so we result with $n+1$ (as we started with $n+1$ terms. $\\blacksquare$\n\nFrom the Lemma and the problem condition that $\\gcd{(x,n+1)}=1$, $\\gcd{(y^n+y^{n-1}+...+y+1,y-1)}=1$. We see that $y-1\\mid x^n$, so unless $y-1=1$ or $y^n+y^{n-1}+...+y+1=1$, the latter of which is absurd, there does not exist a solution. So we conclude that $y=2,$ if any solution exists. Now we wish to find solutions to \n$$x^n+1=2^{n+1}.$$ It's obvious that $x$ must be odd for this to be true. But then note that $x\\geq 3\\implies x^n\\geq 2^{n+1}$ for $n\\geq 2$ and $x\\geq 5\\implies x^n\\geq 2^{n+1}$ for $n\\geq 1$, so if any solution exists then (by the problem condition for $n$) $n=2$ and $x=3$ or $x=1$, both of which clearly do not work, and we may conclude." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "area of a triangle", "Heron\\u0027s formula", "geometry proposed" ], "Problem": "Heron's formula expresses the area of a triangle in terms of the lengths of its sides.\r\nI) Is there an analogus formula for the volume of a tetrahedron in terms of area of its faces?\r\nII) Is there an analogus formula for the volume of a tetrahedron in terms of the lengths of its edges that is symmetric in the edges?\r\nTHe Cayley formula\r\n\\[\r\nV=\\frac{1}{2^{3/2}3!}\r\n\\left (\\begin{array}{ccccc}\r\n0 & 1 & 1 & 1 & 1\\\\\r\n1 & 0 & AB^2 & AC^2 & AD^2\\\\\r\n1 & AB^2 & 0 & BC^2 & BD^2 \\\\\r\n1 & AC^2 & BC^2 & 0 & CD^2\\\\\r\n1 & AD^2 & BD^2 & CD^2 & 0\r\n\\end{array}\\right)^{1/2}\r\n\\]\r\ngives the volume of a tetrahedron with vertices A, B, C, D, but is not symmetric in those lengths.", "Solution_1": "For (1), for any triangle there is a tetrahedron with all faces congruent to the triangle, but I don't think these all have same volume.\r\n\r\nFor (2), what do you mean by symmetrical in the edges? The formula should be symmetrical in the vertices, but some permutations of the edges change the volume.", "Solution_2": "There is no such formula in either case. A tetrahedron with one face equilateral with side of lenght 2 and the other three side of length 3 has volume$\\sqrt{23}/3$.Changing the length ( one side 3 and the other 2) changes the volume.The new volume is $3\\sqrt{3}/4$. Then there is not symmetric formula in terms of edge lengths." } { "Tag": [ "geometry", "circumcircle", "angle bisector" ], "Problem": "[color=darkred]Let $ABC$ be a triangle. Denote the points $\\{\\begin{array}{c}B'\\in (AC)\\\\\\ C'\\in (AB)\\end{array}$ so that $BC'=CB'$.\n\nProve that the second intersection $X$ between the circumcircles of the triangles\n\n$ABB'$ and $ACC'$ belongs to the bisector of the angle $\\widehat{BAC}$.[/color]\r\n\r\n[b][u]Please, I wish and another proofs ! Thank you.[/u][/b]\r\n\r\n[hide=\"The my proof.\"]\n[color=darkblue]The quadrilaterals $\\{\\begin{array}{c}ABXB'\\\\\\ ACXC'\\end{array}$ are cyclically $\\Longrightarrow$ $\\{\\begin{array}{c}\\widehat{XB'C}\\equiv\\widehat{XBC'}\\\\\\ \\widehat{XCB'}\\equiv\\widehat{XC'B}\\end{array}\\|$ $\\Longrightarrow$\n\n$\\{\\begin{array}{c}XB'C\\sim XBC'\\\\\\ B'C=BC'\\end{array}\\|$ $\\Longrightarrow$ $XB'C\\equiv XBC'$ $\\Longrightarrow$ $XB'=XB$ $\\Longrightarrow$\n\n$\\widehat{XB'B}\\equiv\\widehat{XBB'}$ $\\Longrightarrow$ $\\boxed{\\ \\widehat{XAB}\\equiv\\widehat{XAC}\\ }$.\n\n[b][u]Remark.[/u][/b] The line which passes by the midpoints of the segments $[BC]$, $[B'C']$\n\nis parallelly with the bisector of the angle $\\widehat{BAC}$ ![/color][/hide]", "Solution_1": "Your proof is very beautiful, after seeing it I can't think of any other way to do this problem... it would be so ugly :( \r\nHere is a diagram so everyone can enjoy your proof:\r\n[img]6919[/img]", "Solution_2": "[color=blue]my proof[/color][hide]\nLet $s1$ the perp. bisector of $CC'$ and $s2$ the perp. bisector of $BB'$ intersect at $X$. \nAs $XC=XC' (X \\in s1)$ and $XB=XB' (X \\in s2)$ and $CB'=BC'$ $\\Delta XCB' \\equiv \\Delta XC'B$. So perpendiculars from $X$ to $CB'$ and to $BC'$ are equal. This means $X$ [b]lies on the angle bisector[/b] of $\\angle BAC$.\nThe perp. bisector of $AX$ intersects $s1$ in $O1$. Clearly $O1$ is equidistant from $C,X,C',A$ and is the center of the circumscribed circle. The same reasoning shows that $O2$ the intersection of perp. bisector of $AX$ with $s2$ is the center of the circumscribed circle around $B,X,B',A$. [/hide]" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find all $ x\\in\\mathbb R$ such that $ \\frac{3}{2}(5^{x}\\plus{}1)\\equal{}2^{x}\\plus{}3^{x}\\plus{}4^{x}$ .", "Solution_1": "Only $ x\\equal{}0$ and $ x\\equal{}1$.", "Solution_2": "[quote=\"Rust\"]Only $ x \\equal{} 0$ and $ x \\equal{} 1$.[/quote]\r\nCan you prove it? :)", "Solution_3": "Consider $ f(x)\\equal{}\\frac{3}{2}(5^{x}\\plus{}1)\\minus{}2^{x}\\minus{}3^{x}\\minus{}5^{x}$ and $ f'(x), (5^{\\minus{}x}f'(x))'.$", "Solution_4": "How do you prove it?I don't know!!! :?:", "Solution_5": "Let \r\n$ f(x) \\equal{}\\frac{3}{2}(5^{x}\\plus{}1)\\minus{}2^{x}\\minus{}3^{x}\\minus{}4^{x},$\r\n$ g(x) \\equal{} 5^{\\minus{}x}f'(x) \\equal{}\\frac{3}{2}ln5\\minus{}ln2 (\\frac{2}{5})^{x}\\minus{}ln3(\\frac{3}{5})^{x}\\minus{}ln4(\\frac{4}{5})^{x}.$\r\nObviosly $ g'(x) > 0\\ \\forall x\\in R$ and $ g(0) < 0,g(\\infty ) \\equal{}\\frac{3}{2}ln 5 > 0$. Therefore exist $ x_{0}$, suth that\r\n$ g(x) > 0\\to f'(x) > 0$ if $ x > x_{0}$ and $ g(x) < 0\\to f'(x) < 0$ if $ x < x_{0}$.\r\nTherefore $ f(x) \\equal{} 0$ have no more 2 solution $ x_{1}< x_{0}$ and $ x_{2}> x_{0}$.\r\nWe have solutions $ x_{1}\\equal{} 0, x_{2}\\equal{} 1$, therefore they are all solutions.", "Solution_6": "$ 3^{x}\\plus{}4^{x}\\plus{}8^{x}\\equal{}7^{x}\\plus{}2^{x}\\plus{}6^{x}$", "Solution_7": "Only $ x\\equal{}0,x\\equal{}1,x\\equal{}2$.", "Solution_8": "[quote=\"Rust\"]Only $ x \\equal{} 0,x \\equal{} 1,x \\equal{} 2$.[/quote]\r\n\r\n??? Why ?\r\nYou ????", "Solution_9": "Solve by anologue. But these case is longer." } { "Tag": [], "Problem": "sooorry,people if i am dumb, but i aint gettin it, will someone please help me \r\n\r\n2 identical buggies each of mass $150 kg$ move one after the other without friction with same velocity $4 m/sec$. A man of mass $m$ rides the rear buggy . At a certain moment the man jumps into the front buggy with a velocity $v$ relative to his buggy. As a result of this process rear buggy stops .If the sum of kinetic energies of man and front buggy just after collison , differs from that just before collison by $2700 joules$,calculate values of $m$ and $v$", "Solution_1": "What you've got here is a system of three nonlinear equations in three unknowns. Two of the unknowns are $m$ and $v.$ I'm not sure what the most efficient choice of the third unknown, but let's say that the final speed of the front buggy is $4+u.$ There are three states of the system:\r\n\r\nState 1: Man riding in back buggy.\r\n\r\nMomentum = $(150+m)\\cdot4+150\\cdot4$\r\n(Energy irrelevant)\r\n\r\nState 2: Man in mid-air between buggies.\r\n\r\nMomentum = $0+m(4+v)+150\\cdot4$\r\nEnergy = $(0 + m(4+v)^2 + 150\\cdot4^2)/2$\r\n\r\nState 3: Man riding in front buggy.\r\n\r\nMomentum = $0+(150+m)(4+u)$\r\nEnergy = $(0+(150+m)(4+u)^2)/2$\r\n\r\nMomentum is conserved; energy is not. Two of our three equations come from conservation of momentum:\r\n\r\n$(150+m)\\cdot4+150\\cdot4=0+m(4+v)+150\\cdot4=0+(150+m)(4+u)$\r\n\r\nwhich can be simplified to $mv=600=(150+m)u.$\r\n\r\nThe energy condition says that\r\n\r\n$(0 + m(4+v)^2 + 150\\cdot4^2)/2=(0+(150+m)(4+u)^2)/2+2700$ or\r\n\r\n$m(4+v)^2+150\\cdot16=(150+m)(4+u)^2+5400.$\r\n\r\nI'll let you wrestle with these three equations, but that's the setup." } { "Tag": [], "Problem": "I've translated the Chinese TST for 2003. Thanks to zhaoli for providing the source.", "Solution_1": "Thanks for your excellent work.\r\nHave you received the TST problems from 1986-2002?\r\nIf you had time, would you please translate them into English and post them in Mathlinks?", "Solution_2": "Hmm. They are already translated. I will post them when I have time. \r\n\r\nThe credit for translating goes to Charmaine Sia and Yue Yang, both extremely successful and experienced IMO contestants. :)", "Solution_3": "If I have spare time, I am going to translate the other problems of China TST 2003.", "Solution_4": "In China training camp, there are 8~9 sets of problems to select IMO contestants.\r\nUsually, only the last two sets having six problems is officially called as TST, the others are called quizes.\r\n :)", "Solution_5": "[quote=\"zhaoli\"]In China training camp, there are 8~9 sets of problems to select IMO contestants.\nUsually, only the last two sets having six problems is officially called as TST, the others are called quizes.\n :)[/quote]\r\n\r\nYou can take your time and translate the TSTs for 2004 and after that. The problem sets I have mostly 6 problems a year. If you sent a more comprehensive problem set to Billzhao he can translate his time. Otherwise it is not worth going for it as I already have it in English. ;)", "Solution_6": "Did you post them? I can't find them...", "Solution_7": "[quote=\"filletwho\"]Did you post them? I can't find them...[/quote]\r\n\r\nNo, not yet. Please be patient. :)", "Solution_8": "I can be patient, very patient. so are they up now? howabout now? jk :D" } { "Tag": [ "induction", "number theory unsolved", "number theory" ], "Problem": "Let {$ a_n$}={$ 2,4,8,1,3,6,...$} be the infinite integer such that $ a_n$ ids the leftmost digit in the decimal of $ 2^n$ and {$ b_n$} ={$ 5,2,1,6,3,1,...$} be the infinite inte integer such that $ b_n$ ids the leftmost digit in the decimal of $ 5^n$ .Prove that foer any block of consecutive terms in {$ a_n$} there is a block of consecutive terms in {$ b_n$} in the reverse order.", "Solution_1": "[hide]It is a well known theorem of Jordan, that given any $ t$ such that $ t$ is not a power of 10, for any finite sequence of digits $ \\{d_i\\}$, there exists an $ n$, such that $ t^n$ begins with that sequence of digits.\n\nUsing this theorem, we assert that if the given sequence is $ \\{a_i,a_{i\\plus{}1},...,a_k\\}$, then there exists an $ n$ such that $ 5^n$ begins with the sequence of digits forming $ 2^k$. Thus, let $ 5^n\\equal{}2^k\\cdot10^m\\plus{}c$, where $ c<10^m$. We claim that the desired sequence is $ \\{b_n,b_{n\\plus{}1},...,b_{n\\plus{}k\\minus{}i}\\}$. We prove that $ b_{n\\plus{}j}$ begins with the sequence of digits forming $ 2^{k\\minus{}j}$ by induction on $ j$. The claim is obvious for $ j\\equal{}0$ since $ b_n\\equal{}2^k\\cdot10^m\\plus{}c$. Now assume that the claim is true for $ j\\equal{}j_0$. Then we have that $ b_{n\\plus{}j_0}\\equal{}2^{k\\minus{}j_0}\\cdot10^{m_0}\\plus{}c_0$ for $ c_0<10^{m_0}$. Notice that \n\n$ b_{n\\plus{}j_0\\plus{}1}\\equal{}5b_{n\\plus{}j_0}\\equal{}\\frac{10}{2}b_{n\\plus{}j_0}\\equal{}2^{k\\minus{}j_0\\minus{}1}\\cdot10^{m_0\\plus{}1}\\plus{}\\frac{10c_0}{2}$\n\nSince $ c_0<10^{m_0}$, we have that $ \\frac{10c_0}{2}<10^{m_0\\plus{}1}$, thus $ b_{n\\plus{}j_0\\plus{}1}$ begins with the sequence of digits forming $ 2^{k\\minus{}j_0\\minus{}1}$ and the claim is proven.[/hide]" } { "Tag": [ "inequalities", "induction", "limit", "factorial", "inequalities proposed" ], "Problem": "Prove that:\r\n\r\n\\[ \\boxed{\\left(1\\plus{}\\frac{1}{1^2}\\right)\\left(2\\plus{}\\frac{1}{2^2}\\right)\\left(3\\plus{}\\frac{1}{3^2}\\right)...\\left(n\\plus{}\\frac{1}{n^2}\\right)<\\sqrt{6} \\cdot n!}\\]", "Solution_1": "Any ideas? It seems to be a hard inequality. I don't think that induction is the best choice.", "Solution_2": "If we divide by $ n!$, we get $ (1 \\plus{} \\frac {1}{1^3})(1 \\plus{} \\frac {1}{2^3})...(1 \\plus{} \\frac {1}{n^3}) < \\sqrt {6}$. Now I think that $ lim_{n \\rightarrow \\plus{} \\infty}LHS \\equal{} 2 < \\sqrt {6}$...\r\nI may be wrong, please correct me in such a case...", "Solution_3": "hello, the searched limit of your product is\r\n$ \\lim_{n \\to \\infty}\\prod_{i\\equal{}1}^{n}1\\plus{}\\frac{1}{i^3}\\equal{}\\frac{1}{(\\frac{1}{2}\\minus{}\\frac{\\sqrt{3}}{2}i)!(\\frac{1}{2}\\plus{}\\frac{\\sqrt{3}}{2}i)!}$\r\nSonnhard.", "Solution_4": "Is the factorial defined for complex numbers? Please provide details.", "Solution_5": "If $ a_n={\\left(2+\\frac{1}{2^{2}}\\right)\\left(3+\\frac{1}{3^{2}}\\right)...\\left(n+\\frac{1}{n^{2}}\\right)}$ and $ b_n={\\left(2-\\frac{1}{2^{2}}\\right)\\left(3-\\frac{1}{3^{2}}\\right)...\\left(n-\\frac{1}{n^{2}}\\right)}$ then the following inequalities are true:\r\n\r\n$ \\boxed{\\star}$ $ a_n \\cdot b_n <(n!)^2$\r\n\r\n$ \\boxed{\\star}$ $ \\frac{a_n}{b_n}<\\frac{3}{2}$" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Hi everybody,\r\n\r\nI'm working on a game.\r\nWe have a board 3x3.\r\nWe want to color (using one color) 5 squares\r\nWe have 126 possibilities but....\r\nI want to eliminate all the symetrical colored squares (3x3). The \"cross\" have no symetry, you can put it in any shape you want you will have only one. But the \"L\" you can shape it four ways, so there is 3 ways to eliminate. \r\nI can do it by checking all the squares (126).\r\nWhat if I have a square with n x n (where n is bigger as 1000)?\r\nIs there any formula to know it?\r\nThnk you for any help", "Solution_1": "Here is a picture of 2 different shapes\r\nEach one have 4 ways to color it" } { "Tag": [ "topology", "real analysis", "real analysis theorems" ], "Problem": "I am afraid that I am somehow misunderstanding the proof of Urysohn's Lemma. The particular book I am using is Gamelin and Greene's \"Introduction to Topology\". \r\n\r\nLet $E,F$ be disjoint closed sets of our normal topological space $X$. The strategy in the text is to construct for each dyadic rational $r \\in (0,1)$ an open set $U_{r}$ such that\r\n$\\overline{U_{r}}\\subset U_{s}\\; \\; \\; 0 < r < s < 1$\r\n$E \\subset U_{r}$\r\n$U_{r}\\subset X\\setminus F$\r\nAnd then we let $f(x) = \\sup\\{r \\, | \\, x \\,\\text{not belonging to}\\, U_{r}\\}$\r\n\r\nThe way this is done is by first choosing $U_{1/2}$ such that \r\n$E \\subset U_{1/2}\\subset \\overline{U_{1/2}}\\subset X\\setminus F$\r\nThen we continue with dyadics of the form\u00a0$\\frac{p}{2^{2}}$ and we get\r\n$E \\subset U_{1/4}\\subset \\overline{U_{1/4}}\\subset U_{1/2}\\subset \\overline{U_{1/2}}\\subset U_{3/4}\\subset \\overline{U_{3/4}}\\subset X\\setminus F$\r\nAnd so on.\r\n\r\nI am trying to understand why we choose dyadics. I read some comment on Wikipedia that the dyadics are somehow a \"small\" set and that's why they are used. But I don't see why the proof doesn't work exactly the same if we just index all rationals (or any countable dense set) in $(0,1)$ and do the exact same thing with one number at a time. It seems to me that everything works out fine if you do this. But then it seems very weird to use dyadics instead of just the rational numbers. Am I misunderstanding something?", "Solution_1": "Dyadic rationals appear naturally when at each step you are inserting a new number between two numbers that are already there." } { "Tag": [ "trigonometry" ], "Problem": "Solve the following equation: $(ax-i\\sqrt{1-x^2})^n+(ax+i\\sqrt{1-x^2})^n=0, \\ a\\in\\mathbb{R}$", "Solution_1": ":? Shouldn't it work for any x?", "Solution_2": "sorry, it was $a\\in\\mathbb{R}$", "Solution_3": "Actually, I was wrong the first time. But by DeMoivre's we get that x = 0.", "Solution_4": "Can you be more explicit :? How I use DeMoivre?", "Solution_5": "Are we solving for x for all n,a?", "Solution_6": "yes", "Solution_7": "So just put $n=1$, and $a \\neq 0$ and pop", "Solution_8": "Well I was actually wrong, but sub $x = \\cos \\theta$ and you have\r\n\r\n$(a \\cos \\theta - i \\sin \\theta )^n + (a \\cos \\theta + i \\sin \\theta )^n = 0$\r\n\r\nShouldn't be hard from there.", "Solution_9": "wow. That's nice!!! Thanks ;)", "Solution_10": "Errrr, problem. That $a$ messes everything up....", "Solution_11": "[quote=\"seamusoboyle\"]Errrr, problem. That $a$ messes everything up....[/quote]\r\n\r\nGood point. Doing too much MOP homework has fried my brain. I think binomial expansion might help..." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "function", "inequalities", "calculus", "derivative" ], "Problem": "I saw this problem in Engel, USAMO 1980; \r\n\r\n$If a,b,c$ are 3 reals that can take any value from 0 to 1 (both inclusive); prove that $(a/(b+c+1))+(b/(c+a+1))+(c/(a+b+1))+(1-a)(1-b)(1-c)$ can take a maximum value of 1.\r\nHe has proved it by the following method:\r\n\r\n\" LHS = f(a,b,c) ; f is strictly convex in each variable, and so is defined on a closed convex cube. \"\r\nI understand this. But then he says, \"thus, the function assumes extremal values at extremal points\". How is this ? \r\n\r\nHe cites some theorem of Weierstrass. Please help. I need to get this cleared quick. Thanks, in advance.", "Solution_1": "If a function is strictly convex then its extrema are not saddle points. In other words, the gradient of the graph doesn't go from positive to zero and then back to positive, it must always go from negative to positive because it is convex.\r\n\r\nTherefore any stationary point will be extremal.", "Solution_2": "can anyone provide a solution which doesnt use function,jensen inequality :maybe:", "Solution_3": "Something nicer posted here:\r\nhttp://www.mathlinks.ro/viewtopic.php?p=758089#758089\r\n\r\nBut am still confused about the double derivative $ f(a,b,c)$\r\n\r\nCan anyone post the complete derivation process? :maybe:" } { "Tag": [ "geometry", "geometric transformation", "reflection", "incenter", "geometry unsolved" ], "Problem": "Let $ O$ be the incenter of triangle ABC. Let $ O_1$, $ O_2$, and $ O_3$ be the reflections of O across $ AB$, $ BC$, and $ CA$ respectively. Also, let $ A'$ be the centre of the excircle of ABC tangent to $ BC$, $ B'$ be the centre of the excircle of ABC tangent to $ CA$, and $ C'$ be the centre of the excircle of ABC tangent to $ AB$.\r\n\r\n1. Prove that $ C'O_1$, $ A'O_2$, and $ B'O_3$ meet at a point.\r\n2. Let the point determined in question 1 be $ H$. Prove that triangles $ A'B'C'$ and $ O_1O_2O_3$ are similar, and that $ \\frac{C'O_1}{O_1H} \\equal{} \\frac{A'O_2}{O_2H} \\equal{} \\frac{B'O_3}{O_3H}$.\r\n\r\nThis looks like it could have been a problem on one of the older Olympiads, but I don't know the solution. It seems trivial at first, but I still couldn't find it.", "Solution_1": "DE//B'C'//I2I3 \r\nthey are concurrent at their homothetic center", "Solution_2": "[quote=\"plane geometry\"]they are concurrent at their homothetic center[/quote]\r\n\r\nSo wait, explain the diagram...? I don't see any concurrent lines, nor do I see the homothetic center...", "Solution_3": "Note that $ I_2I_3//B'C', I_1I_3//A'C', I_1I_2//A'B'$ and we are done!", "Solution_4": "[quote=\"livetolove212\"]Note that $ I_2I_3//B'C', I_1I_3//A'C', I_1I_2//A'B'$ and we are done![/quote]\r\n\r\nAh. I thought that might have been it once you said that...\r\n\r\n...but how is that obvious, besides seeing it through the diagram?\r\n\r\nI know that EF is parallel to I2I3 by definition, but how do you show that EF is parallel to BC? Unless it's something really obvious that I'm missing here.", "Solution_5": "I mean, EF is parallel to B'C'.", "Solution_6": "Never mind, I got it.\r\n\r\n*EDIT* ...wait. no, I [i]didn't[/i]! Argh.\r\n\r\n*EDIT2* Okay, now I have it for real." } { "Tag": [], "Problem": "A cheetah runs at a speed of 70 mph, while a snail crawls at a rate of 0.005 mph. How many miles will a cheetah run in the same amount of time that it takes a snail to crawl 5.28 feet?", "Solution_1": "A cheetah runs at 70/.005 = 14000 times the rate of the snail.\r\n\r\nWhen the snail travels 5.28 feet, the cheetah will travel 14*5.28*1000=14*5280 feet. Since there are 5280 feet in one mile, the cheetah will run 14 miles." } { "Tag": [ "analytic geometry" ], "Problem": "When will the admission test B be posted? I thought it was supposed to be today?\r\n\r\nAlso, is there a way we could postpone getting items (for example recommendations; those will take a while) but still send in the test by the due date? I'm on spring break right now and it will be hard to coordinate everything.", "Solution_1": "I think it is supposed to be posted the 30th...or at least that is what the website said 11:20 AM ET.", "Solution_2": "It is currently online, so don't worry. Also, it looks to be about the same difficulty as the first (which was not hard).", "Solution_3": "yeah it is now up:\r\nhttp://www.awesomemath.org/tests/TestB.pdf", "Solution_4": "nice test...i might apply for this...i ono i have a few other things going on...but awesome math is a strong possibility...\r\n\r\nEDIT: it is 4:35 am right now...but i am pretty sure i [talked about the admission test]!", "Solution_5": "zacacox says: Side effects of talking about the Admission Test before it's over include swelling, muscle soreness, and dryness of mouth. Please consult your doctor before talking about any new Admission Test before it's over.", "Solution_6": "I think I figured out where they [zacacox 1, premature Admission Test banter 0]\r\n\r\nBut I do want to get in...", "Solution_7": "LMAO :rotfl: \r\nyou're hilarious zac\r\n\r\ni think our standards for your start of camp speech have greatly risen. \"DO NOT START A FIRE FOR ANY REASON WHATSOEVER\" never gets old :D", "Solution_8": "[quote=\"Max300\"]LMAO :rotfl: \nyou're hilarious zac\n\ni think our standards for your start of camp speech have greatly risen. \"DO NOT START A FIRE FOR ANY REASON WHATSOEVER\" never gets old :D[/quote]\r\n\r\nrofl i agree!\r\n\r\nGO ZAC!! (even though you were never there at the dorm)", "Solution_9": "lol haha..i love how it reads now", "Solution_10": "[quote=\"PenguinIntegral\"]\nAlso, is there a way we could postpone getting items (for example recommendations; those will take a while) but still send in the test by the due date? I'm on spring break right now and it will be hard to coordinate everything.[/quote]", "Solution_11": "how many problems do we solve for awesome math?", "Solution_12": "[quote=\"not_trig\"][quote=\"Max300\"]LMAO :rotfl: \nyou're hilarious zac\n\ni think our standards for your start of camp speech have greatly risen. \"DO NOT START A FIRE FOR ANY REASON WHATSOEVER\" never gets old :D[/quote]\n\nrofl i agree!\n\nGO ZAC!! (even though you were never there at the dorm)[/quote]\r\n\r\nlol...was anyone else here in zac's cabin?", "Solution_13": "I was - for 1 day. Then I moved to Dvince's place. hey all. nice to talk to you again.\r\n\r\n[hide=\":P\"]\nBest line of AMP 2006: Daniel Vitek to Damien Jiang\n\nDamien, make like a tree and GET THE F*** OUT OF HERE!!!\n\nSecond-best line of AMP 2006: Daniel Vitek to David Vincent\nDude, there's like s*** on the floor. (which there was - I had to clean it up though it wasn't mine :stretcher: :mad: :bomb: :cursing: :censored: :furious: :cleaning: )\n\nokay, so maybe these weren't the best, but they have to be up there...\n[/hide]" } { "Tag": [ "induction", "ratio" ], "Problem": "Let n be a positive integer. A bit string of length n is a sequence of n numbers consisting of 0\u2019s and 1\u2019s. Let f (n) denote the number of bit strings of length n in which every 0 is surrounded by 1\u2019s. (Thus for n = 5, 11101 is allowed, but 10011 and 10110 are not allowed, and we have f (3) = 2,\r\nf (4) = 3.) Prove that f (n) < (1.7)^n for all n.", "Solution_1": "[quote=\"mdk\"]Let n be a positive integer. A bit string of length n is a sequence of n numbers consisting of 0\u2019s and 1\u2019s. Let f (n) denote the number of bit strings of length n in which every 0 is surrounded by 1\u2019s. (Thus for n = 5, 11101 is allowed, but 10011 and 10110 are not allowed, and we have f (3) = 2,\nf (4) = 3.) Prove that f (n) < (1.7)^n for all n.[/quote]\r\n\r\n[hide=\"Hint\"]Establish an induction formula between $ f(n)$, $ f(n\\minus{}1)$ and $ f(n\\minus{}2)$[/hide]", "Solution_2": "Hope this is correct...\r\n[hide]\nLet $ A_{n}$ be the number of bit strings length n that end in a 1.\nLet $ B_{n}$ be the number of bit strings length n that end in a 0.\n\nIt is easy to see that $ A_{n}\\equal{} A_{n\\minus{}1}\\plus{}B_{n\\minus{}1}$ and $ B_{n}\\equal{} A_{n\\minus{}1}$, so:\n\n$ A_{n}\\plus{}B_{n}\\equal{} (A_{n\\minus{}1}\\plus{}B_{n\\minus{}1})\\plus{}A_{n\\minus{}1}$\n$ F(n) \\equal{} F(n\\minus{}1)\\plus{}F(n\\minus{}2)$.\n\nThis implies that F(n) is the $ n^{th}$ fibonacci number, which can be explicitly written as $ \\frac{\\phi^{n}\\minus{}\\overline{\\phi}^{n}}{\\sqrt{5}}$ with $ \\phi$ being the golden ratio $ \\frac{1\\plus{}\\sqrt{5}}{2}$ and $ \\overline{\\phi}$ its conjugate $ \\frac{1\\minus{}\\sqrt{5}}{2}$.\n\n$ \\frac{\\phi^{n}\\minus{}\\overline{\\phi}^{n}}{\\sqrt{5}}<\\phi^{n}$, and we know that $ \\phi < 1.7\\Leftrightarrow\\phi^{n}< 1.7^{n}$, thus $ \\frac{\\phi^{n}\\minus{}\\overline{\\phi}^{n}}{\\sqrt{5}}< 1.7^{n}$.\n\n[/hide]" } { "Tag": [ "least common multiple" ], "Problem": "Determine the number of ordered pairs of positive integers $ (a,b)$ such that the least common multiple of $ a$ and $ b$ is $ 2^{3}5^{7}11^{13}$.", "Solution_1": "[hide]One of $ a$ and $ b$ must be divisible by each of $ 2^{3}$, $ 5^{7}$, and $ 11^{1}3$. The other number can have an exponent less than or equal to the first.\n\nThe factor combinations with powers of 2 are (disregarding order):\n\n$ (2^{0},2^{3})$\n$ (2^{1},2^{3})$\n$ (2^{2},2^{3})$\n$ (2^{3},2^{3})$\n\nThey can be distributed among $ a$ and $ b$ in 7 ways. Similarly, the powers of 5 can be split in 13 ways, and the powers of 11 in 25 ways, for a total of $ 125\\times13\\equal{}1625$ ordered pairs.[/hide]" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c>0$ s.t. $ a\\plus{}b\\plus{}c\\equal{}\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}$. If $ a\\leq b\\leq c$, then:\r\n$ ab^2c^3\\geq1$", "Solution_1": "No solution to this one? :?: :(", "Solution_2": "[quote=\"Inequalities Master\"]Let $ a,b,c>0$ s.t. $ a\\plus{}b\\plus{}c\\equal{}\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}$. If $ a\\leq b\\leq c$, then:\n$ ab^2c^3\\geq1$[/quote]\nSee here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=358621" } { "Tag": [ "trigonometry", "geometry", "3D geometry", "prism", "geometric transformation", "reflection" ], "Problem": "Does anyone here have AP Physics B with the \"College Physics\" text by Serway and Faughn 5th edition? There are 2 problems that i am really struggling with but in order for someone to be able to help me, they'd need to see the diagrams...", "Solution_1": "[quote=\"ROLEX4life\"]Does anyone here have AP Physics B with the \"College Physics\" text by Serway and Faughn 5th edition? There are 2 problems that i am really struggling with but in order for someone to be able to help me, they'd need to see the diagrams...[/quote]I taught freshman physics from the 5th edition of Serway and Faughn. What's your question?", "Solution_2": "I have problems with Questions 38 and 41 from Chapter 22. I don't know how to figure out the indices of refraction and/or the incident angle. Both those problems refer to the diagrams on page 742 just for quick reference. Thanks for your help in advance.", "Solution_3": "[quote=\"ROLEX4life\"]I have problems with Questions 38 and 41 from Chapter 22. I don't know how to figure out the indices of refraction and/or the incident angle. Both those problems refer to the diagrams on page 742 just for quick reference. Thanks for your help in advance.[/quote]Both problems deal with the same phenomenon: Total internal reflection.\r\n\r\nKeep in mind Snell's law: $n_i\\sin\\theta_i = n_r\\sin\\theta_r$\r\n\r\nIf you solve this for $\\sin\\theta_r$, then $\\sin\\theta_r =\\frac{n_i\\sin\\theta_i}{n_r}$\r\n\r\n$\\sin\\theta_r \\not > 1$; therefore, if this condition is violated the light rays cannot pass into the second medium, whatever it might be.\r\n\r\nFor question # 38 Serway and Faughn ask you to calculate the minimum index of refraction for the prism.\r\n\r\nFor which we find $\\sin\\theta_{air} = \\frac{n_g \\sin\\theta_g}{n_{air}} > 1$\r\n\r\nIt is immediately apparent then that $n_g > \\frac{n_{air}}{\\sin\\theta_g}$.\r\n\r\nIn question #41 Serway & Faughn ask you to find the maximum index of the fluid surrounding a prism having $n_g = 1.80$.\r\n\r\nSame problem $\\frac{1.80 \\sin\\theta_g}{n_f} > 1$\r\n\r\nFrom which we see $1.80 \\sin\\theta_g > n_f$\r\n\r\nWhat's left, and I have to leave something for you to do, is a geometry problem.", "Solution_4": "don't u mean that n < (air)/ sin 0(g) ?\r\n\r\nI'm still confused (sorry, i'm stupid). \r\n\r\nI end up with 1.8 = sin (incident angle)/ sin (exit angle) but wouldn't the incident angle be 90 degrees since the diagram shows the ray entering parallel to the normal? I can't figure out the geometry set up. Also, how did you find that n(g) = 1.8? Do you always assume a prism has n = 1.8?\r\n\r\nAlso, could you help me with number 45 part B and number 51? I'm really sorry. I truly want to understand this stuff but it's just too hard. I get the concepts but i can't apply them to problems. Also, my stupid teacher doesn't even go over any problems in class. He ONLY explains the most BASIC concepts. All he has actually taught us is how to draw ray diagrams for mirrors and lenses, the law of reflection, and snell's law. I'm not kidding. You have to figure out the problems for yourselves....\r\n\r\nman i'm so screwed in this class....:(", "Solution_5": "[quote=\"ROLEX4life\"]don't u mean that n < (air)/ sin 0(g) ?\n\nI'm still confused (sorry, i'm stupid). \n\nI end up with 1.8 = sin (incident angle)/ sin (exit angle) but wouldn't the incident angle be 90 degrees since the diagram shows the ray entering parallel to the normal? I can't figure out the geometry set up. Also, how did you find that n(g) = 1.8? Do you always assume a prism has n = 1.8?\n\nAlso, could you help me with number 45 part B and number 51? I'm really sorry. I truly want to understand this stuff but it's just too hard. I get the concepts but i can't apply them to problems. Also, my stupid teacher doesn't even go over any problems in class. He ONLY explains the most BASIC concepts. All he has actually taught us is how to draw ray diagrams for mirrors and lenses, the law of reflection, and snell's law. I'm not kidding. You have to figure out the problems for yourselves....\n\nman i'm so screwed in this class....:([/quote]First, you're not stupid; you're a student. These things are difficult. If you have trouble with them it comes from lack of experience, not dullness of mind.\r\n\r\nSecond, I meant what I wrote. In both cases, the glass is the incident medium. In the first problem the light tries to refract from the glass into air, and in the second problem, from glass into some unknown fluid.\r\n\r\nThe result is that in the first problem $n_g$ greater than some number (its minimum value is that number) and in the second problem you get $n_f$ less than some other number (its maximum value is this other number).\r\n\r\nNo, you don't assume that a prism always has $n = 1.8$. Remember, in the first problem the index of the prism is what you're trying to solve for. Index of refraction is a material property. If the index of a prism was always 1.8 that would mean that all prisms were made of the same material, which obviously isn't true.\r\n\r\nFor your second problem, #41 I believe, the fact that $n_g = 1.8$ was given in the problem. You were supposed to find the maximum index of the surrounding fluid.\r\n\r\nOther problems will have to wait as my copy of S & F is at home, and I'm quite busy at work right now.", "Solution_6": "Ok, thanks for your help and patience with me :P \r\n\r\nbtw i figured out number 51 so you don't need to explain that...", "Solution_7": "[quote=\"ROLEX4life\"]Ok, thanks for your help and patience with me :P \n\nbtw i figured out number 51 so you don't need to explain that...[/quote]S & F, Problem 22.45b, at what value of $n_2$ does total internal reflection cease?\r\n\r\n$n_1\\sin\\theta_1 = n_2 \\sin\\theta_r$\r\n\r\nWhen refraction actually occurs $\\sin\\theta_r \\le 1$\r\n\r\n$\\Rightarrow n_1\\sin\\theta_1 \\le n_2$.\r\n\r\nTotal internal reflection ceases when $n_2 = n_1 \\sin\\theta_1$." } { "Tag": [ "Gauss", "function", "integration", "trigonometry", "complex numbers", "complex analysis", "complex analysis unsolved" ], "Problem": "Find the complex numbers $ z$ for which the series \r\n\\[ 1 \\plus{} \\frac {z}{2!} \\plus{} \\frac {z(z \\plus{} 1)}{3!} \\plus{} \\frac {z(z \\plus{} 1)(z \\plus{} 2)}{4!} \\plus{} \\cdots \\plus{} \\frac {z(z \\plus{} 1)\\cdots(z \\plus{} n)}{(n \\plus{} 2)!} \\plus{} \\cdots\\]\r\nconverges and find its sum.", "Solution_1": "there is many way how to prove that series converges for $ |z| < 1$\r\n for example one can use Gauss formula for Gamma function and obtain that $ z(z + 1)(z + 2)...(z + n) \\approx C(z) n^{z}n!$\r\n and when $ n \\to \\infty$ we have that $ C(z) \\to \\Gamma(z)$\r\n (you can easily make all this a proper proof, I think..)\r\nThe main result is to find sum .\r\nLets denote your sum by $ S(z)$ and consider $ S(z) - 1$\r\nwe can write it as \r\n$ S(z) - 1 = \\sum_{n\\geq 0} \\frac {\\prod_{i = 0}^{n}(z + i)}{(n + 2)!}$\r\nuse Gamma function property that $ \\Gamma(z + 1) = z\\Gamma(z)$ and we see that :\r\n${ \\Gamma(z)\\prod_{i = 0}^{n}(z + i)} = \\Gamma(z + n + 1)$\r\nthus we obtain :\r\n\\[ \\Gamma(z)\\Gamma(2 - z)(S(z) - 1) = \\sum_{n\\geq 0} \\frac {\\Gamma(z + n + 1)\\Gamma(2 - z)}{\\Gamma(n + 3)} = \\sum_{n\\geq 0}\\beta (z + n + 1,2 - z)\r\n\\]\r\n\r\n\\[ = 2\\sum_{n\\geq 0} \\int_{0}^{\\pi/2}\\sin^{2z + 2n + 1}\\alpha \\cos^{3 - 2z}\\alpha d\\alpha\r\n\\]\r\n\r\n\\[ = 2\\int_{0}^{\\pi/2}\\cos^{3 - 2z}\\alpha \\sin^{2z + 1}\\alpha \\sum_{n\\geq 0}\\sin^{2n}\\alpha d\\alpha = 2\\int_{0}^{\\pi/2}\\cos^{1 - 2z}\\alpha \\sin^{2z + 1}\\alpha d\\alpha = \\Gamma(1 + z)\\Gamma(1 - z)\r\n\\]\r\nthus :\r\n\\[ S(z) = \\frac {\\Gamma(1 + z)\\Gamma(1 - z)}{\\Gamma(z)\\Gamma(2 - z)} + 1 = (1 - z)^{ - 1}\r\n\\]", "Solution_2": "It converges at negative integers too... ;)", "Solution_3": "Sure :blush: , not only it converges also for all $ Re z \\leq \\minus{}1$" } { "Tag": [ "vector", "linear algebra" ], "Problem": "I need help in understanding these proofs. Can someone please explain?\r\n\r\nDefine $T: V\\rightarrow V$, a linear transformation mapping $V$ to $V$, where $V$ is a vector space of finite dimension.\r\n\r\na) If $T$ is one-to-one, prove that $T$ is also onto.\r\n\r\nb) If $T$ is onto, prove that $T$ is one-to-one.", "Solution_1": "This is a theorem about matrices. Choose a basis for the $n$-dimensional vector space $V.$ Expressed in terms of that basis, a linear transformation is given by an $n\\times n$ matrix.\r\n\r\nThen the result follows from rank + nullity = dimension. This fundamental theorem of matrices can be proved using the mechanics of gaussian elimination and by counting pivot elements and non-pivot columns.\r\n\r\nThere are some other ways of going through this. You could use the following assertions:\r\n\r\n$T$ is one-to-one iff it maps every independent set to an independent set.\r\n\r\n$T$ is onto iff it maps every spanning set to a spanning set.\r\n\r\nAny two bases for $V$ have the same number of elements.\r\n\r\nTwo side notes:\r\n\r\nIn your statement of the problem, you said that $V$ was a vector space but didn't say over what field. You didn't have to; the theorem is the same, and with the same proof, over all fields.\r\n\r\nYou did say that $V$ was finite dimensional. This is vital, as the statement would be false in an infinite-dimensional space." } { "Tag": [], "Problem": "Solve that equation of system:\r\n\r\n$ \\left\\{ \\begin{array}{l}\r\n x \\plus{} \\frac{{2xy}}{{\\sqrt[3]{{x^2 \\minus{} 2x \\plus{} 9}}}} \\equal{} x^2 \\plus{} y \\\\ \r\n y \\plus{} \\frac{{2xy}}{{\\sqrt[3]{{y^2 \\minus{} 2y \\plus{} 9}}}} \\equal{} y^2 \\plus{} x \\\\ \r\n \\end{array} \\right.$", "Solution_1": "hello, i have only found the solutions $ x\\equal{}0,y\\equal{}0$ or $ x\\equal{}1,y\\equal{}1$.\r\nSonnhard." } { "Tag": [ "number theory", "greatest common divisor" ], "Problem": "if the $ gcd (a,b) \\equal{} 1$, prove that the $ gcd (a\\plus{}b, a\\minus{}b) \\equal{} 1$ or $ 2$.", "Solution_1": "I am no math wiz or anything close to it. I have my own post on here I am waiting for an answer to, Here is as far as I get. Someone wanna pick up where I left off?\r\n\r\n\r\nLet a = 80, b = 83, c = a+b, d = a-b\r\ngcd(a,b) = 1\r\nlet e = gcd(c,a), e = 1\r\nlet f = gcd(d,b), f = 1\r\n\r\ngcd(e,f) = 1, therefore, gcd(c,d) = 1 or restated, gcd(a+b,a-b) = 1\r\n\r\nagain, correct me if I am wrong.", "Solution_2": "Lemma: $ (a,kb)|k*(a,b)$ where $ a$, $ b$, and $ k$ are integers.\r\n\r\nSo $ (a\\plus{}b,a\\minus{}b)\\equal{}(a\\plus{}b,2a)|2*(a\\plus{}b,a)\\equal{}2*(b,a)\\equal{}2$. Then since the gcd divides two, it is equal to one or two.", "Solution_3": "Let the GCD be $ k$\r\nthen $ k|2a$ and $ k|2b$ \r\nas $ gcd(a,b)\\equal{}1$\r\n$ k\\equal{}1 or 2$", "Solution_4": "[hide=\"Another approach\"] [b]Lemma:[/b] If $ ax \\plus{} by \\equal{} d$ then $ \\gcd(x, y) | d$. \n\n[b]Bezout's Lemma:[/b] $ \\forall x, y \\in \\mathbb{Z} : \\exists a, b \\in \\mathbb{Z} : ax \\plus{} by \\equal{} \\gcd(x, y)$.\n\nBut of course,\n\n$ (a \\plus{} b) \\plus{} (a \\minus{} b) \\equal{} 2a$\n$ (a \\plus{} b) \\minus{} (a \\minus{} b) \\equal{} 2b$\n\nSo if $ ac \\plus{} bd \\equal{} \\gcd(a, b) \\equal{} 1$ then\n\n$ (a \\plus{} b)(c \\plus{} d) \\plus{} (a \\minus{} b)(c \\minus{} d) \\equal{} 2$\n\nHence $ \\gcd(a \\plus{} b, a \\minus{} b) | 2$. [/hide]" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Fix two integers $ n > m > 1$, and $ m$ integers $ 1 \\le x_1 \\le x_2 \\le \\ldots \\le x_{m} \\le n \\minus{} 1$ such that $ n \\mid x_1 \\plus{} x_2 \\plus{} \\ldots \\plus{} x_m$.\r\n\r\nProve, or disprove, that there exist always a permutation $ \\{\\sigma(x_1),\\ldots,\\sigma(x_m)\\}$ of $ \\{x_1,\\ldots,x_m\\}$ such that the numbers $ y_j: \\equal{} j \\plus{} \\sigma(x_j)$ [color=orange](edited)[/color] for $ j \\equal{} 1,2,\\ldots,m$ are all different modulo $ n$ [color=orange]and $ n \\nmid y_j$ for all $ j\\equal{}1,2,\\ldots,m$ (edited).[/color]\r\n\r\n\r\n[size=75]Ps. This problem arises from a discussion from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=308048]here[/url] :)[/size]", "Solution_1": "Pick $ n \\equal{} 5$, $ m \\equal{} 4$, and $ \\{x_i\\} \\equal{} 1,1,4,4$. WLOG $ \\sigma(x_1)$ is a $ 1$, then $ \\sigma(x_2)$ must be $ 1$ (otherwise the $ \\{y_i\\}$ hit $ 0$ too early), then $ \\sigma(x_3)$ has to be a $ 4$ and $ y_3 \\equal{} 1$ is repeated. Perhaps the sequence should be strictly increasing?", "Solution_2": "Omg, very sorry. I have wrong copying the definition of $ y_j$ :o \r\n\r\nThe correct statement is:\r\n\r\n[i]\"Fix two integers $ n > m > 1$, and $ m$ integers $ 1 \\le x_1 \\le x_2 \\le \\ldots \\le x_{m} \\le n \\minus{} 1$ such that $ n \\mid x_1 \\plus{} x_2 \\plus{} \\ldots \\plus{} x_m$.\n\nProve, or disprove, that there exist always a permutation $ \\{\\sigma(x_1),\\ldots,\\sigma(x_m)\\}$ of $ \\{x_1,\\ldots,x_m\\}$ such that the numbers $ y_j: \\equal{} j \\plus{} \\sigma(x_j)$ for $ j \\equal{} 1,2,\\ldots,m$ are all different modulo $ n$, and $ n \\nmid y_j$ for all $ j \\equal{} 1,2,\\ldots,m$.\"[/i] :P" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "prove that : $ \\forall n>2: \\exists x_n,y_n: 2^n\\equal{}7x_n^2\\plus{}y_n^2$\r\n\r\n\r\n\r\nsorry if it posted before...", "Solution_1": "See PEN H37" } { "Tag": [], "Problem": "(1) Find the equation of the circle with centered $ (5,\\ 2)$, which touches the line $ 3x\\minus{}4y\\minus{}22\\equal{}0.$ \r\n\r\n(2) Find the range of $ m$ for which a circle $ x^2\\plus{}y^2\\minus{}5\\equal{}0$ has distinct points of intersection with the line $ 2x\\plus{}y\\plus{}m\\equal{}0.$", "Solution_1": "[quote=\"kunny\"](1) Find the equation of the circle with centered $ (5,\\ 2)$, which touches the line $ 3x \\minus{} 4y \\minus{} 22 \\equal{} 0.$ \n\n(2) Find the range of $ m$ for which a circle $ x^2 \\plus{} y^2 \\minus{} 5 \\equal{} 0$ has distinct points of intersection with the line $ 2x \\plus{} y \\plus{} m \\equal{} 0.$[/quote]\r\n\r\n[b](1)[/b] Distance from the center to the line $ \\equal{}|\\frac{3(5)\\minus{}4(2)\\minus{}22}{\\sqrt{3^2\\plus{}4^2}}|\\equal{}3$\r\n\r\nSo the equation : $ (x\\minus{}5)^2\\plus{}(y\\minus{}2)^2\\equal{}3^2$\r\n\r\n$ x^2\\plus{}y^2\\minus{}10x\\minus{}4y\\plus{}20\\equal{}0$\r\n\r\n\r\n[b](2)[/b] Put $ y\\equal{}\\minus{}2x\\minus{}m$ into the equation of the circle.\r\n\r\n$ 5x^2\\plus{}4mx\\plus{}(m^2\\minus{}5)\\equal{}0$\r\n\r\n$ Discrimant \\equal{}(4m)^2\\minus{}4(5)(m^2\\minus{}5)>0$\r\n$ \\Leftrightarrow 25\\minus{}m^2>0$\r\n$ \\Leftrightarrow \\minus{}5 0 \u03ba\u03b1\u03b9 \u03b9\u03c3\u03cd\u03b5\u03b9 [b][u]\u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 x,y >0[/u][/b], ... $ a(x^2 \\plus{} y^2 ) \\plus{} \\frac {b}{x \\plus{} y}\\ge c \\minus{} 2xya$, \r\n\r\n\u03bd\u03b1 \u03b4.\u03bf. ... $ 27ab^2\\ge4c^3$", "Solution_1": "\u039a\u03b1\u03bd\u03ad\u03bd\u03b1 \u03ac\u03bb\u03bb\u03bf \u03c0\u03b5\u03c1\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03cc \u03b3\u03b9\u03b1 \u03c4\u03b1 x,y \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9; :ninja:", "Solution_2": "\u0397 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2...vangelismak \u03b5\u03b9\u03c3\u03b1\u03b9 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2?", "Solution_3": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03bc\u03bf\u03bd\u03bf \u03b1\u03c5\u03c4\u03cc\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03c0\u03b5\u03c1\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 ! \u0394\u03b5\u03bd \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03bb\u03ac\u03b8\u03bf\u03c2 :( \r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03bc\u03b5 \u03c3\u03c5\u03bc\u03b2\u03b1\u03c4\u03b9\u03ba\u03ad\u03c2 \u03bc\u03b5\u03b8\u03cc\u03b4\u03bf\u03c5\u03c2 \u03b4\u03b5\u03bd \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c4\u03b9. \u039c\u03ac\u03bb\u03bb\u03bf\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c6\u03cd\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7\u03c2.", "Solution_4": "\u039d\u03b1 \u03b7 \u03b4\u03b9\u03ba\u03ae \u03bc\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7 \u03c7\u03c1\u1f20\u03c3\u03b7 \u03b1\u03bd\u03ac\u03bb\u03c5\u03c3\u03b7\u03c2 . \u0397 \u03b4\u03bf\u03b8\u03b5\u03af\u03c3\u03b1 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \r\n\r\n$ a(x\\plus{}y)^3\\minus{}c(x\\plus{}y)\\plus{}b\\geq 0$ \r\n\r\n\u0398\u03b5\u03c9\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 $ f(u)\\equal{}au^3\\minus{}cu\\plus{}b$ . \u03a4\u03cc\u03c4\u03b5 $ f(u)\\geq 0$ \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 u>0 . \u0391\u03c5\u03c4\u03ae \u03b7 \u03c3\u03c5\u03bd\u03ac\u03c1\u03c4\u03b7\u03c3\u03b7 \u03b3\u03b9\u03b1 u>0 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03b9\u03ac\u03b6\u03b5\u03b9 \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf \u03c3\u03c4\u03bf $ u_1\\equal{}\\sqrt{\\frac{c}{3a}}$ . \u038c\u03bc\u03c9\u03c2 \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 $ f(u_1)\\geq 0$ \u03ac\u03c1\u03b1 \r\n\r\n$ \\sqrt{\\frac{c}{3a}}(c\\minus{}\\frac{c}{3})\\leq b$ \u03ae \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b1 $ 27ab^2\\geq 4c^3$ :) \r\n\r\n\r\n***\r\nOne line solution by Michael Savvas uses only AM-GM :wink:", "Solution_5": "One line solution ? :wow: \r\n\r\n\u0398\u03b1 \u03b4\u03ce \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c4\u03b9 ... :mad: \r\n\r\n :P" } { "Tag": [ "AMC", "AIME", "geometry", "topology", "inequalities", "Ross Mathematics Program", "Harvard" ], "Problem": "From what I heard, all people who scored over 50 in COMC gets a pamphlet to 'try out' for the Math Camp in Maine.\r\n\r\nDid anyone actually go there? I'm curious about the level of math they teach there.", "Solution_1": "Got it last year because of AIME I think, but I didn't go. Anyways, I know someone (Chen Huang) who knows someone (Andrew Mao) who went there. lol\r\nHe didn't recommand people to go there.\r\nPlus, as far as I know, the camp costs like 3000 bucks. So....ya... ;) .", "Solution_2": "Er...is this is the big one in the states? http://www.mathcamp.org? Or are you refering to the one that Mr. Griffiths talks about sometimes?", "Solution_3": "I actually went there last summer, I mean the big one in the states :P \r\nIt's actually a pretty cool math camp which lasts for over a month and has a variety of activities like hiking and others.\r\nThey teach quite a lot of topics too including abstract algebra, projective geometry, number theory, topology and many others with cool sounding names. I'm sure these stuff never get taught in school, (Canadian schools anyways), and they are more university oriented. But there's also introductory and olympiad problem solving that teach you mostly how to solve equations and medium-hard inequalities respectively.", "Solution_4": "Yeah, if you're looking at the big one in the states, Ti (remember her Jenn? ) told me it was totally worth it. \r\n\r\nHaven't heard anything about the other one though.\r\n\r\nMyself, I might apply to the big one, we'll see.", "Solution_5": "[quote]Got it last year because of AIME I think, but I didn't go. Anyways, I know someone (Chen Huang) who knows someone (Andrew Mao) who went there. lol \nHe didn't recommand people to go there. \nPlus, as far as I know, the camp costs like 3000 bucks. So....ya... .[/quote]\r\n\r\nI think Andrew went to the math camp in Maine, while the big US math camp is in Colorado. The one in Colorado costs like $\\$3000, while the one in Maine costs much less. I remember Andrew said he applied for financial assistance, and only paid something like $\\$500. Even without financial assitance, I don't think it will cost over $\\$1500", "Solution_6": "[quote=\"weileaf\"]I think Andrew went to the math camp in Maine, while the big US math camp is in Colorado. The one in Colorado costs like $\\$3000, while the one in Maine costs much less. I remember Andrew said he applied for financial assistance, and only paid something like $\\$500. Even without financial assitance, I don't think it will cost over $\\$1500[/quote]\r\n\r\nUmm.. Do you know the reason Andrew doesn't recommend it to others?", "Solution_7": "I got a flyer for one in Portland, I assumed for my AIME. circa $3000 for a month. It looks interesting, but I would assume it would be more for grade 11s? After all, I am going off to university to do math in 5 months, and if it is a choice between spending 3000 [even though it could work out to more like 1000] and earning 2000 and getting some work experiance, I would pick the latter.", "Solution_8": "[quote=\"jpark\"][quote=\"weileaf\"]I think Andrew went to the math camp in Maine, while the big US math camp is in Colorado. The one in Colorado costs like $\\$3000, while the one in Maine costs much less. I remember Andrew said he applied for financial assistance, and only paid something like $\\$500. Even without financial assitance, I don't think it will cost over $\\$1500[/quote]\n\nUmm.. Do you know the reason Andrew doesn't recommend it to others?[/quote]\r\n\r\nOh yeah, forgot to mention. Dr. Liu told me that at this camp, they often have famous people fly in, but they leave during that same day, and the camp is not as organized as it should be. He told me no one really leaves having learnt anything, he told me definitely not to apply.", "Solution_9": "Ah, it's decided then.. I guess we won't see each other in Maine Sarah.. Oh well.. :D", "Solution_10": "I recently received in the mail a brochure for some math camp at The Ohio State University. From the looks of the brochure, it's 8 weeks of solid number theory. That seems rather unusual... Has anyone heard of it?", "Solution_11": "[quote=\"geoffreynham\"]I recently received in the mail a brochure for some math camp at The Ohio State University. From the looks of the brochure, it's 8 weeks of solid number theory. That seems rather unusual... Has anyone heard of it?[/quote]\r\n\r\nI assume that you are referring to [url=http://www.math.ohio-state.edu/ross/]Ross[/url]. In any case, I have heard that it is a very good program. However, while I did not attend Ross, a senior at CC said that she attended it and she liked it enough to recommend it. I suppose that I should note that she will be attending Harvard next year and did RSI last year. I hope that's a good enough reference for you :D" } { "Tag": [], "Problem": "let, there be $ \\triangle ABC$.\r\nlet, $ \\triangle A'B'C'$ be another triangle. with $ B' \\in AB$ ext. if necessary and similarly $ C' \\in AC$. Also we have that $ AB' \\equal{} AB.k$ and $ AC' \\equal{} AC.k$. prove or disprove if $ B'C' \\equal{} BC.k$", "Solution_1": "[quote=\"communist\"]Let there be $ \\triangle ABC$.\n\nLet $ \\triangle A'B'C'$ be another triangle with $ B' \\in AB$, extended if necessary, and similarly $ C' \\in AC$.\n\nAlso, we have that $ AB' \\equal{} k(AB)$, and $ AC' \\equal{} k(AC)$. Prove or disprove: $ B'C' \\equal{} k(BC).$[/quote]\r\n\r\nInstead of trying to use periods for multiplication, for abbreviations, \r\nand for ending sentences, I did a suggested redo above. This and \r\nsome other changes should help someone read your content\r\nthrough more readily." } { "Tag": [ "email", "\\/closed" ], "Problem": "For the last month or so, my PMs that I sent are never received by anybody, although they are notified that I have PMed them.\r\nAlso, I receive nobody's PMs except cognos599's and undefined117's (although maybe I can receive other people's, too).\r\nFor example, logic812 and MustItAlwaysBeMe have complained about my PM system; apparently they have PMed me, but I did not receive them.\r\nAlso, I've PMed cognos599 (multiple times) and undefined117 (once), but they say they never got my PMs.\r\n\r\nSo, I guess this topic is basically directed to Valentin, or anyone else who can help me. I know it may not be an easy task to fix this; and it's strange how I can only receive some PMs but not others.", "Solution_1": "is your inbox full??\r\nhmm i'll pm you and see if you receive it\r\n\r\nand try making a poll to see if other people have the same problem you do.\r\ngood luck.", "Solution_2": "No, my inbox is completely empty (although I wish it wasn't!).\r\n\r\nAnd I didn't receive your PM, and I don't think that I will.\r\n\r\nEDIT: I received PMs from xoangieexo and levans, and I replied.\r\nHopefully, you two have received those replies.\r\n\r\nCould any admin help out here (don't worry levans, I know that you're an admin too)?", "Solution_3": "[quote=\"MathAndKnowledge\"]So, I guess this topic is basically directed to Valentin, or anyone else who can help me. I know it may not be an easy task to fix this; and it's strange how I can only receive some PMs but not others.[/quote]PM me. We might be on something strange here (that is in case you're not wasting our time and deleting the message from outbox, after the email notification has been sent ...).", "Solution_4": "I was going to edit my previous posts, but for some reason it won't let me.\r\n\r\nAnyways, I have breaking news.\r\n\r\nI am not very smart. I didn't know about the whole outbox thing; I just (for the last month or so, coincidentally [note the sarcasm]) decided to delete outbox messages, because I thought they weren't necessary; the post had been sent, hadn't it? But I guess that I was wrong, and now I know why I haven't been able to PM for the last month or so.\r\nI've already told this to Valentin and levans, so I don't think much further help is necessary.\r\nAll I need to do now is figure out why people like logic812 and MustItAlwaysBeMe can't PM me, but others can. I'll work it out between them and me, so thanks for the concern and willingness to help... now this topic should die." } { "Tag": [ "calculus", "inequalities", "limit", "function" ], "Problem": "prove that a^a > 1/2 with a is a positive real number", "Solution_1": "You can use calculus, it's pretty straightforward.", "Solution_2": "indeed,min( x.ln(x) ) = ?\r\nderivate: ln(x) + 1 = 0 so x = e^-1\r\nx^x = 0,6922 which is clearly greater than 1/2", "Solution_3": "I can easily use calculas to show that the minimum is (1/e)^(1/e), for which the numerical value is that given by Peter, but is there anyway to get the result without pulling out a calculator? My impression (I may be wrong, I have very limited experience in Mathematics Olympiads) is that often you are not allowed to use a calculator.", "Solution_4": "you are NEVER allowed to use a calculator I think :? not in any round\r\n\r\nbut you don't need a calculator to prove that (1/e)^(1/e) is larger than 1/2 do you? :D", "Solution_5": "Hmm... I might, actually. I'm REALLY bad at inequalities stuff.\r\n\r\nI honestly haven't a clue how to go about it.", "Solution_6": "well i just know the value of e by heart, but you can easily see it, as elseway the thing to prove won't be correct :D I know it's weak but it works", "Solution_7": "I have seen many literatures that start out with the following inequality regarding the value of e:\r\n\r\nProve that 2n as n->oo and\r\n\r\n(2) e=1+(1/1!)+(1/2!)+(1/3!)+...\r\n\r\nLet the sequence {en} be defined by en=(1+1/n)n , for all naturals n>0. You can easily show that this sequence is increasing.\r\n\r\nThe first definition can be shown easily (I will only post it if asked).\r\n\r\nUsing (1), 2=e1n2)+... =2+((1/2)/(1-(1/2)))=2+1=3.\r\n\r\nSo we have shown 21/e>1/2 ? Simple.\r\nSince e>2, 2e>22=4>3>e. So 1/e>1/2e. Both sides of the inequality are positive numbers,\r\nthen (1/e)1/e>1/2. No calculators, as I promised. :D", "Solution_9": "[quote=\"3X.lich\"] What is so special about e? How is e defined? [/quote]\r\nmy idea is that e is so special because it is the only positive number for which the function ex derivated remains the same. This can also hold for a definition. Try to prove using this definition the other properties of e (ie being the limit of the sequence en described above by 3x.lich for example)." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c\\geq 0$, no two of which are zero.Prove that:\r\n$ \\frac{1}{a\\plus{}b}\\plus{}\\frac{1}{b\\plus{}c}\\plus{}\\frac{1}{c\\plus{}a}\\geq \\frac{2a}{3a^2\\plus{}bc}\\plus{}\\frac{2b}{3b^2\\plus{}ca}\\plus{}\\frac{2c}{3c^2\\plus{}ab}$.", "Solution_1": "No one can solve? :( :?: \r\n\r\n[hide=\"to arqady\"]arqady, please post your's or Vasc's solution. [/hide]", "Solution_2": "I think it can be killed by SOS", "Solution_3": "[quote=\"Allnames\"]I think it can be killed by SOS[/quote]\r\nAre you sure? Maybe after expanding? :wink:", "Solution_4": "[quote=\"Inequalities Master\"]Let $ a,b,c\\geq 0$, no two of which are zero.Prove that:\n$ \\frac {1}{a \\plus{} b} \\plus{} \\frac {1}{b \\plus{} c} \\plus{} \\frac {1}{c \\plus{} a}\\geq \\frac {2a}{3a^2 \\plus{} bc} \\plus{} \\frac {2b}{3b^2 \\plus{} ca} \\plus{} \\frac {2c}{3c^2 \\plus{} ab}$.[/quote]\r\n\r\nWell, this form is not that difficult, with some efforts, I can do it through expanding.\r\nI also prove this one, which is much fun to expand. \r\n$ a,b,c$ are positive number,\r\n$ \\frac{c^3}{a\\plus{}b}\\plus{}\\frac{a^3}{b\\plus{}c}\\plus{}\\frac{b^3}{a\\plus{}c} \\geq \\frac{2a^4}{3a^2\\plus{}bc}\\plus{}\\frac{2b^4}{3b^2\\plus{}ac}\\plus{}\\frac{2c^4}{3c^2\\plus{}ab}$" } { "Tag": [ "quadratics", "number theory unsolved", "number theory" ], "Problem": "Find all real numbers a such that all solutions to the quadratic equation $ x^2 \\minus{} ax \\plus{} a \\equal{} 0$ are integers.", "Solution_1": "Let $ m, n$ be the roots of the original equation (not necessarily distinct) WLOG, let $ m\\geq n$\r\nThen $ m\\equal{}\\frac{a\\plus{}\\sqrt{a^2\\minus{}4a}}{2}, n\\equal{}\\frac{a\\minus{}\\sqrt{a^2\\minus{}4a}}{2}$.\r\nSince $ m, n$ are integers, $ m\\minus{}n \\equal{} \\sqrt{a^2\\minus{}4a}$ is also a positive integer, name it as $ k$.\r\nSo $ a\\equal{}2m\\minus{}k$ is also an integer.\r\nNow, $ a^2\\minus{}4a \\equal{} k^2$\r\n$ (a\\minus{}2)^2 \\minus{} k^2 \\equal{} 4$\r\n$ (a\\minus{}2\\plus{}k)(a\\minus{}2\\minus{}k) \\equal{} 4$,\r\nas $ a,k$ are integers, $ (a\\minus{}2\\plus{}k)$, $ (a\\minus{}2\\minus{}k)$ have the same parity.\r\nSo (a-2+k) = 2, (a-2-k) = 2 or (a-2+k) = -2, (a-2-k) = -2.\r\nSolve the two simultaneous equations we can get $ (a,k) \\equal{} (4,0)$ or $ (0,0)$.\r\nSo $ a \\equal{} 0, 4$" } { "Tag": [ "algebra", "polynomial", "algebra open" ], "Problem": "Let n > 1 be a positive integer, let E={+1,-1}^n, let s=(s_1,...,s_n) in E,\r\nlet f_s:=s_1 sqrt(x+1)+...+s_n sqrt(x+n) and let N(x)=prod_{s in E} f_s. \r\n\r\nProve that N(x) is a polynomial in Z[x] and determine its degree in x.\r\n\r\nGood luck !\r\n\r\n_________________\r\nThe Owls Are Not What They Seem", "Solution_1": "is this an open question, or a own questions to which you have the answer to?", "Solution_2": "> is this an open question, or a own questions to which you have the answer to?\r\n\r\nI have a partial answer to this question. I did not post it to the\r\n\"Unsolved Problems\" section, because I do not consider that it matches\r\nthe definition \"Problem you couldn't solve and to which you know that\r\nthere is a solution (i.e. a problem from a contest, etc.) but you\r\ndon't know it\". I had the same doubt while thinking to send it to the \"Proposed & Own\r\nProblem\" section, whose definition is \"Problems that you have already\r\nsolved and you are interested in second opinions or solutions\"...\r\nBy the way, what exactly your definition of an \"open question\" is ?\r\n\r\nAlin.", "Solution_3": "an open question is obvioulsy a question that is not closed :)\r\n\r\nhowever a more detalied description could be the following: \r\n\r\n[i]An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not. [/i]", "Solution_4": "I see. Maybe you should put down this definition at\r\nhttp://www.mathlinks.ro/viewforum.php?f=3\r\nas you did for the other themes.\r\n\r\n_________________\r\nThe Owls Are Not What They Seem" } { "Tag": [ "Alcumus", "AMC", "AIME", "MATHCOUNTS", "Support" ], "Problem": "Does Alcumus get harder and harder as you progress?\r\nAs in, does it get to about AIME / just below olympiad level difficulty?", "Solution_1": "It gets to easy-medium AIME level but no harder than that.", "Solution_2": "I thought that Alcumus problems didn't have a specific level :huh:", "Solution_3": "All problems have a level :)\r\n\r\nThe problems start out at easy MATHCOUNTS level and progress to easy-medium AIME level.", "Solution_4": "How hard is easy AIME?", "Solution_5": "Thanks for the replies. \r\n\r\nBtw, has anybody tried out Mildorf's mock AIME's?\r\nShould check out his [url=http://web.mit.edu/~tmildorf/www/]site[/url]\r\n\r\nThe questions on his mock exams are really hard.. supposedly even three times harder than regular AIME questions.. is it a good idea to prepare for contests by solving problems that are harder than the ones on the actual contest??", "Solution_6": "i'd say yes\r\nif you know how to solve problems that are harder than problems you'll be encountering, most likely, you'll know how to solve the problems you will encounter\r\n\r\nbut right before the competition, i suggest doing a little of the competition problems, to get to know about how hard it is", "Solution_7": "alrighty, thanks a bunch :D" } { "Tag": [ "calculus", "derivative", "function", "calculus computations" ], "Problem": "I'll begin with the the lemniscate, who's function is otherwise known as $ (x^2 \\plus{} y^2)^2 \\equal{} x^2 \\minus{} y^2$.\r\n\r\nNow, once I begin the differentiation process, I am given this:\r\n\r\n$ 2(x^2 \\plus{} y^2)(2x \\plus{} 2y\\frac {dy}{dx}) \\equal{} 2x \\minus{} 2y \\frac {dy}{dx}$\r\n\r\nHow would one go about trying to remove the other dy/dx? How should I treat it as?", "Solution_1": "hello, i have $ 2(x^2 + y^2)(2x + 2y\\cdot y^') = 2x - 2y\\cdot y^'$.\r\nSonnhard.", "Solution_2": "You have not differentiated correctly. Using chain rule, you should get $ 2(x^2+y^2)(2x+2yy')=2x-2yy'$...", "Solution_3": "[quote]You have not differentiated correctly. Using chain rule, you should get $ 2(x^2+y^2)(2x+2yy')=2x-2yy'$[/quote]\r\n\r\nUh, how is that different from what I posted?", "Solution_4": "The $ \\minus{}$ sign.", "Solution_5": "[quote=\"Tyl\"]The $ \\minus{}$ sign.[/quote]\r\n\r\nSorry for the typo. However, I still don't know how to treat the dy/dx parts.", "Solution_6": "hello, your answer is still not ok.\r\nSonnhard.", "Solution_7": "[quote=\"Dr Sonnhard Graubner\"]hello, you answer is still not ok.\nSonnhard.[/quote]\r\n\r\nWould you mind explaining how?", "Solution_8": "hello, when you differentiate $ (x^2+y(x))^2$ with respect to $ x$ you must use the chain rule, namely\r\n$ 2(x^2+y(x)^2) \\cdot(2x+2y(x)\\cdot y^'(x))$\r\nSonnhard.", "Solution_9": "I changed my post accordingly. Now, how would I treat the dy/dx parts of the equation?", "Solution_10": "hello, no it isn't, your left hand-side is\r\n$ 2(x^2\\plus{}y^2)\\cdot 2x\\plus{}2y\\frac{dy}{dx}$\r\nand it must be\r\n$ 2(x^2\\plus{}y^2)(2x\\plus{}2y\\frac{dy}{dx})$\r\nSonnhard.", "Solution_11": "I changed my post accordingly. Now, how would I treat the dy/dx parts of the equation?", "Solution_12": "hello, we have\r\n$ 4x(x^2+y^2)-2x=-4(x^2+y^2)yy^'-2yy^'$\r\n$ 2x(x^2+y^2)-x=-2(x^2+y^2)yy^'-yy^'$\r\n$ x-2x(x^2+y^2)=y^'\\left(2(x^2+y^2)y+y\\right)$\r\n$ y^'=\\frac{x-2x(x^2+y^2)}{2(x^2+y^2)y+y}$\r\nSonnhard.", "Solution_13": "let $ y'$ denote $ \\frac {dy}{dx}$\r\n\r\ndifferentiating both sides of the equation w.r.t x using the chain rule, we get\r\n\r\n$ (x^2 \\plus{} y^2)^2 \\equal{} x^2 \\minus{} y^2$\r\n\r\n$ \\implies 2(x^2 \\plus{} y^2)(2x \\plus{} 2yy') \\equal{} 2x \\minus{} 2yy'$\r\n\r\n$ \\implies 2(x^2 \\plus{} y^2)(x \\plus{} yy') \\equal{} x \\minus{} yy'$\r\n\r\nrearrangement\r\n\r\n$ yy'(2x^2 \\plus{} 2y^2 \\plus{} 1) \\equal{} x \\minus{} 2x(x^2 \\plus{} y^2)$\r\n\r\n$ y' \\equal{} \\frac {x(1 \\minus{} 2(x^2 \\plus{} y^2)}{y(1 \\plus{} 2(x^2 \\plus{} y^2)}$ :)" } { "Tag": [ "inequalities", "geometry", "function", "number theory", "greatest common divisor", "least common multiple", "trigonometry" ], "Problem": "[quote]I think the cut-off was pretty high this time. Delhi seniors have performed extremely well because all 6 selected crossed 71. However, I expect the same kind of performance from seniors from other states like TN. \n\nAgr_94 and nuclear alchemist bought up the topic of IMOTC attendees writing RMO. I completely agree with you guys. It is not nice at all that they write RMO and block seats. People who attend IMOTC are given the chance to write INMO directly even if they don't clear postals. Some people write RMO with the aim of getting perfect score, which is not good because in the process other 12th standard guys who worked so hard are being deprived of a chance to write INMO. \n\nSome observations about the RMO and INMO exams. \n1. The Sophie Germain identity was used even in INMO 2008 number theory question. \n2. The geometry question on the INMO or RMO has almost always involved a construction of some sort. \n3. The greatest integer function was used in INMO 2007 and INMO 2009. \n4. Any inequality in the past three INMO's has been solvable by normalization. \n\nSomeone should start a new thread to discuss INMO strategy or something like that. I can't think of anything else right now. But the INMO questions are following a definite pattern over the past three years. So keep that in mind while preparing. All the best for INMO to all selected Better luck next time for others. \n\nAnd one last thing about cut-offs before signing off: \nThe INMO 2007 paper was easy compared to a usual standard INMO paper. As a result the cut-off was high, something like 35 or 40. Because this RMO was easier it completely makes sense that the cut-off be higher.\n[/quote]\r\n\r\nhere is the new topic.... my observations:\r\n\r\n1)INMO 2009 Q1 was very ingenius(at least their proof)...i dont mind Q's involving ingenius proofs..those which involve trival and knowledge-based identities worry me....\r\n2)i dont know what normalization is ...and i've started preparing for INMo today itself...so i guess i'm not ought to..so can some1 PLZ tell me what it its??\r\n3) INMO 2007 inequality Q6 maybe solved using normalization(which i dunno) but i am sure its solvable using simple convertion (taking xy=a , yz=b , zx=c etc) and then using AM-HM inequality...\r\n4) i would love to have a INMO 2005 alpha beta one like Q in INMO'10...it was soo very awesm and required ingenuity rather than knowledge....i wud LOVE to use my brain rather than any concepts pre-stored in it...", "Solution_1": "Well from experience of having written INMO 2009, I would like to say you many things:\r\n\r\n1. Do not go by any kind of pattern oriented preparation. befuddlers is a guy more experienced than me (He wrote INMO 2007 and 2008) but I disagree in this regard. Geometry problems in INMO are very nice and not the same construction to look for etc. Construction is done using intuition at that moment and that hugely depends on your temperament and skill at that time. Just try to solve many geometry problems and you will be in a good mind-set. Don't try to hunt for pattern in the kind of constructions the offiical solutions use in the INMOs. Same goes for the greatest integer function problems. I think Ashwath Rabindranath got an excellent solution for the INMO 2007 gif (Sahil it was 2007 not 2005). The idea is completely different from what was required in INMO 2009. It depends and varies from problems to problems. So does it go for the inequalities. Befuddlers, I think the INMO 2007 inequality was actually made more round about by the normalisation which was not required after all. Fermat point or full expanding and killing it by AM GM was enough I think (Believe me I did the 2nd one just for fun-- it took around 25 minutes that is all). INMO 2009 was a good example of a normalisation but even then the problem was not very obvious. And solutions without normalisations were more elegant too. \r\nAs for Sophie Germain Identity, well, it is indeed a prowerful factorisation but not the most reliable for tough problems. You see , if it is not a generalised one, at times you have got to make use of the numbers and not the identity alone. The numbers can play a role in facotrisation. There are several examples . If you want, I can give you some. \r\n\r\n2. You have to try and get exposure to tougher problems. Try to make a strategy on the basis of which areas you are good at and which you are not so strong. Since in the INMO, you get problems from all, (not like this year's RMO where they did not have a single algebra or a good geometry ).., try to strengthen your strong areas. For me, it appears that you are quite good in algebra -so try to make yourself strong even in number theory (mostly requires algebraic tricks). \r\n\r\n3. Try to gain as much experience by trying problems than seeing solutions. Helps you more.\r\n\r\n4. I also did not like one comment of yours somewhere when you said :\" refer to challenge and thrill Pg 63 some exercise problem: Know one thing that it is not possible to remember each and every problem from every book and resources you see. Instead try to recollect ideas. In that case, I shall tell you that any geometry problem can be solved using some problem of challenge and thrill but the problems in challenge and thrill are very much harder than INMO level. \r\n\r\n5. Try to solve more and more problems instead of learning more things(which you may feel like). Learn some necessary things (that you may already be knowing: if not, it is like the basic things that people use in olympiad-defitnely not at the level of mathlinks users. You may ask if you want in PM what they are. ).\r\n\r\n6. Develop your thinking by solving some tough ones too. \r\n\r\nAll the best and congratulations for clearing RMO.", "Solution_2": "[quote=\"Agr_94_Math\"]Well from experience of having written INMO 2009, I would like to say you many things:\n\n1. Do not go by any kind of pattern oriented preparation. befuddlers is a guy more experienced than me (He wrote INMO 2007 and 2008) but I disagree in this regard. Geometry problems in INMO are very nice and not the same construction to look for etc. Construction is done using intuition at that moment and that hugely depends on your temperament and skill at that time. Just try to solve many geometry problems and you will be in a good mind-set. Don't try to hunt for pattern in the kind of constructions the offiical solutions use in the INMOs. Same goes for the greatest integer function problems. I think Ashwath Rabindranath got an excellent solution for the INMO 2007 gif (Sahil it was 2007 not 2005). The idea is completely different from what was required in INMO 2009. It depends and varies from problems to problems. So does it go for the inequalities. Befuddlers, I think the INMO 2007 inequality was actually made more round about by the normalisation which was not required after all. Fermat point or full expanding and killing it by AM GM was enough I think (Believe me I did the 2nd one just for fun-- it took around 25 minutes that is all). INMO 2009 was a good example of a normalisation but even then the problem was not very obvious. And solutions without normalisations were more elegant too. \nAs for Sophie Germain Identity, well, it is indeed a prowerful factorisation but not the most reliable for tough problems. You see , if it is not a generalised one, at times you have got to make use of the numbers and not the identity alone. The numbers can play a role in facotrisation. There are several examples . If you want, I can give you some. \n\n2. You have to try and get exposure to tougher problems. Try to make a strategy on the basis of which areas you are good at and which you are not so strong. Since in the INMO, you get problems from all, (not like this year's RMO where they did not have a single algebra or a good geometry ).., try to strengthen your strong areas. For me, it appears that you are quite good in algebra -so try to make yourself strong even in number theory (mostly requires algebraic tricks). \n\n3. Try to gain as much experience by trying problems than seeing solutions. Helps you more.\n\n4. I also did not like one comment of yours somewhere when you said :\" refer to challenge and thrill Pg 63 some exercise problem: Know one thing that it is not possible to remember each and every problem from every book and resources you see. Instead try to recollect ideas. In that case, I shall tell you that any geometry problem can be solved using some problem of challenge and thrill but the problems in challenge and thrill are very much harder than INMO level. \n\n5. Try to solve more and more problems instead of learning more things(which you may feel like). Learn some necessary things (that you may already be knowing: if not, it is like the basic things that people use in olympiad-defitnely not at the level of mathlinks users. You may ask if you want in PM what they are. ).\n\n6. Develop your thinking by solving some tough ones too. \n\nAll the best and congratulations for clearing RMO.[/quote]\r\n\r\nseriously arvind...i was just saying see the prob in CTPCM and commenting on the ~ty with RMO prob..i would be the last one to mug or remember....and i follow ditto.recollect ideas... :D ..and sure i was talking of inmo 2005 Q2 alpha beta one...n BTW i dont have any strongs and weak....just depends whether it clicks or not..yaa sure i have most experience in inequalities...and sure i love geometry (evn if i cnt do every problem)...so i wont put one over another..btw NT is as awesm as others..BTW i'm happy u say CTPCM is above inmo :) ;)..that makes me a relieved man...and BTW i think inmo 2007 Q1..was seriously above any CTPCM ....look at the way they have used f(t) in the end ..it cud NEVER come to my mind :oops: :oops: ..and arvind PLZ tell me what the real basics u talk about (in ineq i think they mean strengthing AM-GM , rearrangement , jensen cauchy etc)", "Solution_3": "As far as inequalities are concerned, don't go by the level of the people here in mathlinks forum of Inequalities Olympiad Section. It is too too too good level compared to INMO, infact too much higher than IMOTC's and IMOs. Several people are so adept at solving inequalities around the world, they have strong techniques and ideas that are gained by experience and intuition. As regards inequalities, know stuff like Cauchy Schwatrz, Means, Rearrangement , Applying AM-GM by weights, etc).", "Solution_4": "and for NT , geometry(this is the real problem point)", "Solution_5": "Geometry: Ceva, Menelau, Ptolemy, and a few basic properties of triangle knowledge,quadrilaterals, and some theorems on tangents and secants is enough. (Tis may sound too much like a portion but not like that). More than knowledge, in geometry, trying more and more problems and doing soem things yourself will give you a better edge.\r\n\r\nNT: Basic stuff like Modulos, GCD and LCM, Fermat's little theorem, Euler's Theorem, Wilson's Theorem, Chinese Remainder Theorem.", "Solution_6": "[quote=\"Agr_94_Math\"]Geometry: Ceva, Menelau, Ptolemy, and a few basic properties of triangle knowledge,quadrilaterals, and some theorems on tangents and secants is enough. (Tis may sound too much like a portion but not like that). More than knowledge, in geometry, trying more and more problems and doing soem things yourself will give you a better edge.\n\nNT: Basic stuff like Modulos, GCD and LCM, Fermat's little theorem, Euler's Theorem, Wilson's Theorem, Chinese Remainder Theorem.[/quote]\r\n\r\nPlz tell me 1 thing... I have never seen application of menelaus or ceva's ...in any inmo..what i HAVE seen is application of sine / cosine rule...pythagorus etc etc..can u plz give me an example frm inmo where these theorems may help (it is a humble request ..no offences :roll: :roll: :lol: ....", "Solution_7": "[quote=\"sahilsharma94\"]\n\nPlz tell me 1 thing... I have never seen application of menelaus or ceva's ...in any inmo..what i HAVE seen is application of sine / cosine rule...pythagorus etc etc..can u plz give me an example frm inmo where these theorems may help (it is a humble request ..no offences :roll: :roll: :lol: ....[/quote]\r\nWell then you probably havent looked too hard....just see the official solution of q1 of INMO 2003", "Solution_8": "can any1 briefly and briskly explain invariance principle?", "Solution_9": "Oops! Sorry guys, looks like I didn't convey what I wanted to say in the right spirit. \r\n\r\nOkay, just to clear things up:\r\n1. When I said \"look for a strategy,\" I meant something more like prepare and practice problems from the specific areas I mentioned. For example, people say greatest integer function problems are very easy, but I don't think anyone except Akashnil in this forum solved the GIF problem in INMO 2009.(I am only talking about the people who wrote INMO 2009 officially here). And of course constructions will be based on the flavor of the problem, but constructions I think are extremely important to solve some geometry problems. \r\n\r\nPerhaps also knowing \"higher concepts\" like homothety and transformations in geometry will help. However, INMO tends to focus on angle chasing and other relatively simple methods to solve geometry problems. \r\n\r\nAnd as for inequalities, there are definitely more elegant methods than normalization, but normalization is almost a sure success yielder in the inequalities on INMO because INMO inequalities are mostly symmetric (thus can be normalized).\r\n\r\nCombinatorics on INMO 2008 and 2009 were combinatorial geometry problems( I think). But knowing bijections and recursions should help. \r\n\r\nINMO problems tend not to ask NT problems with straight out applications of Euler's theorem etc, but gcd and other related stuff do play a part in the problem(almost always)\r\n\r\n@sahil: As the name suggests \"Invariance\" means that some parameter remains constant/doesn't change during a specific operation/transformation. That parameter is said to be \"invariant.\" You could look into Problem solving strategies by Engel for some good problems. \r\n\r\nIn general INMO problems don't require tremendous preparation. Good angle chasing skills and basic knowledge in other areas should be enough.\r\n\r\nSo, good luck guys! :D", "Solution_10": "[quote=\"befuddlers\"]\nPerhaps also knowing \"higher concepts\" like homothety and transformations in geometry will help. However, INMO tends to focus on angle chasing and other relatively simple methods to solve geometry problems. \n[/quote]\nI don't think just \"knowing\" these concepts will help....you may need some practice....(but I think that's what you meant in the first place...on second thoughts :) )\n\n[quote=\"befuddlers\"]\nAnd as for inequalities, there are definitely more elegant methods than normalization, but normalization is almost a sure success yielder in the inequalities on INMO because INMO inequalities are mostly symmetric (thus can be normalized).\n[/quote]\nYou may want to replace \"symmetric\" by \"homogenous\"\n\n[quote=\"befuddlers\"]\nIn general INMO problems don't require tremendous preparation. Good angle chasing skills and basic knowledge in other areas should be enough.\n[/quote]\r\nSorry....but I don't think just basic knowledge will suffice...", "Solution_11": "aur yeh hai rijul ka ek aur successful katta :rotfl: :rotfl: :roll:", "Solution_12": "[quote=\"sahilsharma94\"]aur yeh hai rijul ka ek aur successful katta :rotfl: :rotfl: :roll:[/quote]\r\n :P HEHEHE...!!!", "Solution_13": "Been long since we had a aops fight :P", "Solution_14": "[quote=\"nuclear_alchemist\"]Been long since we had a aops fight :P[/quote]\r\nBhadka mat be.....", "Solution_15": "[quote=\"Rijul saini\"][quote=\"nuclear_alchemist\"]Been long since we had a aops fight :P[/quote]\nBhadka mat be.....[/quote]\r\n\r\nunfortunately all ppl out here wud know and agree that the pen is mightier than the sword..so no fights...sorry [size=150]nuke DEAL[/size]", "Solution_16": "Sorry for the typo. Normalization can be used for homogenous equations.\r\n\r\nSome techniques of normalization can include restraining x,y,z in an inequality with the condition \r\n$ x \\plus{} y \\plus{} z$ = 1 and so on", "Solution_17": "Hey all,\r\n\r\nHappy New Year! :) \r\n\r\nHope your INMO preparation is going well. \r\n\r\nHere are a couple of websites which are really good. \r\n\r\nhttp://mathcircle.berkeley.edu/\r\nGo to the archives. \r\n\r\nThe IMO Compendium website is also really helpful.\r\n\r\nGood luck!", "Solution_18": "hey is arthur engel required for inmo\n", "Solution_19": "No not really.\nI mean I didn't refer to it much.\nMaybe for starting complex geometry and the idea of extremal and invariance but that is all.\nI believe PSS is now old for contemporary olympiads.\n\nOthers' opinions may differ of course.", "Solution_20": "Can someone please explain what normalisation is?\n", "Solution_21": "[quote=Hrithik007]Can someone please explain what normalisation is?[/quote]\n\nYou can find it in any standard inequality text, for example 'Secrets in Inequalities'..but put briefly and rougly it is the following:\n\nSay you have a homogenous inequality in 3 variables, i.e., you have $f(a, b, c) \\ge 0$ and $f$ is such that $f(ka, kb, kc)=k^{d}f(a, b, c)$....for some constant $d$....Now if you are given that $a, b, c$ are positive.. then the given inequality is true iff $\\left( \\frac{1}{a+b+c}\\right)^df(a, b, c)) \\ge 0$ which by the homogenous condition is same as $f(\\frac{a}{a+b+c}, \\frac{b}{a+b+c}+\\frac{c}{a+b+c}) \\ge 0$...since $\\frac{a}{a+b+c}+\\frac{b}{a+b+c}+\\frac{c}{a+b+c}=1$..All this means that you could have assumed from the outset that $a+b+c=1$....\n\nNow what you did with $a+b+c$ can be done with any homogenous function..meaning you can impose a condition like $g(a,b,c)=1$ for any homogenous function $g$ of degree say $r$ (Do you see why?)..eg. you can assume $ab+bc+ca=1$ and so on..but one at a time of course(it is possible to apply multiple conditions in some cases..say if the sets of variables are disjoint..e.g. in one proof of Cauchy's inequality..)...\n\nNow, try and prove your favourite homogenous inequality by normalisation.." } { "Tag": [ "LaTeX", "trigonometry" ], "Problem": "Buna ziua, \r\n\r\nCum se demonstreaza ca sinn nu are limita cand n tinde la +infinit? \r\n\r\nMai general, daca x(n) este un sir care are limita +infint cand n tinde la +infinit, si f(x) este periodica de perioada principala T, cum se demonstreaza ca sirul f(x(n)) nu are limita cand n tinde la +infinit? \r\n\r\nVa multumesc si cer iertare dar nu am stiut sa bat in Latex textul", "Solution_1": "Daca $\\sin n\\rightarrow L$, atunci $\\sin (n+1)-\\sin (n-1) =2\\sin 1\\cos n \\rightarrow L-L=0$, deci $\\cos n \\rightarrow 0.$ Atunci $\\sin 2n =2\\sin n \\cos n \\rightarrow 0,$ de unde rezulta $L=0.$ Contradictia se obtine observand ca\r\n\\[1=\\sin^{2}n+\\cos^{2}n \\rightarrow 0.\\]\r\n(demonstratia e din cartea prof. Siretchi)\r\n\r\nRezultatul \"mai general\" nu este adevarat.", "Solution_2": "Domnule Bogdan Enescu, va multumesc pentru solutie! \r\n\r\nNumai bine,", "Solution_3": "[quote=\"enescu\"]Daca $\\sin n\\rightarrow L$, atunci $\\sin (n+1)-\\sin (n-1) =2\\sin 1\\cos n \\rightarrow L-L=0$, deci $\\cos n \\rightarrow 0.$ Atunci $\\sin 2n =2\\sin n \\cos n \\rightarrow 0,$ de unde rezulta $L=0.$ Contradictia se obtine observand ca\n\\[1=\\sin^{2}n+\\cos^{2}n \\rightarrow 0. \\]\n(demonstratia e din cartea prof. Siretchi)\n\nRezultatul \"mai general\" nu este adevarat.[/quote]\r\n\r\n[color=red]This a great and inspirated solution(!!!) , which I see for first time.Untill now I knew only the solution based on the negation of the definition of the limit ![/color]\r\n[u] Babis[/u]" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $ ABC$ be a right angled triangle where $ \\angle C=90^{\\circ}$ and $ CD$ is the altitude. Let $ K$ be a point in plane. If $ O$ is the circumcenter of triangle $ ABK$, prove that $ AO$ is perpendicular to $ DK$ if and only if $ AC=AK$. :)", "Solution_1": "we know that $ KD \\bot AO \\iff KA^{2}-KO^{2}=DA^{2}-DO^{2}$\r\ni.e. $ KA^{2}-DA^{2}=KO^{2}-DO^{2}$\r\nbut according to the power of point $ D$ to circle $ (O)$ we have:\r\n$ |DA.DB|=|DO^{2}-OK^{2}|$\r\n$ \\Rightarrow \\boxed{DA.DB=OK^{2}-DO^{2}(I)}$\r\nand also we have $ \\triangle ADC \\sim \\triangle CDB$ so we have $ DA.DB=DC^{2}$\r\nthus according to $ (I)$ we get that:\r\n$ DC^{2}=KO^{2}-DO^{2}$\r\nthus we have:\r\n$ KD \\bot AO \\iff KA^{2}-DA^{2}=DC^{2}$\r\nbut in the right angled $ \\triangle ACD$ we have $ DC^{2}+DA^{2}=AC^{2}$\r\nthus we have:\r\n$ KA^{2}-DA^{2}=DC^{2}\\iff KA^{2}=DC^{2}+DA^{2}=AC^{2}\\iff KA=AC$\r\nthus we have:\r\n$ KD \\bot AO \\iff KA=AC$ as wanted...", "Solution_2": "Again a nice solution Babak :) \r\nMy solution for this one was:\r\n\r\nLet $ DK\\cap AO=H$ and let $ M$ be the midpoint of $ BK$. we'll prove that $ OMHK$ is a cyclic quadrilateral. Note that because $ ABC$ is right angled so triangles $ ABC$ and $ ACD$ are similar. so we have: $ AC^{2}=AD\\cdot AB$ but $ AK=AC$ so $ \\boxed{AK^{2}=AD\\cdot AB}$ but this shows the similarity of triangles $ AKD$ and $ ABK$ so we have: \r\n$ \\angle AKD=\\angle ABK$, so now we can write:\r\n$ \\angle DKM=\\boxed{\\angle HKM=\\angle AKB-\\angle ABK}$\r\n\r\nbut we also have:\r\n$ \\angle AOM=\\angle KOM-\\angle KOA=\\frac{1}{2}\\angle KOB-2\\angle ABK$\r\n$ =\\frac{1}{2}\\left(2\\angle ABK+2\\angle AKB\\right)-2\\angle ABK=\\boxed{\\angle AKB-\\angle ABK}$\r\n\r\n as a consequence $ \\angle AOM=\\angle HOM=\\angle HKM$ and the quadrilateral $ OMHK$ is cyclic and $ \\angle KHM=\\angle KMO=90^{\\circ}$ which means $ AO$ and $ KD$ are perpendicular to each other. :)\r\n\r\nand now its easy to prove that if $ AO$ and $ DK$ are perpendicular then $ AK=AC$.", "Solution_3": "I read carefully both solutions. They are really very nice and entirely natural. Congrats to both of you !\r\n\r\n [u]Babis[/u]", "Solution_4": "midpoint of $ AK$ is $ M$. $ N$ is intersection of $ AO,DK$. $ AO$ is perpenticular $ DK$ if and only if $ >This part I got, \r\n dy/dt = ky (1-y)\r\nb) Solve the differential equation\r\n\r\n>>I tried integrating the above equation but in the end what I got was y = (Ce^kt)+1\r\n\r\nbut the book says that the answer is y = y0/[y0 + (1-y0)e^(-kt)] \r\nhow did they get this?\r\n\r\nThanks", "Solution_1": "Just seperate the variables\r\n$\\frac{dy}{dt}=ky(1-y)$\r\n$\\frac{dy}{y(1-y)}=kdt$\r\n$\\int \\frac{dy}{y(1-y)} = \\int kdt$\r\nUse partial fractions on the LHS.\r\n$\\frac{1}{y(1-y)} = \\frac{1}{y} - \\frac{1}{y-1}$\r\n$\\int \\frac{1}{y} dy - \\int \\frac{1}{y-1} dy = \\int k dt$\r\n$\\ln(y) - \\ln(y-1) = kt+C$\r\n$\\ln{\\left(\\frac{y}{y-1}\\right) = kt +C}$\r\n$\\frac{y}{y-1} = Ce^{kt}$\r\n$y=\\frac{Ce^{kt}}{Ce^{kt}-1}$\r\nSince $C$ is just a constant we can divided the numberator and denominator by $C$ to get:\r\n$\\boxed{y=\\frac{e^{kt}}{e^{kt}+C}}$. \r\nWere there initial conditions given in the problem?\r\nAnd what does $y_0$ represent?", "Solution_2": "that was all that was given, no more information was given... so i do not understand how they got \r\n\r\ny = y0/[y0 + (1-y0)e^(-kt)]", "Solution_3": "The usual thing that $y_0$ would mean would be an initial condition: $y_0=y(0).$\r\n\r\nLook a couple of lines up from the bottom of Jimmy's post: he says\r\n\r\n$\\frac{y}{y-1} = Ce^{kt}$\r\n\r\nLet $t=0$ and $y=y_0$. Then we get $C=\\frac{y_0}{y_0-1}.$\r\n\r\nGoing to the next line in Jimmy's answer (where it's still the same $C$):\r\n\r\n$y=\\frac{Ce^{kt}}{Ce^{kt}-1} = \\frac{\\frac{y_0}{y_0-1}e^{kt}} {\\frac{y_0}{y_0-1}e^{kt}-1} = \\frac{y_0e^{kt}}{y_0e^{kt}-y_0+1} = \\frac{y_0}{y_0+(1-y_0)e^{-kt}}$" } { "Tag": [ "Euler" ], "Problem": "\u039f\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03bf\u03bc\u03bf\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bc\u03b5 \u03c0\u03c1\u03cc\u03c3\u03c9\u03c0\u03b1 \u03ae \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03c4\u03c5\u03c7\u03b1\u03af\u03b1 ? :wink: \r\n\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03bf\u03c0\u03bf\u03b8\u03b5\u03c4\u03ce \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 G3 IMO SL 1996 \u03c3\u03c4\u03b7\u03bd official version \u03c4\u03bf\u03c5 (\u03c0\u03b9\u03c3\u03c4\u03ad\u03c8\u03c4\u03b5 \u03bc\u03b5 \u03b4\u03b5\u03bd \u03b8\u03ad\u03bb\u03b5\u03c4\u03b5 \u03bd\u03b1 \u03b4\u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03bd official solution .\u03a0\u03ad\u03c1\u03bd\u03b5\u03b9 \u03c0\u03bf\u03bb\u03bb\u03ac \u03cc\u03bc\u03bf\u03b9\u03b1 \u03bf \u03ac\u03c4\u03b9\u03bc\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03c9 )\r\n(\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c3\u03b1\u03c2 \u03c3\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b5\u03b9 \u03c4\u03bf \u03b2\u03b1\u03c3\u03b9\u03ba\u03cc \u03af\u03c7\u03bd\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 \u03bc\u03b5 F \u03c3\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 .\u0391\u03bb\u03bb\u03ac \u03bd\u03b1 \u03c4\u03ce\u03c1\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03c3\u03ba\u03ad\u03c6\u03c4\u03bf\u03bc\u03b1\u03b9 \u03c0\u03c1\u03b9\u03bd \u03b1\u03c0\u03cc \u03ba\u03b1\u03bc\u03b9\u03ac \u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b1 \u03ad\u03bb\u03c5\u03bd\u03b1 \u03bc\u03af\u03b1 \u03bc\u03b5 F \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 \u03c0\u03bf\u03c5 :roll: )\r\n\r\n\u0388\u03c3\u03c4\u03c9 ABC \u03bf\u03be\u03c5\u03b3\u03ce\u03bd\u03b9\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 $ BC > CA$ . \u0388\u03c3\u03c4\u03c9 $ O$ \u03c4\u03bf \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf , $ H$ \u03c4\u03bf \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03ba\u03b1\u03b9 $ F$ \u03c4\u03bf \u03af\u03c7\u03bd\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae $ C$ .\u0388\u03c3\u03c4\u03c9 \u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c3\u03c4\u03b7\u03bd $ OF$ \u03c3\u03c4\u03bf $ F$ \u03cc\u03c4\u03b9 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ CA$ \u03c3\u03c4\u03bf $ P$ . \u0394\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\angle{FHP} \\equal{} \\angle{BAC}$ \r\n\r\n(\u039a\u03b1\u03b9 \u03c4\u03ce\u03c1\u03b1 \u03b5\u03be\u03b7\u03b3\u03b5\u03af\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03c0\u03c1\u03b9\u03bd \u03bc\u03b9\u03b1 \u03b5\u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b1 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03bb\u03c5\u03bd\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ae\u03c4\u03b1\u03bd \u03ac\u03c7\u03c1\u03b7\u03c3\u03c4\u03b1 :huh: )\r\n\u0395\u03c5\u03c4\u03c5\u03c7\u03ce\u03c2 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c4\u03bf \u03b2\u03c1\u03ae\u03ba\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03bd\u03b1 \u03b2\u03c1\u03ac\u03b4\u03c5 \u03b5\u03af\u03c7\u03b1 \u03c0\u03b5\u03c4\u03b1\u03c7\u03c4\u03b5\u03af \u03ba\u03b1\u03b9 \u03b5\u03af\u03c7\u03b1 \u03c0\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03c9 \u03be\u03b1\u03bd\u03b1\u03b4\u03b5\u03af \u03bb\u03c5\u03bc\u03bc\u03ad\u03bd\u03bf \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c0\u03b5\u03c4\u03b1\u03bb\u03bf\u03cd\u03b4\u03b1\r\n(see IMO compenium 3rd solution)\r\n\r\n***\u039a\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ae\u03ba\u03b7 . \u0397 \u03c0\u03b9\u03bf \u03c9\u03c1\u03b1\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03b4\u03b5\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03c4\u03bf\u03c5 Euler (\u03bd\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03c5\u03bd\u03cc\u03c4\u03b1\u03bd \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 .\u0386\u03c1\u03b5 \u0386\u03bb\u03ba\u03b7\u03c3\u03c4\u03b7 \u03b1\u03c4\u03c5\u03c7\u03b9\u03b1...\u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9 \u03cc\u03bc\u03c9\u03c2 keep going)", "Solution_1": "[quote=\"silouan\"]\u039f\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03bf\u03bc\u03bf\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bc\u03b5 \u03c0\u03c1\u03cc\u03c3\u03c9\u03c0\u03b1 \u03ae \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03bd\u03c4\u03b5\u03bb\u03ce\u03c2 \u03c4\u03c5\u03c7\u03b1\u03af\u03b1 ? :wink: \n\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03c4\u03bf\u03c0\u03bf\u03b8\u03b5\u03c4\u03ce \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 G3 IMO SL 1996 \u03c3\u03c4\u03b7\u03bd official version \u03c4\u03bf\u03c5 (\u03c0\u03b9\u03c3\u03c4\u03ad\u03c8\u03c4\u03b5 \u03bc\u03b5 \u03b4\u03b5\u03bd \u03b8\u03ad\u03bb\u03b5\u03c4\u03b5 \u03bd\u03b1 \u03b4\u03b5\u03af\u03c4\u03b5 \u03c4\u03b7\u03bd official solution .\u03a0\u03ad\u03c1\u03bd\u03b5\u03b9 \u03c0\u03bf\u03bb\u03bb\u03ac \u03cc\u03bc\u03bf\u03b9\u03b1 \u03bf \u03ac\u03c4\u03b9\u03bc\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c4\u03b7\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03ac\u03b2\u03c9 )\n(\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03b4\u03b5\u03bd \u03be\u03ad\u03c1\u03c9 \u03b1\u03bd \u03ba\u03b1\u03bd\u03b5\u03af\u03c2 \u03c3\u03b1\u03c2 \u03c3\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ad\u03c7\u03b5\u03b9 \u03b4\u03b5\u03b9 \u03c4\u03bf \u03b2\u03b1\u03c3\u03b9\u03ba\u03cc \u03af\u03c7\u03bd\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 \u03bc\u03b5 F \u03c3\u03b5 \u03ac\u03bb\u03bb\u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 .\u0391\u03bb\u03bb\u03ac \u03bd\u03b1 \u03c4\u03ce\u03c1\u03b1 \u03c0\u03bf\u03c5 \u03c4\u03bf \u03c3\u03ba\u03ad\u03c6\u03c4\u03bf\u03bc\u03b1\u03b9 \u03c0\u03c1\u03b9\u03bd \u03b1\u03c0\u03cc \u03ba\u03b1\u03bc\u03b9\u03ac \u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b1 \u03ad\u03bb\u03c5\u03bd\u03b1 \u03bc\u03af\u03b1 \u03bc\u03b5 F \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03ac\u03bc\u03b1\u03b9 \u03c0\u03bf\u03c5 :roll: )\n\n\u0388\u03c3\u03c4\u03c9 ABC \u03bf\u03be\u03c5\u03b3\u03ce\u03bd\u03b9\u03bf \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u03bc\u03b5 $ BC > CA$ . \u0388\u03c3\u03c4\u03c9 $ O$ \u03c4\u03bf \u03c0\u03b5\u03c1\u03af\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf , $ H$ \u03c4\u03bf \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03ba\u03b1\u03b9 $ F$ \u03c4\u03bf \u03af\u03c7\u03bd\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03cd\u03c8\u03bf\u03c5\u03c2 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03ae $ C$ .\u0388\u03c3\u03c4\u03c9 \u03b7 \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c3\u03c4\u03b7\u03bd $ OF$ \u03c3\u03c4\u03bf $ F$ \u03cc\u03c4\u03b9 \u03c4\u03ad\u03bc\u03bd\u03b5\u03b9 \u03c4\u03b7\u03bd $ CA$ \u03c3\u03c4\u03bf $ P$ . \u0394\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\angle{FHP} \\equal{} \\angle{BAC}$ \n\n(\u039a\u03b1\u03b9 \u03c4\u03ce\u03c1\u03b1 \u03b5\u03be\u03b7\u03b3\u03b5\u03af\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03c0\u03c1\u03b9\u03bd \u03bc\u03b9\u03b1 \u03b5\u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b1 \u03c3\u03b5 \u03b1\u03c5\u03c4\u03cc \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03bb\u03c5\u03bd\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ae\u03c4\u03b1\u03bd \u03ac\u03c7\u03c1\u03b7\u03c3\u03c4\u03b1 :huh: )\n\u0395\u03c5\u03c4\u03c5\u03c7\u03ce\u03c2 \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c4\u03bf \u03b2\u03c1\u03ae\u03ba\u03b1 \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03bd\u03b1 \u03b2\u03c1\u03ac\u03b4\u03c5 \u03b5\u03af\u03c7\u03b1 \u03c0\u03b5\u03c4\u03b1\u03c7\u03c4\u03b5\u03af \u03ba\u03b1\u03b9 \u03b5\u03af\u03c7\u03b1 \u03c0\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03ad\u03c7\u03c9 \u03be\u03b1\u03bd\u03b1\u03b4\u03b5\u03af \u03bb\u03c5\u03bc\u03bc\u03ad\u03bd\u03bf \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03bc\u03b5 \u03c0\u03b5\u03c4\u03b1\u03bb\u03bf\u03cd\u03b4\u03b1\n(see IMO compenium 3rd solution)\n\n***\u039a\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c4\u03b5\u03bb\u03b5\u03c5\u03c4\u03b1\u03af\u03b1 \u03c0\u03c1\u03bf\u03c3\u03b8\u03ae\u03ba\u03b7 . \u0397 \u03c0\u03b9\u03bf \u03c9\u03c1\u03b1\u03af\u03b1 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03b4\u03b5\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03ba\u03cd\u03ba\u03bb\u03bf \u03c4\u03bf\u03c5 Euler (\u03bd\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03c5\u03bd\u03cc\u03c4\u03b1\u03bd \u03ba\u03b1\u03b9 \u03ad\u03c4\u03c3\u03b9 .\u0386\u03c1\u03b5 \u0386\u03bb\u03ba\u03b7\u03c3\u03c4\u03b7 \u03b1\u03c4\u03c5\u03c7\u03b9\u03b1...\u03b4\u03b5\u03bd \u03c0\u03b5\u03b9\u03c1\u03ac\u03b6\u03b5\u03b9 \u03cc\u03bc\u03c9\u03c2 keep going)[/quote]\r\n\r\n\r\n\r\n\u0391\u03bd\u03b5\u03be\u03ac\u03bd\u03c4\u03b7\u03bb\u03b7\u03c4\u03bf\u03c2!!! \u0391\u03c5\u03c4\u03cc \u03b5\u03be\u03b7\u03b3\u03b5\u03af \u03b2\u03ad\u03b2\u03b1\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c0\u03ce\u03c2 \u03c4\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03ad\u03c6\u03c4\u03b1\u03c3\u03b5 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c4\u03bf\u03c5 Cosmin Pohoata \u03ba\u03b1\u03b9 \u03c4\u03bf\u03c5 Virgil Nicula \u03b3\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03b1\u03c1\u03bc\u03bf\u03bd\u03b9\u03ba\u03ad\u03c2 \u03c4\u03b5\u03c4\u03c1\u03ac\u03b4\u03b5\u03c2. \u03a0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03c0\u03b9\u03b2\u03b5\u03b2\u03b1\u03b9\u03ce\u03c3\u03b5\u03b9\u03c2 silouan \u03cc\u03c4\u03b9 \u03c4\u03bf \u03b5\u03af\u03c7\u03b1 \u03c5\u03c0\u03bf\u03c0\u03c4\u03b5\u03c4\u03c5\u03b8\u03b5\u03af \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c0\u03c1\u03ce\u03c4\u03b7 \u03c3\u03c4\u03b9\u03b3\u03bc\u03ae, \u03ba\u03b1\u03b9 \u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1 \u03b5\u03af\u03c7\u03b1 \u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03bb\u03cc\u03b3\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03bf\u03c0\u03bf\u03af\u03bf\u03c5\u03c2 \u03c4\u03bf \u03c0\u03af\u03c3\u03c4\u03b5\u03c5\u03b1...\u03ad\u03c4\u03c3\u03b9 \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9?\r\n\r\n\u0395\u03c0\u03af\u03c3\u03b7\u03c2 \u03bc\u03bf\u03c5 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03b5\u03bd\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 \u03b1\u03c5\u03c4\u03cc \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd Euler...\u03bc\u03cc\u03bb\u03b9\u03c2 \u03c0\u03b1\u03c4\u03ae\u03c3\u03c9 submit \u03b8\u03b1 \u03c0\u03ac\u03c9 \u03bd\u03b1 \u03c4\u03bf \u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03c9. \u03a1\u03b5 \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd \u03b1\u03bd \u03c4\u03bf \u03b5\u03b9\u03c7\u03b5\u03c2 \u03b8\u03c5\u03bc\u03b7\u03b8\u03b5\u03af \u03bd\u03c9\u03c1\u03af\u03c4\u03b5\u03c1\u03b1, \u03b8\u03b1 \u03b5\u03b9\u03c7\u03b1\u03bc\u03b5 \u03c0\u03ac\u03c1\u03b5\u03b9 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03cc\u03bd\u03c4\u03bf\u03c5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u0386\u03bb\u03ba\u03b7\u03c3\u03c4\u03b7...\u03b1\u03bd \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03ae\u03c4\u03b1\u03bd \u03bb\u03af\u03b3\u03bf \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf \u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03a3\u03ba\u03bf\u03c0\u03b9\u03b1\u03bd\u03bf\u03cd\u03c2...\r\n\r\np.s. Bravo re silouan!kali douleia." } { "Tag": [ "number theory", "prime numbers" ], "Problem": "Here is a picture as attachment.\r\nThe game is simple : find all the digits such as if you read vertically, horizontally and diagonally (from the top) you have always a prime number.\r\nThe prime numbers created have to be distincts.", "Solution_1": "Ahh, a game I can understand! \r\n\r\nSorta...\r\n\r\nWhen we put numbers in the 3x3 grid, do they have to be digits? Or can they be numbers like 10573?" } { "Tag": [ "MATHCOUNTS", "complex numbers" ], "Problem": "Hmm.. I haven't actually seen this written down officially anywhere, but I suppose someone has defined it.\r\nFor what real a is (-2)^a defined?", "Solution_1": "This question needs concepts that are beyond mathcouts level, I believe. Like converging sequences.", "Solution_2": "It's defined for all real (and complex) a. Uniquely defined and real definitions are another story.", "Solution_3": "Okay, I was thinking reals. But either way, it's still not for mathcounts.", "Solution_4": "Actually it's not that difficult and the definition of exponents are very important in MATHCOUNTS.\n\nAnswer: [hide]When a is the multiplicative inverse of an even number[/hide]", "Solution_5": "That's not right, Tare.", "Solution_6": "Er, I don't know why it's wrong...Let me restate it: [hide]For all a element of real number (-2)^a is an element of real number if and only if a is the multiplicative inverse of an even number, excluding 0.[/hide]", "Solution_7": "It's wrong because it fails in every case. (-2)^(1/2), the simplest case of those you suggested, is not a real number.", "Solution_8": "Hmm.. I was thinking of non complex-stuff. What I thought was that a had to be rational with its denominator odd (in lowest terms) if a is positive or numerator odd if a is negative.. maybe I was wrong..", "Solution_9": "But how do you prove, say, if (-2)^:rt2: is real or not? On the rational numbers, you're correct, but it's not clear what to do on irrational numbers.", "Solution_10": "Oh whoops, I meant it's [b]not[/b] defined for the cases I mentioned, so the answer would be \"For all a element of real number (-2)^a is not an element of real number if and only if a is the multiplicative inverse of an even number.\"\r\n\r\nTeehee :oops:\r\n\r\nI'm still incorrect anyways since irrational power of a negative number is undefined no matter what.", "Solution_11": "Yes. But proving that irrational powers are undefined for negative numbers requires knowledge of sequences and such that is definitely not mathcouts level.", "Solution_12": "That is true...oh, so that's why you said this is beyond MATHCOUNTS level...\r\n\r\n...this is harder than I thought :?", "Solution_13": "[quote=\"Tare\"]\nI'm still incorrect anyways since irrational power of a negative number is undefined no matter what.[/quote]\r\n\r\nLike 0 of what you said here is correct. Assume w is a complex constant and z != 0, then z^w is defined for all complex z. It's singlevalued when w is an integer, takes finitely many values when w is rational, takes infinitely many numbers in all other cases.", "Solution_14": "The problem only wants real numbers, not comlex numbers. I know it isn't explicitly stated, but in general when you ask a question of this sort, the answers are taken over the field of the real numbers. So for this question, what Tare said was correct: (-2)^[i]a[/i] will only be defined in real numbers if [i]a[/i] is a rational number with odd denominator.", "Solution_15": "Mmm, I agree. After all this is just MATHCOUNTS level, or close to it, so I doubt complex numbers come into play.\r\n\r\nI finally got it right! Whoopy! :)", "Solution_16": "[quote=\"JBL\"]The problem only wants real numbers, not comlex numbers. I know it isn't explicitly stated, but in general when you ask a question of this sort, the answers are taken over the field of the real numbers. So for this question, what Tare said was correct: (-2)^[i]a[/i] will only be defined in real numbers if [i]a[/i] is a rational number with odd denominator.[/quote]\r\n\r\nDon't think thats quite right if a is negative :)\r\n\r\nIt just came up when, like, you are asked to solve something like x^(2/3) = 4, then you go x = 4^(3/2) but then you may miss the negative solution.. or is there a negative solution? Well thats what I was thinking about that particular moment of the day anyway..", "Solution_17": "You're going to have to explain to me why having a negative [i]a [/i]changes anything here.", "Solution_18": "It doesn't. oops:oops:", "Solution_19": "[quote=\"ComplexZeta\"]It's defined for all real (and complex) a. Uniquely defined and real definitions are another story.[/quote]\n\n\nwhat do ya mean by defined ?" } { "Tag": [ "ratio", "AMC", "AIME", "probability" ], "Problem": "The sequence $ z_{n}$ is defined with $ z_{n+2}z_{n-2}= z_{n}^{2}+2007$ for all $ n \\geq 4$. It is given that $ z_{2}= 1$ and $ z_{4}= 3$. Find the largest integer less than or equal to $ \\frac{3 z_{2008}^{2}+3 z_{2006}^{2}-z_{2008}z_{2006}}{3 z_{2008}z_{2006}}$", "Solution_1": "[hide=\"does this work?\"]\nlet $ f(n)=z_{2n}$\nlet x=f(n-1),ax=f(n) and abx=f(n+1)\nthen $ b=a+\\frac{2007}{ax^{2}}$\nthe part on the right rapidly approaches zero, so f(x) should be approximatly geometric\nsince we are finding the greatest integer less than or equal to \n$ \\frac{f(1004)}{f(1003)}+\\frac{f(1003)}{f(1004)}-\\frac{1}{3}$ we just have to find the approximate ratio\n\nfor n=2, we have x=1, and a=3, so b=672\nthen for n=3, x=3, a=672, so b=672.331...\nthe ratio should be a bit over 672.3, so\n$ \\frac{f(1004)}{f(1003)}+\\frac{f(1003)}{f(1004)}-\\frac{1}{3}\\approx672.3+\\frac{1}{672.3}-\\frac{1}{3}$\nthe greatest integer less than or equal to that is 671\n\nAddmitably, this is the same reasoning that made me miss this problem on the real AIME\n[/hide]", "Solution_2": "Well done. This is essentially SAME problem as AIME 2007 #14. However, there was one slight variation. As previous poster noted, the answer was actually 1/3 less than what AIME asked (in the format that is). \r\n\r\nNow just a thought that came to my mind... Is it possible to answer same question for $ \\frac{z^{2}_{2008}+z^{2}_{2006}}{z_{2008}z_{2006}}$?\r\n\r\nAll we know for that is that it's less than 672.33333333333... I haven't been able to figure this \"additional\" problem out so any great idea will be appreciated.", "Solution_3": "wouldn't that be even easier? It would just be 672", "Solution_4": "Hmm. That's what I think but how do you know how small the difference is? In original AIME problem, we knew that the answer was slightly less than 225 so we choose 224 but like in this case, is there anyway to confirm that the answer is at least 672?\r\n\r\nLike... By any chance, can it be something like 671.99999999999999999997?\r\n\r\nThen the answer is 671. \r\n\r\nI DO think that 672 is right (because I'm assuming the difference is very small from 672+1/3) but this is just a thought... which is not 100%.", "Solution_5": "what exactly are you asking?\r\n\r\nbeucase if the question is $ 672.3+\\frac{1}{672.3}$, then the answer is obivously 672", "Solution_6": "I realized something now that I looked into your solution more carefully. It's a lot different from way I solved it.\r\n\r\nI solved method that probability used in this thread http://www.artofproblemsolving.com/Forum/viewtopic.php?p=784670&search_id=259307746#784670\r\n\r\nGoing with what he did, in this problem, my work was:\r\n\r\n$ G_{2008}+1/G_{2008}< G_{2008}+1/G_{2006}= 672+1/3$\r\n\r\nThus, that's where my idea came. Honestly, I'm not sure if your idea is 100% accurate.. I don't know, I don't see any definite fault on it but it just feels like it's just \"idea\" rather than \"proved.\"", "Solution_7": "well my solution isn't exactly rigorous\r\nFor example, I never proved that $ \\frac{z_{n+2}}{z_{n}}$ converged" } { "Tag": [ "algebra theorems", "algebra" ], "Problem": "Could anyone tell me what Lambek-moser Theorem is, and how to prove it?\r\nI would be very very thankful if anyone could answer my question.", "Solution_1": "Look at the following link\r\n[url]http://www.mathsoft.com/mathresources/constants/discretestructures/article/0,,2267,00.html[/url]", "Solution_2": "Too thanks! :D", "Solution_3": "link is broken,", "Solution_4": "Actually, quite a nice and deep result; see [url]http://en.wikipedia.org/wiki/Lambek%E2%80%93Moser_theorem[/url]." } { "Tag": [ "vector", "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $X$ be a vector space of dimension $n$, $B(X)$ be the space of all endomorphisms of $X$, $T \\in B(X)$. Prove that the following statements are equivalent:\r\n\r\n(1) all eigenvalues of $T$ are simple;\r\n(2) $(T^{k}x)_{0 \\le k \\le n-1}$ are linearly independent in $X$ for some $x \\in X$;\r\n(3) $(T^{k})_{0 \\le k \\le n-1}$ are linearly independent in $B(X)$.", "Solution_1": "\\[Let \\;T=\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix}\\]\r\n$T$ satisfies (2),(3) but doesn't satisfy (1).", "Solution_2": "Indeed. :blush: Here is the right one: prove that (1) implies (2) and (3).", "Solution_3": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=minimal&t=84742]Here[/url]'s something a little stronger." } { "Tag": [ "calculus", "derivative", "Pythagorean Theorem", "geometry", "calculus computations" ], "Problem": "I started to learn calculus following a book but I don\u2019t understand the explanation of a derivative example. Here is the excerpt of the book:\r\n\r\n[color=darkblue]Then the Pythagorean theorem tells us that\n\nh(t)^2 + b(t)^2 = 13^2\n\nWe may differentiate both sides of this equation with respect to the variable t\n(which is time in minutes) to obtain\n\n2 \u2022 h(t) \u2022 h\u2019(t) + 2 \u2022 b(t) \u2022 b\u2019(t) = 0[/color]\r\n\r\nIt seems that the author applied the Chain Rule but I am not sure about the intermediary steps of this differentiation. Could someone show me intermediary steps and explain them ?", "Solution_1": "Yes, the author applied the chain rule to the expressions $ h(t)^{2}$ and $ b(t)^{2}$.\r\n\r\nLet $ g(x) = x^{2}$. Then\r\n\r\n$ h(t)^{2}= g(h(t))$\r\n\r\nSo that\r\n\r\n$ \\frac{d}{dt}h(t)^{2}= g'(h) h'(t)$\r\n\r\nWhere\r\n\r\n$ g'(x) = 2x$\r\n\r\nSo that\r\n\r\n$ g'(h) = 2h$\r\n\r\nAnd\r\n\r\n$ g'(h) h'(t) = 2h(t) h'(t)$" } { "Tag": [ "function", "number theory", "greatest common divisor", "modular arithmetic", "prime numbers", "number theory proposed" ], "Problem": "Hi,\r\n\r\nI have some difficulty to prove this statement (see the picture) \r\nIs it true?\r\nThank you for any help or comment.", "Solution_1": "I checked the statement for some hundreds values it holds.\r\nEven if it is not true for all values p,q there is a way to check numbers by a set of primes.\r\nI tried with 561 (Carmichael number) checking with 89 it shows that 561 is composite (2^560 mod 561=1)\r\nI tried with 341 (Carmichael number) checking it with 89 it shows that 341 is composite. (2^340 mod 341=1)\r\nMy goal now it to find the set of prime numbers to use as witness.", "Solution_2": "If I haven't misunderstood the question , try $ p\\equal{}23,q\\equal{}11$, the statement will be wrong.", "Solution_3": "[quote=\"stephencheng\"]If I haven't misunderstood the question , try $ p \\equal{} 23,q \\equal{} 11$, the statement will be wrong.[/quote]\r\n\r\nThank you for your comment.\r\nI checked 11 and 23.\r\nThe statement is wrong. Because 23=2*11 +1 and p=11.\r\nSo I have to introduce a condition of co-primality between s and p.\r\n[u]My idea is to check the primality using other known certified prime.[/u] I have found that [b]the twin primes [/b] as numbers for checking out the primality of any dubious number are efficient. Why? I do not know. \r\nIt works often.\r\nI think that there is a way to solve the primality if we exploit this path.\r\nWhat is sure is that if we find a function f(p,q) easily computable giving an output t [u]if and if only if[/u] p and q are both prime then the primality problem is done.\r\nI have found another function f(p,q) I'm testing it now.", "Solution_4": "Is this one working under the conditions that (p,s)=(q,t)=1 ?\r\nThank you for any comment.", "Solution_5": "Here is the rectified statement \r\n\r\np=2t+1 >3\r\nq=2s+1 >3\r\n\r\ngcd(s,p)=1\r\ngcd(t,q)=1\r\n\r\na=(t*s)^(t*s) mod (p*q)\r\n\r\nIf p and q are both prime then \r\n\r\na mod p = 1 or p-1\r\na mod q = 1 or q-1\r\n\r\nBut you have to understand that the statement is the basis of the test because you have to check Carmichael numbers.\r\nThat means that if you find the 1 or p-1 and 1 or q-1 you have to check any dubious number by using a set of known certified primes.\r\nThat is the new test.\r\n341 is going to pass but if you test with 89 which is a certified prime it is not going to pass.", "Solution_6": "Sorry the simplified statement I erased now is wrong.\r\nI made a mistake on my calculations.", "Solution_7": "There's such a thing as an edit button...", "Solution_8": "[quote=\"Matharith\"]p=2t+1 >3\nq=2s+1 >3\n\ngcd(s,p)=1\ngcd(t,q)=1\n\na=(t*s)^(t*s) mod (p*q)\n\nIf p and q are both prime then \n\na mod p = 1 or p-1\na mod q = 1 or q-1[/quote]\r\nThe statement is correct.\r\n\r\nIn the first place, we are going to prove that\r\n\\[ a^2 \\equal{} (ts)^{\\frac {(p \\minus{} 1)(q \\minus{} 1)}2}\\equiv 1\\pmod{pq}.\r\n\\]In order to do this, it is enough to prove that $ (ts)^{\\frac {(p \\minus{} 1)(q \\minus{} 1)}2}\\equiv 1\\pmod p$ and $ (ts)^{\\frac {(p \\minus{} 1)(q \\minus{} 1)}2}\\equiv 1\\pmod q$. Of course, these are analogous to each other, and therefore it suffices to prove the first one. Since $ p$ does not divide neither $ t$ nor $ s$, it follows $ (ts)^{p \\minus{} 1}\\equiv 1\\pmod p$ (by Fermat's little theorem), and therefore $ \\left((ts)^{p \\minus{} 1}\\right)^{\\frac {q \\minus{} 1}2}\\equiv 1\\pmod p$, what was to be proven.\r\n\r\nTherefore, $ a$ is a square root of $ 1$ modulo $ pq$. Since $ pq$ is the product of two different prime numbers, there are four different square roots of $ 1$ modulo $ pq$, all of which could be found by solving a system of the form\r\n\\[ \\left.\\begin{array}{c}a\\equiv\\pm 1\\pmod p \\\\\r\na\\equiv\\pm 1\\pmod q\\end{array}\\right\\},\r\n\\]what is precisely your conclusion." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "We have non-negative numbers such as: $ w\\plus{}x\\plus{}y\\plus{}z\\equal{}2$\r\n\r\nFind the maximum value of: $ A\\equal{}\\frac{w^2}{(w^2\\plus{}1)^2}\\plus{}\\frac{x^2}{(x^2\\plus{}1)^2}\\plus{}\\frac{y^2}{(y^2\\plus{}1)^2}\\plus{}\\frac{z^2}{(z^2\\plus{}1)^2}$\r\n\r\n\r\nI have only:\r\n* $ f(x)\\equal{}\\frac{x^2}{(x^2\\plus{}1)^2}\\equal{}\\frac{1}{(x\\plus{}\\frac{1}{x})^2}$ rises iff $ x \\in \\langle 0, 1 \\rangle$ and comes down iff $ x \\in \\langle 1,\\plus{}\\infty \\rangle$\r\n\r\n** it's easy to get that A is maximum iff $ w,x,y,z \\in \\langle 0,1 \\rangle$ (because $ f(x)\\equal{}f(\\frac{1}{x})$)\r\n\r\n*** so: f is convex iff $ x \\in \\langle 0, \\sqrt{\\frac{4\\minus{}\\sqrt{13}}{3}} \\rangle$ and concave iff $ x \\in \\langle \\sqrt{\\frac{4\\minus{}\\sqrt{13}}{3}}, 1 \\rangle$", "Solution_1": "[quote=\"Sylwek\"]We have non-negative numbers such as: $ w \\plus{} x \\plus{} y \\plus{} z \\equal{} 2$\n\nFind the maximum value of: $ A \\equal{} \\frac {w^2}{(w^2 \\plus{} 1)^2} \\plus{} \\frac {x^2}{(x^2 \\plus{} 1)^2} \\plus{} \\frac {y^2}{(y^2 \\plus{} 1)^2} \\plus{} \\frac {z^2}{(z^2 \\plus{} 1)^2}$\n\n\nI have only:\n* $ f(x) \\equal{} \\frac {x^2}{(x^2 \\plus{} 1)^2} \\equal{} \\frac {1}{(x \\plus{} \\frac {1}{x})^2}$ rises iff $ x \\in \\langle 0, 1 \\rangle$ and comes down iff $ x \\in \\langle 1, \\plus{} \\infty \\rangle$\n\n** it's easy to get that A is maximum iff $ w,x,y,z \\in \\langle 0,1 \\rangle$ (because $ f(x) \\equal{} f(\\frac {1}{x})$)\n\n*** so: f is convex iff $ x \\in \\langle 0, \\sqrt {\\frac {4 \\minus{} \\sqrt {13}}{3}} \\rangle$ and concave iff $ x \\in \\langle \\sqrt {\\frac {4 \\minus{} \\sqrt {13}}{3}}, 1 \\rangle$[/quote]\r\nWe have : \r\n$ \\frac {t^2}{(t^2 \\plus{} 1)^2} \\le\\ \\frac{48}{125}t\\minus{}\\frac{4}{125}$\r\n\r\n$ \\rightarrow A \\le\\ \\frac{16}{25}$", "Solution_2": "Thanks :) , I solved it, but I had to check four conditions, and your solution is simpler (we have only to check: $ t < 0,12$, which is easier than my sollution (I checked: if three, two, one and none of numbers w,x,y,z are smaller than $ \\sqrt{\\frac{4\\minus{}\\sqrt{13}}{3}}$), because your inequality is true f.e. for $ t \\in \\approx \\langle 0.119455591306, 1 \\rangle$ ).\r\n\r\nP.S. $ 4f(\\frac{1}{2})\\minus{}3f(\\frac{2}{3}) \\approx 0,001$, so we can't pass over this condition.", "Solution_3": "[quote=\"onlylove_math\"][quote=\"Sylwek\"]We have non-negative numbers such as: $ w \\plus{} x \\plus{} y \\plus{} z \\equal{} 2$\n\nFind the maximum value of: $ A \\equal{} \\frac {w^2}{(w^2 \\plus{} 1)^2} \\plus{} \\frac {x^2}{(x^2 \\plus{} 1)^2} \\plus{} \\frac {y^2}{(y^2 \\plus{} 1)^2} \\plus{} \\frac {z^2}{(z^2 \\plus{} 1)^2}$\n\n\nI have only:\n* $ f(x) \\equal{} \\frac {x^2}{(x^2 \\plus{} 1)^2} \\equal{} \\frac {1}{(x \\plus{} \\frac {1}{x})^2}$ rises iff $ x \\in \\langle 0, 1 \\rangle$ and comes down iff $ x \\in \\langle 1, \\plus{} \\infty \\rangle$\n\n** it's easy to get that A is maximum iff $ w,x,y,z \\in \\langle 0,1 \\rangle$ (because $ f(x) \\equal{} f(\\frac {1}{x})$)\n\n*** so: f is convex iff $ x \\in \\langle 0, \\sqrt {\\frac {4 \\minus{} \\sqrt {13}}{3}} \\rangle$ and concave iff $ x \\in \\langle \\sqrt {\\frac {4 \\minus{} \\sqrt {13}}{3}}, 1 \\rangle$[/quote]\nWe have : \n$ \\frac {t^2}{(t^2 \\plus{} 1)^2} \\le\\ \\frac {48}{125}t \\minus{} \\frac {4}{125}$\n\n$ \\rightarrow A \\le\\ \\frac {16}{25}$[/quote]\r\nAre you sure this inequality always true?\r\nBy the way, the max is $ \\frac{16}{25}$.\r\nAnd the exists a strange proof for it." } { "Tag": [ "geometry", "geometric transformation", "reflection", "inequalities unsolved", "inequalities" ], "Problem": "In which country is the magazine [i]Mathematical Reflections [/i]published?", "Solution_1": "This journal is published in USA (Texas university)\r\nBut it is free. You can look any issues you want with the following website:\r\n\r\n[url]http://reflections.awesomemath.org/[/url]" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c$ be positive reals. Prove that\r\n\r\n$ \\frac{3a^2\\minus{}2c^2\\plus{}bc}{b\\plus{}c}\\plus{}\\frac{3b^2\\minus{}2a^2\\plus{}ca}{c\\plus{}a}\\plus{}\\frac{3c^2\\minus{}2b^2\\plus{}ab}{a\\plus{}b}\\geq a\\plus{}b\\plus{}c$", "Solution_1": "taking a,b,c from RHS o LHS , we get\r\n\r\n$ 3\\sum{\\frac{c^{2}\\minus{}b^{2}}{a\\plus{}b}}$ which is greater than or equal to 0.\r\n\r\nhence proved" } { "Tag": [ "function" ], "Problem": "prove that the equation:\r\n\r\n$x(x+1)(x+2)...(x+n)=1$ has a postive solution that is less then $\\frac{1}{n!}$", "Solution_1": "[hide]\nWell let $f(x) = x(x+1)\\cdots(x+n)$. We have $f(0) = 0$ and $f\\left(\\frac{1}{n!}\\right) = \\frac{\\left(1+\\frac{1}{n!}\\right)\\left(2+\\frac{1}{n!}\\right)\\cdots\\left(n+\\frac{1}{n!}\\right)}{n!} > 1$ so there must be some solution between them (since $f$ is continuous).[/hide]", "Solution_2": "[quote=\"paladin8\"][hide]\nWell let $f(x) = x(x+1)\\cdots(x+n)$. We have $f(0) = 0$ and $f\\left(\\frac{1}{n!}\\right) = \\frac{\\left(1+\\frac{1}{n!}\\right)\\left(2+\\frac{1}{n!}\\right)\\cdots\\left(n+\\frac{1}{n!}\\right)}{n!} > 1$ so there must be some solution between them (since $f$ is continuous).[/hide][/quote]\r\n\r\nis this the mean value theorem or the intermediate value theorem?", "Solution_3": "[quote=\"xxreddevilzxx\"][quote=\"paladin8\"][hide]\nWell let $f(x) = x(x+1)\\cdots(x+n)$. We have $f(0) = 0$ and $f\\left(\\frac{1}{n!}\\right) = \\frac{\\left(1+\\frac{1}{n!}\\right)\\left(2+\\frac{1}{n!}\\right)\\cdots\\left(n+\\frac{1}{n!}\\right)}{n!} > 1$ so there must be some solution between them (since $f$ is continuous).[/hide][/quote]\n\nis this the mean value theorem or the intermediate value theorem?[/quote]\r\n\r\nDidn't know it had a name, but it's the Intermediate Value Theorem. The Mean Value Theorem is for calculus.", "Solution_4": "[quote=\"paladin8\"]\n $f\\left(\\frac{1}{n!}\\right) = \\frac{\\left(1+\\frac{1}{n!}\\right)\\left(2+\\frac{1}{n!}\\right)\\cdots\\left(n+\\frac{1}{n!}\\right)}{n!} > 1$ [/quote]\r\n\r\nprove this ;)", "Solution_5": "[quote=\"m509\"]prove that the equation:\n\n$x(x+1)(x+2)...(x+n)=1$ has a postive solution that is less then $\\frac{1}{n!}$[/quote]\r\n\r\n\r\nLet x=1: we get on the LHS $(n+1)!$; so divide that by n! and we get $1 1$ [/quote]\n\nprove this ;)[/quote]\r\n\r\nThe $LHS =\\frac{\\left(n!+1\\right) \\left(2\\cdot n!+1\\right) \\left(3 \\cdot n!+1\\right) ... \\left(n \\cdot n!+1\\right)}{\\left(n!\\right)^2}$\r\n\r\nThen $LHS > \\frac{n!(n!)^n}{\\left(n!\\right)^2} = (n!)^{n-1} \\ge 1$\r\n\r\nSo $LHS > 1$", "Solution_7": "[quote=\"m509\"][quote=\"paladin8\"]\n $f\\left(\\frac{1}{n!}\\right) = \\frac{\\left(1+\\frac{1}{n!}\\right)\\left(2+\\frac{1}{n!}\\right)\\cdots\\left(n+\\frac{1}{n!}\\right)}{n!} > 1$ [/quote]\n\nprove this ;)[/quote]\r\n\r\n****************************************************\r\nI think $(\\frac{1+\\frac{1}{n!}}{1})(\\frac{2+\\frac{1}{n!}}{2})....(\\frac{n+\\frac{1}{n!}}{n})$ is clear enough.", "Solution_8": "[hide]\nLet $f(x) = x(x + 1)(x + 2) \\dots (x + n) - 1$. Clearly, $f(x)$ is an increasing function for $x \\ge 0$, $f(0) = -1 < 0$, and $f(1) = (n + 1)! > 0$, so $f(x)$ has exactly one positive root $r$.\n\nFurthermore,\n\\[ r = \\frac{1}{(r + 1)(r + 2) \\dots (r + n)} < \\frac{1}{1 \\cdot 2 \\dots n} = \\frac{1}{n!}. \\]\n[/hide]" } { "Tag": [ "trigonometry", "inequalities", "geometry open", "geometry" ], "Problem": "Prove to all the triangle ABC, we have: $ \\tan^3\\frac{A}{2}\\plus{}\\tan^3\\frac{B}{2}\\plus{}\\tan^3\\frac{C}{2}\\ge\\frac{1}{\\sqrt3}$", "Solution_1": "$ \\tan^3\\frac{A}{2}\\plus{}\\tan^3\\frac{B}{2}\\plus{}\\tan^3\\frac{C}{2}\\ge\\frac{(tan\\frac{A}{2}\\plus{}\\tan\\frac{B}{2}\\plus{}\\tan\\frac{C}{2})}{9}^3$\r\nAnd by using Jensen inequality we have \r\n$ tan\\frac{A}{2}\\plus{}\\tan\\frac{B}{2}\\plus{}\\tan\\frac{C}{2}\\ge\\sqrt{3}$ :lol:", "Solution_2": "[quote=\"TRAN THAI HUNG\"]$ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2}\\ge\\frac {(tan\\frac {A}{2} \\plus{} \\tan\\frac {B}{2} \\plus{} \\tan\\frac {C}{2})}{9}^3$\nAnd by using Jensen inequality we have \n$ tan\\frac {A}{2} \\plus{} \\tan\\frac {B}{2} \\plus{} \\tan\\frac {C}{2}\\ge\\sqrt {3}$ :lol:[/quote]\r\nDifferentiate $ tan(x)$ twice and get: $ tan''(x) \\equal{} \\frac{2 \\cdot sin(x)}{(cos(x))^3}$, which is not always positive!\r\n\r\nThe proof should be more like: $ (tan^3(x))'' \\equal{} \\frac{6 \\cdot sin(x) \\cdot (sin^2(x)\\plus{}1)}{cos^5(x)}$. So we have $ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2} \\ge \\frac{1}{\\sqrt{3}}$ when $ A,B,C < \\pi$.\r\nNow assume $ A \\ge \\pi \\ge B \\ge C$ wlog because of symmetry.\r\nNow $ \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2} \\ge 2\\tan^3\\frac{B\\plus{}C}{4} \\equal{} 2\\tan^3\\frac{180\\minus{}A}{4}$.\r\nSo it is necessary to prove:\r\n$ \\tan^3\\frac{A}{2} \\plus{} 2\\tan^3\\frac{180\\minus{}A}{4} \\ge \\frac{1}{\\sqrt{3}}$.\r\nBut that's obvious since $ \\tan^3\\frac{A}{2} \\ge \\tan^3\\frac{90}{2} \\equal{} 1$.", "Solution_3": "[quote=\"Mathias_DK\"][quote=\"TRAN THAI HUNG\"]$ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2}\\ge\\frac {(tan\\frac {A}{2} \\plus{} \\tan\\frac {B}{2} \\plus{} \\tan\\frac {C}{2})}{9}^3$\nAnd by using Jensen inequality we have \n$ tan\\frac {A}{2} \\plus{} \\tan\\frac {B}{2} \\plus{} \\tan\\frac {C}{2}\\ge\\sqrt {3}$ :lol:[/quote]\nDifferentiate $ tan(x)$ twice and get: $ tan''(x) \\equal{} \\frac {2 \\cdot sin(x)}{(cos(x))^3}$, which is not always positive!\n\nThe proof should be more like: $ (tan^3(x))'' \\equal{} \\frac {6 \\cdot sin(x) \\cdot (sin^2(x) \\plus{} 1)}{cos^5(x)}$. So we have $ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2} \\ge \\frac {1}{\\sqrt {3}}$ when $ A,B,C < \\pi$.\nNow assume $ A \\ge \\pi \\ge B \\ge C$ wlog because of symmetry.\nNow $ \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2} \\ge 2\\tan^3\\frac {B \\plus{} C}{4} \\equal{} 2\\tan^3\\frac {180 \\minus{} A}{4}$.\nSo it is necessary to prove:\n$ \\tan^3\\frac {A}{2} \\plus{} 2\\tan^3\\frac {180 \\minus{} A}{4} \\ge \\frac {1}{\\sqrt {3}}$.\nBut that's obvious since $ \\tan^3\\frac {A}{2} \\ge \\tan^3\\frac {90}{2} \\equal{} 1$.[/quote]\r\n \r\n at first for 0 \\frac{A}{2},\\frac{B}{2},\\frac{C}{2} > 0$ - becouse A,B,C - are angles of triangle)", "Solution_4": "[quote=\"laisac\"]Prove to all the triangle ABC, we have: $ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2}\\ge\\frac {1}{\\sqrt3}$[/quote]\r\nDo not use a powerful tool :lol: \r\nWe have:$ \\tan^3\\frac{A}{2}\\plus{}\\tan^3\\frac{B}{2}\\ge 2\\sqrt{\\tan^3\\frac{A}{2}.\\tan^3\\frac{B}{2}}$ (*)\r\nBecause: $ \\tan\\frac{A}{2}.\\tan\\frac{B}{2}\\equal{}\\frac{\\sin\\frac{A}{2}.\\sin\\frac{B}{2}}{\\cos\\frac{A}{2}.\\cos\\frac{B}{2}}$$ \\equal{}\\frac{\\cos\\frac{A\\minus{}B}{2}\\minus{}\\cos\\frac{A\\plus{}B}{2}}{\\cos\\frac{A\\minus{}B}{2}\\plus{}\\cos\\frac{A\\plus{}B}{2}}$\r\n$ \\equal{}1\\minus{}\\frac{2\\cos\\frac{A\\plus{}B}{2}}{\\cos\\frac{A\\minus{}B}{2}\\plus{}\\cos\\frac{A\\plus{}B}{2}}$\r\n$ \\ge 1\\minus{}\\frac{2\\cos\\frac{A\\plus{}B}{2}}{1\\plus{}\\cos\\frac{A\\plus{}B}{2}}\\equal{}\\frac{1\\minus{}\\cos\\frac{A\\plus{}B}{2}}{1\\plus{}\\cos\\frac{A\\plus{}B}{2}}\\equal{}\\tan^2\\frac{A\\plus{}B}{4}$\r\n Inequalities (*) Write another:$ \\tan^3\\frac{A}{2}\\plus{}\\tan^3\\frac{B}{2}\\ge 2\\tan^3\\frac{A\\plus{}B}{4} (1)$\r\n Marked \"=\" when: A=B.\r\nSame $ \\tan^3\\frac{C}{2}\\plus{}\\tan^3\\frac{60^0}{2}\\ge 2\\tan^3\\frac{C\\plus{}60^0}{4} (2)$\r\n Marked \"=\" when:$ C\\equal{}60^0$\r\n(1)+(2):\r\n$ \\tan^3\\frac{A}{2}\\plus{}\\tan^3\\frac{B}{2}\\plus{}\\tan^3\\frac{C}{2}\\plus{}\\tan^3\\frac{60^0}{2}\\ge 2(\\tan^3\\frac{A\\plus{}B}{4} \\plus{}\\tan^3\\frac{C\\plus{}60^0}{4} )\\ge 4\\tan^3\\frac{A\\plus{}B\\plus{}C\\plus{}60^0}{8}\\equal{}4\\tan^330^0$\r\nOr $ \\tan^3\\frac{A}{2}\\plus{}\\tan^3\\frac{B}{2}\\plus{}\\tan^3\\frac{C}{2}\\ge 3\\tan^330^0\\equal{}\\frac{1}{\\sqrt3}$\r\n Marked \"=\" when: $ A\\equal{}B\\equal{}C\\equal{}60^0$.\r\nHelp!\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=240992", "Solution_5": "[quote=\"laisac\"]Do not use a powerful tool :lol: \nWe have:$ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2}\\ge 2\\sqrt {\\tan^3\\frac {A}{2}.\\tan^3\\frac {B}{2}}$ (*)\nBecause: $ \\tan\\frac {A}{2}.\\tan\\frac {B}{2} \\equal{} \\frac {\\sin\\frac {A}{2}.\\sin\\frac {B}{2}}{\\cos\\frac {A}{2}.\\cos\\frac {B}{2}}$$ \\equal{} \\frac {\\cos\\frac {A \\minus{} B}{2} \\minus{} \\cos\\frac {A \\plus{} B}{2}}{\\cos\\frac {A \\minus{} B}{2} \\plus{} \\cos\\frac {A \\plus{} B}{2}}$\n$ \\equal{} 1 \\minus{} \\frac {2\\cos\\frac {A \\plus{} B}{2}}{\\cos\\frac {A \\minus{} B}{2} \\plus{} \\cos\\frac {A \\plus{} B}{2}}$\n$ \\ge 1 \\minus{} \\frac {2\\cos\\frac {A \\plus{} B}{2}}{1 \\plus{} \\cos\\frac {A \\plus{} B}{2}} \\equal{} \\frac {1 \\minus{} \\cos\\frac {A \\plus{} B}{2}}{1 \\plus{} \\cos\\frac {A \\plus{} B}{2}} \\equal{} \\tan^2\\frac {A \\plus{} B}{4}$\n [/quote]\r\ni'm sorry you're wrong here but it should be:\r\n$ 1 \\minus{} \\frac {2\\cos\\frac {A \\plus{} B}{2}}{1 \\plus{} \\cos\\frac {A \\plus{} B}{2}}$\r\n$ \\ge1 \\minus{} \\frac {2\\cos\\frac {A \\plus{} B}{2}}{\\cos\\frac {A \\minus{} B}{2} \\plus{} \\cos\\frac {A \\plus{} B}{2}}$", "Solution_6": "[quote=\"lizhi\"]\n\ni'm sorry you're wrong here but it should be:\n$ 1 \\minus{} \\frac {2\\cos\\frac {A \\plus{} B}{2}}{1 \\plus{} \\cos\\frac {A \\plus{} B}{2}}$\n$ \\ge1 \\minus{} \\frac {2\\cos\\frac {A \\plus{} B}{2}}{\\cos\\frac {A \\minus{} B}{2} \\plus{} \\cos\\frac {A \\plus{} B}{2}}$[/quote]\r\nSorry, I was mistaken. Thank you.\r\nIn other :lol: \r\nWe have $ \\frac{{tg^3 \\frac{A}{2} \\plus{} tg^3 \\frac{B}{2}}}{2} \\ge \\left( {\\frac{{tg\\frac{A}{2} \\plus{} tg\\frac{B}{2}}}{2}} \\right)^3$(1)\t\r\nWe have $ tg\\frac{A}{2} \\plus{} tg\\frac{B}{2}\\equal{}\\frac{{\\sin \\frac{{A \\plus{} B}}{2}}}{{\\cos \\frac{A}{2}\\cos \\frac{B}{2}}} \\equal{} \\frac{{2\\sin \\frac{{A \\plus{} B}}{2}}}{{\\cos \\frac{{A \\minus{} B}}{2} \\plus{} \\cos \\frac{{A \\plus{} B}}{2}}}$\r\n$ \\ge \\frac{{4\\sin \\frac{{A \\plus{} B}}{4}\\cos \\frac{{A \\plus{} B}}{4}}}{{1 \\plus{} \\cos \\frac{{A \\plus{} B}}{2}}} \\equal{} 2tg\\frac{{A \\plus{} B}}{4}$\r\nSo: $ tg^3 \\frac{A}{2} \\plus{} tg^3 \\frac{B}{2} \\ge 2tg^3 \\frac{{A \\plus{} B}}{4}$\t(2)\r\nSimilarly, we have:$ tg^3 \\frac{C}{2} \\plus{} tg^3 \\frac{{60^0 }}{2} \\ge 2tg^3 \\frac{{C \\plus{} 60^0}}{4}$ \t(3)\r\nThe Community (2) and (3) we have \r\n$ tg^3 \\frac{A}{2} \\plus{} tg^3 \\frac{B}{2} \\plus{} tg^3 \\frac{C}{2} \\plus{} tg^3 \\frac{{60^0 }}{2} \\equal{} 2(tg^3 \\frac{{A \\plus{} B}}{4} \\plus{} tg^3 \\frac{{C \\plus{} 60^0 }}{4})$\r\n$ \\ge 4tg^3 \\frac{{A \\plus{} B \\plus{} C \\plus{} 60^0 }}{8} \\equal{} 4tg^3 \\frac{{60^0 }}{2}$\r\nOr : $ tg^3 \\frac{A}{2} \\plus{} tg^3 \\frac{B}{2} \\plus{} tg^3 \\frac{C}{2} \\ge 3tg^3 30^0 \\equal{}\\frac{1}{{\\sqrt 3 }}$(*)\t\r\nEquality, when $ A \\equal{} B \\equal{} C \\equal{} 60^0$", "Solution_7": "[quote=\"laisac\"] [color=darkred]Prove that for any triangle ABC, we have $ \\tan^3\\frac {A}{2} \\plus{} \\tan^3\\frac {B}{2} \\plus{} \\tan^3\\frac {C}{2}\\ge\\frac {1}{\\sqrt3}$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof[/u].[/b] $ \\sum\\tan^3\\frac A2 \\equal{} \\sum\\left(\\frac {r_a}{p}\\right)^3 \\equal{} \\frac {\\sum r_a^3}{p^3}\\ge\\frac {\\left(\\sum r_a\\right)^3}{9p^3} \\equal{} \\frac {(4R \\plus{} r)^3}{9p^3}\\ge \\frac {\\left(p\\sqrt 3\\right)^3}{9p^3} \\equal{} \\frac {1}{\\sqrt 3}$ .\n\n[b][u]Remark[/u].[/b] Il used the following well-known relations : \n\n$ \\odot\\ \\boxed {\\ r_a \\equal{} p\\tan\\frac A2\\ }$ a.s.o. ; $ \\begin{array}{c} \n\\boxed {\\ S\\equal{}pr\\equal{}(p\\minus{}a)r_a\\equal{}(p\\minus{}b)r_b\\equal{}(p\\minus{}c)r_c\\ }\\\\\\\\\n \\boxed{\\ \\begin{array}{ccc}\nr_a \\plus{} r_b \\plus{} r_c & \\equal{} & 4R \\plus{} r\\\\\\\nr_ar_b \\plus{} r_br_c \\plus{} r_cr_a & \\equal{} & p^2 \\\\\n\\ r_ar_br_c & \\equal{} & p^2r\\end{array}\\ }\\end{array}$ .\n\n$ \\odot\\ \\boxed {\\ \\{x,y,z\\}\\subset (0,\\infty )\\ \\implies\\ (x \\plus{} y \\plus{} z)^3\\le 9\\left(x^3 \\plus{} y^3 \\plus{} z^3\\right)\\ }\\ (*)$ .\n\nIndeed, $ \\left\\|\\ \\begin{array}{ccc} \\left(x \\plus{} y \\plus{} z\\right)\\left(x \\plus{} y \\plus{} z\\right) & \\le & 3\\left(x^2 \\plus{} y^2 \\plus{} z^2\\right) \\\\\n \\\\\n\\left(x \\plus{} y \\plus{} z\\right)\\left(x^2 \\plus{} y^2 \\plus{} z^2\\right) & \\le & 3\\left(x^3 \\plus{} y^3 \\plus{} z^3\\right)\\end{array}\\ \\right\\|\\ \\bigodot\\ \\implies\\ (*)$ .\n\n$ \\odot\\ \\boxed {\\ p\\sqrt 3\\ \\le\\ 4R \\plus{} r\\ }\\ (**)$ .\n\nIndeed, $ 3\\left(r_ar_b \\plus{} r_br_c \\plus{} r_cr_a\\right)\\le\\left(r_a \\plus{} r_b \\plus{} r_c\\right)^2\\ \\implies\\ 3p^2\\le (4R \\plus{} r)^2\\ \\implies\\ (**)$ .[/color]" } { "Tag": [], "Problem": "Prove that\r\n\r\nIn any integer sided right triangle, the product of the altitude and base is a multiple of 12.", "Solution_1": "let the sides be $ a,b,c$\r\n$ a^{2} \\plus{} b^{2} \\equal{} c^{2}$ by Pythagoras theorem\r\nwhen a perfect square is divided by 12 the possible remainders are $ 0,1,9,4$\r\nout of these 4 remainders \r\nonly when $ a^{2}$ leaves remainder 4 and $ b^{2}$ leaves remainder 9 ,when $ a^{2}$ leaves 0 and $ b^{2}$ leaves 0,when $ a^{2}$ leaves 0 and $ b^{2}$ leaves 1 ,only out of these three cases $ c^{2}$ has the required form when a perfect square is divided by 12\r\nand for the remaining combinations of remainders of a and b , $ c^{2}$ has a remainder which a perfect sqauare cant have when it is divided by12.\r\nand out of the three valid cases in all $ a*b \\equiv 0\\mod12$.it is obvious that a can be the base or altitude ,b can be base or altitude, and c is the hypotenuse\r\nhence proved", "Solution_2": "also you can use [url=http://en.wikipedia.org/wiki/Pythagorean_triple]primitive pythagorean triplets[/url]\r\n\r\none of $ 2mn$ or $ m^2\\minus{}n^2$ is divisible by $ 3$ and one is divisible by $ 4$ for all $ m,n$" } { "Tag": [ "geometry", "circumcircle", "inradius", "trigonometry", "incenter", "Pythagorean Theorem", "geometry proposed" ], "Problem": "Given triangle $ABC$ circumcircle $(O),A',B',C'$ are circumceter of triangle $OBC,OCA,OAB$ and $A'',B'',C''$ are symetric point of $A',B',C'$ with lines $BC,CA,AB$ resp prove that in four circles (three excircles and incircle of triangle $A''B''C''$) always exist a circle pass through $E$ - ninepoint center of $ABC$ and in four number (three exradius and inradius of triangle $A''B''C''$) always exist a number equal to $\\frac{1}{2}OH$ with $H$ is orthocenter of $ABC$.", "Solution_1": "Compare with the 1st solution of the Furhmann triangle problem at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101687[/url] (no hint was given about Fuhrmann at the time).\r\n\r\nAssume $\\triangle ABC$ is acute. B'C', C'A', A'B' are perpendicular bisectors of OA, OB, OC, $\\triangle A'B'C'$ is similar to the tangential triangle with similarity center O and coefficient 1/2. With usual notation,\r\n\r\n$A'B' = R\\ \\frac{\\tan A+\\tan B}{2}= R\\ \\frac{\\sin C}{2 \\cos A \\cos B}$\r\n\r\n$OA'' = \\frac{R}{2 \\cos A}-2R \\cos A = R\\ \\frac{1-4 \\cos^{2}A}{2\\cos A}$\r\n\r\n$OB'' = R\\ \\frac{1-4 \\cos^{2}B}{2\\cos B}$\r\n\r\nBy the cosine law for the $\\triangle OA''B'',$\r\n\r\n$\\frac{A''B''^{2}}{A'B'^{2}}= \\frac{OA''^{2}+OB''^{2}+2\\ OA'' \\cdot OB''\\ \\cos C}{A'B'^{2}}=$\r\n\r\n$\\frac{(1-4 \\cos^{2}A) \\cos^{2}B+(1-4 \\cos^{2}B) \\cos^{2}A+2 (1-4 \\cos^{2}A)(1-4 \\cos^{2}B) \\cos A \\cos B \\cos C}{\\sin^{2}C}=$\r\n\r\n(substituting $2 \\cos A \\cos B \\cos C = 1-(\\cos^{2}A+\\cos^{2}B+\\cos^{2}C)$)\r\n\r\n$= \\frac{(1-\\cos^{2}C)-4 \\cos^{2}A (1-\\cos^{2}C)-4 \\cos^{2}B (1-\\cos^{2}C)+16 \\cos^{2}A \\cos^{2}B (1-\\cos^{2}C)+4 (\\cos^{2}A-\\cos^{2}B)^{2}}{\\sin^{2}C}=$\r\n\r\n$= 1-4 \\cos^{2}A-4 \\cos^{2}B+16 \\cos^{2}A \\cos^{2}B+4 \\sin^{2}(A-B) =$\r\n\r\n$= 1+8 \\cos A \\cos B (\\cos A \\cos B-\\sin A \\sin B) = 1-8 \\cos A \\cos B \\cos C$\r\n\r\nIf D is foot of the A-altitude, power of H to (O) is\r\n\r\n$R^{2}-HO^{2}= 2 HD \\cdot HA = 8R^{2}\\cos A \\cos B \\cos C$\r\n\r\nTherefore $\\frac{A''B''}{A'B'}= \\frac{HO}{R}$ and similarly, $\\frac{B''C''}{B'C'}= \\frac{C''A''}{C'A'}= \\frac{HO}{R},$ $\\triangle A''B''C'' \\sim \\triangle A'B'C'$ with coefficient $\\frac{HO}{R}.$ If parallel to BC through H meet A'A'' at X,\r\n\r\n$XH = \\frac{|b^{2}-c^{2}|}{2a}= R\\ \\frac{|\\cos^{2}B-\\cos^{2}C|}{\\sin A}= R |\\sin (B-C)|$\r\n\r\n$XA'' = \\frac{R}{2 \\cos A}-R \\cos A-2R \\cos B \\cos C = R\\ \\frac{1-2 \\cos^{2}A-4 \\cos A \\cos B \\cos C}{2\\cos A}$\r\n\r\n$OA' = \\frac{R}{2 \\cos A}$\r\n\r\nBy Pythagorean theorem for the $\\triangle HXA'',$\r\n\r\n$\\frac{HA''^{2}}{OA'^{2}}= \\frac{XH^{2}+XA''^{2}}{OA'^{2}}=$\r\n\r\n${= (1-2 \\cos^{2}A-4 \\cos A \\cos B \\cos C)^{2}+4 \\sin^{2}(B-C)}\\cos^{2}A=$\r\n \r\n$= 1-8 \\cos A \\cos B \\cos C-4 \\cos^{2}A (1-\\cos^{2}A)+16 \\cos^{2}A \\cos^{2}B \\cos^{2}C+$\r\n$+4 \\cos^{2}A (2 \\cos A \\cos B \\cos C)+4 \\cos A (1-\\cos^{2}A-\\cos^{2}B-\\cos^{2}C)+$\r\n$+4\\cos^{2}A \\cos^{2}B+4 \\cos^{2}A \\cos^{2}C-8 \\cos^{2}A \\cos B^{2}\\cos C^{2}-$\r\n$-8 \\cos^{2}A \\sin B \\sin C \\cos B \\cos C=$\r\n\r\n$= 1-8 \\cos A \\cos B \\cos C+4 \\cos^{2}A (2 \\cos A \\cos B \\cos C)+$\r\n$+8 \\cos^{2}A \\cos B \\cos C (\\cos B \\cos C-\\sin B \\sin C) =$\r\n\r\n$= 1-8 \\cos A \\cos B \\cos C = \\frac{HO^{2}}{R^{2}}$\r\n\r\nTherefore $\\frac{HA''}{OA'}= \\frac{HO}{R}$ and similarly, $\\frac{HB''}{OB'}= \\frac{HC''}{OC'}= \\frac{HO}{R}.$ Since O is the incenter of $\\triangle A'B'C',$ H is the incenter of the $\\triangle A''B''C''.$ R is inradius of the tangential triangle, R/2 inradius of the $\\triangle A'B'C'$ and OH/2 inradius of the $\\triangle A'' B''C''.$ Incircle of the last triangle cuts OH in half, passing through the 9-point center of the original $\\triangle ABC.$ If the $\\triangle ABC$ is obtuse, O is the appropriate excenter of the tangential triangle and of the $\\triangle A'B'C'$ and H is the corresponding extenter of the $\\triangle A''B''C''.$" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let the roots of the cubic equation $ x^3 \\plus{} ax^2 \\plus{} bx \\plus{} c \\equal{} 0$ be all real.\r\nShow that the difference between the greatest and last of them is not less than $ \\sqrt {a^2 \\minus{} 3b}$ and not greater than $ 2\\sqrt {a^2 \\minus{} 3b}$.", "Solution_1": "Let $ x_1, x_2, x_3$ be roots of this polynomial and $ x_1 \\leq x_2 \\leq x_3$. Applying Viete'a we have\r\n \\[ \\sqrt{a^2\\minus{}3b}\\equal{}\\sqrt{x_1^2\\plus{}x_2^2\\plus{}x_3^2\\minus{}x_1x_2 \\minus{} x_2x_3 \\minus{} x_3 x_1} \\le x_3 \\minus{} x_1\\]\r\nafter squaring we get $ (x_2 \\minus{} x_1 )(x_2 \\minus{} x_3 ) \\le 0$\r\n\r\n\\[ x_3 \\minus{} x_1 \\le 2\\sqrt{a^2\\minus{}3b}\\equal{}2\\sqrt{x_1^2\\plus{}x_2^2\\plus{}x_3^2\\minus{}x_1x_2 \\minus{} x_2x_3 \\minus{} x_3 x_1}\\]\r\n\r\nafter aquaring we get $ 0\\leq 2(x_1\\minus{}x_2)^2 \\plus{} \\left(2x_2 \\minus{}x_1 \\minus{} x_3 \\right)^2$" } { "Tag": [ "calculus", "integration", "number theory", "prime factorization" ], "Problem": "What is the smallest number that contains precisely 18 integral factors other than one or the number itself.", "Solution_1": "[hide] 18 factors not including 1 and itself gives 20 factors including everything. So, if the number was $p_{1}^{e_{1}}+p_{2}^{e_{2}}...p_{n}^{e_{n}}$, the number of factors would be $(e_{1}+1)(e_{2}+1)(e_{3}+1)...(e_{n}+1)=20$, since there are 20 factors. $20=2^{2}*5$, so the lowest number would be $2^{5-1}*3^{2-1}*5^{2-1}=16*3*5=\\boxed{240}$ [/hide]", "Solution_2": "English please?\r\n :blush:", "Solution_3": "[quote=\"happyme\"]English please?\n :blush:[/quote]\r\nOkay, here is a specific case.\r\nConsider how many factors 100 has.\r\n100 is prime factorized as $2^{2}\\cdot 5^{2}$.\r\nAny factor of 100 can be constructed with 2s and 5s.\r\nYou can have either 0, 1, or 2 2s. You can have either 0, 1, or 2 5s.\r\n3*3=9\r\nSo 100 has 9 factors.", "Solution_4": "[hide]\ni think the lowest would be something to the fourth times two other primes\n\n16*3*5=240\n\nis that it?\n\n[/hide]\r\njorian", "Solution_5": "My answer and complete explanation:\r\n\r\n[hide]\nI got $240$.\n\nThe number of factors a number has including 1 and itself is found by taking the prime factorization to a number, adding one to each of the exponents, and multiplying the exponents together.\n\nHowever, here we are not including 1 and itself, so add 2 to 18, making up for 1 and itself.\n\nNow we need to find numbers that multiply to get 18.\nI got: 2 x 2 x 5\n\nNow subtract 1 from each.\n\nWe have 1, 1, and 4.\n\nNow what is the smallest prime? 2.\nSo lets give 2 the power of 4, since we are aiming for the smallest possible number.\n\nThen what are the next to primes? 3 and 5. They both get a power of 1.\n\n$2^{4}* 3^{1}* 5^{1}= 240$\n\n[/hide]\r\n\r\nI hope this is right :)", "Solution_6": "[hide]\n$x^{x_{1}}\\cdot y^{y_{1}}\\cdot z^{z_{1}}=n$ where x, y, and z are prime factors. The number of prime factors is $(x_{1}+1) \\cdot (y_{1}+1) \\cdot (z_{1}+1)=20$, since we have to add two to 18 for one and the number. The possible values are 1, 1, 20; 1, 2, 10; 2, 2, 5; 1, 4, 5.\n\nThe one that has the closest terms is the smallest one. \n$2^{4}\\cdot 3^{1}\\cdot 5^{1}= \\boxed{240}$[/hide]", "Solution_7": "[quote=\"bpms\"][quote=\"happyme\"]English please?\n :blush:[/quote]\nOkay, here is a specific case.\nConsider how many factors 100 has.\n100 is prime factorized as $2^{2}\\cdot 5^{2}$.\nAny factor of 100 can be constructed with 2s and 5s.\nYou can have either 0, 1, or 2 2s. You can have either 0, 1, or 2 5s.\n3*3=9\nSo 100 has 9 factors.[/quote]\r\n\r\nI see. Thanks." } { "Tag": [ "inequalities", "trigonometry", "algebra unsolved", "algebra" ], "Problem": "$0 \\leqslant x \\leqslant \\frac{\\pi}{2}$\r\n$\\frac{\\pi}{6} \\leqslant \\alpha \\leqslant \\frac{\\pi}{3}$\r\nProve that:\r\n$\\tan({\\frac{\\pi\\sin{x}}{4\\sin{\\alpha}}) + \\tan({\\frac{\\pi\\cos{x}}{4\\cos{\\alpha}}) > 1}}$", "Solution_1": "anyone ?" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "search", "AMC", "USA(J)MO" ], "Problem": "Let $ a$, $ b$, and $ c$ denote three distinct integers and let $ P$ denote a polynomial having all integral coefficients. Show that it is impossible that $ P(a)\\equal{}b$, $ P(b)\\equal{}c$, and $ P(c)\\equal{}a$.", "Solution_1": "Posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1931887680&t=103046]here[/url].", "Solution_2": "Suppose $ a$, $ b$, and $ c$ are distinct integers such that $ P(a)\\equal{}b$, $ P(b)\\equal{}c$, and $ P(c)\\equal{}a$. Then each of $ a$, $ b$, and $ c$ are (distinct) roots of the polynomial $ P(P(P(x))) \\minus{} x$, and are not roots of the polynomial $ P(x)\\minus{}x$. Since each value of $ y$ with $ P(y)\\equal{}y$ also clearly has $ P(P(P(y)))\\equal{}y$, the polynomial $ P(P(P((x)))\\minus{}x$ is divisible by the polynomial $ P(x)\\minus{}x$. They have the same constant coefficient by inspection, and hence the quotient $ Q(x) \\equal{} \\frac{P(P(P((x)))\\minus{}x}{P(x)\\minus{}x}$ will have a constant coefficient of 1.\r\n\r\nBy assumption, each of $ a$, $ b$, and $ c$ are integers and are roots of the polynomial $ Q(x)$ which has integral coefficients by the division algorithm. [It is also possible to verify directly that $ Q(x)$ has integral coefficients by writing it out in terms of $ P(x)$.] The rational root theorem implies that each of $ a$, $ b$, and $ c$ divides the constant coefficient of $ R(x)$, which is 1. But since there are only two distinct integers which divide 1, two of $ a$, $ b$, and $ c$ must be equal. This is a contradiction, and therefore no such $ a$, $ b$, and $ c$ can exist.", "Solution_3": "My first USAMO proof!!! :D \r\n\r\nAll the terms of a given polynomial are divisible by x except for the constant term, so we can write the polynomial as $ P(x)\\equal{}nx\\plus{}r$, where n is an expression in terms of x and r is the constant term. By the problem, n and r are both integers. So, we are trying to prove that $ P(P(P(a)))$ cannot equal a. $ P(x)$ is $ nx\\plus{}r$, so $ P(P(P(a)))\\equal{}n(n(na\\plus{}r)\\plus{}r)\\plus{}r\\equal{}n^3a\\plus{}[n(n\\plus{}1)\\plus{}1]r\\equal{}n^3a\\plus{}(n^2\\plus{}n\\plus{}1)r$.\r\n\r\nSetting that equal for a and looking for a contradiction, if $ n^3a\\plus{}(n^2\\plus{}n\\plus{}1)r\\equal{}a, then (n^3\\minus{}1)a\\plus{}(n^2\\plus{}n\\plus{}1)r\\equal{}0$. This factors to $ (n\\minus{}1)(n^2\\plus{}n\\plus{}1)a\\plus{}(n^2\\plus{}n\\plus{}1)r\\equal{}0$, or $ [(n\\minus{}1)a\\plus{}r](n^2\\plus{}n\\plus{}1)\\equal{}0$. This implies that either $ n^2\\plus{}n\\plus{}1\\equal{}0$ or $ (n\\minus{}1)a\\plus{}r\\equal{}0$. Checking the roots of $ n^2\\plus{}n\\plus{}1\\equal{}0$, we find that neither is an integer, so that is eliminated. Working with $ (n\\minus{}1)a\\plus{}r\\equal{}0$, this expands to $ na\\minus{}a\\plus{}r\\equal{}0$, or $ na\\plus{}r\\equal{}a$. That means, $ P(a)\\equal{}a$. But wait, $ P(a)\\equal{}b$, and that would mean $ b\\equal{}a$. But the problem states that a, b, and c are distinct, which is a contradiction.", "Solution_4": "[quote=\"Textangle\"]$ P(P(P(a))) \\equal{} n(n(na \\plus{} r) \\plus{} r) \\plus{} r \\equal{} n^3a \\plus{} [n(n \\plus{} 1) \\plus{} 1]r \\equal{} n^3a \\plus{} (n^2 \\plus{} n \\plus{} 1)r$.[/quote]\r\nThis is valid if and only if $ n$ is a constant; otherwise, because $ n$ is a polynomial, substituting $ P(a)$ into $ P(x)$ changes $ n$ itself. In other words, you've only proven the result if $ P$ is a linear polynomial.", "Solution_5": "@chaucerchu: Your claim that $ P(P(P(x))) \\minus{} x$ and $ P(x) \\minus{} x$ have the same constant term is invalid -- take $ P(x) \\equal{} x \\plus{} 1$.", "Solution_6": "yes it only true if $ P(P(P(0))) \\equal{} P(0)$" } { "Tag": [ "inequalities", "algorithm", "inequalities proposed" ], "Problem": "Suppose x1,x2,x3,x4,x5,x6 and x7 are non-negative integers and S denotes the sum of these seven integers. Then we have the following inequality:\r\n (x4+x7)/(S-x3)+(x5+x7)/(S-x1)-(x6+x7)/(S-x2)<=1.\r\n My question is: can we prove this inequality mechanically? i.e., an algorithm can be given to test whether this inequality holds.", "Solution_1": "[quote=\"yao\"]Suppose x1,x2,x3,x4,x5,x6 and x7 are non-negative integers and S denotes the sum of these seven integers. Then we have the following inequality:\n (x4+x7)/(S-x3)+(x5+x7)/(S-x1)-(x6+x7)/(S-x2)<=1.\n My question is: can we prove this inequality mechanically? i.e., an algorithm can be given to test whether this inequality holds.[/quote]\r\n\r\n$\\frac{x_{4}+x_{7}}{S-x_{3}}+\\frac{x_{5}+x_{7}}{S-x_{1}}-\\frac{x_{6}+x_{7}}{S-x_{2}}\\leq$\r\n$\\leq\\frac{x_{4}+x_{3}+x_{7}}{S}+\\frac{x_{1}+x_{5}+x_{7}}{S}+\\frac{x_{2}+x_{6}-x_{7}}{S}=1.$ :)", "Solution_2": "Thank arqady. But I know this proof, my question is if I dot not know how to prove, whether can I verify this inequality only using computer?" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove or disprove:\r\nFor positive reals $a_1,\\ldots,a_n$ and $x_1,\\ldots,x_n$ the inequality\r\n\r\n\\[ \\frac{a_1}{x_1}+\\ldots+\\frac{a_n}{x_n}\\geq \\frac{n(a_1+ \\ldots + a_n)}{x_1 + \\ldots +x_n} \\]\r\n\r\nholds.", "Solution_1": "No, this is not true. Say $a_i=x_iy_i$ and think about the condition for Chebyshev Inequality." } { "Tag": [ "calculus", "integration", "trigonometry", "function", "complex analysis", "real analysis", "real analysis unsolved" ], "Problem": "1.) The integral of sin(tx)/(pi*(1+x^2)) dx from negative infinity to positive infinity. (should be 0 but don't know how to get it.)\r\n\r\n2.) The integral of cos(tx)/(pi*(1+x^2)) dx from negative infinity to positive infinity. (should be e^-abs.(t), but again don't know how to get there.)", "Solution_1": "(1) is $ \\int_{ \\minus{} \\infty}^{\\infty}\\frac {\\sin tx}{\\pi(1 \\plus{} x^2)}\\,dx$ and (2) is $ \\int_{ \\minus{} \\infty}^{\\infty}\\frac {\\cos tx}{\\pi(1 \\plus{} x^2)}\\,dx$\r\n\r\nBoth integrals converge absolutely by comparison to $ \\int_{ \\minus{} \\infty}^{\\infty}\\frac {1}{\\pi(1 \\plus{} x^2)}\\,dx,$ so we may take that for granted.\r\n\r\nWe can do (1) very easily: the integrand is an odd function, and the interval of integration is symmetric. In such a case, as long as the integral converges, the integral is zero.\r\n\r\nWe need to do more work for (2). This is, of course, a Fourier transform. One pathway is to assume that we know the appropriate theorems and formulas for the inversion of Fourier transforms and simply compute the inverse Fourier transform of $ e^{ \\minus{} |t|}.$ I regard that as a perfectly legitimate way to do the problem - but it does rely on two things: knowing the formula for Fourier inversion, and knowing the answer to this particular problem ahead of time.\r\n\r\nMy second suggestion is to use the residue theorem from complex analysis to compute the integral along the real axis of $ \\frac {e^{itz}}{\\pi(1 \\plus{} z^2)}.$ Why do we get one answer for $ t > 0$ and a different answer for $ t < 0?$ Because we must take the return path through whichever of the upper or lower half planes that exponential is small in." } { "Tag": [ "linear algebra", "matrix" ], "Problem": "1)Let $ A,B$ be complex matrices of size $ 3$x$ 3$ .Prove that :\r\n $ \\det(AB\\minus{}BA)\\equal{}Tr(AB(AB\\minus{}BA)BA)$\r\n 2) Let $ A$ and B be 2 by 2 matrices with integer entries such that $ AB\\equal{}BA$ and $ \\det(B)\\equal{}1$\r\nProve that if $ \\det(A^3\\plus{}B^3)\\equal{}1$ then $ A^2\\equal{}I$\r\n 3) Let $ A$ be a $ 2$x$ 2$ matrix with complex entries and $ C(A)\\equal{}\\{B \\in M_2{(\\mathbb{C})}\\}$\r\n Show that $ |\\det(A\\plus{}B)| \\geq |\\det(B)| , \\forall B \\in C(A)$\r\n If and only if $ A^2\\equal{}O$", "Solution_1": "For problem number $ 2$, note that, if $ A \\equal{} 0$, $ B \\equal{} I$, then $ A$ and $ B$ are matrix with integer entries, $ AB \\equal{} BA$, $ det(B) \\equal{} 1$ and $ det(A^3 \\plus{} B^3) \\equal{} 1$, but $ A^2$ is not an identity matrix, so maybe we should add some statement, such as \"... $ A$ is an invertible matrix\", or other statement.\r\nI am sorry if I am wrong...", "Solution_2": "[quote=\"Qruni\"]For problem number $ 2$, note that, if $ A \\equal{} 0$, $ B \\equal{} I$, then $ A$ and $ B$ are matrix with integer entries, $ AB \\equal{} BA$, $ det(B) \\equal{} 1$ and $ det(A^3 \\plus{} B^3) \\equal{} 1$, but $ A^2$ is not an identity matrix, so maybe we should add some statement, such as \"... $ A$ is an invertible matrix\", or other statement.\nI am sorry if I am wrong...[/quote]\r\n Sorry Prove that $ A^2 \\equal{} O_2$ :oops:\r\n [color=red]I Edited [/color]: 2) Let $ A$ and $ B$ be $ 2$ x $ 2$ matrices with integer entries such that $ AB \\equal{} BA$ and $ det(B) \\equal{} 1$ \r\nProve that if $ det(A^3 \\plus{} B^3) \\equal{} 1$ then $ A^2\\equal{}O_2$", "Solution_3": "[quote=\"QuyBac\"] 2) Let $ A$ and $ B$ be $ 2$ x $ 2$ matrices with integer entries such that $ AB \\equal{} BA$ and $ det(B) \\equal{} 1$ \nProve that if $ det(A^3 \\plus{} B^3) \\equal{} 1$ then $ A^2 \\equal{} O_2$[/quote]\r\n\r\nThe \"edited\" second:\r\n\r\nFirst we notice that $ det(B^{ \\minus{} 1}) \\equal{} 1$\r\n\r\nThen we have:\r\n\r\n$ det(A^3 \\plus{} B^3) \\equal{} det(B^{ \\minus{} 1})^3 det((AB^{ \\minus{} 1})^3 \\plus{} I) \\equal{} 1 \\rightarrow det((AB^{ \\minus{} 1})^3 \\plus{} I) \\equal{} 1$\r\n\r\nSetting $ X \\equal{} AB^{ \\minus{} 1}$, we have $ det(X^3 \\plus{} I) \\equal{} det(X \\plus{} I)det(X \\plus{} wI)det(X \\plus{} w^2I) \\equal{} 1$ where w is a third root of unity. \r\n\r\nWe know that: \r\n$ det(X \\minus{} sI) \\equal{} s^2 \\minus{} tr(X)s \\plus{} det(X)$ \r\n\r\nSetting det(X)=D, tr(X)=T we see that T is an integer since $ B^{ \\minus{} 1} \\equal{} \\frac {1} {detB} Adj(B)$ and therefore has integer entries... (obviously D is an integer as well)\r\n\r\n$ det(X \\plus{} wI) \\equal{} w^2 \\plus{} Tw \\plus{} D \\equal{} \\overline{w} \\plus{} T \\overline {w^2} \\plus{} D \\equal{} \\overline{det(X \\plus{} w^2I)} \\\\\r\n\\rightarrow det(X \\plus{} wI)det(X \\plus{} w^2I) \\in R_ \\plus{}$\r\n\r\nThen $ det(X \\plus{} I) \\equal{} 1 \\equal{} det(X \\plus{} wI)det(X \\plus{} w^2I)$ \r\n\r\nand we have:\r\n\r\n$ 1 \\plus{} T \\plus{} D \\equal{} 1 \\rightarrow D \\equal{} \\minus{} T \\\\\r\n(w^2 \\plus{} Tw \\plus{} D)(w \\plus{} Tw^2 \\plus{} D) \\equal{} 1 \\rightarrow D^2(2 \\minus{} w \\plus{} w^2) \\equal{} 0 \\rightarrow D \\equal{} T \\equal{} 0$\r\n\r\nThen from Caley-Hamilton: \r\n$ X^2 \\minus{} X \\cdot trX \\plus{} det(X) \\equal{} 0 \\rightarrow X^2 \\equal{} 0 \\rightarrow (AB^{ \\minus{} 1})^2 \\equal{} 0$\r\n\r\nBut $ AB \\equal{} BA \\rightarrow B^{ \\minus{} 1}A \\equal{} AB^{ \\minus{} 1}$ yielding A^2B^{-2}=0\r\n\r\nMultiplying on the right by B we get $ A^2 \\equal{} 0_2$", "Solution_4": "For 2) new version.\r\n$ A^3\\plus{}B^3\\equal{}(A\\plus{}B)(A^2\\plus{}B^2\\minus{}AB)$, then $ det(A\\plus{}B)\\equal{}\\epsilon,det((B\\plus{}jA)(B\\plus{}j^2A))\\equal{}\\epsilon$ where $ j^2\\plus{}j\\plus{}1\\equal{}0$ and $ \\epsilon\\equal{}\\pm{1}$. $ \\epsilon\\equal{}1$ because the last det is $ >0$.\r\nLet $ f: \\lambda\\rightarrow{d}et(B\\plus{}\\lambda{A})\\equal{}a\\lambda^2\\plus{}b\\lambda\\plus{}1$ where $ a\\equal{}det(A),b$ are real numbers. $ f(1)\\equal{}1,f(j)f(j^2)\\equal{}1$. Therefore $ b\\equal{}\\minus{}a,3a^2\\plus{}1\\equal{}1$ and $ f(\\lambda)\\equal{}1$. $ A,B$ are simultaneously triangularizable over $ \\mathbb{C}$; then $ spectrum(A)\\equal{}\\{0,0\\}$ and $ A^2\\equal{}0$.\r\nMore precisely: if $ A\\not\\equal{}0$ then we may assume that (over $ \\mathbb{C}$):\r\n$ A\\equal{}\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix},B\\equal{}\\begin{pmatrix}u&v\\\\0&u\\end{pmatrix}$ with $ u^2\\equal{}1$, that is $ u\\equal{}\\epsilon$. Finally $ B\\equal{}\\epsilon{I}_2\\plus{}vA$, where $ A\\not\\equal{}{0},A^2\\equal{}0$.", "Solution_5": "About the first...\r\n\r\nI think there is a mistake in this as well (unless i am wrong somewhere)\r\n\r\nAnyways:\r\n\r\nCaley-Hamilton for C=AB-BA gives:\r\n\r\n$ C^3\\minus{}tr(C) C^2 \\plus{}aC\\minus{}det(C)I\\equal{}0$\r\n\r\nTracing this with tr(C)=0 we get:\r\n\r\n$ tr((AB\\minus{}BA)^3)\\minus{}3det(AB\\minus{}BA)\\equal{}0$\r\n\r\n$ (AB\\minus{}BA)^3\\equal{}(AB)^3 \\minus{} ABABBA\\minus{}ABBAAB\\plus{}ABBABA\\minus{}BAABAB\\plus{}BAABBA\\plus{}BABAAB\\minus{}(BA)^3$\r\n\r\nNow if i have done the multiplications correctly, tracing this and using the commutativity inside the trace we get $ tr((AB\\minus{}BA)^3)\\equal{}3tr(AB(BA\\minus{}AB)BA)$\r\n\r\nTherefore i think the result needed (and proved) is:\r\n\r\n$ det(AB\\minus{}BA)\\equal{}tr(AB(BA\\minus{}AB)BA)$", "Solution_6": "For 1), lemonidas is right.\r\nFor 3), I assume that $ C(A)\\equal{}\\{B;BA\\equal{}AB\\}$.\r\n-Quybac, your 3 exercises were incorrectly stated, that is remarkable-\r\nThe result is true in dimension $ n$:\r\nAssume that $ A$ is not a nilpotent matrix. Then $ A$ admits a non zero eigenvalue $ \\lambda$. We take $ B\\equal{}\\minus{}\\lambda{I_n}$. Therefore $ 0\\geq|\\lambda|^n$, a contradiction.\r\nConversely, if $ A$ is a nilpotent matrix anf if $ AB\\equal{}BA$, then we may assume that $ A$ is a strictly upper triangular matrix and $ B$ is an upper triangular matrix. Therefore $ det(A\\plus{}B)\\equal{}det(B)$.", "Solution_7": "[quote=\"loup blanc\"]\nFor 3), I assume that $ C(A) \\equal{} \\{B;BA \\equal{} AB\\}$.\n-[/quote]\r\n Oke ! :blush:" } { "Tag": [ "Vieta" ], "Problem": "What is the sum of the solutions of the equation $ x(2x\\minus{}9) \\equal{} 0$?\nExpress your answer as a common fraction.", "Solution_1": "hello, this equation has exactly two solutions, $ x_1\\equal{}0$ or $ x_2\\equal{}\\frac{9}{2}$, so the sum is given by\r\n$ x_1\\plus{}x_2\\equal{}\\frac{9}{2}$.\r\nSonnhard.", "Solution_2": "Or you can quickly expand in your head, and use Vieta's to get $ \\frac{9}{2}$, but seeing as its already factored for you, yours is a bit faster.", "Solution_3": "How do you get 9/2?", "Solution_4": "We expand and get $2x^2 - 9x = 0$. Hence, by Vieta's, the sum of the roots of $ax^2 + bx + c$ is $-\\frac{b}{a}$, so in this case the sum would be $-\\frac{-9}{2} = \\boxed{\\frac{9}{2}}$." } { "Tag": [ "integration", "logarithms", "calculus", "calculus computations" ], "Problem": "$ \\int_{a}^{3a} (e^{3x} \\plus{} e^{x})\\, dx \\equal{} 100$.\r\n\r\n[b](i)[/b] Show that $ a \\equal{} \\frac {1}{9}\\ln (300 \\plus{} 3e^{a} \\minus{} 2e^{3a})$.\r\n\r\n[b](ii)[/b] Use an iterative process, based on the equation in part [b](i)[/b], to find the value of $ a$ correct to 4 decimal places. Use a starting value of $ 0.6$ and show the result of each step of the process.\r\n\r\n[I make it $ a\\equal{}0.6309 \\, (4dp)$.]", "Solution_1": "hello, we have $ \\int e^{3x}\\plus{}e^x\\,dx\\equal{}\\frac{e^{3a}}{3}\\plus{}e^x$ and the definite integral is\r\n$ \\frac{e^{9a}}{3}\\plus{}e^{3a}\\minus{}\\frac{e^{3a}}{3}\\minus{}e^a\\equal{}100$ from here we get by multiplying with 3\r\n$ e^{9a}\\plus{}2e^{3a}\\minus{}3e^a\\equal{}300$ or $ e^{3a}\\equal{}300\\plus{}3e^a\\minus{}2e^{3a}$ or $ a\\equal{}\\frac{1}{9}\\ln(300\\plus{}3e^a\\minus{}2e^{3a})$\r\nSonnhard.", "Solution_2": "I agree, Sonnhard. But is my answer to part (ii) correct?" } { "Tag": [ "calculus", "integration", "function", "trigonometry", "LaTeX", "calculus computations" ], "Problem": "$ \\int\\sqrt(1\\plus{} cos^ {2} {x}) dx$", "Solution_1": "Hi :) !!\r\n\r\n$ \\int \\sqrt {1 + cos^2(x)}dx = \\int \\sqrt {2 - sin^2(x)} = \\sqrt {2} \\int \\sqrt {1 - \\frac {1}{2}sin^2(x)}dx = \\sqrt {2}E (x | \\frac {1}{2})$\r\n\r\nwith $ E(x| h) = \\int \\sqrt {1 - hsin^2(x)}dx$ is a function $ Elliptic \\ integral$\r\n\r\nand thx :)", "Solution_2": "That doesnt make sense what u wrote mathema :wink: \r\nAlso check the definition of the elliptic integral of the second kind again... :wink:", "Solution_3": "[quote=\"yeahnigga\"]That doesnt make sense what u wrote mathema :wink: \nAlso check the definition of the elliptic integral of the second kind again... :wink:[/quote]\r\n\r\nHi \r\n\r\nyou can check that the function $ x \\rightarrow 1 \\plus{} cos^2(x)$ is a derivate of$ x \\rightarrow \\sqrt {2} E(x;1/2)$\r\n\r\nBUT i don't understand what do you want to say !!!\r\n\r\nthx", "Solution_4": "Thanks mathema*: the guy who asked me the question said it was really easy which sent me on some wild goose chases.", "Solution_5": "hello, your integral looks so better $ \\int\\sqrt{1\\plus{}\\cos^2(x)}\\,dx$.\r\nSonnhard.", "Solution_6": "Lieber Sonnhard, Ihr Integral sieht gut aus aber Sie wollten wohl sagen \"your integral would look better as \".", "Solution_7": "[quote=\"AndrewTom\"]Thanks mathema*: the guy who asked me the question said it was really easy which sent me on some wild goose chases.[/quote]\r\n\r\nYOU ARE WELCOME :) ;)", "Solution_8": "hello, i wanted to say the integral would look so in $ \\text{\\LaTeX}$ better as yours.\r\nThank your for your hint.\r\nSonnhard.", "Solution_9": "Lieber Sonnhard, \r\n\r\ndanke f\u00fcr den Ratschlag; er war sehr n\u00fctzlich und ich habe ihn gern angenommen. (\u00dcbrigens ist \"as\" in Deutsch \"wie\" und \"than\" das deutsche \"als\" im Komparativ.)" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ f : \\mathbb{R} \\rightarrow \\mathbb{R}$ be a twice differentiable, $ 2 \\pi$-periodic even function. Prove that if \\[ f''(x)\\plus{}f(x)\\equal{}\\frac{1}{f(x\\plus{} 3 \\pi /2 )}\\] holds for every $ x$, then $ f$ is $ \\pi /2$-periodic. \r\n\r\n\r\n[i]Z. Szabo, J. Terjeki[/i]", "Solution_1": "Since $f$ is even, we get that \\[\\frac{1}{f(x+3\\pi/2)}=f''(x)+f(x) = f''(-x)+f(-x)=\\frac{1}{f(-x+3\\pi/2)}=\\frac{1}{f(x-3\\pi/2)}\\]\n\nTherefore $f$ is $3\\pi$-periodic and $2\\pi$-periodic, therefore $f$ must be $\\pi$-periodic\nThat means \n\\begin{align*}\n&f''(x)+f(x)=\\frac{1}{f(x+\\pi/2)}\\\\\n&f''(x+\\pi/2)+f(x+\\pi/2)=\\frac{1}{f(x+\\pi)}=\\frac{1}{f(x)}\\\n\\end{align*}\nThat means that \\[f''(x)f(x+\\pi/2)-f''(x+\\pi/2)f(x)=0 \\Rightarrow (f'(x)f(x+\\pi/2)-f(x)f'(x+\\pi/2))'=0\\]\nTherefore \\[f'(x)f(x+\\pi/2)-f(x)f'(x+\\pi/2) = c\\]\nSince $f(x+\\pi/2)=f(x-\\pi/2)$ ($f$ is even and has period $\\pi$), by setting $x\\rightarrow -x$ we get that \\[-f'(x)f(x+\\pi/2)+f(x)f'(x+\\pi/2) = c\\]\n\nSo, we got that $c=0$.\n\\begin{align*}\n&f'(x)f(x+\\pi/2)-f(x)f'(x+\\pi/2) = 0 \\\\\\\\\n\\Rightarrow &(f(x)/f(x+\\pi/2))'=0 \\\\\\\\\n\\Rightarrow &f(x)=kf(x+\\pi/2)=k^2f(x+\\pi)=k^2f(x) \\\\\\\\\n\\Rightarrow &k^2=1 \\Rightarrow k\\in\\{1,-1\\}\\\n\\end{align*}\nSince $f$ doesn't vanish and is continuous, $f$ must have the same sign on $\\mathbb{R}$. That forces $k$ to be $1$, which means $f$ has period $\\pi/2$.\n\n\n\n", "Solution_2": "[quote=ehsan2004]Let $ f : \\mathbb{R} \\rightarrow \\mathbb{R}$ be a twice differentiable, $ 2 \\pi$-periodic even function. Prove that if \\[ f''(x)\\plus{}f(x)\\equal{}\\frac{1}{f(x\\plus{} 3 \\pi /2 )}\\] holds for every $ x$, [/quote]\n\n\nCan one give an example $f$ function wich satisfy above ", "Solution_3": "Since $f$ is $\\pi$ periodic \n$ f : \\mathbb{R} \\rightarrow \\mathbb{R}$ be a twice differentiable, $ \\pi$-periodic even function $\\[ f''(x)\\plus{}f(x)\\equal{}\\frac{1}{f(x)}\\]$ holds for every $ x$\n\n\nCan one give an example $f$ function wich satisfy this condition", "Solution_4": "Probably not what you are looking for but $f(x) = 1$ works...", "Solution_5": "Non constant function ", "Solution_6": "A function satisfying $\\frac{df}{dx}^2=C+2ln(f)-f^2, where C is a constant", "Solution_7": "[quote=SassTramm]A function satisfying $\\frac{df}{dx}^2=C+2ln(f)-f^2$where C is a constant[/quote]\n\nDo you an explicit form, I mean with usual functions ? Does it involve trigonometry functions since f is pi- periodic?" } { "Tag": [ "LaTeX" ], "Problem": "Hello,\r\nI am having a problem with the following problem:\r\n\r\n\"Find an equation of the tangent line to the circle $x^{2}+y^{2}=25$ at the point $(4,-3)$.\"\r\n\r\nPlease give some pointers - thank you.\r\n\r\nedit - couldn't find latex button", "Solution_1": "This is way too hard for Classroom. This should be moved to Getting Started.", "Solution_2": "Oh? I am sorry. It was in my Equations of Lines packet.\r\nApologies." } { "Tag": [ "analytic geometry", "MATHCOUNTS", "email" ], "Problem": "What was the highest score in your chapter?", "Solution_1": "3 perfects in Santa Clara Valley, California. I got ONE sprint problem wrong and I ended up tied for 6th. :? Stupid tiebreakers.\r\n\r\n2nd in Countdown due to sudden panic after the guy I was against in the final beat me in the first question.", "Solution_2": "In the Central Florida Chapter, there were 5 perfect scores, mine being one of them, hehe! :) :) I came up 2nd because this guy went from 6th to first.", "Solution_3": "I got full score, as the only one. So it was :winner_first::winner_first::winner_first::winner_first::winner_first::winner_first::winner_first::winner_first:!!!", "Solution_4": "One of 3 perfect scores in Lake Washington Chapter (in Washington, of course). I won the first ever tiebreaker in our chapter! YAY!", "Solution_5": "sorry boys... In our chapter.. I got first with a 35 =)", "Solution_6": "The first place in our chapter got a 35 as well.", "Solution_7": "wow in my chapter some guys got 40s and 42s, but the 40s were 5th to 6th", "Solution_8": "i got 42 and a :winner_first:", "Solution_9": "Aren't you not allowd to post your score?\r\n\r\nEDIT: whoops, new announcement sry.\r\n\r\nI got 1st with 44 in my chapter.", "Solution_10": "I got first with a 43. Next was like 20 points down, :rotfl: :rotfl:", "Solution_11": "Sounds like Delaware state will be this year\r\n\r\n[hide]\n[hide]\n[hide]\n\n\n\n\n\n\n\n\n\njust kidding\n\n\n\n\n\n\n\n\n\n[/hide]\n[/hide]\n[/hide]", "Solution_12": "I was 4th w/ 45. The lowest score at our Seattle chapter was 41. But I think at the Lake Washington chapter, the lowest score was 42.", "Solution_13": "Woah thats just incredible.", "Solution_14": "Why? What happened at your chapter? I was in a 3 way tie, but a girl beat everyone and went from 10th to 1st in the countdown round. At the Lake Washington Chapter, there were 3 perfect scores.", "Solution_15": "First a 29? \r\nOur 10th got 26 though.", "Solution_16": "our 10th got 40 (can't remember what the others got).", "Solution_17": "Our 83rd place got 3 :rotfl: :rotfl:", "Solution_18": "there are people out there who are...mathmatically challenged.", "Solution_19": "[quote=\"robinhe\"]Our 83rd place got 3 :rotfl: :rotfl:[/quote]\r\nA 3!!!\r\nIf someone is that bad my coach wouldn't have even taken them. She'd rather not take the maximum if it means taking someone who isn't good at these problems.", "Solution_20": "keep in miond it mioght have been an inexperienced coach who didn't know what the competition was like...or the kid was having a really bad day!", "Solution_21": "I'm sure there were multiple 3- at my chapter. Its amazing how low some scores are.", "Solution_22": "Our 10th was either 45 or 44.\r\n83rd would've been somewhere around 32.\r\n\r\nA 3 would've gotten you a solid 190th place.", "Solution_23": "There were only like 50 people at my chapter. 190 is a lot for just one chapter.", "Solution_24": "Yeah, our chapter has like 60 people. I got a :first: with a 43 though, but last year i won it with only 37. I can't believe that people even got perfect scores. That's crazy that more than one person would from one chapter. Our 10th place about 30. I'm scared of state, but really want to go to nationals. Good luck to everyone else just as screwed as i am :lol:", "Solution_25": "I do Mathcounts! It Rules! NEW YORK, BABY! :lol:\r\n\r\n\r\nEmbarassed I only came in third individually in my chapter cause I made four retarded mistakes\r\nI knew the answers, just wrote down in wrong unit Sad", "Solution_26": "My coach got the results then took them while he's on a bussiness trip and he still hasn't come back. His son told me that I got top ten but I want to know which place. Anyone who went to Georgia Tech's chapter with the results, please email me at jingesu9@gmail.com \r\nthanks.\r\nOh yeah, my team got first in chapter.", "Solution_27": "Which chapter were you in? There were 5 different chapters at the competition at GA Tech. :o I think GA's the only place with multiple competitions at the same place at the same time.....", "Solution_28": "actually no. my uncle proctors a chapter, (actually two) on the border of illinois and iowa. there's one chapter from each state there, so they have to do countdown rounds seperately.", "Solution_29": "Colorado Northern Chapter, 45 by me!\r\nI know it's a bit late to be saying this (State was today), but it's true!" } { "Tag": [ "puzzles" ], "Problem": "Please put $ 6$ coins on $ 5 \\times 5$ grid points so that all each distance is different.\r\n\r\n[asy]for(int n=0; n <= 5; ++n)\ndraw((n,0)--(n,5)^^(0,n)--(5,n));\n\nfilldraw(circle((1,2),0.25));\nfilldraw(circle((1,4),0.25));\nfilldraw(circle((3,1),0.25));\nfilldraw(circle((3,3),0.25));\nfilldraw(circle((4,2),0.25));\nfilldraw(circle((5,4),0.25));[/asy]\r\n\r\n\r\nThis problem is questions from \"Takeshi Kitano presents Comaneci University Mathematics\".\r\nIt is a Japanese TV program.", "Solution_1": "oh nvm 5x5\r\n6 points means you need 15 distances i believe and on a 5x5 thats possible....\r\nhm.....", "Solution_2": "[quote=\"bobma99\"]oh nvm 5x5\n6 points means you need 15 distances i believe and on a 5x5 thats possible....\nhm.....[/quote]\r\n\r\nYes, that's right :) \r\n$ \\binom{6}{2}\\equal{}15$.\r\n\r\nAnd, \uff54he existing distances is $ \\sqrt{1}$, $ \\sqrt{2}$, $ \\sqrt{4}$, $ \\sqrt{5}$, $ \\sqrt{8}$, $ \\sqrt{9}$, $ \\sqrt{10}$, $ \\sqrt{13}$, $ \\sqrt{16}$, $ \\sqrt{17}$, $ \\sqrt{18}$, $ \\sqrt{20}$, $ \\sqrt{25}$, $ \\sqrt{26}$, $ \\sqrt{29}$, $ \\sqrt{32}$, $ \\sqrt{34}$, $ \\sqrt{41}$, $ \\sqrt{50}$.\r\n($ 19$ distances.)" } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "LaTeX", "function", "absolute value" ], "Problem": "Solve for $ x$:\r\n\r\n$ \\begin{pmatrix}2 & 1 & x \\\\\r\n0 & 1 & 1 & \\\\ 1 & x & 0\\end{pmatrix} \\minus{} \\begin{pmatrix}5 & 1 \\\\ x & 2\\end{pmatrix} \\equal{} 8$.", "Solution_1": "As stated, that is nonsense. One cannot add, or equate, matrices of different sizes.\r\n\r\nDo you perhaps mean to take the determinant of each of those matrices? That would make this a polynomial equation.", "Solution_2": "I was looking for a latex to put \"vertical lines\" (like absolute value) instead of the \"parentheses\"... I am not familiar with the difference between matrices and determinants.", "Solution_3": "[quote=\"jeez123\"]$ \\begin{vmatrix}2 & 1 & x \\\\ 0 & 1 & 1 & \\\\ 1 & x & 0\\end{vmatrix} \\minus{} \\begin{vmatrix}5 & 1 \\\\ x & 2\\end{vmatrix} \\equal{} 8.$[/quote]\r\n\r\nIt's vmatrix.\r\nI usually use the notation $ \\det(\\cdot)$; it makes it clear that the determinant is a function from matrices to numbers in the base field.", "Solution_4": "At this point, it's really just a grind-through-it calculation. The equation turns out to be linear.\r\n\r\n$ (1\\minus{}3x)\\minus{}(10\\minus{}x)\\equal{}8$ yields $ x\\equal{}\\minus{}\\frac{17}{2}.$\r\n\r\nFor some reason, I kept making computational errors, so it took a while for me to get this to check." } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "Let $ A_1,A_2,...,A_n$ be finite sets. If $ |A_i \\cap A_j| \\equal{} i \\plus{} j$, $ (\\forall) i\\neq j$, find $ min|A_1 \\cup A_2 \\cup ... \\cup A_n|$.\r\n \r\nWhere $ |A|$ is the number of elements of A. \r\n\r\n\r\nSorin Dumitrica", "Solution_1": "[hide]One needs to craft a careful inductive argument with the following ingredients:\n1. Lets call the set $ \\{A_1,A_2,\\dots,A_n\\}$ optimal if $ |A_1\\cup A_2\\cup \\dots \\cup A_n|$ is minimum. For an optimal set $ A_{n\\minus{}1}\\subset A_n$.\n\n2. For a set $ \\{A_1, \\dots, A_{n\\plus{}1}\\}$, the number $ |A_1\\cup A_2\\cup \\dots \\cup A_{n\\plus{}1}|$ is at least $ n\\plus{}1$ more than $ min\\{|A_1\\cup A_2\\cup \\dots \\cup A_n|\\}$[/hide]" } { "Tag": [ "inequalities", "Cauchy Inequality", "inequalities proposed" ], "Problem": "Let $ a, b, c$ positive real numbers. Prove or disprove that\r\n\r\n$ \\sum[\\frac{ab \\plus{} 3c}{a \\plus{} b \\plus{} 2c}]\\le \\frac{a \\plus{} b \\plus{} c \\plus{} 9}{4}$.", "Solution_1": "My answer for your inequality is absolutely true. Well, firstly, the inequality can be divided into 2 parts. \r\nThe first part is to prove \r\n$ \\sum_{cyc}\\frac{ab}{a\\plus{}b\\plus{}2c} \\le \\frac{a\\plus{}b\\plus{}c}{4}$\r\nand the second part need proving like \r\n$ \\sum_{cyc}\\frac{3c}{a\\plus{}b\\plus{}2c} \\le \\frac{9}{4}$\r\nThey are both trivial after using Cauchy inequality according to the standard form\r\n$ \\frac{1}{a} \\plus{}\\frac{1}{b} \\ge \\frac{4}{a\\plus{}b}$" } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "Evaluate the limit :\r\n\\[ \\lim_{n \\to \\infty}{\\sqrt[3]{n\\plus{}1} \\cdot \\cos{\\sqrt{n\\plus{}1}}}\\minus{} \\sqrt[3]{n} \\cdot \\cos{\\sqrt{n}}.\\]", "Solution_1": "$ {\\sqrt[3]{n+1} \\cdot \\cos{\\sqrt{n+1}}}- \\sqrt[3]{n} \\cdot \\cos{\\sqrt{n}} = \\sqrt[3]{n}(\\sqrt[3]{1+\\frac{1}{n}}\\cos\\sqrt{n+1}-\\cos\\sqrt{n})$\r\n\r\n$ \\boxed{\\sqrt[3]{1+\\frac{1}{n}}=1+\\frac{1}{3n}+o(n^{-1})}$\r\n\r\n$ \\sqrt[3]{n}(\\sqrt[3]{1+\\frac{1}{n}}\\cos\\sqrt{n+1}-\\cos\\sqrt{n})\\approx\\sqrt[3]{n}((1+\\frac{1}{3n})\\cos\\sqrt{n+1}-\\cos\\sqrt{n})=$\r\n$ =\\sqrt[3]{n}(\\cos\\sqrt{n+1} - \\cos\\sqrt{n} + \\frac{1}{3n}\\cos\\sqrt{n+1})=$\r\n$ =\\sqrt[3]{n}(-2\\sin\\frac{\\sqrt{n+1}+\\sqrt{n}}{2}\\sin\\frac{\\sqrt{n+1}-\\sqrt{n}}{2}+ \\frac{1}{3n}\\cos\\sqrt{n+1})=$\r\n$ =\\sqrt[3]{n}(-2\\sin\\frac{\\sqrt{n+1}+\\sqrt{n}}{2}\\sin\\frac{1}{2(\\sqrt{n+1}+\\sqrt{n})}+ \\frac{1}{3n}\\cos\\sqrt{n+1})$\r\n\r\n$ \\boxed{\\sin\\frac{1}{2(\\sqrt{n+1}+\\sqrt{n})}=\\frac{1}{2(\\sqrt{n+1}+\\sqrt{n})}+o(n^{-0.5})\\approx\\frac{1}{4\\sqrt{n}}+o(n^{-0.5})}$\r\n\r\n$ \\sqrt[3]{n}(-2\\sin\\frac{\\sqrt{n+1}+\\sqrt{n}}{2}\\sin\\frac{1}{2(\\sqrt{n+1}+\\sqrt{n})}+ \\frac{1}{3n}\\cos\\sqrt{n+1})\\approx$\r\n$ \\approx -\\frac{1}{2\\sqrt[6]{n}}\\sin\\frac{\\sqrt{n+1}+\\sqrt{n}}{2}+\\frac{1}{3\\sqrt[3]{n^2}}\\cos\\sqrt{n+1})$\r\n\r\nobviously $ \\lim_{n\\to\\infty}(-\\frac{1}{2\\sqrt[6]{n}}\\sin\\frac{\\sqrt{n+1}+\\sqrt{n}}{2}+\\frac{1}{3\\sqrt[3]{n^2}}\\cos\\sqrt{n+1}))=0$", "Solution_2": "Stolz-Cesaro does the trick.", "Solution_3": "or can't we use M.V.T.?", "Solution_4": "[quote=\"\u00a7outh\u00a7tar\"]Stolz-Cesaro does the trick.[/quote]\r\nStolz-Cesaro doesn't do the trick.\r\nIn this case it only shows that the limit is 0, IF the limit exists." } { "Tag": [ "modular arithmetic", "number theory", "least common multiple", "greatest common divisor", "function" ], "Problem": "Find $ 3^{2004}\\pmod{2004}$", "Solution_1": "hello, we have $ 3^{12} \\equiv 381 \\mod 2004$ and $ 381^{167}\\equiv 381 \\mod 2004$ so your result is\r\n$ 3^{2004}\\equiv 381\\ mod 2004$.\r\nSonnhard.", "Solution_2": "$ 3^{2004}\\equiv(\\minus{}1)^{2004}\\equiv1\\pmod4$, $ 3^{2004}\\equiv0\\pmod3$, and from FLT, $ 3^{166}\\equiv1\\pmod{167}\\implies3^{2004}\\equiv3^{12}\\equiv47\\pmod{167}$, so solving the congruence gives $ 3^{2004}\\equiv\\boxed{381}\\pmod{2004}$.", "Solution_3": "[hide=\"Alternatively...\"]\n$ 3^{2004}\\equiv 3x\\pmod{2004}\\implies 3^{2003}\\equiv x\\pmod{668}$.\n$ \\varphi(668)\\equal{}332$\n$ 2003\\equiv 11\\pmod{332}\\implies 3^{2003}\\equiv 3^{11}\\pmod{668}$.\n$ 3^{11}\\equiv 3\\cdot 9\\cdot 81^2\\equiv \\minus{}119\\cdot 27\\equiv 127\\pmod {668}$.\n$ x\\equal{}127\\implies 3x\\equal{}\\boxed{381}$.\n[/hide]", "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, we have $ 3^{12} \\equiv 381 \\mod 2004$ and $ 381^{167}\\equiv 381 \\mod 2004$[/quote]\r\n\r\nWhy is the last one true?", "Solution_5": "$ 381^{167}\\equiv 3x\\pmod{2004}\\implies 381^{166}\\cdot 127\\equiv x\\pmod{668}$.\r\n\r\nWe conveniently have $ 668\\equal{}2^2\\cdot 167\\implies\\lambda(668)\\equal{}\\text{lcm}(\\lambda(2^2),\\lambda(167))\\equal{}166$. Therefore, because $ \\text{gcd}(381,668)\\equal{}1$, we have $ 381^{166}\\equiv 1\\pmod{668}$, so $ x\\equiv 127\\pmod{668}\\implies 3x\\equiv 381\\pmod{2004}$.\r\n\r\nI suppose Dr Sonnhard Graubner simply left that computation to the reader :D.\r\n\r\n(If anyone doesn't know, $ \\lambda(n)$ denotes the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael Function[/url])." } { "Tag": [ "calculus", "integration", "geometry", "trigonometry", "calculus computations" ], "Problem": "Find the values of $ a,\\ b$ such that the area bounded by the curve $ y \\equal{} \\cos x\\ \\left(0\\leq x\\leq \\frac {\\pi}{2}\\right)$, the $ x$-axis and $ y$-axis are devided into three equal parts by the curves $ y \\equal{} a\\sin x,\\ y \\equal{} b\\sin x\\ (a > b)$.", "Solution_1": "$ \\int_0^{\\frac{\\pi}{2}} \\cos x dx= [\\sin x]_0^{\\frac{\\pi}{2}}=1$\r\n\r\n$ c\\sin x = \\cos x \\to \\cot x = c \\to x=\\cot^{-1} c=r_c$\r\n\r\n$ 1=3\\int_0^{r_a} \\cos x - a\\sin x dx = 3[\\sin x +a\\cos x]_0^{r_a}$\r\n\r\n$ =3[\\sin r_a(1+ a \\cot r_a) - a]=3(\\frac{1+a^2}{\\sqrt{1+a^2}}-a)=3(\\sqrt{1+a^2}-a)$\r\n\r\n$ 0=(1+3a)^2-9(1+a^2)=1+6a+9a^2-9-9a^2=6a-8 \\to a=\\frac{4}{3}$\r\n\r\n$ 1=3\\left( \\int_0^{r_b} b\\sin x dx +\\int_{r_b}^{\\frac{\\pi}{2}} \\cos xdx\\right)=3\\left([-b\\cos x]_0^{r_b}+[\\sin x]_{r_b}^{\\frac{\\pi}{2}\\right)}$\r\n\r\n$ 3b(1-\\cos r_b)+3(1-\\sin r_b)=3(b-\\frac{b^2}{\\sqrt{1+b^2}})+3(1-\\frac{1}{\\sqrt{1+b^2}})=3b+3 - 3\\sqrt{1+b^2}$\r\n\r\n$ 0=(3b+2)^2-(3\\sqrt{1+b^2})^2=9b^2+12b+4-9-9b^2=12b-5 \\to b=\\frac{5}{12}$", "Solution_2": "That's correct answer." } { "Tag": [ "function", "inequalities", "symmetry" ], "Problem": "$a, b, c$ are real numbers. Given two functions: $f(x)=ax^2+bx+c$ and $g(x)=ax+b$. When $|x| \\leq 1$ we have $|f(x)|\\leq 1$.\r\n(1) Prove: $|c| \\leq 1$\r\n(2) Prove: When $|x| \\leq 1$, prove $|g(x)| \\leq 2$\r\n(3) Let $a>0$. When $|x| \\leq 1$, the maximum value of $g(x)$ is $2$. Find $f(x)$.", "Solution_1": "For 1:\r\nAssume the contrary that $|c|>1$ then $f(0)>1$ contradiction.\r\nFor 2:\r\n\\[1\\geq |f(x)|\\geq |ax^2+bx|+c\\geq |g(x)|+c\\]\r\n\\[1-c\\geq |g(x)|\\]\r\n\\[2\\geq |g(x)|\\]\r\nfrom 1", "Solution_2": "For 3:$f(x)=2x^2-1$. This problem's background is Tchevichev's polynomial.", "Solution_3": "[quote=\"amd2357\"]$1\\geq |f(x)|\\geq |ax^2+bx|+c\\geq |g(x)|+c$[/quote]\r\nHowever, the inequality $|ax^2+bx+c|\\geq |ax^2+bx|+c$ is not always true. It is true only when $c$ is negative.\r\n[hide=\"solution to part 2\"]Since $\\begin{cases}f(0)=c\\\\f(-1)=a-b+c\\\\f(1)=a+b+c\\end{cases}$, we have $a=\\frac{1}{2}[f(1)+f(-1)-2f(0)]$, $b=\\frac{1}{2}[f(1)-f(-1)]$.\nSo when $-1\\leq x\\leq 1$,\n$|g(x)|=|ax+b|$\n$=|\\frac{1}{2}[f(1)+f(-1)-2f(0)]x+\\frac{1}{2}[f(1)-f(-1)]|$\n$=\\frac{1}{2}|(x+1)f(1)+(x-1)f(-1)-2xf(0)|$\n$\\leq\\frac{1}{2}(|x+1||f(1)|+|x-1||f(-1)|+|2x||f(0)|)$\n$\\leq\\frac{1}{2}(|x+1|+|x-1|+2)$\n$=\\frac{1}{2}((x+1)+(1-x)+2)$\n$=2$[/hide]\n[hide=\"solution to part 3\"]Since $a>0$, $g(x)=ax+b$ is increasing on the interval $[-1,1]$. When $x=1$, $g(x)$ attains its maximum. From the conditions,\n$1=g(1)=a+b=f(1)-f(0)$\nTherefore $-1\\leq f(0)=f(1)-2\\leq 1-2\\leq -1$\nSo $c=f(0)=-1$\nWhen $|x|\\leq 1$, $|f(x)|\\leq 1$, so\n$f(x)\\geq -1=c=f(0)$\nThis shows that on the interval $[-1,1]$, $f(0)$ is the minimum of $f(x)$. $x=0$ must be the axis of symmetry of $f(x)$. Therefore, $b=0$, $a=2$ and $f(x)=2x^2-1$.[/hide]", "Solution_4": "Actually, $g_{max}(x)=2 \\Longleftrightarrow |f(1)|=|f(-1)|=|f(0)|=1$. From here we can easily get $f(x)$." } { "Tag": [ "geometry", "perpendicular bisector", "geometry unsolved" ], "Problem": "let triangle $ ABC$ be a triangle and construct squares $ ABB_1A_2$, $ BCC_1B_2$, $ CAA_1C_2$ externally on its sides.prove that perpendicular bisector of $ A_1A_2,B_1B_2,C_1C_2$ are concurrent", "Solution_1": "See [url]http://www.mathlinks.ro/viewtopic.php?t=22917[/url]." } { "Tag": [ "calculus", "integration", "number theory", "prime factorization", "geometric sequence" ], "Problem": "Prove that if p is a prime number, then $p^n$ is never a perfect number.", "Solution_1": "It comes directly if you write :\r\np^n = p*p....*p (n times)\r\n\r\n:)", "Solution_2": "[hide]since $p$ is prime, then we have that the prime factorization is $p^n$\n\nso the definition of a perfect number is that the sum of the divisors is double the number,\n\nclearly the divisors are 1,p,p^2,...p^n, and that is a geometric sequence so\n\n$\\sum_{k|p^n}k=\\frac{p^{n+1}-1}{p-1}$\n\nthen using the definition of a perfect number, we have that \n\n$\\frac{p^{n+1}-1}{p-1}=2p^n$\n$p^{n+1}-1=2p^{n+1}-2p^n$\n$p^{n+1}-2p^n+1=0$\nand the only integral answer we get for that is 1, and that is not prime, so there are no perfect numbers in that form[/hide]", "Solution_3": "[quote=\"Altheman\"][hide]since $p$ is prime, then we have that the prime factorization is $p^n$\n\nso the definition of a perfect number is that the sum of the divisors is double the number,\n\nclearly the divisors are 1,p,p^2,...p^n, and that is a geometric sequence so\n\n$\\sum_{k|p^n}k=\\frac{p^{n+1}-1}{p-1}$\n\nthen using the definition of a perfect number, we have that \n\n$\\frac{p^{n+1}-1}{p-1}=2p^n$\n$p^{n+1}-1=2p^{n+1}-2p^n$\n$p^{n+1}-2p^n+1=0$\nand the only integral answer we get for that is 1, and that is not prime, so there are no perfect numbers in that form[/hide][/quote]\r\n\r\nI'm not entirely convinced by your proof. Why can't primes be roots of your equation? Your final equation can be obtained by multiplying $p^n=p^{n-1}+p^{n-2}+\\cdots+1$ (which is what we want to prove) by $p-1$. Basically, it's restating the given.", "Solution_4": "That's rational root theorem.. Basically you can write, his last line as $p-2+ \\frac 1{p^n}=0$ now $p-2$i integer and last part can not be integer." } { "Tag": [ "polynomial", "algebra", "function", "domain", "Ring Theory", "superior algebra", "superior algebra theorems" ], "Problem": "Through some research, I found that for any N, there can only be up to one Gaussian prime with a norm of N, up to associates and their conjugates. Is this true for Eisenstein primes as well? How can this be proven?\r\n\r\nMore specifically, any rational prime congruent to 3 modulo 4 is a Gaussian prime, as are its associates. Any Gaussian integer with a rational prime norm congruent to 1 modulo 4 is a Gaussian prime, as are its associates and their conjugates - and for any rational prime congruent to 1 modulo 4, there is always [i]gauranteed[/i] to be a Gaussian integer with a norm equal to it. With Eisenstein integers, all is in terms of congruence to 1 modulo 3 versus to 2 modulo 3. Does the same existence and uniqueness apply to Eisenstein primes that does to Gaussian ones?\r\n\r\nLastly, can any of this be extended to arbitrary Unique Factorization Domains?[/i]", "Solution_1": "Yes; this is a consequence of unique factorization, which holds in both the Gaussian and Eisenstein integers. \r\n\r\nYour second question doesn't quite make sense as stated. For one thing, not every UFD has a natural notion of norm.", "Solution_2": "Generally a number ring is a UFD iff PID iff all irreducibles have prime-power norms (i.e. norms divisible by only one rational prime)." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Find \"all\" the Sylow subgroups of S5 and S6\r\n\r\nI don't know how to find 1.Sylow 2-subgroups of S5\r\n 2.Sylow 2-subgroups of S6\r\n 3.Sylow 3-subgroups of S6\r\n 4.Sylow 5-subgroups of S6\r\n\r\nI use Sylow thms to get the possible numbers of Sylow subgroups but don't know how to find the right one.\r\nEven I know the right number.\r\nHow to find them all?", "Solution_1": "You may want to work out the Sylow 5-subgroups of S6 first. Make sure you know how to find the Sylow 3-subgroups of S3, S4, and S5 too. You should be able to find the Sylow 5-subgroups of S5, S6, S7, S8, and S9.", "Solution_2": "Let n_5 = #Sylow 5-subgroups of S5 ; \r\nthen n_5|5!/5 = 24 and is congruent to 1 mod 5, so n_5 = 1 or 6.\r\nBut a (sub)group of order 5 is cyclic, so generated by some 5-cycle;\r\nand the 5-cycles are of the form (abcde), of which there are 5!/5 = 24.\r\nSince the Sylow 5-subgroups intersect trivially,\r\nthere are n_5 (5 - 1) = 24 elements of order 5 in S_5,\r\nso we must have n_5 = 6.\r\n\r\nSimilarly for S6.\r\nLet n_5 = #Sylow 5-subgroups of S6 ; \r\nthen n_5|6!/5 = 144 and is congruent to 1 mod 5, so n_5 = 1 or 6 or 36.\r\nAgain,a (sub)group of order 5 is cyclic, so generated by some 5-cycle;\r\nand there are 24*6=144 5-cycles.\r\nSince the Sylow 5-subgroups intersect trivially,\r\nthere are n_5 (5 - 1) = 144 elements of order 5 in S_5,\r\nso we must have n_5 = 36.\r\n\r\nSimiarly for Sylow 3-subgroups of S5.\r\n\r\nHow to proceed to find \r\n1.Sylow 2-subgroups of S5 \r\n2.Sylow 2-subgroups of S6\r\n3.Sylow 3-subgroups of S6 ??", "Solution_3": "Yup.\r\n\r\nThe next easiest is probably the Sylow 3-subgroups of S6.\r\n\r\n< (1,2,3), (4,5,6) > has order 9, and 9 is the highest power of 3 dividing 6!.\r\n\r\nAny conjugate < (a,b,c), (d,e,f) > of < (1,2,3), (4,5,6) > is also a Sylow 3-subgroup. How many conjugates are there? Well choosing the conjugate amounts to choosing the set of orbits {{a,b,c},{d,e,f}}. There are Binomial(6,3)/2=10 ways to do this (choosing a,b,c is Binomial(6,3) ways, then {d,e,f} is determined, but instead of choosing {a,b,c} it is the same to choose {d,e,f}, so one has double counted).\r\n\r\nOne can also check that the normalizer of < (1,2,3), (4,5,6) > is < (1,2,3), (4,5,6), (2,3), (5,6), (1,4)(2,5)(3,6) > of order 72 and index 10.\r\n\r\nThis should also describe how to find the Sylow 3-subgroups of S7 and S8.\r\n\r\n\r\nDescribing the Sylow 2-subgroups of S5 is the same as describing the Sylow 2 subgroups of S4.\r\n\r\n< (1,2), (3,4), (1,3)(2,4) > has order 8, so it is a Sylow 2-subgroup of S4 and S5.\r\n\r\nSo is any conjugate < (a,b), (c,d), (a,c)(b,d) >. Again choosing such a conjugate comes down to choosing {{a,b},{c,d}}, and so there are Binomial(4,2)/2=3 ways to do this in S4 and Binomial(5,2)/2*Binomial(3,2)=15 ways to do this in S5. In S4 {a,b} determines {c,d}, but in S5 you still have three points left, so there is the factor of Binomial(3,2).\r\n\r\nAlternatively one can calculate that the Sylow 2-subgroup is its own normalizer, and so has index 3 in S4 and and index 15 in S5.\r\n\r\n\r\nThe Sylow 2-subgrous of S6 and S7 are somewhat similar. They are all < (a,b), (c,d), (a,c)(b,d), (e,f) >. There are Binomial(6,2)*Binomial(4,2)/2=45 ways of choosing {{a,b},{c,d}} for S6 and Binomial(7,2)*Binomial(5,2)/2*Binomial(3,2)=315 ways of choosing ({{a,b},{c,d}},{e,d}) for S7. Notice that in S6 {e,f} is determined by {{a,b},{c,d}}. Again you can also calculate that the Sylow 2-subgroups are self-normalizing, and so have index 45 and 315.", "Solution_4": "I know the groups of the form < (a,b,c),(d,e,f)> are all Sylow 3-subgroups of S6.\r\nBut why all Sylow 3-subgroups of S6 are of the form < (a,b,c),(d,e,f)> ?\r\nI'm confused.\r\n\r\nSimilar problems in the case Sylow 2-subgroups of S5 and S6.", "Solution_5": "[quote=\"laipou\"]I know the groups of the form < (a,b,c),(d,e,f)> are all Sylow 3-subgroups of S6.\nBut why all Sylow 3-subgroups of S6 are of the form < (a,b,c),(d,e,f)> ?[/quote]\r\n\r\nThis follows from the \"conjugacy\" part of Sylow's theorem and a description of conjugacy in Sn.\r\n\r\nAll Sylow 3-subgroups of S6 are conjugate to < (1,2,3),(4,5,6) > by Sylow's theorem. In other words, if P is a Sylow 3-subgroup of S6, then there is a g in S6 such that P = $ \\langle (1,2,3),(4,5,6) \\rangle^g$.\r\n\r\nIn S6, (1,2,3)^g = ( 1^g, 2^g, 3^g ) and (4,5,6)^g = ( 4^g, 5^g, 6^g ) where g is the permutation that takes 1 to 1^g, 2 to 2^g, 3 to 3^g, 4 to 4^g, 5 to 5^g, and 6 to 6^g. In other words, P = < (a,b,c), (d,e,f) > where a,b,c,d,e,f are the images under the permutation g of 1,2,3,4,5,6." } { "Tag": [ "\\/closed" ], "Problem": "Hi! Mathlinkers.\r\n\r\nHow many people are there those who can understand or speak Japanese,in this site?\r\n\r\nkunny", "Solution_1": "Hi kunny! :) \r\n\r\nI've lived in Japan for 15 yrs since I was born, and came to the U.S. 2 yrs ago. My English skill has imporved a lot since I came here, but I still have to work on it. Anyway, I can definitely understand Japanese because it's my native language ;) \r\n\r\nI saw your posts everywhere, and it seems like you're really fluent in both English and Japanese... When did you come to the U.S.?\r\n\r\nI'm sure there are a couple of other Japanese AoPSers, but maybe they haven't found this topic yet.\r\n\r\nBTW, I'm new to the AoPS, and I really feel like I should've found this web site earlier. This is definately one of the most excellent math web sites that I've ever seen!!! :)", "Solution_2": "Hello,frt.\r\n\r\nThank you for your reply.I'm very glad to receive that.\r\n\r\n[quote=\"frt\"]\nI've lived in Japan for 15 yrs since I was born, and came to the U.S. 2 yrs ago. My English skill has imporved a lot since I came here, but I still have to work on it. Anyway, I can definitely understand Japanese because it's my native language ;) [/quote]\n\nMe,too. :) ;) :lol: \n\n[quote=\"frt\"]\nI saw your posts everywhere, and it seems like you're really fluent in both English and Japanese... When did you \ncome to the U.S.?[/quote]\n\nNo, my English is not good. I can speak English a little bit. I live in Tokyo now.If there is something wrong about my English, please correct for me without restraint, because I want to learn plain English. :) \n\n[quote=\"frt\"]\nI'm sure there are a couple of other Japanese AoPSers, but maybe they haven't found this topic yet.\n[/quote]\n\n :( Come on, every one! :) \n\n[quote=\"frt\"]\nBTW, I'm new to the AoPS, and I really feel like I should've found this web site earlier. This is definately one of the most excellent math web sites that I've ever seen!!! :)[/quote]\r\n\r\nYes, I think this site is very excellent. It's a pity that there are few site such as this in Japan.\r\n\r\nBest regards.\r\n\r\nkunny", "Solution_3": "[quote]No, my English is not good. I can speak English a little bit. I live in Tokyo now.[/quote]\n\nYour English is much better than me ;) Just curious, are you a high school student? (or...)\n\n[quote]It's a pity that there are few site such as this in Japan. \n[/quote]\r\n\r\nYeh, I agree. We need more math-related web sites in Japan.\r\n\r\nAlso, why is it that there's no Japanese forum in AoPS while there are Korean and Chinese ones? Is it just because there are very few Japanese people here?\r\n\r\n\r\nYuichi (my real first name) :)", "Solution_4": "I think Kunny's purpose of making this post is to have a Japanese community built in ML.", "Solution_5": "[quote=\"frt\"]Also, why is it that there's no Japanese forum in AoPS while there are Korean and Chinese ones? Is it just because there are very few Japanese people here?\n\nYuichi (my real first name) :)[/quote]Yup, that's the reason. As soon as you guys gather (3-4 persons at least) we can easily make one for you :)", "Solution_6": "I can speak.. lets see... a few words/phrases and i can count in Japanese! I visited Japan for a month when i was 5 and thats all I remember...\r\n\r\nO'renkideskaw = How are you?", "Solution_7": "Konnichiwa, ToBe. :) \r\n\r\n[color=blue]Ogenkidesuka?[/color] Message [color=red]arigatou[/color]. Where did you visit in Japan?\r\n\r\nI visited Canada several years ago.\r\n\r\nkunny", "Solution_8": "Sugoi...\r\n\r\nI used to live in Japan for four years, but I'm nowhere near fluent. I haven't spoken Japanese with someone else in a REALLY LONG time so my skill with it kind of faded away :-(. And my accent sucks.", "Solution_9": "I know how to count to five in Japanese, sort of :blush: , and know how to say hello and goodbye. Are japanese symbols related in anyway to chinese characters?", "Solution_10": "[quote=\"236factorial\"]Are japanese symbols related in anyway to chinese characters?[/quote]\r\nYeah, Japanese letters originally came from Chinese letters. A long time ago, Japanese adopted Chinese letters and started using them. That's the origin of one of the three Japanese characters ([i]kanji[/i]). Then, for convenience, Chinese letters were simplified to create the second type of Japanese characters ([i]hiragana[/i]). The third type of Japanese characters, [i]katakana[/i], was also created from Chinese letters.", "Solution_11": "Katakana is soooo easy to learn though, then hiragana, then kanji...well...I won't even go there :-o. 2000+ characters...", "Solution_12": "My friend speak Japanese, and i know a little bit as well." } { "Tag": [ "ratio", "trigonometry", "quadratics" ], "Problem": "cos(x)+4sin(x)=19/5\r\n\r\ndetermine the value of sec(x)\r\n\r\nand a similiar \r\n2cos(x)+sin(x)=2\r\n\r\ndetermine value of tan(x)\r\n\r\ngive me some hints please i want to actually figure this out", "Solution_1": "remember that tan(x)=sin(x) divided by cos(x)", "Solution_2": "[quote=\"Farcus\"]cos(x)+4sin(x)=19/5\n\ndetermine the value of sec(x)\n\nand a similiar \n2cos(x)+sin(x)=2\n\ndetermine value of tan(x)\n\ngive me some hints please i want to actually figure this out[/quote]\r\nWhat well-known ratios could be the values for $\\cos x$ and $\\sin x$ (Hint: they're very well-known :D)\r\n[hide=\"Answers (kinda)\"] $\\cos x=\\frac{3}{5}$\n$\\cos x =1$\n:D [/hide]", "Solution_3": "How can we do this with identities/algebraically?", "Solution_4": "[quote=\"mysmartmouth\"]How can we do this with identities/algebraically?[/quote]\r\nUse $\\sin x =\\pm\\sqrt{1-\\cos^{2}x}$. Isolate the square root, then square both sides. Solve the quadratic for $\\cos x$. :D" } { "Tag": [ "AMC", "AIME", "complex numbers" ], "Problem": "This problem is from AIME 1983.\r\nw and z are complex numbers such that w^2 + z^2=7, and w^3 + z^3=10, what is the LARGEST REAL VALUE OF w + z.\r\n\r\nThe solution was posted, along with lots of other problems, at http://www.kalva.demon.co.uk but I dont see how to come up with it.\r\nPost EXPLANATIONS.", "Solution_1": "A Try : \r\n\r\n\r\nX = w + z\r\n\r\nX^3 = w^3 + z^3 + wz * X = 10 + wz * X\r\n\r\nX^2 = w^2 + z^2 + 2 wz = 7 + 2 wz\r\n\r\n\r\nReplacing wz in the first equation by (X^2 - 7)/2 gives a real cubic equation with (presumably, too lazy to compute it) obvious real roots.", "Solution_2": "Let x = w + z, and y = wz.\r\n\r\n(w + z)^2 = w^2 + 2wz + z^2 = 7 + 2wz\r\n--> x^2 = 7 + 2y, y = (x^2 - 7)/2\r\n\r\n(w + z)^3 = w^3 + 3w^2z + 3wz^2 + z^3 = 10 + 3wz(w + z)\r\n--> x^3 = 10 + 3xy\r\n\r\nSo 2x^3 = 20 + 3x(x^2 - 7)\r\nx^3 - 21x + 20 = 0\r\n(x-1)(x^2 + x -20) = 0\r\n(x-1)((x+5)(x-4) = 0\r\n--> x = 1, 4, or -5\r\n\r\nSo the largest possible real value for x = w + z is 4.\r\n\r\nIn general whenever you see a problem involving symmetric terms and sums of powers, think of looking at the expansions of their powers. The sums usually appear somewhere in there.", "Solution_3": "This is similar to i/3:\r\n\r\nLet X=w+z and Y=xz. Then\r\n\r\nX^2-2Y=7\r\nX^3-3XY=X(X^2-2Y)-XY=7X-XY=10.\r\n\r\nY=(X^2-7)/2\r\n\r\n7X-X^3/2+7X/2=10, or\r\n\r\nX^3-21X+20=0.\r\n\r\n1 is a root, so dividing gives X^2+X-20, which has roots at 4 and -5, so the maximum for w+z is 4." } { "Tag": [], "Problem": "Let $ U \\equal{} 2\\times2004^{2005}$, $ V \\equal{} 2004^{2005}$, $ W \\equal{} 2003\\times2004^{2004}$, $ X \\equal{} 2\\times2004^{2004}$, $ Y \\equal{} 2004^{2004}$ and $ Z \\equal{} 2004^{2003}$. Which of the following is the largest?\r\n\r\n$ \\textbf{(A)}\\ U \\minus{} V \\qquad \\textbf{(B)}\\ V \\minus{} W \\qquad \\textbf{(C)}\\ W \\minus{} X \\qquad \\textbf{(D)}\\ X \\minus{} Y \\qquad \\textbf{(E)}\\ Y \\minus{} Z$", "Solution_1": "[hide=\"Solution\"]A.\n\nSome quick factoring shows us that A is largest. :) [/hide]", "Solution_2": "[hide]W-X is my quick guess looking at the numbers...\nedit: should stop making quick guesses, hehehe[/hide]", "Solution_3": "[hide]$U-V=2004^{2005}$\n$V-W=2004^{2004}$\n$W-X=2001\\times2004^{2004}$\n$X-Y=2004^{2004}$\n$Y-Z=2003\\times2004^{2003}$\n\n$U-V$ is the largest, so $\\boxed A$[/hide]", "Solution_4": "[hide]A. U-V is the largest.[/hide]\r\n\r\n\r\nDid you guys use a calculator or is there an easier way to do it?", "Solution_5": "[hide]$Y=1 *2004^{2004}$\n$X=2 * 2004^{2004}$\n$W=2003 * 2004^{2004}$\n$V=2004 * 2004^{2004}$\n$U=4008 * 2004^{2004}$\nwe can now forget about the $2004^{2004}$ part\nand $4008-2004=U-V$ is obviously the largest of those differences\n\n$\\boxed{A: U-V}$[/hide]", "Solution_6": "[quote=\"XCgirl\"]Did you guys use a calculator or is there an easier way to do it?[/quote]\r\nYou can factor them. These numbers would be too big for most calculators anyway. ;)", "Solution_7": "[quote=\"catcurio\"]These numbers would be too big for most calculators anyway. ;)[/quote]Make that \"for all calculators.\" :D" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let be $ a,b,c\\in \\mathbb{R}_\\plus{}$ such that $ \\sqrt{a}\\plus{}\\sqrt{b}\\plus{}\\sqrt{c}\\equal{}1$ . Prove that :\r\n\r\n$ a(\\sqrt{b}\\plus{}\\sqrt{c})\\plus{}b(\\sqrt{c}\\plus{}\\sqrt{a})\\plus{}c(\\sqrt{a}\\plus{}\\sqrt{b})\\le \\frac{1}{4}$", "Solution_1": "[quote=\"alex2008\"]Let be $ a,b,c\\in \\mathbb{R}_ \\plus{}$ such that $ \\sqrt {a} \\plus{} \\sqrt {b} \\plus{} \\sqrt {c} \\equal{} 1$ . Prove that :\n\n$ a(\\sqrt {b} \\plus{} \\sqrt {c}) \\plus{} b(\\sqrt {c} \\plus{} \\sqrt {a}) \\plus{} c(\\sqrt {a} \\plus{} \\sqrt {b})\\le \\frac {1}{4}$[/quote]\r\nPut $ \\sqrt{a}\\equal{}x,,\\sqrt{b}\\equal{}y,,\\sqrt{c}\\equal{}z$.This inequality becom:\r\n$ x^3\\plus{}y^3\\plus{}z^3\\plus{}6xyz \\ge x^2(y\\plus{}z)\\plus{}y^2(z\\plus{}x)\\plus{}z^2(x\\plus{}y)$\r\nWhich is obvious true :)" } { "Tag": [ "LaTeX", "function", "complex numbers" ], "Problem": "Either the book I am using right now (NOT aops) has made a very big typo, or something is wrong with me... :(\r\n\r\n\"However, since there are n distinct nth roots of each complex number, the answer is not unique. Thus {complex numbers} is not closed under the operation [nth root of - I don't know how to LaTeX that, sorry]\"\r\n\r\nI thought complex numbers were closed...", "Solution_1": "It's more that it is not a function than that the complex numbers aren't closed under it.\r\n\r\nCheers,\r\n\r\nRofler", "Solution_2": "Without more context it's not clear what is being said. The map \"find all complex $ n$th roots of $ z$\" is a map from complex numbers to [i]sets[/i] of complex numbers -- it looks to me that this may be the (not very interesting) point. It is true that every algebraic equation with complex coefficients has the \"right\" number of complex solutions (i.e., the Fundamental Theorem of Algebra)." } { "Tag": [ "More Sequences" ], "Problem": "Let $\\{a_{n}\\}$ be a strictly increasing positive integers sequence such that $\\gcd(a_{i}, a_{j})=1$ and $a_{i+2}-a_{i+1}>a_{i+1}-a_{i}$. Show that the infinite series \\[\\sum^{\\infty}_{i=1}\\frac{1}{a_{i}}\\] converges.", "Solution_1": "maybe the problem is not exatly same writen here...because the condition $ \\gcd(a_i,a_j) \\equal{} 1$ is extra...\r\n$ a_n \\minus{} a_{n \\minus{} 1} > a_{n \\minus{} 1} \\minus{} a_{n \\minus{} 2} > .... > a_2 \\minus{} a_1\\geq1$ therefore $ a_n \\minus{} a_{n \\minus{} 1}\\geq{n \\minus{} 1}$...and also:\r\n\r\n$ a_k \\equal{} (a_k \\minus{} a_{k \\minus{} 1}) \\plus{} (a_{k \\minus{} 1} \\minus{} a_{k \\minus{} 2}) \\plus{} .... \\plus{} (a_2 \\minus{} a_1) \\plus{} a_1\\geq{(k \\minus{} 1) \\plus{} (k \\minus{} 2) \\plus{} .... \\plus{} 1 \\plus{} 1}\\Rightarrow{a_k > \\frac {k(k \\minus{} 1)}{2}} \\\\\r\n\\Longrightarrow\\sum^{\\infty}_{i \\equal{} 1}\\frac {1}{a_{i}} < \\frac {1}{a_1} \\plus{} \\sum^{\\infty}_{i \\equal{} 2}\\frac {2}{i(i \\minus{} 1)} \\equal{} a_1 \\plus{} 2\\sum^{\\infty}_{i \\equal{} 2}{\\frac {1}{i \\minus{} 1} \\minus{} \\frac {1}{i}} \\equal{} a_1 \\plus{} 2$", "Solution_2": "Very good point, and nice solution. I have no idea what $ \\gcd(a_i,a_j)\\equal{}1$ is doing there. :maybe:", "Solution_3": "Maybe the $ >$ should be a $ \\geq$ (and the $ \\gcd(a_i,a_j)\\equal{}1$ still be there). Otherwise it isn't number theory at all.", "Solution_4": "[quote=\"ZetaX\"]Maybe the $ >$ should be a $ \\geq$ (and the $ \\gcd(a_i,a_j) \\equal{} 1$ still be there). Otherwise it isn't number theory at all.[/quote]That looks more like a NT problem indeed. Do you have a solution for that problem?", "Solution_5": "A solution for when we only assume $a_{i+2} - a_{i+1} \\ge a_{i+1} - a_i$ for all $i \\ge 1$, as suggested by ZetaX almost 15 years ago:\n\nFor a constant $c \\in \\mathbb{N}$ let $S_c = \\{a_k, a_{k+1}, \\ldots, a_{k+l-1}\\}$ be the set of elements such that the difference $a_{k+i} - a_{k+i-1}$ is exactly equal to $c$ for all $i$ with $1 \\le i \\le l$. I claim that $|S_c| = l < 2c-3$ for all $c \\ge 3$. Otherwise, note that $a_k, a_{k+1}, \\ldots, a_{k+c-2}$ are all distinct modulo $c-1$, and so are $a_{k+c-1}, a_{k+c}, \\ldots, a_{k+2c-3}$. That means that in each of these two subsequences there exists an element divisible by $c-1$, which is not allowed by the assumption $\\gcd(a_i, a_j) = 1$. Likewise, $|S_1| < 3$ and $|S_2| < 5$, as otherwise there must be two integers divisible by two or three respectively. \\\\\n\nThe inequalities $|S_1| < 3$ and $|S_2| < 5$ imply $a_5 - a_4 \\ge 2$ and $a_{10} - a_9 \\ge 3$. The inequality $|S_c| < 2c-3$ for $c \\ge 3$ can now be used to show via induction, that $a_{i+1} - a_i \\ge n$ holds for all $i \\ge n^2$ in general. Or, stated differently, $a_{i+1} - a_i \\ge \\left \\lfloor \\sqrt{i} \\right \\rfloor$. This, in turn, implies that $\\displaystyle a_{i+1} = a_1 + \\sum_{j=1}^{i} (a_{j+1} - a_j) \\ge 1 + \\sum_{j=1}^i \\left \\lfloor \\sqrt{j} \\right \\rfloor > \\sum_{j=1}^i \\frac{1}{2} \\sqrt{j} > \\int_0^i \\frac{1}{2}\\sqrt{j} \\, dj = \\frac{i^{3/2}}{3}$. \nNow the convergence of $\\displaystyle \\sum_{i=1}^{\\infty} \\frac{1}{a_i}$ is clear." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c>0$. Prove that\r\n\r\n$ \\frac{a^3\\plus{}4a\\plus{}2}{b\\plus{}c}\\plus{}\\frac{b^3\\plus{}4b\\plus{}2}{c\\plus{}a}\\plus{}\\frac{c^3\\plus{}4c\\plus{}2}{a\\plus{}b}\\geq\\frac{21}{2}$", "Solution_1": "${ \\{{a}^{3} + 1 + 1 > = 3a }$\r\n\r\nand it becomes well-known equality.\\" } { "Tag": [ "floor function", "inequalities", "algebra", "polynomial", "function", "number theory unsolved", "number theory" ], "Problem": "Find all real numbers $x$ such that $\\lfloor x^3 \\rfloor = 4x + 3$.", "Solution_1": "This is rather boring \"go through all cases\" type solution for this\r\nproblem, but it is correct and totaly doable in about 10 minutes, so\r\nnever mind. First what we can see that $4x$ is an integer, in other\r\nwords $x\\in \\{\\frac{k}4|k\\in Z\\}$.\r\n\r\nSuppose that $[x^3]=0$ then $x=-\\frac34$. If $x>0$ then $x>1$ (or\r\nelse $[x^3]=0$) so we have $4x+3=[x^3]>x^3-1$ so $4x+4>x^3$. From\r\nthis we easily get $x<3$ so $1[x^3]=4x+3$ so $x^3>4x+2$. Again $-1[/code]\r\n\r\nor something like that?", "Solution_4": "The alt attribute is supposed to do that. Apparently it doesn't work on Firefox :D Ok, I'll see what I can do." } { "Tag": [], "Problem": "What is the greatest prime factor of $12!+14!$?\r\n\r\n[hide=\"solution because I just figured it out\"]\nFactoring is your friend :) \n\nFactor:\n$12!(1+14\\cdot13)$\nClearly on the left side the biggest prime factor is 11.\n\n$14\\cdot13+1=182+1=183$.\nIt is $3\\cdot61$\n61 is prime, and thus it is the largest prime factor[/hide]", "Solution_1": "[hide]You can factor that expression into $12!(1+14*13)$, which equals $12!(183)$. The Greatest prime factor in the $12!$ portion is 11, but in the 183 portion, 183 is 3*61. Sixty-one is prime. Therefore, the greatest prime factor of $12!+14!$ is $61$.[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all $(m,n)$ such that $m,n \\in N$ and $(5+3\\sqrt{2})^m=(3+5\\sqrt{2})^n$", "Solution_1": "I think it was already posted...\r\nIndeed:\r\nTake conjugates to get the equivalent equation $(5-3\\sqrt{2})^m=(3-5\\sqrt{2})^n$. But $0<5-3\\sqrt{2}<1$ and $-5<3-5\\sqrt{2}<-4$, so equality is not possible.", "Solution_2": "I think the following argument is valid as well:\r\n\r\nWorking in $\\mathbb{Z}_7$ if $\\sqrt{2} \\equiv 3$ so $5+3\\sqrt{2} \\equiv 5+9 \\equiv 0$ whereas $3+5\\sqrt{2} \\equiv 4$ so if the equality holds we will have $4^n \\equiv 0$ which is not possible.\r\n\r\nif $\\sqrt{2} \\equiv -3$ we take conjugations to both side as ZetaX showed above and arrive at contradiction.", "Solution_3": "Fools! $(0,0)$ !", "Solution_4": "I've already said before that the definitions of $\\mathbb{N}$ are different. But the same applies when $0$ is allowed.", "Solution_5": "I have something that is similar to ZetaX's solution, but a little different.\r\n\r\nWork in $\\mathbb{Q}[\\sqrt{2}]$. Every element can be unique written in the form $a + b \\sqrt{2}$ for $a, b \\in \\mathbb{Q}$ and $f: a + b \\sqrt{2} \\rightarrow a - b\\sqrt{2}$ is an automorphism (the conjugate).\r\n\r\nSince $f$ is an automorphism, we know that $f^m(3+5\\sqrt{2}) = f^n(5+3\\sqrt{2}) \\Leftrightarrow (3+5\\sqrt{2})^m=(5+3\\sqrt{2})^n$.\r\nSo we may multiply both sides and get\r\n$f^m(3+5\\sqrt{2})(3+5\\sqrt{2})^m = f^n(5+3\\sqrt{2})(5+3\\sqrt{2})^n$\r\n$(3-5\\sqrt{2})^m(3+5\\sqrt{2})^m = (5-3\\sqrt{2})^n(5+3\\sqrt{2})^n$\r\n$((3-5\\sqrt{2})(3+5\\sqrt{2}))^m = ((5-3\\sqrt{2})(5+3\\sqrt{2}))^n$\r\n$(-41)^m = 7^n$\r\n\r\nSince $7$ and $41$ are prime, this equation cannot hold for positive integers $m$ and $n$." } { "Tag": [ "invariant", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ G$ be a finitely generated group that acts on a tree $ T$ with no inversions (no element of $ G$ interchanges the vertices of an edge). We say that such a tree $ T$ is a $ G$-tree. Show that any $ G$-tree contains a minimal (with respect to inclusion) $ G$ sub-tree (subtree with $ G$ action).\r\n\r\nAlso, the minimal $ G$-subtree is unique unless $ G$ fixes at least two vertices.\r\n\r\nGive and example when this fails for an infinitely generated group.", "Solution_1": "First the uniqueness assertion:\r\n\r\nLet $ A,B$ be distinct minimal subtrees of $ T$. They cannot intersect, because then their intersection would be a minimal subtree strictly smaller than one of them. The geodesic connecting the two must be fixed pointwise by $ G$, and it contains at least two vertices, since $ A$ and $ B$ are disjoint. \r\n\r\n\r\nExistence:\r\n\r\nCase 1 - there is an element $ s\\in G$ which fixes no vertices of $ T$. In this case there is a unique bi-infinite path $ P\\subset T$ on which $ s$ acts as a non-trivial translation. I claim that the smallest tree containing $ \\bigcup_{g\\in G}gP$, which is obviously a $ G$-tree, is minimal. It suffices to prove that every $ G$-subtree contains $ P$. Indeed, let $ a$ be a vertex of some subtree. Then the path connecting $ a$ and $ sa$ contains a segment of the path $ P$, and this segments iterates by $ s$ (both positive and negative iterates) make up the whole of $ P$; this means that every subtree invariant under the group generated by $ s$ contains $ P$, and we're done.\r\n\r\nCase 2 - there is a vertex $ a$ fixed by all the elements of $ G$. Here, there's nothing to prove: $ a$ is a minimal $ G$-subtree.\r\n\r\nWhen $ G$ is finitely-generated cases 1 and 2 cover all the possibilities: if a finitely generated group acts (without inversion) on a tree such that every element fixes some vertex, then the whole group fixes a vertex. I won't prove this. It can be found, for instance, in Serre's book \"Trees, amalgams and $ SL_2$\". \r\n\r\n\r\nThe counterexample (this was the hardest part, actually :)):\r\n\r\nThe vertices of $ T$ will be pairs of the form $ (n,\\ \\omega)$, where $ n$ is a natural number and $ \\omega \\equal{} (\\omega_n,\\ \\omega_{n \\plus{} 1}, \\ldots)$ is a binary sequence which is eventually zero. Notice that $ \\omega$ is indexed beginning with $ n$; that's the point of the notation: the $ n$ in $ (n,\\ \\omega)$ is supposed to indicate that we regard the sequence $ \\omega$ as starting from position $ n$. For all $ n\\ge 0$ and all $ \\omega \\equal{} (\\omega_n,\\ \\omega_{n \\plus{} 1},\\ \\ldots)$, the corresponding point $ a \\equal{} (n,\\ \\omega)$ is connected to $ (n \\plus{} 1,\\ \\omega)$, where this second $ \\omega$ stands for $ (\\omega_{n \\plus{} 1},\\ \\omega_{n \\plus{} 2},\\ \\ldots)$. We call the vertices of the form $ (n,\\ \\omega)$ \"vertices at level $ n$\". This describes a tree in which the vertices at level $ 0$ are leaves (i.e. are connected only to one other vertex) and those at higher levels have degree $ 3$ (they are connected to two vertices at a lower level and to one vertex at a higher level). \r\n\r\nThe group $ G$ is the group of automorphisms of $ T$ which preserve the levels of the vertices and, for some $ n\\ge 0$, fix the set of vertices at levels $ \\ge n$ pointwise. Now, if a $ G$-subtree has a vertex $ (n,\\ \\omega)$, then it contains all the vertices at the level $ n$, as can easily be seen. It follows that the invariant subtrees are precisely those consisting of vertices at levels $ \\ge n$ for $ n\\in\\mathbb N$, and so that there can be no [i]minimal[/i] $ G$-subtree." } { "Tag": [ "symmetry", "geometry", "geometric transformation", "reflection", "perimeter" ], "Problem": "Prove that the quadrilateral with symetric center is parallelogram.", "Solution_1": "sorry...what do you mean by symetric center...", "Solution_2": "You know...... the intersection between the diagonals is the symetric center of the parallelogram.", "Solution_3": "I hope I understand corectly.Let the quadrilateral be $ ABCD$ and $ O$ the symetric center( If I am right $ AO\\equal{}OC$ and $ BO\\equal{}BD$).Then \r\n$ \\Delta AOB \\equiv \\Delta DOC$ and $ \\Delta AOD \\equiv \\Delta BOC$ then it results that $ AB\\parallel{}DC$ and $ BC\\parallel{}ACD$.\r\n\r\nI repeat, I don't think I understand corectly what is the symetric center.It may be also the altitudes from $ O$ to $ AB$ and $ DC$(are equal) and from $ O$ to $ AD$ and $ BC$ are equal(please corect me!)", "Solution_4": "[color=darkblue][b]Center of symmetry[/b] of a polygon $ P$ is a point, lets call it $ O$, such that for any point $ X \\in P$, the reflection of $ X$ wrt $ O$ also lies on the perimeter of $ P$.[/color]", "Solution_5": "Hi \r\n\r\nAre you sure yuor definition is correct . I am trying to translate what is sad in my book : A polygon $ P$ has a center of symetry if there exist a point $ O$ such that every point of this polygon has each own reflaction according to $ O$ in the plane limited by the polygon. \r\n\r\nIt is also a large possibility that my book may be wrong.", "Solution_6": "Can somebody insure me about the definition of center of symetry of a polygon" } { "Tag": [], "Problem": "What is the least amount a number can have if it is divided by 3,4, and 5 and yields a remainder of 2 with the exception of 2?", "Solution_1": "[hide=\"answer\"]Since it leaves a remainder of $2$, it simply is $3 \\cdot 4 \\cdot 5+2=\\boxed{62}$[/hide]\r\n\r\nOh, and by the way, please refrain from using \"pretty obvious\" or \"really easy\" etc, because it might make other people uncomfortable if they are learning and trying to get better", "Solution_2": "i'll keep that in mind for next time. :?", "Solution_3": "No Problem :)" } { "Tag": [ "quadratics" ], "Problem": "It is given that one root of $2x^2+rx+s=0,$ with $r$ and $s$ real numbers, is $3+2i$ ($i=\\sqrt{-1}$). The value of $s$ is:\r\n(A) undetermined\r\n(B) 5\r\n(C) 6\r\n(D) -13\r\n(E) 26", "Solution_1": "[hide]\nFor $r$ and $s$ to be real numbers, the second root must be $3 - 2i$\nExpanding, we get the quadratic to be $x^2 - 6x + 13$, then to make the lead coefficient 2, $2x^2 - 12x + 26$. $s = 26$. $D$. Tricky there at the end...\n[/hide]", "Solution_2": "I got the same answer as you but a different way,\r\n$\\frac{-r+\\sqrt{r^2-8s}}{4}=3+2i$\r\nSo equating coefficients we get,\r\n$\\frac{-r}{4}=3$\r\n$r=-12$\r\nsimilarily,\r\n$\\frac{\\sqrt{r^2-8s}}{4}=2\\sqrt{-1}$\r\nsquaring both sides we get,\r\n$\\frac{r^2-8s}{16}=-4$\r\nSolving we get,\r\n$\\boxed{s=26}$\r\nSo it is $\\boxed{E}$\r\nCorrect?\r\nMy method is not elegant though :? :(", "Solution_3": "[hide]the two roots must be 3+2i and 3-2i for coefficients to be real, so the expansion we have to do is (x-3-2i)(x-3+2i), the i's will disappear. \nwhich equals $x^2 - 6x + 13$. \n\nNote that $-(2i)^2 = 4$. \n\nDoubling that, $2x^2 - 12x + 26$. \n\ns = 26. E. [/hide]", "Solution_4": "Just use the product of two roots = $\\frac {s}{2}$ and you are done ." } { "Tag": [], "Problem": "How many 5 digit integers are there consisting of the digits 1, 2 and 3 each appearing at least once?", "Solution_1": "[hide]We count all numbers with digits limited to 1, 2, 3. Then we subtract those that don't have a 1, 2, or 3.\n\nThe first is equal to $ 3^5\\equal{}243$.\n\nThe second is equal to $ \\{\\text{digits are 1 and 2}\\}\\plus{}\\{\\text{digits are 1 and 3}\\}\\plus{}\\{\\text{digits are 2 and 3}\\}\\minus{}\\{\\text{digits all 1}\\}\\minus{}\\{\\text{digits all 2}\\}\\minus{}\\{\\text{digits all 3}\\}\\equal{}2^5\\plus{}2^5\\plus{}2^5\\minus{}1\\minus{}1\\minus{}1\\equal{}93$.\n\nThe answer is $ 243\\minus{}93\\equal{}\\boxed{150}$.[/hide]", "Solution_2": "@lingomaniac88, what about a number like $ 99123$? It works...\r\n\r\n[hide]A five-digit number has five digits. We need to have $ 1, 2, 3$ each appear at least once, so we need to select $ 3$ of the $ 5$ digits to represent $ 1, 2, 3$. We split it into two cases:\nThere are $ \\binom{5}{3}=10$ ways to do so.\nOf these $ 10$ ways, $ 6$ of them fill the first digit.\nThen for $ 6$ of the choices, there are $ 10^2$ ways to fill the two remaining digits.\nFor the $ 10-6=4$ choices, there are $ 9\\cdot 10$ ways to fill the two remaining digits.\n\nIn total, the answer is $ 6 \\cdot 10^2+4 \\cdot 9\\cdot 10=600+360=\\boxed{960}$.[/hide]", "Solution_3": "No lingomaniac's answer is correct. The number consists of only the digits 1,2,3. Sorry if my statement is troubling.", "Solution_4": "[quote=\"nayel\"]No lingomaniac's answer is correct. The number consists of only the digits 1,2,3. Sorry if my statement is troubling.[/quote]\r\n\r\nOh sorry. Ignore my solution then. lol", "Solution_5": "No it's okay. There's no harm in discussing more problems. :roll:" } { "Tag": [ "Euler" ], "Problem": "Find the sum of the digits of $ 2^{1000}$.", "Solution_1": "Is there a formula or pattern to find this?\r\n\r\nObviously, finding all digits of $ 2^1000$ doesn't seem like a good idea", "Solution_2": "$ 2^{1000} \\equal{} 16^{250}$\r\n[hide=\"Observation\"]Observe:\n[list]$ 16^0 \\equal{} 1 \\rightarrow 1$\n$ 16^1 \\equal{} 16 \\rightarrow 7$\n$ 16^2 \\equal{} 256 \\rightarrow 13$\n$ 16^3 \\equal{} 4096 \\rightarrow 19$\n$ 16^4 \\equal{} 65536 \\rightarrow 25$\n...\nSee a pattern?[/list][/hide]\n[hide=\"Possible Solution\"]\nFrom the observation, we can conclude that the sum of the digits of $ 16^n$ for $ n >\\equal{}0$ is $ 6n \\plus{} 1$ (there yet has to be a proof or disproof). The sum of the digits of $ 2^{1000}$, or $ 16^{250}$, is $ 6(250) \\plus{} 1 \\equal{} \\boxed{1501}$.[/hide]", "Solution_3": "Perhaps taking it modulo 9 would help? Also, this is a project Euler problem!", "Solution_4": "[quote=\"Patterns_34\"]$ 2^{1000} \\equal{} 16^{250}$\n[hide=\"Observation\"]Observe:\n[list]$ 16^0 \\equal{} 1 \\rightarrow 1$\n$ 16^1 \\equal{} 16 \\rightarrow 7$\n$ 16^2 \\equal{} 256 \\rightarrow 13$\n$ 16^3 \\equal{} 4096 \\rightarrow 19$\n$ 16^4 \\equal{} 65536 \\rightarrow 25$\n...\nSee a pattern?[/list][/hide]\n[hide=\"Possible Solution\"]\nFrom the observation, we can conclude that the sum of the digits of $ 16^n$ for $ n > \\equal{} 0$ is $ 6n \\plus{} 1$ (there yet has to be a proof or disproof). The sum of the digits of $ 2^{1000}$, or $ 16^{250}$, is $ 6(250) \\plus{} 1 \\equal{} \\boxed{1501}$.[/hide][/quote]\r\n\r\nFails for $ n \\equal{} 8$. You shouldn't expect patterns like this to hold. \r\n\r\nIf this is a Project Euler problem, it isn't meant to be solved by hand.", "Solution_5": "Thank you for the insight t0rajir0u.\r\n\r\nJust a question, why have a [url=http://projecteuler.net/]Project Euler[/url] [url=http://projecteuler.net/index.php?section=problems]problem[/url] be posted in High School Basics?", "Solution_6": "[quote=\"math92\"]Find the sum of the digits of $ 2^{1000}$.[/quote]\r\n\r\n1 - 2^0\r\n\r\n7 - 2^10\r\n\r\n31 - 2^20\r\n\r\n37 - 2^30\r\n\r\n61 - 2^40\r\n\r\ni'm not too experienced with euler...but i think you can find it out that way\r\n\r\n\r\ni found a cool thing from a wiki page somewhere", "Solution_7": "[quote=\"math92\"]Find the sum of the digits of $ 2^{1000}$.[/quote]\r\n\r\nA quick basic computation with Excel gives a value of $ 1366$\r\n\r\nThe max value for $ n<1000$ is obtained for $ 2^{996}$ which gives $ 1468$\r\n\r\nBut I have no idea about how to find these values except just computing $ 2^n$", "Solution_8": "[quote=\"pco\"]A quick basic computation with Excel gives a value of $ 1366$[/quote]\r\nPco, how did you get Excel to do that kind of computation?\r\n\r\nWhen I imput $ 16^{13}$, or $ 2^{52}$, or greater my Excel program doesn't return an accurate number.", "Solution_9": "[quote=\"Patterns_34\"][quote=\"pco\"]A quick basic computation with Excel gives a value of $ 1366$[/quote]\nPco, how did you get Excel to do that kind of computation?\n\nWhen I imput $ 16^{13}$, or $ 2^{52}$, or greater my Excel program doesn't return an accurate number.[/quote]\r\n\r\nYou just have to use slices of 6 digits, for example :\r\nPut in A0 value 1 and in the remainder of line A values 0.\r\nPut in B0 the formule \"=\\mod(2*A0;1000000)\"\r\nPut in the remainder of line B the formula \"=\\mod(2*uppercell+ent(upperleftcell/500000);1000000)\"\r\n\r\nYou just have to extend right and down and you get values of $ 2^n$ by slices of 6 digits." } { "Tag": [ "LaTeX", "function", "integration" ], "Problem": "Is there a way to prevent functions in LaTeX to work. For example you just want to show the function \r\n\\frac{}{}.", "Solution_1": "For a small amunt of text use \\verb using a delimiter such as =[code] \\verb=\\frac{}{}= [/code] which gives $ \\verb= \\frac{ }{ }=$. \nFor longer text use the verbatim environment instead.\n[code]\\begin{verbatim}\n\\textbf{bold text} \\\\\n\\frac{}{}\n\\int_0^1 x\\;dx\n\\end{verbatim}[/code] which gives $ \\par\\begin{verbatim}\r\n\\textbf{bold text} \\\\\r\n\\frac{1}{2}\r\n\\int_0^1 x\\;dx\r\n\\end{verbatim}$\r\n(ignore the \\par which is only to make it work on the forum. In a document the above will show as 3 lines of text).\r\nThe starred versions show a $ \\verb*= =$ for spaces.", "Solution_2": "Thank you!" } { "Tag": [], "Problem": "King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 53 and an angle of 38. A cannonball that was accidentally dropped from the top of the castle wall hit the moat below in 1.6 .", "Solution_1": "What's the question? (and I think you forgot the units: 53 m/s, 30 degrees, 1.6 s?)", "Solution_2": "hey man where's the question??? :rotfl: :rotfl: :P", "Solution_3": "Same doubt here!!!!\r\nAlso the units........" } { "Tag": [ "function", "Support", "real analysis", "real analysis unsolved" ], "Problem": "This is, probably, a classical one but I do not remember it posted. Let $x_{k}$ be a sequence of pairwise distinct real numbers and let $d_{k}$ be any sequence of reals tending to $0$. Prove or disprove that then there always exists a function $f : \\mathbb R\\to\\mathbb R$ such that it is continuous at every point except the points $x_{k}$ and at each point $x_{k}$ the left and the right limits exist and the difference between them (the \"jump\") is exactly $d_{k}$.", "Solution_1": "Yes, there is such a function. We can construct a sequence of functions $f_{n}$ converging uniformly to $f$ with the property that $f_{n}$ is continuous except at $x_{1},x_{2},\\dots,x_{n}$ with jumps $d_{1},d_{2},\\dots,d_{n}$ at these points. Verifying that the limit has the right jumps and is continuous elsewhere given these properties is a routine exercise.\r\n\r\nLet $g(x)=\\begin{cases}0& x\\le 0\\\\ 1-x&01\\end{cases}$. Let $g_{k}(x)=d_{k}g(\\frac{x-x_{k}}{c_{k}})$, where $c_{k}$ are positive constants to be chosen later. Then we will have $f_{n}(x)=g_{1}(x)+g_{2}(x)+\\dots+g_{n}(x)$.\r\n\r\nLet $N_{0}=0$, and choose an increasing sequence $N_{i}$ such that $d_{k}<\\frac1{2^{i}}$ for $k>N_{i}$. If $N_{i-1}0$, and $ 1\\le a\\in\\mathbb N<2$, $ a\\equal{}1\\implies b\\plus{}\\frac1{c\\plus{}\\frac1d}\\equal{}\\frac{25}{22}$\r\n$ \\implies b\\equal{}1\\implies c\\plus{}\\frac1d\\equal{}\\frac{22}3\\implies c\\equal{}7\\implies d\\equal{}3$\r\n$ \\implies3a\\minus{}2b\\plus{}c^2\\equal{}3(1)\\minus{}2(1)\\plus{}(7)^2\\equal{}\\boxed{50}$.", "Solution_2": "Yes, You're Right. :lol:" } { "Tag": [], "Problem": "How did everybody do?", "Solution_1": "I got a 42, but I know somebody who got a 45. I beat him in the Countdown Round though :) .", "Solution_2": "Please just refer to the other topic that is up for Chapters, since it already has a lot of posts. Thanks.", "Solution_3": "I got a 36, sad I know. I didn't even get to Countdown. Yet out of the ashes of this tradegy our team made it to state and so I can still compete in the team and individual again. Luck is with me.", "Solution_4": "I missed out (barely) on countdown with a 34. I'm not too ashamed of that as a sixth grader :).\r\n\r\nI really messed up on 4 points worth of problems. I got swamped on the sprint, and the target was like, the easiest target i've ever taken chapter level (I only got 5 in shcool lol, and I got 7 here with 1 stupid mistake)", "Solution_5": "The target was too easy, but the sprint was really hard. I got 18/7(32), and got 9 on the team round.", "Solution_6": "I do agree that the target was easy, but the sprint round I thought was harder than the target round!!! \r\nMy score: 15/30 (sprint round), 12/16 (target) = 27/46, getting seventh inmy chapter. I was expecting higher...", "Solution_7": "About the target round... I think I solved all but one by hand... b/c they reminded me of the first pages of a State Sprint... ya but I think the fact that the CT competitions are so early in the morning makes it a bit tough to be fresh for the Sprints... in hindsight even the spring (except for 24) was pretty easy.. but it had too much combinatorics for my own good. ;)" } { "Tag": [ "\\/closed" ], "Problem": "I want to activate HTML in my posts, and I did so in my profile. However, when I am posting, under the Options box it still says HTML is off. This is what I get when I post HTML:\r\n\r\nMy School", "Solution_1": "You have to do it every time you post. Did you press Save in your profile? (sometimes I don't and my selections don't save)", "Solution_2": "yes, I pressed save, even before my last post...", "Solution_3": "HTML is not allowed on th site. You can use BBCode to make the same thing: \r\n[code]\n[url=http://ligon.wcpss.net/]My School[/url]\n[/code]\r\ngives you\r\n\r\n[url=http://ligon.wcpss.net/]My School[/url]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Solve the following equation over rationals\r\n$ a^b \\equal{} b^a$", "Solution_1": "Well, it has surely infinitely many solutions since $ f(x) \\equal{} x^{\\frac {1}{x}}$ is continous and is strictly monotonic (rising) on $ (0,e)$ and monotonic (deacreasing) on $ e, \\plus{} \\infty)$, thus using the Darboux property $ \\forall_{a\\in (0,e)} \\ \\exists_{b\\in (e, \\plus{} \\infty)} \\ f(a) \\equal{} f(b)$ :wink:\r\nAlso look here:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=150682", "Solution_2": "Similarly, one finds that the set of real solutions is given by $ (x, y) \\equal{} \\left( k^{ \\frac{1}{k\\minus{}1} }, k^{ \\frac{k}{k\\minus{}1} } \\right)$.", "Solution_3": "thanks polskimisiek for the reference." } { "Tag": [ "function" ], "Problem": "A checkerboard of 13 rows and 17 columns has a number written in each square, beginning in the upper-left corner, so that the first row is numbered $1, 2, \\ldots , 17$, the second row $18, 19, \\ldots , 34$, and so on down the board. If the board is renumbered so that the left column, top to bottom is $1,2, \\ldots , 13$, the second column $14, 15, \\ldots , 26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).\r\n\r\n\\[ \\text{(A) } 222 \\qquad \\text{(B) } 333 \\qquad \\text{(C) } 444 \\qquad \\text{(D) } 555 \\qquad \\text{(E) } 666 \\]", "Solution_1": "[hide]\nIf we label the rows x and the columns y, we can create expressions for each number which end up being $13(y-1)+x$ and $17(x-1)+y$. \nEquating these:\n$13y - 13 + x = 17x - 17 + y$\n$12y - 16x = -4$\n$4x - 3y = 1$\n\nNext, we look for x and y values which satisfy this equation which can be easily done by graphing the function (and going to the table to look for integer values). These end up being (1,1); (4,5); (7,9); (10,13); (13,17).\nWe get a sum:\n$1 + 56 + 111 + 166 + 221 = 555$\n$\\mbox{D}$\n[/hide]", "Solution_2": "[hide=\"Answer\"]Any numbers of this form are such that $(n-1)x+m=(m-1)y+n$, so $13(n-1)+m=17(m-1)+n\\Rightarrow 13n+m-13=17m+n-17\\Rightarrow$\n$12n+4=16m\\Rightarrow 3n+1=4m$. Therefore, our possible answers are $(1,1)$, $(5,4)$, $(9,7)$, $(13,10)$, and $(17,13)$. Substituting those into the original equation, we have $1$, $56$, $111$, $166$, and $221$, which sum to $555\\Rightarrow \\boxed{D}$.[/hide]" } { "Tag": [], "Problem": "Let $p_{i}$ be the $i^{th}$ prime.\r\n\r\n$p_{1}p_{2}p_{3}\\cdots p_{n}=p_{x}$\r\n\r\nwhere $x>n$\r\n\r\nDoes $x=2n$?", "Solution_1": "Huh? By definition a prime can't be expressed as a product of primes.", "Solution_2": "[quote=\"matt276eagles\"]Huh? By definition a prime can't be expressed as a product of primes.[/quote]\r\n\r\nI think he meant to add $1$ to the product.", "Solution_3": "[hide=\"Uh?\"]\n$(2)(3)(5)+1=31\\neq p_{6}=13$.\n[/hide]", "Solution_4": "Rofl. I didn't look at the second part of the question. \r\n\r\nI guess not...", "Solution_5": "sorry. I meant p_1p_2...p_n +1 = p_x.", "Solution_6": "[quote=\"1=2\"]sorry. I meant p_1p_2...p_n +1 = p_x.[/quote]\r\nUhh, there's no guarantee that the LHS is prime\r\n$2\\cdot 3 \\cdot 5\\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19+1=1487\\cdot 593$", "Solution_7": "diophantient, check your work", "Solution_8": "[quote=\"Derek\"]diophantient, check your work[/quote]\r\noops, thanks for pointing that out.\r\n2*3*5*11*13+1=7*613" } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "Number of posts: 11182\r\nPosts per day: 29.81 \r\n\r\nNumber of topics: 2793 \r\nTopics per day: 7.45 \r\n\r\nNumber of users: 432 \r\nUsers per day: 1.15 \r\n\r\nBoard started: 03 Feb 2003 11:16 pm \r\nDatabase size: 11.92 MB", "Solution_1": "if latex is available, i think, the size of database will grow faster.\r\nYou need to store many gif files.\r\nand all latex-to-html software converts duplicated formulae into different pictures - that's bad.", "Solution_2": "[quote]all latex-to-html software converts duplicated formulae into different pictures - that's bad[/quote]That's not quite true. Latexrender is very efficient and caches images so duplicated formulae re-use the same images.", "Solution_3": "Whoops, forgot to log in before posting that message :?", "Solution_4": "one month has passed, so I will update the stats\r\n\r\nNumber of posts: 14080 \r\nPosts per day: 34.81 \r\n\r\nNumber of topics: 3576\r\nTopics per day: 8.84 \r\n\r\nNumber of users: 583\r\nUsers per day: 1.44 \r\n\r\nDatabase size: 15.24 MB", "Solution_5": "another month has passed, so I will again update the stats\r\n\r\nNumber of posts: 16038\r\nPosts per day: 36.72 \r\n\r\nNumber of topics: 3959\r\nTopics per day: 9.06 \r\n\r\nNumber of users: 748\r\nUsers per day: 1.71 \r\n\r\nDatabase size: 17.73 MB", "Solution_6": "more stats this month :) \r\n\r\nNumber of posts: 18090\r\nPosts per day: 38.83 \r\n\r\nNumber of topics: 4453\r\nTopics per day: 9.56 \r\n\r\nNumber of users: 878\r\nUsers per day: 1.88 \r\n\r\nDatabase size: 20.65 MB", "Solution_7": "Number of posts: 21016\r\nPosts per day: 41.84 \r\n\r\nNumber of topics: 5050\r\nTopics per day: 10.05 \r\n\r\nNumber of users: 1065\r\nUsers per day: 2.12 \r\n\r\nDatabase size: 24.77 MB", "Solution_8": "now some nice graphs and we'll all clap our hands :D\r\n\r\nwell seems like the site is nicely growing! Now too bad there's no statistic to see how much of the post are really mathematical and how much are just gossip :P" } { "Tag": [ "inequalities", "induction", "inequalities proposed" ], "Problem": "I'm interested in finding out some new proofs for the next inequality.\r\n$ \\frac{a^{n\\plus{}1}}{b^n}(a\\minus{}b)\\plus{}\\frac{b^{n\\plus{}1}}{c^n}(b\\minus{}c)\\plus{}\\frac{c^{n\\plus{}1}}{a^n}(c\\minus{}a)\\geq0$, where: $ a,b,c>0$ and $ n\\in N$.", "Solution_1": "I'd like to give you a huge hint:\r\n$ LHS\\equal{}\\frac{(a^2)^{n\\plus{}1}}{(ab)^n}\\plus{}\\frac{(b^2)^{n\\plus{}1}}{(bc)^n}\\plus{}\\frac{(c^2)^{n\\plus{}1}}{(ca)^n}\\stackrel{(*)}{\\geq}\\underbrace{\\frac{a^2\\plus{}b^2\\plus{}c^2}{ab\\plus{}bc\\plus{}ca}}_{\\geq1}\\cdot\\left(\\frac{(a^2)^n}{(ab)^{n\\minus{}1}}\\plus{}\\frac{(b^2)^n}{(bc)^{n\\minus{}1}}\\plus{}\\frac{(c^2)^n}{(ca)^{n\\minus{}1}}\\right)\\geq RHS$\r\nAny remarks? I'm interested to find out your oppinions related to the inequality $ (*)$.", "Solution_2": "I am trying not to be offensive but a little bit more constructive here. How can you get the inequality (*) ? Please restate the used inequality and where you applied it. \r\n\r\nRegards,", "Solution_3": "$ \\frac{a^{n\\plus{}1}}{b^{n}}(a\\minus{}b)\\plus{}\\frac{b^{n\\plus{}1}}{c^{n}}(b\\minus{}c)\\plus{}\\frac{c^{n\\plus{}1}}{a^{n}}(c\\minus{}a)\\geq0 \\Longleftrightarrow$\r\n$ \\Longleftrightarrow\\ \\frac{a^{n\\plus{}2}}{b^n}\\plus{}\\frac{b^{n\\plus{}2}}{c^n}\\plus{}\\frac{c^{n\\plus{}2}}{a^n}\\geq\\frac{a^{n\\plus{}1}}{b^{n\\minus{}1}}\\plus{}\\frac{b^{n\\plus{}1}}{c^{n\\minus{}1}}\\plus{}\\frac{c^{n\\plus{}1}}{a^{n\\minus{}1}}$\r\nWe obtain:\r\n$ LHS\\equal{}\\frac{(a^{2})^{n\\plus{}1}}{(ab)^{n}}\\plus{}\\frac{(b^{2})^{n\\plus{}1}}{(bc)^{n}}\\plus{}\\frac{(c^{2})^{n\\plus{}1}}{(ca)^{n}}\\stackrel{(*)}{\\geq}\\underbrace{\\frac{a^{2}\\plus{}b^{2}\\plus{}c^{2}}{ab\\plus{}bc\\plus{}ca}}_{\\geq1}\\cdot\\left(\\frac{(a^{2})^{n}}{(ab)^{n\\minus{}1}}\\plus{}\\frac{(b^{2})^{n}}{(bc)^{n\\minus{}1}}\\plus{}\\frac{(c^{2})^{n}}{(ca)^{n\\minus{}1}}\\right)\\geq\\frac{(a^{2})^{n}}{(ab)^{n\\minus{}1}}\\plus{}\\frac{(b^{2})^{n}}{(bc)^{n\\minus{}1}}\\plus{}\\frac{(c^{2})^{n}}{(ca)^{n\\minus{}1}}\\equal{} RHS$\r\nTry to prove the following result, and use it in $ (*)$ for $ n\\equal{}3$. I haven't seen this result before anywhere. I think and hope it's something new. I have already a new article concerning this inequality, and its derived results.\r\n[b]MY LEMMA:[/b]\r\n\r\nIf: $ a_i\\geq0,b_i>0,i\\equal{}1,2,\\dots,k$ and $ k,n\\in N,k\\geq1$, then:\r\n$ \\frac{{a_1}^{n\\plus{}1}}{b_1^n}\\plus{}\\frac{a_2^{n\\plus{}1}}{b_2^n}\\plus{}\\cdots\\plus{}\\frac{{a_k}^{n\\plus{}1}}{b_k^n}\\geq\\frac{a_1\\plus{}a_2\\plus{}\\cdots\\plus{}a_k}{b_1\\plus{}b_2\\plus{}\\cdots\\plus{}b_k}\\cdot\\left(\\frac{{a_1}^n}{b_1^{n\\minus{}1}}\\plus{}\\frac{{a_2}^n}{b_2^{n\\minus{}1}}\\plus{}\\cdots\\plus{}\\frac{{a_k}^n}{b_k^{n\\minus{}1}}\\right)$\r\nThis is my solution. I tought that someone could give me a different solution. I'm sure that everything is OK now, [b]great math[/b]!", "Solution_4": "Thanks for the lemma. I haven't proven it yet but I have another proof for the given inequality\r\n\r\nUsing mathematical induction, it suffices to prove that\r\n\r\n\\[ \\frac{a^{n\\plus{}2}}{b^n} \\plus{}\\frac{a^n}{b^{n\\minus{}2}} \\ge 2\\sum_{cyc} \\frac{a^{n\\plus{}1}}{b^{n\\minus{}1}}\\]\r\n\r\nand \r\n\r\n\\[ \\sum_{cyc}\\frac{a^{n\\plus{}1}}{b^{n\\minus{}1}} \\ge \\sum_{cyc}\\frac{a^n}{b^{n\\minus{}2}}\\] (due to mathematical induction)\r\n\r\nHence, we complete our proof here.", "Solution_5": "[quote=\"marin.bancos\"]I'm interested in finding out some new proofs for the next inequality.\n$ \\frac {a^{n \\plus{} 1}}{b^n}(a \\minus{} b) \\plus{} \\frac {b^{n \\plus{} 1}}{c^n}(b \\minus{} c) \\plus{} \\frac {c^{n \\plus{} 1}}{a^n}(c \\minus{} a)\\geq0$, where: $ a,b,c > 0$ and $ n\\in N$.[/quote]\r\nThis inequality is true according to the obvious inequality\r\n$ \\frac{2a^{n\\plus{}1}}{b^n} (a\\minus{}b) \\ge a^2\\minus{}b^2.$", "Solution_6": "Thank you, [b]great math[/b]! Thank you, [b]can_hang2007[/b]! Very nice ideas! Using your solution, this inequality could be generalized:\r\n$ \\frac{a_1^{n\\plus{}1}}{a_2^{n}}(a_1\\minus{}a_2)\\plus{}\\frac{a_2^{n\\plus{}1}}{a_3^{n}}(a_2\\minus{}a_3)\\plus{}\\cdots\\plus{}\\frac{a_k^{n\\plus{}1}}{a_1^{n}}(a_k\\minus{}a_1)\\geq0$, where: $ a_1,a_2,\\dots,a_k\\geq0$ and $ n,k\\in N, k\\geq2.$\r\nA very beautiful expression.\r\nI'm sure that my lemma is also a valuable inequality because of its well known and interesting derived results.", "Solution_7": "Some friends of MathLinks asked me to post my proof for MY LEMMA. You can see it here. I'm very interested in your oppinions. What do you think about it, [b]Manlio[/b]? Owing to [b]can_hang2007[/b]'s idea, we can give a new generalization for my initial inequality. Try to prove it. \r\nProve that: $ \\frac{a_{1}^{n}(a_{1}\\minus{}a_{2})}{a_{2}^{n\\minus{}k\\plus{}1}}\\plus{}\\frac{a_{2}^{n}(a_{2}\\minus{}a_{3})}{a_{3}^{n\\minus{}k\\plus{}1}}\\plus{}\\cdots\\plus{}\\frac{a_{p}^{n}(a_{p}\\minus{}a_{1})}{a_{1}^{n\\minus{}k\\plus{}1}}\\geq0,$ where: $ a_{1},a_{2},\\dots,a_{p}\\geq0$ and $ n,k\\in N, n\\geq k\\geq1.$", "Solution_8": "You can see MY LEMMA here.", "Solution_9": "First, let me recall your lemma:\r\n$ \\frac {{a_{1}}^{n \\plus{} 1}}{b_{1}^{n}} \\plus{} \\frac {a_{2}^{n \\plus{} 1}}{b_{2}^{n}} \\plus{} \\cdots \\plus{} \\frac {{a_{k}}^{n \\plus{} 1}}{b_{k}^{n}}\\geq\\frac {a_{1} \\plus{} a_{2} \\plus{} \\cdots \\plus{} a_{k}}{b_{1} \\plus{} b_{2} \\plus{} \\cdots \\plus{} b_{k}}\\cdot\\left(\\frac {{a_{1}}^{n}}{b_{1}^{n \\minus{} 1}} \\plus{} \\frac {{a_{2}}^{n}}{b_{2}^{n \\minus{} 1}} \\plus{} \\cdots \\plus{} \\frac {{a_{k}}^{n}}{b_{k}^{n \\minus{} 1}}\\right)$\r\n\r\nI see your proof is nice, but I think we can get a simpler proof as follows:\r\n\r\nBecause the inequality is homogeneous in $ a_1,a_2,\\ldots, a_k$ and $ b_1,b_2,\\ldots, b_k,$ we may assume that $ a_1 \\plus{} a_2 \\plus{} \\cdots\\plus{} a_k \\equal{} b_1 \\plus{} b_2 \\plus{} \\cdots \\plus{} b_k \\equal{} k.$ Then, we can write our inequality as\r\n$ \\sum^k_{i \\equal{} 1}\\frac {a^{n \\plus{} 1}_i}{b^n_i} \\ge \\sum^k_{i \\equal{} 1}\\frac {a^n_i}{b^{n \\minus{} 1}_i},$\r\nor\r\n$ \\sum^k_{i \\equal{} 1}\\frac {a^{n \\plus{} 1}_i}{b^n_i} \\plus{} \\sum^k_{i \\equal{} 1}(b_i \\minus{} a_i)\\ge \\sum^k_{i \\equal{} 1}\\frac {a^n_i}{b^{n \\minus{} 1}_i}.$\r\nThis inequality is true because \r\n$ \\frac {a^{n \\plus{} 1}_i}{b^n_i} \\plus{} b_i \\minus{} a_i \\minus{} \\frac {a^n_i}{b^{n \\minus{} 1}_i} \\equal{} \\frac {a^n_i(a_i \\minus{} b_i)}{b^n_i} \\plus{} b_i \\minus{} a_i \\equal{} \\frac {(a^n_i \\minus{} b^n_i)(a_i \\minus{} b_i)}{b^n_i} \\ge 0.$", "Solution_10": "Very nice proof, [b]can_hang2007.[/b] Thanks for it. What do you think about the final generalization of my initial inequality?\r\nI'm waiting for new proofs or new results, my friends!" } { "Tag": [ "trigonometry" ], "Problem": "Find all x such that:\r\n\r\n$ \\sqrt{cos2x\\minus{}sin4x}\\equal{}sinx\\minus{}cosx$", "Solution_1": "[quote=\"MR.129\"]Find all x such that:\n\n$ \\sqrt {\\cos2x \\minus{} \\sin4x} \\equal{} \\sin x \\minus{} \\cos x$\n [/quote]\r\n[hide=\"solution\"]After squaring we attain:\n$ \\cos 2x \\minus{} 2 \\sin 2x \\cdot \\cos 2x \\equal{} (\\sin x \\minus{} \\cos x)^2 \\implies \\sin 2x\\plus{}\\cos2x\\equal{}2\\sin 2x \\cdot \\cos 2x$\n\nSubstituting $ \\cos^2 2x\\equal{}1\\minus{}\\sin^2 2x$ and calculating we get:\n\\[ 4 \\sin^4 2x \\cdot (\\sin^2 2x\\minus{}1)\\equal{}0\\]\n\nand therefore the solutions are:\n$ x_1\\equal{}k\\cdot \\pi$\n\n$ x_2\\equal{}\\frac{\\pi}{4}\\plus{}k\\cdot \\pi$\n\n$ x_3\\equal{}\\frac{3\\pi}{4}\\plus{}k\\cdot \\pi$\n\nwhere $ k \\in Z$[/hide]", "Solution_2": "hello, $ x\\equal{}2\\cdot\\pi$ is not a solution of your equation.\r\nSonnhard.", "Solution_3": "[quote=\"Dr Sonnhard Graubner\"]hello, $ x \\equal{} 2\\cdot\\pi$ is not a solution of your equation.\nSonnhard.[/quote]\r\n\r\nI didn't actually check solutions properly but let's see\r\n\r\n$ \\sqrt{\\cos2 \\cdot 2\\pi \\minus{} \\sin 4\\cdot 2\\pi}\\equal{}\\sin 4\\pi \\minus{} \\cos 4\\pi$\r\n\r\n$ \\sqrt{1}\\equal{}\\minus{}1$ .. what;s wrong ?", "Solution_4": "hello, $ \\sqrt{1}\\equal{}\\minus{}1$ is definitively wrong.\r\nSonnhard.", "Solution_5": "[quote=\"Dr Sonnhard Graubner\"]hello, $ \\sqrt {1} \\equal{} \\minus{} 1$ is definitively wrong.\nSonnhard.[/quote]\r\n\r\nwhy ?", "Solution_6": "By definition, $ \\sqrt{x}$ denotes the nonnegative real number $ y$ such that $ y^2 \\equal{} x$ (if it exists). Therefore, $ \\sqrt{x} \\ge 0$ for all $ x$ and hence $ \\sqrt{1} \\not\\equal{} \\minus{}1$.", "Solution_7": "hello, see here\r\nhttp://en.wikipedia.org/wiki/Square_root\r\nSonnhard.", "Solution_8": "Got it.\r\n\r\nSo $ x = k\\cdot \\pi$ it's correct for the equation\r\n\\[{ \\cos2x - \\sin 4x} = (\\sin x - \\cos x)^2\\]\r\nbut not for\r\n\\[ \\sqrt {\\cos2x - \\sin4x} = \\sin x - \\cos x\\]\r\nthat's cool :ninja:", "Solution_9": "hello, ok, one more problem is solved.\r\nSonnhard.", "Solution_10": "I think the most important thing to solve this trigonometric equation is all real numbers x that cause expression under the root is less than 0 must be eliminated from the set of possible solutions.\r\n\r\nI have another solution that is quite clearly and completed.\r\n\r\n[b]Answer.[/b]\r\n\r\n$ \\sqrt {cos2x \\minus{} sin4x} \\equal{} sinx \\minus{} cosx$\r\n\r\n$ \\Leftrightarrow \\left\\{\\begin{array}{l} \\sin x \\minus{} \\cos x \\ge 0 \\\\\r\n\\cos 2x \\minus{} \\sin 4x \\equal{} 1 \\minus{} \\sin 2x \\\\\r\n\\end{array} \\right.$\r\n\r\n$ \\Leftrightarrow \\left\\{\\begin{array}{l}\\sin (x \\minus{} \\frac {\\pi }{4}) \\ge 0 \\\\\r\n(\\cos 2x \\plus{} \\sin 2x)(\\cos 2x \\plus{} \\sin 2x \\minus{} 1) \\equal{} 0 \\\\\r\n\\end{array} \\right.$\r\n\r\n$ \\Leftrightarrow \\left\\{\\begin{array}{l}k2\\pi \\le x \\minus{} \\frac {\\pi }{4} \\le \\pi \\plus{} k2\\pi \\,(k \\in Z)\\,\\,\\,\\,\\,\\,\\,(*) \\\\\r\n\\left[ \\begin{array}{l} \\sin (2x \\plus{} \\frac {\\pi }{4}) \\equal{} 0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1) \\\\\r\n\\sin (2x \\plus{} \\frac {\\pi }{4}) \\equal{} \\frac {{\\sqrt 2 }}{2} \\equal{} \\sin \\frac {\\pi }{4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(2) \\\\\r\n\\end{array} \\right. \\\\\r\n\\end{array} \\right.$\r\n\r\nWe have\r\n\r\n$ (1): \\sin (2x \\plus{} \\frac {\\pi }{4}) \\equal{} 0$\r\n\r\n$ \\Leftrightarrow x \\equal{} \\minus{} \\frac {\\pi }{8} \\plus{} \\frac {{l\\pi }}{2}\\,\\,\\,(l \\in Z)$ \r\n\r\nBecause the condition (*) must be satisfied by x, therefore :\r\n\r\n$ k2\\pi \\leq \\minus{} \\frac {\\pi }{8} \\plus{} \\frac {{l\\pi }}{2} \\minus{} \\frac {\\pi}{4} \\leq \\pi \\plus{} k2\\pi$ $ (l,k \\in Z)$\r\n\r\n$ \\Rightarrow \\frac {3\\pi}{8} \\leq \\frac {l \\pi}{2} \\minus{} k2\\pi \\leq \\frac {11\\pi}{8}$\r\n\r\n$ \\Rightarrow \\frac {3}{8} \\leq \\frac {l}{2} \\minus{} 2k \\leq \\frac {11}{8}$\r\n\r\n$ \\Rightarrow l \\equal{} 2(2k \\plus{} 1) \\equal{} 2a (a \\in Z)$\r\n\r\n$ \\Rightarrow x \\equal{} \\minus{} \\frac {\\pi}{8} \\plus{} a\\pi (a \\in Z)$\r\n\r\nWe have\r\n\r\n$ (2): sin(2x \\plus{} \\frac {\\pi}{4}) \\equal{} sin\\frac {\\pi}{4}$\r\n\r\n$ \\Leftrightarrow \\left[ \\begin{array}{l} x \\equal{} m\\pi \\\\\r\nx \\equal{} \\frac {\\pi }{4} \\plus{} m\\pi \\\\\r\n\\end{array} \\right.(m \\in Z)$\r\n\r\nSimilarly, we obtain $ x \\equal{} \\frac {\\pi}{4} \\plus{} m\\pi$ and $ x \\equal{} (2k \\plus{} 1)\\pi$ where $ k, m\\in Z$\r\n\r\nConclusion, the solutions for the given equation are: $ x \\equal{} \\minus{} \\frac {\\pi}{8} \\plus{} a\\pi, x \\equal{} \\frac {\\pi}{4} \\plus{} m\\pi, x \\equal{} (2k \\plus{} 1)\\pi$ where $ a, m, k \\in Z$." } { "Tag": [ "MATHCOUNTS", "analytic geometry", "geometry", "email", "complex numbers", "NSML" ], "Problem": "Any Illinois AoPSers attending the NSML Meet #1 on September 30 at Naperville North?\r\n\r\nIf so, any plans/meetups/thoughts/whatever?\r\n\r\nI'm going, and I'm a (fre)shmen from IMSA. I will be competing in the freshman division (which I have to), but my school's tryouts haven't been complete yet, so I'm not sure fully yet whether or not I'm also doing sophomore or senior division. (Obviously, if I make the team for senior, I'll do senior) I'm pretty excited in meeting everyone there! :D", "Solution_1": "Sadly my school didn't get invited so i can't go.\r\n\r\nmaybe Aryth is going.", "Solution_2": "Yeah, I'm pretty sure he is. But for the first four meets, your school doesn't have to get \"invited\" to go there, it just has to be initially registered. Did your school not register or something?\r\n\r\nIt's just for the [b]NSML All-Conference Meet[/b](that's at Evanston Township High School) after all four meets are done (probably around March, I think) that you have to be invited to in order to go there (and that's based on your scores from the first four regular meets).", "Solution_3": "Our school got a new principal and our main math person is very old (I'd say about 65) so he can't do much.\r\nAll we do is the amc's and mathcounts and also wyse.", "Solution_4": "Wait, since you do Mathcounts, I'm guessing you're in middle school, right?\r\n\r\nFYI, only high school students can compete on a NSML team.", "Solution_5": "My school is pre school-12th grade.", "Solution_6": "dude we should play like fantasy NSML math team :rotfl:", "Solution_7": "hm I'm gonna be NSMLing this year as an alternate for Payton since my school (Northside) doesn't do it. I have no idea what the topics are or what the questions look like, but this could be fun I guess...", "Solution_8": "These are the topics for the first meet:\r\n\r\nFreshmen: Sets and Venn Diagrams\r\nSophomores: Sets, Logic, and Venn Diagrams\r\nJuniors: Coordinate Geometry\r\nSeniors: Algebra of Complex Numbers\r\n\r\nWhat grade levels are you competing for? I'm doing (fre)shmen (sophomores who skipped ninth grade) and senior teams for IMSA.\r\n\r\nalso, we IMSA students are gonna see you, aryth. :D (yes, we're competing at Naperville North, and the schools traveling there are IMSA, Naperville Central, Glenbrook North, and Stevenson)", "Solution_9": "NCHS calls dibs on the chipotle in downtown naperville.", "Solution_10": "Lol... Do IMSA people know if we still have to eat Sodhexo on meet nights?", "Solution_11": "Uh....idk yet. wait until our coaches send an email for more info.\r\n\r\n@Aryth: Yet fortunately I do not believe IMSA is gonna eat at chipotle. If we did that'd be funny. :rotfl: :rotfl: :rotfl: \r\n\r\nrofl.", "Solution_12": "I can see it now.. an epic battle between the Naperville Central and IMSA math teams for who gets the Chipotle...or maybe it could be resolved with an extreme duck, duck, goose tournament.", "Solution_13": "lol...But at least IMSA can hope for dinner at the cafe before leaving for the bus. :rotfl: :rotfl: :rotfl:", "Solution_14": "Is anyone going to the one at Hinsdale Central or is that a different contest?", "Solution_15": "One set of schools is competing at Hinsdale Central today. Wait, echen, aren't you still in middle school? Because NSML is only for high school students (or actually, only high school students can compete as actual school contestants)", "Solution_16": "just FYI guys, AceofDiamonds is saszs :D", "Solution_17": "LOL RLY?\r\n\r\nwow.\r\n\r\na spammer coming back to the forum= :o \r\n\r\nBy the way, the results are up on the NSML homepage:\r\n\r\n[url]http://www.nsml.org/statistics[/url]", "Solution_18": "No i'm in 9th grade but I didn't go to the first meet.", "Solution_19": "i went... hinsdale south meet... soph team... epic failed...", "Solution_20": "you should set your goal to make all conference this year so that you can get a chance to meet me.\r\n\r\nalso, what was the hardest sophomore question that you couldn't do?", "Solution_21": "Is all conference the one at Evanston cause my coach said that everyone goes to that one.", "Solution_22": "dude I am going there. well not. but good luck to people that are.\r\n\r\noh no seriously aceofdiamonds is saszs?", "Solution_23": "[quote=\"echen\"]Is all conference the one at Evanston cause my coach said that everyone goes to that one.[/quote]\r\n\r\n@echen: Yes. Every year, the final fifth meet, the All-Conference meet, is at Evanston Township High School.\r\n\r\nHowever, participation is by invite-only, you have to be in the top scoring students from the first four regular meets to get invited to the Conference meet. And yes, all 50something schools in the meets participate in that final meet, same place, same time.\r\n\r\n@Truffles: yes. I know. it's scary. :o", "Solution_24": "[quote=\"gauss1181\"]you should set your goal to make all conference this year so that you can get a chance to meet me.\n\nalso, what was the hardest sophomore question that you couldn't do?[/quote]\r\n\r\nhaha they were surprisingly really easy. but for the last question it was multi choice and u had to put all of the ones that applied... but i only put one of them >.< and i don't remember what the other one i missed was about..." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find the positive real root of $ x^4 \\minus{} 2x^2 \\minus{} x \\minus{} 1 \\equal{} 0$.", "Solution_1": "hello, this is the solution\r\n$ \\{x\\to \\frac{1}{2 \\sqrt{\\frac{3}{4-8 \\sqrt[3]{\\frac{2}{3 \\sqrt{2193}-133}}+\r\n\\sqrt[3]{\\frac{1}{2} (3\r\n \\sqrt{2193}-133)}}}}+\\frac{1}{2} \\sqrt{\\frac{8}{3}+\\frac{8}{3} \\sqrt[3]{\\frac{2}{3 \\sqrt{2193}-133}}-}$${ \\frac{1}{3}\r\n \\sqrt[3]{\\frac{1}{2} (3 \\sqrt{2193}-133)}+2 \\sqrt{\\frac{3}{4-8 \\sqrt[3]{\\frac{2}{3\r\n \\sqrt{2193}-133}}+\\sqrt[3]{\\frac{1}{2} (3 \\sqrt{2193}-133)}}}}\\}$\r\nSonnhard.", "Solution_2": "Do you meant \r\n$ x = \\frac {1}{2 \\sqrt {\\frac {3}{4 - 8 \\sqrt [3]{\\frac {2}{3 \\sqrt {2193} - 133}} + \\sqrt [3]{\\frac {1}{2} (3 \\sqrt {2193} - 133)}}}} + \\frac {1}{2} \\sqrt {\\frac {8}{3} + \\frac {8}{3} \\sqrt [3]{\\frac {2}{3 \\sqrt {2193} - 133}} -}$${ \\frac {1}{3} \\sqrt [3]{\\frac {1}{2} (3 \\sqrt {2193} - 133)} + 2 \\sqrt {\\frac {3}{4 - 8 \\sqrt [3]{\\frac {2}{3 \\sqrt {2193} - 133}} + \\sqrt [3]{\\frac {1}{2} (3 \\sqrt {2193} - 133)}}}}\\}$?", "Solution_3": "What method did you use?", "Solution_4": "The POWA of Maple 12!" } { "Tag": [ "algebra", "function", "domain", "advanced fields", "advanced fields theorems" ], "Problem": "The following theorem is from Harvey Cohn's \"A Classical Invitation to Algebraic Numbers and Class Fields\" (p.48).\r\n\r\nTheorem 6.18: Let $A$ be a non-zero ideal in a Dedekind domain $R$, and $C|A$. Then there exists $b$ in $R$ such that $C=(A,(b))=A+(b)$.\r\n\r\nProof: Factor $C=P_{1}^{s_{1}}...P_{t}^{s_{t}}$ and $A=P_{1}^{r_{1}}...P_{t}^{r_{t}}$ so $s_{i} \\leq r_{i}$. Let $C_{0}=P_{1}^{s_{1}+1}...P_{t}^{s_{t}+1}$ and $C_{j}=C_{0}/P_{j}$. Now $C_{j}|C_{0}$ (strictly) so there exists $b_{j} \\in C_{j}-C_{0}$. We assert that $b=b_{1}+...+b_{t}$ will satisfy the requirement, because ... Hence $(b)=C.B$ where no $P_{i}|B$. By (6.6c) $A+(b)=C$.\r\n\r\nI believe that the assertion \"Hence $(b)=C.B$ where no $P_{i}|B$\" deserves a non-trivial proof. Can anybody prove that? I have a proof but would like to see yours first.\r\n\r\nBest regards,\r\nMurat Aygen", "Solution_1": "[quote=\"Murat Aygen\"] Hence $(b)=C.B$ where no $P_{i}|B$\" deserves a non-trivial proof.[/quote]\r\n\r\nCould you please give a definition of $B$ ?\r\n\r\n-- \r\nFabrice", "Solution_2": "[quote=\"fmathurin\"]\nCould you please give a definition of $B$ ?[/quote]\r\n\r\n$A,B,C$ are ideals in the Dedekind domain $R$. They need not be principal ideals. Principal ideals are denoted with $(b)$ where $b$ is a generator. In such a domain each ideal is equal to the product of some prime (thus maximal) ideals. Therefore there exists an ideal $B$ such that $(b)=C.B$ whenever $C|(b)$ (This is the contention of Remark 6.8 on page 44).\r\n\r\nBest regards,\r\nMurat Aygen" } { "Tag": [ "induction", "calculus", "geometry", "3D geometry", "arithmetic sequence" ], "Problem": "The set of consecutive odd numbers 1,3,5,7,9...,N has a sum of 400. How many numbers are in the set?\r\n \r\nI'll be posting the answer in about a week", "Solution_1": "you posted it twice :dry:\r\n\r\nI know the solution, but surely I am very old to answer it.\r\n\r\n[hide=\"REALLY strong hint, do not open if you want to do it yourself\"] 20^2=400 [/hide]", "Solution_2": "[hide] since the sum of an odd number sequence is the number of terms squared, 20^2=400.[/hide]", "Solution_3": "[hide=\"Proof using induction\"]\n[b]Base case:[/b] $ n = 1$\nClearly, $ 1 = 1^2$.\n\n[b]Inductive hypothesis:[/b]\n$ \\sum_{i = 1}^n (2i - 1) = n^2$\n\n[b]Inductive step:[/b]\nLet $ k = n$. Then\n$ \\begin{align*} \\sum_{i = 1}^{k + 1} (2i - 1) & = \\left(\\sum_{i = 1}^k (2i - 1)\\right) + (2(k + 1) - 1) \\\\\n& = \\left(\\sum_{i = 1}^k (2i - 1)\\right) + 2k + 1 \\\\\n& = k^2 + 2k + 1 \\\\\n& = (k + 1)^2$\n\nQ.E.D.[/hide]", "Solution_4": "do it like this with the help of arithmetic progression \r\nS=N/2[2A+(N-1)*D]\r\n400=N/2[2+(N-1)*2]\r\n800=N[2+2N-2]\r\n800=2N^2\r\nN^2=400\r\nN=SQRT(400)\r\nN=20\r\nAND I HOPE 20 IS UR ANSWER!", "Solution_5": "[hide]the sum of the first n odd numbers is n^2, so sqrt of 400=20[/hide]", "Solution_6": "hey gogators my method is correct or not!", "Solution_7": "Um no, N is the 20th odd number, not 20.", "Solution_8": "Um, he meant that the number of terms, which was what the problem was asking for, was 20.\r\n\r\nhe was using the formula for the sum of the first N terms of an arithmatic series, with N as n in the formula S=n[2a+(n-1)d]/2, as you should hae seen.\r\n\r\nrt_08: your method IS correct, although somewhat unneccesary as you could simply take the square root of 400.\r\n\r\nas gogators did, and as you did at the end.", "Solution_9": "It's kinda funny that like 50 ppl are responding to this cuz it's an easy question. :rotfl: \r\n\r\n@YongYi, yes, we all know you can use induction. Stop showing off. :)", "Solution_10": "Yes; I recently thought about this. When there's a hard topic, few people reply because most people don't want to look stupid or anything. Then, when there's an easy topic, people always overcomplicate things to show off and stuff because it's easy and they look smart (to the extent where people use calculus to find the cube root of 2...in the MC forum...xD) :rotfl: :rotfl: .", "Solution_11": "[quote=\"Math Geek\"]It's kinda funny that like 50 ppl are responding to this cuz it's an easy question. :rotfl: \n\n@YongYi, yes, we all know you can use induction. Stop showing off. :)[/quote]\r\n\r\nWhat I find funny is that you didn't contribute anything to this topic. :| \r\n\r\nThe reason I took the steps to [i]prove[/i] it was because it was necessary to show that it is true, rather than just \"spoon-feeding\" them the \"formula\" [size=59](Kinda like what they do in middle schools nowadays)[/size]. :) \r\n\r\nAnd if you think this is showing off, then you have no idea of the effort I took to make this as simple as possible. :(", "Solution_12": "i totally agree with yongyi look friends the place where i study we r taught this method \r\ni am getting used to the new method.hope so ill do this pretty quickly \r\nanyways thanx for correcting me !\r\nhope ill learn a lot from u all!", "Solution_13": "$ \\forall$ odd $ L$\r\n$ S\\equal{}1\\plus{}3\\plus{}5\\plus{}7\\cdots\\equal{}\\sum_{L\\equal{}1}^{n}L \\equal{}n^2$\r\n$ \\implies S\\equal{}n^2\\equal{}400$\r\n$ \\implies n\\equal{}20$\r\n\r\ntherefore the answer is $ \\boxed{\\boxed{\\boxed{\\color{red}20}}}$", "Solution_14": "hey thaNKS again \r\nalso \r\nplz answer my question that is \r\ncan u tell me how should i study for olympiads and wich is the best prescribed book for it\r\nalso i submitted an online form for KVPY nd now thay are asking to submit the pay in slip but wen i opened the link to 'pay in slip'\r\nit read the message that \"the file is damaged\" \r\ncould any of u plz help\r\nplz........\r\nthanx!", "Solution_15": "Using induction isn't hard, and using it to prove something such as the formula for the sum of the first $ n$ odd numbers isn't really necessary. I'll prove it in a more leisurely way:\r\n\r\nFirst, I'll prove what the formula for the sum of the first $ n$ natural numbers is:\r\n$ S \\equal{} 1 \\plus{} 2 \\plus{} \\cdots \\plus{} (n \\minus{} 1) \\plus{} n$\r\n$ S \\equal{} n \\plus{} (n \\minus{} 1) \\plus{} \\cdots \\plus{} 2 \\plus{} 1$\r\nWhen we sum these two expressions for the sum of the first $ n$ natural numbers, we can view it as $ n$ pairs of two numbers that sum to $ n \\plus{} 1$:\r\n$ 2S \\equal{} (n \\plus{} 1) \\plus{} (n \\plus{} 1) \\plus{} \\cdots \\plus{} (n \\plus{} 1) \\plus{} (n \\plus{} 1) \\equal{} n(n \\plus{} 1)$\r\nWe seek this sum divided by $ 2$ (which is equal to $ S$):\r\n$ S \\equal{} \\frac {n(n \\plus{} 1)}{2}$\r\n\r\nNow I'll show what the sum of the first $ n$ even numbers is:\r\n$ S \\equal{} 2 \\plus{} 4 \\plus{} \\cdots \\plus{} 2(n \\minus{} 1) \\plus{} 2n$\r\nI divide by $ 2$ to get a familiar expression and then multiply by $ 2$ again:\r\n$ \\frac {S}{2} \\equal{} 1 \\plus{} 2 \\plus{} \\cdots \\plus{} (n \\minus{} 1) \\plus{} n \\equal{} \\frac {n(n \\plus{} 1)}{2}$\r\n$ S \\equal{} n(n \\plus{} 1)$\r\n\r\nAt long last, the sum of the first $ n$ odd numbers can be expressed in a compact form (a formula, in this case):\r\n$ S \\equal{} 1 \\plus{} 3 \\plus{} \\cdots \\plus{} (2n \\minus{} 3) \\plus{} (2n \\minus{} 1)$\r\nI add $ 1$ to each of the $ n$ terms of this sum (which is basically adding $ n$ to the sum) to get a familiar expression and then subtract $ n$ to get back to the original sum:\r\n$ S \\plus{} n \\equal{} 2 \\plus{} 4 \\plus{} \\cdots \\plus{} 2(n \\minus{} 1) \\plus{} 2n \\equal{} n(n \\plus{} 1) \\equal{} n^2 \\plus{} n$\r\n$ S \\equal{} n^2$\r\n\r\nSo, to solve the problem, we know that $ n^2 \\equal{} 400$, and obviously the answer to the problem (the number of members of the set of the first $ n$ odd numbers) is $ n$. Certainly, $ n$ can't be negative, so we can take the positive square root of the equation we made to get $ n$: $ n \\equal{} 20$. The answer is thus $ \\boxed{20}$.\r\n\r\nNotice how, in proving the formulas for the various sums of the first whatever numbers, we built each following formula by the one that preceded it. All it took was a little creativity; you don't need to hustle through induction (and it'll probably help your problem solving skills more in the long run to [i]not[/i] use induction on finding formulas for sums such as these).", "Solution_16": "[hide=\"Another way to come up with this formula\"]\nLet $ S \\equal{} 1 \\plus{} 3 \\plus{} 5 \\plus{} \\dots \\plus{} (2n \\minus{} 1)$. $ S$ can also be rewritten as follows:\n$ S \\equal{} (2n \\minus{} 1) \\plus{} (2n \\minus{} 3) \\plus{} (2n \\minus{} 5) \\plus{} \\dots \\plus{} 1$\n\nNow we pair up (similar to what MathAndKnowledge said). Notice that each pair adds up to $ 2n$, and there are $ \\frac n2$ pairs, so multiplying, we get $ 2n \\cdot \\frac n2 \\equal{} \\boxed {n^2}$.[/hide]", "Solution_17": "[quote=\"Yongyi781\"][quote=\"Math Geek\"]It's kinda funny that like 50 ppl are responding to this cuz it's an easy question. :rotfl: \n\n@YongYi, yes, we all know you can use induction. Stop showing off. :)[/quote]\n\nWhat I find funny is that you didn't contribute anything to this topic. :| \n\nThe reason I took the steps to [i]prove[/i] it was because it was necessary to show that it is true, rather than just \"spoon-feeding\" them the \"formula\" [size=59](Kinda like what they do in middle schools nowadays)[/size]. :) \n\nAnd if you think this is showing off, then you have no idea of the effort I took to make this as simple as possible. :([/quote]\r\n\r\nWell, I wasn't making fun of you for showing off. I was just implying that you're too smart for your own good. :P (Though proving it with induction isn't difficult.)", "Solution_18": "well i dont think yongyi was showing off!\r\n|", "Solution_19": "$ S \\equal{} 1 \\plus{} 3 \\plus{} 5 \\plus{} 7\\cdots \\equal{} \\sum_{L \\equal{} 1}^{n}2L\\minus{}1$\r\n\r\n$ \\implies S\\equal{}\\sum_{L \\equal{} 1}^{n}2L\\minus{}1\\equal{}\\sum_{L \\equal{} 1}^{n}2L\\minus{}\\sum_{L \\equal{} 1}^{n}1$\r\n\r\n$ \\implies S\\equal{}2\\sum_{L \\equal{} 1}^{n}L\\minus{}\\sum_{L \\equal{} 1}^{n}1\\equal{}2\\times\\frac{1}{2}n(n\\plus{}1)\\minus{}n$\r\n\r\n$ \\implies S\\equal{}n^2\\plus{}n\\minus{}n$\r\n$ \\implies S\\equal{} n^2$\r\n$ \\implies S \\equal{} n^2 \\equal{} 400$\r\n$ \\implies n \\equal{} 20$\r\n\r\ntherefore the answer is $ \\boxed{\\boxed{\\boxed{\\color{red}20}}}$", "Solution_20": "all of u got it right...but i dont even know what induction is...so...", "Solution_21": "Induction is basically:\r\n[list=1]\n[*]Showing it works for a starting value, typically $ 1$.\n[*]Showing that [b]if[/b] it works for general $ n$, it works for $ n \\plus{} 1$, by manipulation, etc.[/list]\r\n\r\nThe second step, combined with the first, essentially says that if it works for 1, it works for 2. Since it works for 2, it works for 3. Etc, etc.", "Solution_22": "well thats basically wat mathematic\r\nal induction is?", "Solution_23": "[quote=\"Yongyi781\"]\n\n\nThe reason I took the steps to [i]prove[/i] it was because it was necessary to show that it is true, rather than just \"spoon-feeding\" them the \"formula\" [size=59]([b]Kinda like what they do in middle schools nowadays[/b])[/size]. :) \n\n[/quote]\r\n\r\nYou've been reading the Middle School Teacher Forums! :P", "Solution_24": "to yongyi: thanks! :D :D :D", "Solution_25": "hey i still didnt whether i got my ans in the right way or noT!", "Solution_26": "[quote=\"gogators\"]all of u got it right[/quote]\r\n\r\nu got it right" } { "Tag": [], "Problem": "\u4f60\u597d\r\n\r\n\u4f60\u77e5\u4e0d\u77e5\u9053\u6211\u53ef\u4ee5\u5728\u54ea\u88e1\u627e\u5230\u8da3\u5473\u9ad8\u7b49\u4ee3\u6578 ,\u4e09\u89d2\u554f\u984c?", "Solution_1": "\u662f\u5426\u6709\u9053\u7406\uff1f" } { "Tag": [], "Problem": "Find integers $ m$, $ n$ such that $ m^3\\plus{}m^2\\plus{}m\\plus{}1 \\equal{} n^3$.", "Solution_1": "For $ m>0$, $ m^30$.Prove that :\r\n$ 3(\\sqrt{x(x\\plus{}y)(x\\plus{}z)}\\plus{}\\sqrt{y(y\\plus{}z)(y\\plus{}x)}\\plus{}\\sqrt{z(z\\plus{}x)(z\\plus{}y)})^2\\leq 4(x\\plus{}y\\plus{}z)^3$", "Solution_1": "By Cauchy-Schwarz ineq , we have :\r\n$ LHS \\leq 3(x \\plus{} y \\plus{} z)( \\sum x^2 \\plus{} yz \\plus{} zx \\plus{} xy)$\r\nThen we prove that :\r\n$ 4(x \\plus{} y \\plus{} z)^2 \\geq 3[ (x \\plus{} y \\plus{} z)^2 \\plus{} xy \\plus{} yz \\plus{} zx]$\r\n$ \\leftrightarrow \\sum (x \\minus{} y)^2 \\geq 0$\r\nDone :D" } { "Tag": [ "probability", "calculus", "integration", "search", "Pascal\\u0027s Triangle" ], "Problem": "What are proofs?", "Solution_1": "First of all, what is [b]Intermediate[/b] Counting and Probability doing here?\r\n\r\nWhat are proofs? :o :? Well, it's when you prove something. For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :?", "Solution_2": "[quote=\"chess64\"]What are proofs? :o :? Well, it's when you prove something. [/quote]\n\n :rotfl: duh\n\n[quote=\"chess64\"]For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :?[/quote]\r\n\r\nHow would you prove that?", "Solution_3": "[quote=\"chess64\"]For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :? [/quote]\r\n\r\nhow would you do that? what are integral squares", "Solution_4": "I think integral squares are squares of integers", "Solution_5": "[quote=\"236factorial\"][quote=\"chess64\"]For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :?[/quote]\n\nHow would you prove that?[/quote]\r\n\r\n:mad: this will help on my competition :mad:\r\n\r\n[hide]\nAll perfect squares are $\\equiv 0$ or $1 \\mod {4}$\n\n$2002 \\equiv 2 \\equiv 1+1 \\mod {4}$\nTherefore it is possible. Or you could just find the integer squares.\n[/hide]", "Solution_6": "It will help on your competition :D Thanks :rotfl: \r\n\r\nJust kidding, I'm still a little confused :? NOT :D", "Solution_7": "Aha, but [hide]can you prove that all perfect squares are $\\equiv 0,1 \\mod {4}$?[/hide] :idea: :D", "Solution_8": "[quote=\"chess64\"]Aha, but [hide]can you prove that all perfect squares are $\\equiv 0,1 \\mod {4}$?[/hide] :idea: :D[/quote]\r\n\r\nyeah :roll: \r\nmaybe.", "Solution_9": "Maybe it comes from the following:\r\n\r\nFor all integers $n$, we have $n\\equiv 0,1 \\bmod{2}$. Therefore, $n^2\\equiv 0^2,1^2 \\bmod{2^2}$.\r\n\r\nDoes that work?", "Solution_10": "Questions about classes should be asked in the Classes Information section. And please do a Google define: search before posting questions such as \"what is pascal's triangle\" and 'what is a proof'. Type this into google: define:pascals triangle or define:math proof.\r\n\r\nTopic locked" } { "Tag": [ "induction", "strong induction" ], "Problem": "Here is a problem an interesting problem on Fibonacci Sequences I have tried but have been unable to solve it, any help given would be appreciated since I think its a very interesting problem.\r\n\r\nThe Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...) is defined where $ F_1 \\equal{} 0, F_2 \\equal{} 1$ and $ F_n \\equal{} F_n \\minus{} 1 \\plus{} F_n \\minus{} 2$ for n\u22652.\r\n\r\nShow that $ F_n(F_n \\minus{} 1 \\plus{} F_n \\minus{} 2)$ is a Fibonacci number for all values of n.\r\n\r\nShow that $ (F_n \\minus{} 1)^2 \\plus{} (F_n \\minus{} 2)^2$ is a Fibonacci number for all values of n.", "Solution_1": "[quote=\"The Anomaly\"]Here is a problem I could remember for my Cambridge Interview Exam this past winter, I have tried but have been unable to solve it, any help given would be appreciated since I think its a very interesting problem.\n\nThe Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...) is defined where $ F_1 \\equal{} 0, F_2 \\equal{} 1$ and $ F_n \\equal{} F_{n \\minus{} 1} \\plus{} F_{n \\minus{} 2}$ for $ n\\ge 2$.\n\nShow that $ F_n(F_{n \\minus{} 1} \\plus{} F_{n \\minus{} 2})$ is a Fibonacci number for all values of n.\n\nShow that $ (F_{n \\minus{} 1})^2 \\plus{} (F_{n \\minus{} 2})^2$ is a Fibonacci number for all values of n.[/quote]\n\nI'll assume you meant it this way :) \n\nFirst one doesn't make sense, since $ F_n \\equal{} F_{n \\minus{} 1} \\plus{} F_{n \\minus{} 2}$ so it's basically $ F_n^2$, and that's never a Fibonacci number (except for $ F_n \\equal{} 1$).\nThe second one is correct, and moreover it's $ (F_{n \\minus{} 1})^2 \\plus{} (F_{n \\minus{} 2})^2 \\equal{} F_{2n \\minus{} 3}$ in this setting of indices. Now you know the exact formula, you can (probably) try proving it using strong induction, which would eventually get you to products of Fibonacci numbers. But we can do better than that:\n\n[hide=\"Solution\"]\n[quote=\"matt276eagles\"]\nLemma: $ f_{i} \\equal{} f_{j}f_{i \\minus{} j \\minus{} 1} \\plus{} f_{j \\plus{} 1}f_{i \\minus{} j}$.\n\nProof: We induct on j. For j=1, the lemma becomes $ f_{i} \\equal{} f_{i \\minus{} 2} \\plus{} f_{i \\minus{} 1}$, which is just the definition of Fibonacci numbers. Now assume the lemma is true for j=k, so that $ f_{i} \\equal{} f_{k}f_{i \\minus{} k \\minus{} 1} \\plus{} f_{k \\plus{} 1}f_{i \\minus{} k}$. Then\n\n$ f_{i} \\equal{} f_{k \\plus{} 1}(f_{i \\minus{} k \\minus{} 2} \\plus{} f_{i \\minus{} k \\minus{} 1}) \\plus{} f_{k}f_{i \\minus{} k \\minus{} 1}$\n$ f_{i} \\equal{} f_{k \\plus{} 1}f_{i \\minus{} k \\minus{} 2} \\plus{} (f_{k} \\plus{} f_{k \\plus{} 1})f_{i \\minus{} k \\minus{} 1}$\n$ f_{i} \\equal{} f_{k \\plus{} 1}f_{i \\minus{} k \\minus{} 2} \\plus{} f_{k \\plus{} 2}f_{i \\minus{} k \\minus{} 1}$\n\nThis is our lemma for $ j \\equal{} k \\plus{} 1$, so the induction is complete, and the lemma is proved for all i and j.\n\nNow, let $ i \\equal{} 2n \\minus{} 3$ and $ j \\equal{} n\\minus{}2$ in our lemma. We get $ f_{2n \\minus{}3} \\equal{} f_{n\\minus{}2}^{2} \\plus{} f_{n\\minus{}1}^{2}$[/quote]\n\n[/hide]", "Solution_2": "Sorry my bad, I made an error in writing the first one, I meant to say this:\r\n\r\nShow that $ F_n(F_n \\minus{} 1 \\plus{} F_n \\minus{} 1)$ is a Fibonacci number for all values of n." } { "Tag": [ "vector", "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let $ (X,d)$ be a metric space. We said $ (X,d)$ has the homogeniety property if there exists $ N>0$ such that for any open ball $ B(x,r)\\equal{}\\{y: d(y,x) 0$ is some constant. Show that $ X$ has the homogeniety property defined above.", "Solution_3": "[quote=\"pluricomplex\"][quote=\"Soarer\"]\n3. Any normed vector space should do.[/quote]\nCould you explain this more detains?\n\nNow is the fouth question:\n\n4) Let $ \\mu$ is doubling nontrivial Borel measure on $ X$, ie... for any ball $ B(x,r)$ we must have $ \\mu(B(x,r))\\leq c\\mu(B(x,r/2))$, where $ c > 0$ is some constant. Show that $ X$ has the homogeniety property defined above.[/quote]\r\n\r\nI'm sorry, I meant to say proper normed vector space (those with $ B(x,r)$ compact)\r\n\r\nMay I know what is doubling measure?", "Solution_4": "The definition of doubling measure I gave it above :roll:", "Solution_5": "Let $ x \\in X$ and $ r > 0$. If $ y \\in B(x, r)$ we note that\r\n$ \\mu(B(x,r)) \\le \\mu(B(y, r \\plus{} d(x,y)) \\le \\mu(B(y, 2r)) \\le c^3\\mu(B(y,r/4))$.\r\n\r\nNow let $ x_1, \\ldots, x_N \\in B(x,r)$ so that $ d(x_i, x_j) \\ge r/2$, $ i \\ne j$. We note that the balls\r\n$ B(x_i, r/4)$ are disjoint and $ B(x_i, r/4) \\subset B(x, 5r/4) \\subset B(x, 2r)$. Now we get\r\n$ N\\mu(B(x,r)) \\le c^3\\sum_{i\\equal{}1}^N \\mu(B(x_i, r/4)) \\le c^3\\mu(B(x,2r)) \\le c^4\\mu(B(x,r))$ whence\r\n$ N \\le c^4$.", "Solution_6": "Okey it's time for the fivth question\r\n\r\n5) Give some examples for which $ \\mu$ is not a doubling measure." } { "Tag": [], "Problem": "Eleven teachers run a conference. Every hour, one or more teachers give one-hour presentations, while all of the other teachers observe the presentations. Find the least amount of time during which it is possible for each teacher to observe all other presentations at least once.", "Solution_1": "[quote=\"Phelpedo\"]Eleven teachers run a conference. Every hour, one or more teachers give one-hour presentations, while all of the other teachers observe the presentations. Find the least amount of time during which it is possible for each teacher to observe all other presentations at least once.[/quote]\r\nWe can solve recursively. Divide the eleven teachers into one group of six and one of five. In the first hour the group of five give presentations to the group of six. In the second hour that is reversed. Now each teacher needs to see the presentations of his own group only. Divide the group of five into one group of three and one of two. Divide the group of six into one group of three and another of three. Now we spend two more hours with one group presenting to the other group, and then reversing. Now we have each teacher needing to listen to two other presentations. This can be done in three hours: just have each teacher stand up and present in turn. This takes a total of $7$ hours. I haven't come up with a better method.", "Solution_2": "Hmm, I think this question needs a little explanation. WatsonLadd's solution assumes that the observing teachers can watch more than one presentation at a time, but I interpreted it meaning each hour, each teacher can either present or observe 1 presentation, which would make the answer much larger...", "Solution_3": "That was also how I interpreted it, although I'm not sure \"much larger\" is really accurate. After all, the simplest-looking solution is just to have one person present each hour and have all 10 others watch, for a total of 11 hours. WatsonLad's interpretation seems to be for a more natural question, though.", "Solution_4": "The problem is obviously trivial with that interpretation, so we are probably meant to assume that the non-presenters observe all presentations simultaneously." } { "Tag": [ "calculus", "counting", "derangement", "search", "calculus computations" ], "Problem": "Find the general fomular of $ x_n$:\r\n$ x_n \\equal{} (n \\minus{} 1)(x_{n \\minus{} 1} \\plus{} x_{n \\minus{} 2}), n \\geq 4 \\\\\r\nx_2 \\equal{} 1, x_3 \\equal{} 2$", "Solution_1": "Has got what to do with calculus, exactly? It's counting derangements; the numbers themselves are subfactorial or rencontres numbers; the sequence is [url=http://www.research.att.com/~njas/sequences/A000166]A000166[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url]. A search for any of the key words in this post will find you plenty of information." } { "Tag": [], "Problem": "This one is famous, but neat:\r\n\r\nFind the unique 9 digit number containing each of the digits 1 through 9 such that:\r\n\r\n- the whole number is divisible by 9\r\n- when the last digit (on the right) is removed, the remaining number is divisible by 8\r\n- when the last 2 digits are removed, the remaining number is divisible by 7\r\n- when the last 3 digits are removed, the remaining number is divisible by 6, and so on....", "Solution_1": "number is [size=75][color=darkred](hidden by moderator)[/color][/size]\r\n[hide]\n 381654729\n\ni have find it with lots of efforts.[/hide]", "Solution_2": ":o Hide your answer so others can try it!!!", "Solution_3": "please tell me how to hide my answers.", "Solution_4": "Before you start typing whatever it is you want to hide, click the button that says \"Hide\" just above the text window at the right end of the list of buttons. When you are done with whatever you want to hide click the hide button again.", "Solution_5": "thanks for telling me." } { "Tag": [ "inequalities" ], "Problem": "I'm very bad at inequalities. Is there some elegant way to prove that $5x^{7}+2>3x$ as $0\\leq x\\leq 1$?", "Solution_1": "[hide=\"Hint\"]Apply the AM-GM inequality to\n\\[5 x^{7}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}\\, . \\]\n[/hide]", "Solution_2": "So you mean $\\frac{5x^{7}+1}{7}\\geq \\sqrt[7]{5x^{7}\\left (\\frac{1}{6}\\right )^{6}}$? This implies $5x^{7}+2\\geq 1+7x\\sqrt[7]{\\frac{5}{6^{6}}}$. But $7\\sqrt[7]{\\frac{5}{6^{6}}}\\approx 1.897<3$.", "Solution_3": "I think Ravi B means \r\n\r\n[hide]Apply the AM-GM inequality to\n\\[5 x^{7}+\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}\\, . \\][/hide]", "Solution_4": "Oops, I can't add. SplashD is correct about what I meant.", "Solution_5": "Let $f(x)=5x^{7}-3x+2$ then $f'(x)=35x^{6}-3$. Solving $f'(t)=0$ we get $t=\\sqrt[6]{\\frac{3}{35}}$ Now suffice to prove, that $f(t)>0$\r\n\\[f(t)=\\frac{3}{7}t-3t+2>0 \\iff t < \\frac{7}{9}\\iff \\sqrt[6]{\\frac{3}{35}}< \\frac{7}{9}\\]\r\nBut we have: $t<\\frac{5}{7}<\\frac{7}{9}$. The first inequality is equivalent to: $3\\cdot 7^{5}<5^{7}$" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "can we find the solution of the following equation in positive integers\r\na^2+b^2-ab=2005.\r\nbest regards\r\n(hope this is not too easy.returning to mathematics after quite some time)", "Solution_1": "$a^{2}+b^{2}-ab=.25(a+b)^{2}+.75(a-b)^{2}$\r\nBut $5|x^{2}+3y^{2}$ has no solutions, as:\r\n$x^{2}+3y^{2}=0(mod 5)$\r\niff $(xy^{-1})^{2}=-3(mod 5)$ but this is impossible.\r\n\r\nSo, no solutions ???", "Solution_2": "I suppose I ignored the case where 5|(a+b) and 5|(a-b), but that should be okay because then $25|a^{2}+b^{2}-ab=2005$ a contradiction.", "Solution_3": "Generalisation:\r\nEvery integer $n$ can be written as $n=x^2-xy+y^2$ iff every prime divisor $p,p\\equiv 2 \\mod 3$ occurs an even number of times in the prime factorisation of $n$.\r\nMore generally, there are $ 6 \\cdot \\left( \\sum_{\\substack{ d \\in \\mathbb{N} \\\\ d \\mid n \\\\ d \\equiv 1 \\mod 3}} 1 - \\sum_{\\substack{ d \\in \\mathbb{N} \\\\ d \\mid n \\\\ d \\equiv 2 \\mod 3}} 1 \\right)$ different ways to write $n$ as $n=x^2-xy+y^2$ (the pairs $(x,y),(y,x),(-x,-y)$ etc. are considered as different)" } { "Tag": [ "algebra", "system of equations", "algebra unsolved" ], "Problem": "solve th system of equations :\r\n $\\sqrt{3x}(1+\\frac{1}{x+y})=2$\r\n $\\sqrt{7y}(1-\\frac{1}{x+y})=4\\sqrt2$", "Solution_1": "This is a problem in Vietnam National Olympiad. It is a problem 1 in year 1996", "Solution_2": "\u00fdes ,all right .can you p\u00f3st solution ? :P", "Solution_3": "[quote=\"dieuhuynh\"]solve th system of equations :\n $\\sqrt{3x}(1+\\frac{1}{x+y})=2$\n $\\sqrt{7y}(1-\\frac{1}{x+y})=4\\sqrt2$[/quote]\r\nIt is equavalent to:\r\n$\\frac{2}{\\sqrt{3x}} =1+\\frac{1}{x+y},\\frac{4\\sqrt 2 }{\\sqrt{7y}}=1-\\frac{1}{x+y}\\to$\r\n$\\frac{1}{\\sqrt{3x}}+\\frac{2\\sqrt 2 }{\\sqrt{7y}}=1,\\frac{1}{\\sqrt{3x}}-\\frac{2\\sqrt 2 }{\\sqrt{7y}} =\\frac{1}{x+y}\\to$\r\n$\\frac{1}{3x}-\\frac{8}{x+y}=\\frac{1}{x+y}\\to (7y-24x)(x+y)=21xy\\to y=\\frac{19+\\sqrt{385}}{7}x,$\r\n$x=(\\frac{1}{\\sqrt 3 }+\\frac{2\\sqrt 2 }{\\sqrt{19+\\sqrt{385}}})^2.$" } { "Tag": [ "college contests" ], "Problem": "Vietnamese Undergraduate Mathematical Competition 2004 was held in Quy Nhon last week. My students won 3 gold, 3 silver and 5 bronze. I will post all problems (two sections - Analysis and Algebra) time by time. Enjoy the promlems!\r\n\r\nNamdung", "Solution_1": "Congratulations :) but which of your students are on ML? :) so that we can congrat them too :P" } { "Tag": [ "limit", "trigonometry", "calculus", "calculus computations" ], "Problem": "hi!!\r\nevaluate this limit without using Taylor series or hopital's rule. :\r\n\r\n$ \\lim_{x\\to0}\\frac{x\\minus{}\\sin{x}}{x\\minus{}\\tan{x}}$\r\n\r\nthanx.", "Solution_1": "$ \\tan x\\equal{}\\frac{\\sin x}{\\cos x}$", "Solution_2": "[quote=\"kunny\"]$ \\tan x \\equal{} \\frac {\\sin x}{\\cos x}$[/quote]\r\n\r\nhi! , can you explain more ? :blush: \r\n\r\nthanx again :)", "Solution_3": "divide numerator and denominator by x^3\r\nthen let x = 3y\r\nfirst solve lim (x-sinx)/x^3 . let x=3y\r\nlim [3y-sin(3y)]/3y ,sin3x= 3sinx-4sin^3(x)\r\nalso:lim(x-tanx)/x^3=lim[3y-tan3y]/3y \r\n=lim(3y- [3tany-tan^3y](1-3tan^2y))/27y^3", "Solution_4": "Why not use Taloy series or H'ospital rules ? :?:" } { "Tag": [ "ratio", "geometry", "probability", "3D geometry", "number theory", "prime numbers" ], "Problem": "1. How many positive three-digit integers less than 400 satisfy each of the following characteristics simultaneously?(#7 97 team)\r\nA. Each of the three digits is prime\r\nB. The sum of the three digits is prime\r\nC. The number is prime.\r\n[hide=\"Answer\"]5 numbers[/hide]\n2. Points $ A(0,0), B(9,6)$ and $ C(6,12)$ are vertices of triangle ABC. Point D is on segment $ AB$ such that $ 2(AD)\\equal{}DB$, point $ E$ is on segment $ BC$ such that $ 2(BE)\\equal{}EC$ and point $ F$ is on segment $ CA$ such that $ 2(CF)\\equal{}FA$. What is the ratio of the area of triangle $ DEF$ to the area of triangle $ ABC?$ Express your answer as a common fraction.(#7 03 team)\n[hide=\"Answer\"]$ \\frac13$[/hide]\n3. An integer between 0 and 5000 inclusive is selected at random and is found to be a perfect cube. What is the probability that it is also a perfect fourth power? Express your answer as a common fraction. (#5 97 target)\n[hide=\"Answer\"]$ \\frac16$[/hide]", "Solution_1": "3) Well there are only 18 perfect cubes including 0 and 5000, and of those, the only ones are $ 0$, $ 1$, and $ 4096$.", "Solution_2": "#1:\r\n\r\n[hide]The only prime numbers are 2,3,5, and 7 that are one-digit numbers.\nThe number can't end with 5 either because numbers that end with 5 are multiples of 5.\nThe numbers are :\n223, 227, 337,353,373\nEdit: changed one number to 373\n[/hide]", "Solution_3": "[quote=\"ac-king\"]#1:\n\n[hide]The only prime numbers are 2,3,5, and 7 that are one-digit numbers.\nThe number can't end with 5 either because numbers that end with 5 are multiples of 5.\nThe numbers are :\n223, 227, 337,377, and 353[/hide][/quote]\r\n\r\nHow do you know did you that those are the only ones that are prime, besides the fact that the answer is 5? I can't recognize primes quickly.", "Solution_4": "All the numbers below 400 that follow all the restrictions (except for the one with the number being a prime number) are:\r\n223, 227, 335, 337,353, 373, 377.\r\n335 ends with 5 so it can't be one.\r\n\r\nFor the rest of them, you have to check them out with the divisibility rules.\r\n\r\nThen you have to check the 7,13,17, and 19. \r\nIt isn't necessary to check the ones any higher because for any pair of numbers that multiply up to these numbers, the lower number can't be higher than 19 because all of these numbers are lower than 400.\r\nYou can find that 377 is 29*13 and the rest are prime.", "Solution_5": "[quote=\"JRav\"]3) Well there are only 18 perfect cubes including 0 and 5000, and of those, the only ones are $ 0$, $ 1$, and $ 4096$.[/quote]\r\nhere's a potentially more thorough solution-you can't just check every cube up to 5000 for 4th powered-ness :rotfl: i think this is a bit more practical...i did this problem timed yesterday\r\nif a number is a cube and a 4th power simultaneously, then it is a 12th power(make sure you understand this)\r\n\r\nwell what is probability?\r\nprobability is defined as the number of successes divided by the total number of outcomes.\r\nhow many outcomes are there?\r\nwell, how many cubes couuld we have picked less than 5000? 18(possible outcomes)(please verify)\r\nhow many 12th powers are there(the successes) only 3.\r\nso the answer is successes divided by outcomes=3/18=1/6", "Solution_6": "#2 (03 nat Team 7)\r\nSee [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=816546&search_id=1702538419#816546[/url] for the solution." } { "Tag": [ "MATHCOUNTS", "blogs", "parameterization", "algebra", "system of equations" ], "Problem": "The sides of an equiangular hexagon measure $ 10, 6, 12, 14, x,$ and $ y$\r\nunits, in the order given. Find $ x$ and $ y$.", "Solution_1": "This was actually very similar to a National MATHCOUNTS Sprint round problem. (Actually, it was #28 in the 2008 competition). :)", "Solution_2": "[hide=\"The General Result\"]Suppose the side lengths are a, b, c, d, e, and f.\n\nOrient the hexagon so that sides of length a and d are both horizontal. Then, the \"height\" of the hexagon in terms of b and c is [color=blue][b]sin60*b+sin60*c=sin60*(b+c)[/b][/color]. The height of the hexagon in terms of e and f is, similarly, [b][color=blue]sin60*(e+f)[/color][/b].\n\nSince under the same orientation, the height is the same, we can equate to get [b][color=blue]sin60*(b+c)=sin60*(e+f)[/color][/b], or [b][color=blue]b+c=e+f[/color][/b].\n\nSimilarly, considering the other two orientations, we get the system of equations:\n[b]\n[color=blue]a+b=d+e\nb+c=e+f\nc+d=f+a[/color]\n[/b][/hide]\n\n[hide=\"Particular solution\"]We have the system of equations:\n10+6=14+x\n6+12=x+y\n12+14=y+10\n\nThis system is overdetermined -- we can solve it using the first and third equations only, and use the second to check our answer. We find that [color=green][b](x,y)=(2,16)[/b][/color]\n\n[/hide]", "Solution_3": "Just extend the sides to form an equilateral triangle.", "Solution_4": "[quote=\"The Zuton Force\"][hide=\"The General Result\"]Suppose the side lengths are a, b, c, d, e, and f.\n\nOrient the hexagon so that sides of length a and d are both horizontal. Then, the \"height\" of the hexagon in terms of b and c is [color=blue][b]sin60*b+sin60*c=sin60*(b+c)[/b][/color]. The height of the hexagon in terms of e and f is, similarly, [b][color=blue]sin60*(e+f)[/color][/b].\n\nSince under the same orientation, the height is the same, we can equate to get [b][color=blue]sin60*(b+c)=sin60*(e+f)[/color][/b], or [b][color=blue]b+c=e+f[/color][/b].\n\nSimilarly, considering the other two orientations, we get the system of equations:\n[b]\n[color=blue]a+b=d+e\nb+c=e+f\nc+d=f+a[/color]\n[/b][/hide]\n[/quote]\n[hide=\"More concisely...\"]\n$ |a\\minus{}d|\\equal{}|b\\minus{}e|\\equal{}|c\\minus{}f|$.[/hide]", "Solution_5": "See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=152452]my old blog post[/url]. Verify that the parameterization I gave is equivalent to The Zuton Force's conditions." } { "Tag": [], "Problem": "I just downloaded some movies via torrent, in the dvdrip.xvid format. But the only thing I see, is a bunch of .rar files. \r\nHow can I get the movies to play?\r\n\r\nThanks in advance :)", "Solution_1": "Ok I figured it out.. :D" } { "Tag": [ "search", "trigonometry", "geometry unsolved", "geometry" ], "Problem": "Let $ (O)$ be a circle and a point $ S$ in a plane.Draw $ SA;SB$ is tangent of $ (O)(A;B \\in (O))$ .$ AB$ cuts $ SO$ at $ H$\r\nLet $ d$ be midperpendicular of $ SH$.$ M$ is a point on $ d$.Draw $ MN;MP$ is tagent of $ (O)$$ (N;P \\in (O))$\r\n$ NP$ cuts $ d$ at $ E$\r\nProve that $ \\angle MSE \\equal{} 90^o$", "Solution_1": "http://www.mathlinks.ro/viewtopic.php?search_id=1394237155&t=54479", "Solution_2": "Midpoint of $ [SH]$ is $ K$. $ PMKO$ , $ PKNO$ , $ MKNO$ are cyclic quadrilaterals. \r\n\r\n$ |OH|.|OS|\\equal{}|ON|^2 \\Longrightarrow$ $ m(ONH)\\equal{}m(OSN)\\equal{} \\gamma$ , $ m(KPN)\\equal{}m(KON)\\equal{}\\beta$ , $ m(MPK)\\equal{}m(MNK)\\equal{}m(MOK)\\equal{}m(NEK)\\equal{}\\alpha$.\r\n\r\nWe will prove that $ |MN|\\equal{}|MH|$. Let $ |NO|\\equal{}x$ so $ |MN|\\equal{}xtan(\\alpha\\plus{}\\beta)$ , $ |NH|\\equal{}\\frac{xsin\\beta}{sin(\\beta\\plus{}\\gamma)}$. If we prove that \r\n\r\n$ \\frac{sin\\beta}{sin(\\beta\\plus{}\\gamma)tan(\\alpha\\plus{}\\beta)}\\equal{}2sin\\gamma$ solution is done. \r\n\r\nWe will prove that $ tan(\\alpha\\plus{}\\beta)\\equal{}\\frac{sin\\beta}{2sin\\gamma.sin(\\beta\\plus{}\\gamma)}$\r\n\r\nLemma: Triangle $ ABC$ is given. $ m(ABC)\\equal{}\\alpha$ , $ m(ACB)\\equal{}\\beta$ , $ D$ is the midpoint of $ [BC]$ and $ m(ADC)\\equal{}\\theta$ \r\n\r\n$ cot\\theta\\equal{}\\frac{sin(\\beta\\minus{}\\alpha)}{2sin\\beta.sin\\alpha}$ If we prove this lemma solution is done because in triangle $ SNO$, $ |SK|\\equal{}|KH|$ and \r\n\r\n$ m(NKO)\\equal{}90\\minus{}\\alpha\\minus{}\\beta$ so $ cot(90\\minus{}\\alpha\\minus{}\\beta)\\equal{}\\frac{sin\\beta}{2sin\\gamma.\\sin(\\beta\\plus{}\\gamma)}$.\r\n\r\nLet prove this lemma:\r\n\r\n$ |BD|\\equal{}|DC|$ hence $ \\frac{sin\\alpha}{sin(\\theta\\minus{}\\alpha)}\\equal{}\\frac{sin\\beta}{sin(\\beta\\plus{}\\theta)}$ \r\n\r\n$ \\frac{sin(\\theta \\minus{}\\alpha)}{sin(\\beta \\plus{}\\theta)}\\equal{}\\frac{sin\\alpha}{sin\\beta}$\r\n\r\n$ \\frac{sin\\theta.cos\\alpha\\minus{}cos\\theta.sin\\alpha}{sin\\beta.cos\\theta\\plus{}cos\\beta.sin\\theta}\\equal{}\\frac{sin\\alpha}{sin\\beta}$\r\n\r\n$ cot\\theta\\equal{}\\frac{sin\\beta.cos\\alpha\\minus{}cos\\beta.sin\\alpha}{2sin\\alpha.sin\\beta}\\equal{}\\frac{sin(\\beta\\minus{}\\alpha)}{2sin\\alpha.sin\\beta}$ and solution is done. :wink:" } { "Tag": [], "Problem": "Let $ n_1,n_2,n_3...$ be the odd numbers none of which has a prime divisors greater than 5, prove that\r\n\r\n$ \\frac{1}{n_1}\\plus{}\\frac{1}{n_2}\\plus{}\\frac{1}{n_3}\\plus{}... <\\frac{15}{8}$", "Solution_1": "[hide]Any odd number that has no prime divisor greater than $ 5$ is in the form $ 3^a \\cdot 5^b$ where $ a,b \\in \\mathbb{N}_0$\n\nThus: \n$ \\displaystyle \\sum_{i\\equal{}1}^{\\infty}\\dfrac{1}{n_i} \\equal{} \\sum_{a\\equal{}0}^{\\infty}\\sum_{b\\equal{}0}^{\\infty}\\dfrac{1}{3^a \\cdot 5^b} \\equal{} \\sum_{a\\equal{}0}^{\\infty}\\dfrac{1}{3^a}\\sum_{b\\equal{}0}^{\\infty}\\dfrac{1}{5^b} \\equal{} \\left(\\sum_{a\\equal{}0}^{\\infty}\\dfrac{1}{3^a}\\right)\\left(\\sum_{b\\equal{}0}^{\\infty}\\dfrac{1}{5^b}\\right)$ $ \\equal{} \\dfrac{1}{1\\minus{}\\frac{1}{3}} \\cdot \\dfrac{1}{1\\minus{}\\frac{1}{5}} \\equal{} \\dfrac{3}{2} \\cdot \\dfrac{5}{4} \\equal{} \\dfrac{15}{8}$[/hide]" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $A=12345...20062007$ (Write 2007 numeral from 1 -->2007 )\r\nProve : $A$ not $\\vdots$ 9 \r\n\r\n(We can calculate the sum of numerals from A to prove ?)\r\n ( :maybe: )\r\nLouis Latin and Vicky", "Solution_1": "[quote=\"Vicky and Louis Latin\"]\n(We can calculate the sum of numerals from A to prove ?)\n [/quote]\r\n\r\nYes, you can do that easily... :wink: \r\n\r\nPierre.", "Solution_2": "Oh , no !\r\nIt very dificult because numerals from A not to keep to law !\r\nIt to be different from find the sum of sequence !\r\n( :maybe: )\r\n[i]Louis Latin and Vicky[/i]", "Solution_3": "Let s(A) is sum of digits A. Because $n=s(n)(mod 9)$ we have\r\n$s(A)=\\frac{2007*2008}{2}=0(mod 9)$.", "Solution_4": "Rust you are amazing,but could you proove what you just said ?\r\nmaybe it is obvious to you but not to me", "Solution_5": "Let s(A) is sum of digits A. Because $n=s(n)(mod 9)$ we have\r\n$A=s(A)(mod 9)=\\sum_{k=1}^{2007}s(k)=\\sum_{k=1}^{2007}k (mod 9)=\\frac{2007*2008}{2}=0(mod 9)$.\r\n Therefore $9|A$." } { "Tag": [ "factorial" ], "Problem": "Try this... after some experimenting with factorials I came up with this. perhaps it;'s beginner level, I've never done any American contests.\r\n\r\nSimplify: [((c^d)-1)!((c^d)+1)]/[((c^2d)-1)((c^d)-2)!]\r\n\r\nbtw plz spoiler answers. thankyou.", "Solution_1": "If I understood your question correctly, then\r\n\r\n $\\frac{(c^d - 1)! (c^d + 1)}{(c^{2d} - 1) (c^d - 2)!}$\r\n = $\\frac{(c^d - 1)(c^d + 1)}{(c^{2d} - 1)}$\r\n = $1$\r\n\r\n***********************************************************************************************\r\nSambit" } { "Tag": [], "Problem": "In how many ways can $n$ people be seated in a row of $r$ chairs given that no two people may sit next to one another?", "Solution_1": "[hide]\nNo two people can sit next to each other if nobody sits in the seat immediately after each person. So if we label empty chairs by $E$'s and chairs with a person sitting in them by $S$, we can split a string of $E$'s and $S$'s into $SE$'s and $E$'s...\n\n[b]unless[/b] the last seat is taken, in which case there is no next seat. But we can deal with this anyway by just adding another seat (so now there are $r+1$) and requiring that no person sits in the last seat. Now the string of length $r+1$ can be split into $n$ $SE$'s and $r+1-2n$ $E$'s, for a total of $r+1-n$ sections.\n\nThe number of seatings is just the number of ways of choosing the positions for $n$ $SE$'s out of $r+1-n$ possibilities, which is $\\boxed{\\binom{r+1-n}{n}}$.\n[/hide]", "Solution_2": "Wait, are people distinct?", "Solution_3": "[hide]\nAssuming that each person is indistinct:\nSit each person down in a chair. You have used up $n$ chairs. Put a chair in between each person to make sure they aren't sitting next to each other. You have used up $n-1$ more chairs.\nNow you're left with $r-n-(n-1)=2n-1$ chairs. You can add these chairs in between any two people or on either edge, so there's $n+1$ spots where you can add the rest of the chairs.\nUsing the balls & urns method, you get:\n$\\binom{r-2n+1+n}{n}= \\boxed{\\binom{r-n+1}{n}}$\nIf each person is distinct, you can just permute each of the possibilities to get $\\binom{r-n+1}{n}n!$\n[/hide]", "Solution_4": "[quote=\"scorpius119\"][hide]\nNo two people can sit next to each other if nobody sits in the seat immediately after each person. So if we label empty chairs by $E$'s and chairs with a person sitting in them by $S$, we can split a string of $E$'s and $S$'s into $SE$'s and $E$'s...\n\n[b]unless[/b] the last seat is taken, in which case there is no next seat. But we can deal with this anyway by just adding another seat (so now there are $r+1$) and requiring that no person sits in the last seat. Now the string of length $r+1$ can be split into $n$ $SE$'s and $r+1-2n$ $E$'s, for a total of $r+1-n$ sections.\n\nThe number of seatings is just the number of ways of choosing the positions for $n$ $SE$'s out of $r+1-n$ possibilities, which is $\\boxed{\\binom{r+1-n}{n}}$.\n[/hide][/quote]\r\n\r\nThat's what I got, but note that r>2n-1 for it to be possible to arrange n people into a row of r chairs such that no two people may sit next to each other.", "Solution_5": "[quote=\"computer_nuke\"][quote=\"scorpius119\"][hide]\nNo two people can sit next to each other if nobody sits in the seat immediately after each person. So if we label empty chairs by $E$'s and chairs with a person sitting in them by $S$, we can split a string of $E$'s and $S$'s into $SE$'s and $E$'s...\n\n[b]unless[/b] the last seat is taken, in which case there is no next seat. But we can deal with this anyway by just adding another seat (so now there are $r+1$) and requiring that no person sits in the last seat. Now the string of length $r+1$ can be split into $n$ $SE$'s and $r+1-2n$ $E$'s, for a total of $r+1-n$ sections.\n\nThe number of seatings is just the number of ways of choosing the positions for $n$ $SE$'s out of $r+1-n$ possibilities, which is $\\boxed{\\binom{r+1-n}{n}}$.\n[/hide][/quote]\n\nThat's what I got, but note that r>2n-1 for it to be possible to arrange n people into a row of r chairs such that no two people may sit next to each other.[/quote]\n\nit should be $r \\geq 2n-1$. if the people are distinct then of course this becomes [hide]$\\binom{r-n+1}{n}n!$.[/hide]", "Solution_6": "Whats the balls and urns method? I don't really understand how you got (r-2n+1+n)C(n)." } { "Tag": [ "algebra", "polynomial" ], "Problem": "Consider a real poylnomial $p(x)=a_nx^n+...+a_1x+a_0$.\r\n(a) If $\\deg(p(x))>2$ prove that $\\deg(p(x)) = 2 + deg(p(x+1)+p(x-1)-2p(x))$.\r\n(b) Let $p(x)$ a polynomial for which there are real constants $r,s$ so that for all real $x$ we have \\[ p(x+1)+p(x-1)-rp(x)-s=0 \\]Prove $\\deg(p(x))\\le 2$.\r\n(c) Show, in (b) that $s=0$ implies $a_2=0$.", "Solution_1": "for a) let $n=deg(p(x))$. then let us look at $x^n,x^{n-1},x^{n-2}$ and their coefficents. \r\nthe result of $p(x+1)+p(x-1)-2\\cdot p(x)$ has $a_n+a_n-2a_n=0$ many $x^n$, $({n\\choose 1} \\cdot a_n+a_{n-1})+(-{n\\choose 1}a_n+a_{n-1})-2a_{n-1})=0$ many $x^{n-1}$, and $({n\\choose2}a_n+{n-1\\choose1}a_{n-1}+a_{n-2})+({n\\choose2}a_n-{n-1\\choose1}a_{n-1}+a_{n-2})-2a_{n-2}=2\\cdot {n\\choose2}a_n$ many $x^{n-2}$ in it. since $n(n-1)a_n\\neq 0$ because $n>0$ and $a_n\\neq 0$ it follows that $deg(p(x+1)+p(x-1)-2\\cdot p(x))=n-2$ and hence $deg(p(x))=2+deg(p(x+1)+p(x-1)-2\\cdot p(x))$.", "Solution_2": "for b) let us assume $n=deg(p(x))>2$.\r\nthen we have $a_nx^n+a_nx^n-ra_nx^n=$ so $r=2$.\r\nfrom a) we saw that $deg(p(x+1)+p(x-1)-2p(x))=p(x)-2>0$, so just adding a constant cant result that the sum of plynoms is zero for all real x. contradiction -> the opposite is true\r\n\r\nfor c) \r\n[quote=\"Peter VDD\"]\n(c) Show, in (b) that $s=0$ implies $c=0$.\n[/quote]\r\nwhat is $c$ ?", "Solution_3": "c) has been edited. (also for the post below)", "Solution_4": "Are you sure that is the correct version Peter? \r\n\r\nI think that if $s = 0$, then the polynomial must be constant or linear, you see... :)", "Solution_5": "i guess that makes sense ... (that $a_2=0$, which implies that $p$ is either linear or constant)" } { "Tag": [ "LaTeX", "geometry", "3D geometry", "tetrahedron", "AMC 10", "AMC" ], "Problem": "I know joml88 is doing a mock AMC H tomorrow (which I already did) and other ppl (including myself) did mock AMC, which were mostly on level of AMC-12. Of course, Mildorf did a wonderful job on the mock AIME.\r\n\r\nBut I wonder if anyone can host a mock AMC-10. I did all the past tests and some AHSMEs and I wonder if I can have some extra tests, if possible.\r\n\r\nThank you.", "Solution_1": ":( \r\n\r\nNo reply yet.... :( \r\n\r\nI wonder whether this thread would ever get replies...\r\n\r\nI guess it would be gone soon... :(", "Solution_2": "I collaborated on a Mock AMC with Singthesorrow (who doesn't seem to be around) back in August. I still have the questions but would have to create \"fake\" answers and then put it into LaTEX (which I don't know how to do)... It's about the level of a Mock AMC 10, maybe slightly higher like an 11...\r\n\r\nI have exams coming up though so I'm not too sure I would be able to administer it within January (my last exam is on the 27th, I might be able to do the 29th or 30th)\r\n\r\nWhen are the AMCs?", "Solution_3": "The AMC's are on Febuary 1st and 16th, so having a mock one on January 30th should be okay with me.", "Solution_4": "[quote=\"Lucky707\"]I collaborated on a Mock AMC with Singthesorrow (who doesn't seem to be around) back in August. I still have the questions but would have to create \"fake\" answers and then put it into LaTEX (which I don't know how to do)... It's about the level of a Mock AMC 10, maybe slightly higher like an 11...\n\nI have exams coming up though so I'm not too sure I would be able to administer it within January (my last exam is on the 27th, I might be able to do the 29th or 30th)\n\nWhen are the AMCs?[/quote]\r\n\r\nI'd be willing to put it into LaTeX and proofread it, since I'm not sure it would help me too much if I took it.", "Solution_5": "I'm willing to help as well. I'm a bit old to do the AMC-10 this year.", "Solution_6": "Yeah, just send the test to tetrahedr0n and he will tex it for you.\r\n\r\nAlso, when he tex it, I'm pretty sure he'll check for any errors.\r\n\r\n :)", "Solution_7": "Believe it or not I just finished it...\r\n\r\nIt's on its way for correcting as you read.", "Solution_8": "I sent it to tetrahedron yesterday...\r\n\r\nWould this be the 10th mock AMC? because we just had H and H=8... so would this be Mock AMC 9 or will someone make one before the 30th? Shall I call it Mock AMC 9 because Mock AMC I looks like Mock AMC 1", "Solution_9": "Actually it would be the eleventh (1, 2, A-H)" } { "Tag": [ "number theory", "least common multiple", "function" ], "Problem": "pls help to find all the possible solutions of x & y where\r\n\r\nx - y^3 = LCM(x, y)\r\n\r\nIs there a quick way to solve the quest instead of trial and error?", "Solution_1": "I don't know if it might help, but:\r\n\r\n$ xy \\equal{} LCM (x,y) * GCM (x,y)$.", "Solution_2": "I got this info. Thanks.\r\n\r\nI wrote a vb script in microsoft excel to find the answer since excel provides a lcm function to ease my life. But I am interested in mathematical method to solve it.\r\n\r\ncheers", "Solution_3": "The important observation is that $ y | \\text{lcm}(x, y) \\Rightarrow y | x$. From here it's straightforward.", "Solution_4": "I'm confused, isn't $ lcm(x,y) \\ge x$? Then y would have to be 0.....\r\nAnd what do you mean, t0rajir0u? Doesn't y always divide lcm(x,y)?", "Solution_5": "We are given that $ x \\equal{} y^3 \\plus{} \\text{lcm}(x, y)$. Since $ y | y^3$ and $ y | \\text{lcm}(x, y)$ it follows that $ y | x$. The problem is straightforward from here.", "Solution_6": "Are x and y allowed to be negative (I guess not...)?", "Solution_7": "[quote=\"mathwizarddude\"]Are x and y allowed to be negative (I guess not...)?[/quote]\r\nI don't think so.", "Solution_8": "Hmm. Well, if $ x$ and $ y$ are both non-negative then the question is trivial since $ \\text{lcm}(x, y) \\ge x$, but if either $ x$ or $ y$ is allowed to be negative then the definition of $ \\text{lcm}$ is ambiguous up to sign.", "Solution_9": "Is $ \\text{lcm}(x,0)=0$?\r\nHow about $ \\gcd(x,0)$?\r\nIs lcm allowed to be negative (I guess not...)?", "Solution_10": "$ \\gcd(x, 0)$ is $ x$ because every integer divides $ 0$. $ \\text{lcm}(x, 0)$ is $ 0$ because every multiple of $ 0$ is $ 0$. (By the way, the standard compact notation is $ \\text{lcm}(a, b) \\equal{} [a, b]$; parentheses denote the $ \\gcd$ without further qualification.)" } { "Tag": [ "algebra", "binomial theorem", "number theory unsolved", "number theory" ], "Problem": "prove :\r\n\r\nif $2^n+1$ is prime then $n=2^k$", "Solution_1": "That is easy. You should know the identity that x+y divides x^k+y^k if k is odd. The proof to this identity is a consequence of the binomial theorem: consider x^k=[(x+y)+(-y)]^k and expand.\r\n\r\nThe proof to your question follows from this identity. Suppose n is not a power of 2, then it has an odd factor greater than 1; this is your k. Hence 2^n+1 is composite.\r\n\r\nTherefore if 2^n+1 is prime then n=2^m for some positive integer m. These are called Fermat numbers and there is a conjecture that there are only finite many Fermat numbers which are prime.", "Solution_2": "By the way the nos. of the form 2^n+1 where n=2^k are called Fermat's Numbers", "Solution_3": "Yes, [e^pi], your conjecture is true, but it's an open problem if $2^{2^n}+1>2^{32}$ is always composite." } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let triangle ABC and H is its orthocenter.BH cuts AC at Q, CH cuts AB at P. M is the symmetric point of H wrt AB. K is the symmetric point of B wrt CH. Let point E such that $ \\angle EMA\\equal{}\\angle EHB\\equal{}90^o$. Prove that AE,PQ,CK are concurrent.", "Solution_1": "I prove your problem in few steps:\r\n1.If {$ R$}$ \\equal{} PQ\\cap CK$ and {$ V$}$ \\equal{} AR\\cap KQ$ then $ HV\\parallel{}AB$.\r\n2.If $ BE'\\parallel{}KQ$ $ (E'\\in AR)$, {$ S$}$ \\equal{} HE'\\cap AB$, {$ T$}$ \\equal{} HV\\cap BE'$\r\n then $ \\angle E'HB \\equal{} 90^{\\circ}$.\r\n3.$ \\angle E'MA \\equal{} 90^{\\circ}$\r\n So $ E \\equal{} E'\\Rightarrow AE,PQ,CK$ are concurrent. \r\n \r\n1.It's easy to prove that the quadrilateral $ AQRK$ is cyclic\r\n\r\n$ \\frac {VQ}{VK} \\equal{} \\frac {AQ\\cdot sin\\angle QAR}{KR\\cdot sin\\angle ARK}$ (a)\r\n\r\n$ \\frac {QA}{QC}\\cdot \\frac {PK}{PA}\\cdot \\frac {RC}{RK} \\equal{} 1\\Rightarrow \\frac {AQ}{KR} \\equal{} \\frac {PA}{PK}\\cdot \\frac {QC}{RC}$ (b)\r\n\r\nIt's easy to prove that $ \\frac {sin\\angle QAR}{sin\\angle ARK} \\equal{} \\frac {QR}{AK}$ (c) \r\n\r\n(a),(b),(c)$ \\Rightarrow \\frac {VQ}{VK} \\equal{} \\frac {PA}{PK}\\cdot \\frac {QC}{RC}\\cdot \\frac {QR}{AK}$ (d)\r\n\r\nBut $ \\frac {PA}{PK} \\equal{} \\frac {PA}{PB} \\equal{} \\frac {HQ}{HB}\\cdot \\frac {CA}{CQ}$ (e)\r\n\r\n(d),(e)$ \\Rightarrow \\frac {VQ}{VK} \\equal{} \\frac {HQ}{HB}\\cdot \\frac {AC}{AK}\\cdot \\frac {RQ}{RC} \\equal{} \\frac {HQ}{HB}\\Rightarrow HV\\parallel{}AB$\r\n\r\n2.$ VT\\parallel{}AB\\Rightarrow \\frac {SA}{HV} \\equal{} \\frac {E'S}{E'H} \\equal{} \\frac {SB}{HT}\\Rightarrow \\frac {SA}{SB} \\equal{} \\frac {HV}{HT}$ (f)\r\n\r\nBut $ BT\\parallel{}VQ\\Rightarrow \\frac {HV}{HT} \\equal{} \\frac {HQ}{HB}$ (g)\r\n\r\n(f),(g)$ \\Rightarrow \\frac {SA}{SB} \\equal{} \\frac {HQ}{HB}\\Rightarrow SH\\parallel{}AQ\\Rightarrow HE'\\parallel{}AC$\r\n\r\n$ HE'\\parallel{}AC$ and $ BQ\\bot AC\\Rightarrow BQ\\bot HE'\\Rightarrow \\angle E'HB \\equal{} 90^{\\circ}$\r\n\r\n3.It's easy to prove that $ MA\\bot CK$. {$ D$}$ \\equal{} MA\\cap CK$\r\n\r\n$ \\frac {AE'}{AR} \\equal{} \\frac {AH\\cdot sin\\angle AHE'}{sin\\angle AE'H}\\cdot \\frac {sin\\angle ARK}{AD}$\r\n\r\n$ AH \\equal{} AM$, $ \\angle AHE' \\equal{} \\angle HAC \\equal{} 90^{\\circ} \\minus{} \\angle ACB$ and $ \\angle AE'H \\equal{} \\angle RAQ$\r\n\r\n$ \\Rightarrow \\frac {AE'}{AR} \\equal{} \\frac {AM}{AD}\\cdot \\frac {cos\\angle ACB\\cdot sin\\angle ARK}{sin\\angle RAQ}$\r\n\r\nLet's prove that $ \\frac {cos\\angle ACB\\cdot sin\\angle ARK}{sin\\angle RAQ} \\equal{} 1$.\r\n\r\nIt's easy to prove that $ \\frac {sin\\angle ARK}{sin\\angle RAQ} \\equal{} \\frac {AK}{QR}$ (h)\r\n\r\n$ \\frac {AK}{AP}\\cdot \\frac {QP}{QR}\\cdot \\frac {CR}{CK} \\equal{} 1\\Rightarrow \\frac {AK}{QR} \\equal{} \\frac {AP}{QP}\\cdot \\frac {CK}{CR} \\equal{} \\frac {AC}{BC}\\cdot \\frac {BC}{CR} \\equal{}$\r\n$ \\equal{} \\frac {AC}{CR} \\equal{} \\frac {CK}{CQ} \\equal{} \\frac {CB}{CQ} \\equal{} \\frac {1}{cos\\angle ACB}$ (i)\r\n\r\n(h),(i)$ \\Rightarrow \\frac {cos\\angle ACB\\cdot sin\\angle ARK}{sin\\angle RAQ} \\equal{} 1$\r\n\r\nSo $ \\frac {AE'}{AR} \\equal{} \\frac {AM}{AD}\\Rightarrow ME'\\parallel{}CK$. But $ MA\\bot CK$, so $ MA\\bot ME'\\Rightarrow \\angle E'MA \\equal{} 90^{\\circ}$. :)", "Solution_2": "P.S. $ M$ is the symmetric point of $ H$ wrt $ AB$ $ \\Rightarrow$ \r\n $ \\Rightarrow$ the point $ M$ is on the circumcircle of the triangle $ ABC$.", "Solution_3": "Dear Petry,\r\nYou have a wonderful proof, however the step 3 in your post can be solved by a faster way (download my file).\r\nThis problem is a lemma of another my own problem, it was posted at [url]http://www.mathlinks.ro/viewtopic.php?t=216885[/url]\r\nSincerely, mr.danh" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Find the radius and interval of convergence for the following power series.\r\n\r\n$ \\sum_{n \\equal{} 2}^{\\infty}\\frac {(1 \\plus{} 2cos\\frac {\\pi n}{4})^n}{lnn}x^n$\r\n\r\nAttempt:\r\n\r\n$ R \\equal{} \\frac {1}{\\lim_{n\\rightarrow\\infty}\\sqrt [n]{\\frac {(1 \\plus{} 2cos\\frac {\\pi n}{4})^n}{lnn}}} \\equal{} \\lim_{n\\rightarrow\\infty}\\frac {e^{\\frac {ln(lnn)}{n}}}{(1 \\plus{} 2cos\\frac {\\pi n}{4})}$ In answers $ R \\equal{} \\frac {1}{3}$\r\n\r\n$ \\lim_{n\\rightarrow\\infty}e^{\\frac {ln(lnn)}{n}} \\equal{} 1$.Then is \r\n\r\n$ \\lim_{n\\rightarrow\\infty}(1 \\plus{} 2cos\\frac {\\pi n}{4}) \\equal{} 3$? :|", "Solution_1": "[quote=\"azatkgz\"]Find the radius and interval of convergence for the following power series.\n\n$ \\sum_{n = 2}^{\\infty}\\;\\frac {(1 + 2\\,\\cos\\frac {\\pi n}{4})^n}{\\ln\\,n}\\,x^n$\n\nanswer:\n\n$ R = \\dfrac {1}{\\limsup_{n\\rightarrow\\infty}\\; {\\bigg|\\; \\frac {(1 + 2\\,\\cos\\frac {\\pi n}{4})^n}{\\ln\\,n} \\;\\bigg| }^{\\dfrac{1}{n}} } \\; = \\; \\liminf_{n\\rightarrow\\infty}\\; {\\bigg|\\; \\frac {\\ln\\,n}{(1 + 2\\,\\cos\\frac {\\pi n}{4})^n}\\;\\bigg| }^{\\dfrac{1}{n}}$\n\n$ = \\; \\liminf_{n\\rightarrow\\infty}\\;{\\bigg|\\; \\frac {e^{\\frac {\\ln\\,(\\ln\\,n)}{n}}}{1 + 2\\,\\cos\\frac {\\pi n}{4}} {\\bigg|\\;}}$\n\n\nnow,\n$ \\liminf_{n\\rightarrow\\infty}\\;\\;e^{\\frac {\\ln\\,(\\ln\\,n)}{n}} \\;=\\; \\lim_{n\\rightarrow\\infty}\\;\\;e^{\\frac {\\ln\\,(\\ln\\,n)}{n}} \\;=\\; 1$\n\nand\n$ \\limsup_{n\\rightarrow\\infty}\\;\\;1 + 2\\,\\cos\\frac {\\pi n}{4} = 3$\n\nso, yes, answer is $ R \\;=\\;\\bigg|\\; \\frac {1}{3} \\;\\bigg| \\;= \\frac {1}{3}$ [/quote]\r\n\r\nyou are sort of on the right track, what you really need is \r\n$ R \\; = \\; \\dfrac{1}{\\limsup_{n\\to\\infty} \\;\\left|a_n\\right|^{\\frac {1}{n}}}$\r\n\r\nwhat i mean is your answer is correct, you just needed to justify the steps properly (see what i wrote above)", "Solution_2": "Oh,I see.Thanks :roll:" } { "Tag": [ "geometry", "trigonometry", "inequalities", "power of a point", "radical axis", "geometry solved" ], "Problem": "Let $H$ be the orthocenter of the acute triangle $ABC$. Let $BB'$ and $CC'$ be altitudes of the triangle ($B^{\\prime} \\in AC$, $C^{\\prime} \\in AB$). A variable line $\\ell$ passing through $H$ intersects the segments $[BC']$ and $[CB']$ in $M$ and $N$. The perpendicular lines of $\\ell$ from $M$ and $N$ intersect $BB'$ and $CC'$ in $P$ and $Q$. Determine the locus of the midpoint of the segment $[ PQ]$.\r\n\r\n[i]Gheorghe Szolosy[/i]", "Solution_1": "Note that $\\{ M,P,N,B'\\}$,$\\{ A,C'N,Q\\}$ are concyclic, so $C'H\\cdot CH=MH\\cdot NH=B'H\\cdot BH$, Thus B',C',P,Q are also concyclic. $\\angle HBC=\\angle HC'B'=\\angle HPQ$, which implies PQ//BC. Let D be the midpoint of BC. Then the midpoint of PQ must lie on HD.\r\nHowever, not all the points on HD satisfy the conditions. By simple trigo computation, we obtain $\\frac{2\\cos A}{\\cos A+1}\\le\\frac{HP}{HB}\\le1$. The answer is trivial then.", "Solution_2": "[quote=\"mecrazywong\"]Note that $\\{ M,P,N,B'\\}$,$\\{ A,C'N,Q\\}$ are concyclic, so $C'H\\cdot CH=MH\\cdot NH=B'H\\cdot BH$, Thus B',C',P,Q are also concyclic.[/quote]\r\n\r\nI think you wanted to say:\r\n\r\nNote that the points M, P, N and B' are concyclic, and that the points N, Q, M and C' are concyclic, so that $PH\\cdot B^{\\prime}H=MH\\cdot NH=QH\\cdot C^{\\prime}H$. Thus, the points B', C', P, Q are also concyclic.\r\n\r\nAll the rest of your solution is correct and very nice.\r\n\r\n darij", "Solution_3": "Darij, thanks for reminding me :) and sorry for the typos :blush:", "Solution_4": "[quote=\"mecrazywong\"]However, not all the points on HD satisfy the conditions. By simple trigo computation, we obtain $\\frac{2\\cos A}{\\cos A+1}\\le\\frac{HP}{HB}\\le1$. The answer is trivial then.[/quote]\r\n\r\nI can't understand this part of the solution .... :(", "Solution_5": "This problem is a \"painting\" of a wellknown problem which has the same hypothesis but the conclusion is $PQ\\parallel BC$. The its prove is identically with the Darij's solution. It is a pity for the author of the initial problem!", "Solution_6": "Virgil Nicula, do you know what mecrazywong meant by that trig at the end of his solution?", "Solution_7": "We are finding the locus of the midpoint of PQ. The first part of my solution only proves that the midpoint of PQ lies on HD. Since not every points on HD does belong to the family of the loci, we must eliminate them and the trigo inequality at the end of my solution serves for this purpose.", "Solution_8": "mecrazywong, i also deduced that. but how did you reach that trigo inequality? that was what i was asking", "Solution_9": "[quote=\"perfect_radio\"]mecrazywong, i also deduced that. but how did you reach that trigo inequality? that was what i was asking[/quote]\r\n\r\n?", "Solution_10": "[quote=\"Virgil Nicula\"]This problem is a \"painting\" of a wellknown problem which has the same hypothesis but the conclusion is $PQ\\parallel BC$. The its prove is identically with the Darij's solution. It is a pity for the author $\\boxed {S.\\ J.\\ Weyl}$ of the initial problem![/quote][color=red]This problem (R.M.O. $\\boxed {2005}\\ ,\\ 9^{th}$-grade) appears in my book (only with $PQ\\parallel BC$, without geometrical locus) \" [b]Geometrie plana. Culegere de probleme, Ed. GIL,[/b] $\\boxed {2002}$ (the page 51 with the number 5.17). Am I right ?[/color]\\[ \\boxed {\\ 2005\\ -\\ 2002\\ =\\ 3\\ >\\ 0\\ } \\][b]Question:[/b] [u]Who is the author of this problem from R.M.O., 2005 ?[/u]", "Solution_11": "I don't know who is the author of this problem. My teacher gave me a year ago a \"Gazeta Matematica\", which contained the problems from RMO 2004 and their authors. But he took it back shortly after.\r\n\r\n\r\nDo you know how mecrazywong reached that trig inequality? I can't seem to obtain anything similar. (The other part with $PQ \\| BC$ I understood)", "Solution_12": "The limiting position of midpoint of PQ happens when \r\n\r\n it is colliner with the radical axis of the Droz-Farny cirles.\r\n\r\n\r\n\r\n\r\n T.Y.\r\n\r\n M.T." } { "Tag": [ "inequalities open", "inequalities" ], "Problem": "Let a,b,c be the side of a triangle.Show that:\r\n$\\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\\geq \\frac{17}{8}abc+\\frac{7}{8}\\prod (b+c-a)$", "Solution_1": "[quote=\"NguyenDungTN\"]Let a,b,c be the side of a triangle.Show that:\n$\\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\\geq \\frac{17}{8}abc+\\frac{7}{8}\\prod (b+c-a)$[/quote]\r\n\r\nI think that this is wrong!", "Solution_2": "[quote=\"e.lopes\"][quote=\"NguyenDungTN\"]Let a,b,c be the side of a triangle.Show that:\n$\\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\\geq \\frac{17}{8}abc+\\frac{7}{8}\\prod (b+c-a)$[/quote]\n\nI think that this is wrong![/quote]\r\n\r\nExample?", "Solution_3": "[quote=\"NguyenDungTN\"]\nExample?[/quote]\r\nTry $a=11,$ $b=10$ and $c=2.$ :wink:", "Solution_4": "[quote=\"NguyenDungTN\"]Let a,b,c be the side of a triangle.Show that:\n$\\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\\geq \\frac{17}{8}abc+\\frac{7}{8}\\prod (b+c-a)$[/quote]\r\n$\\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\\geq 2abc+\\prod (b+c-a)$ is true already. :wink:", "Solution_5": "[quote=\"arqady\"]\n$ \\frac {abc(a \\plus{} b \\plus{} c)^{2}}{a^{2} \\plus{} b^{2} \\plus{} c^{2}}\\geq 2abc \\plus{} \\prod (b \\plus{} c \\minus{} a)$ is true already. :wink:[/quote]\r\nWe can suppose that $ abc\\equal{}1$ and $ c\\equal{}min\\{a;b;c\\}$\r\nWe can rewrite this ineq as:\r\n$ (a\\minus{}b)^2M\\plus{}N(a\\minus{}c)(b\\minus{}c) \\geq 0$\r\nwith $ M\\equal{}a\\plus{}b\\minus{}c\\minus{}\\frac{2}{a^2\\plus{}b^2\\plus{}c^2}$\r\n$ N\\equal{}c\\minus{}\\frac{2}{a^2\\plus{}b^2\\plus{}c^2}$\r\n$ M \\geq 0$ <=>$ a^3\\plus{}b^3\\plus{}ab^\\plus{}a^2b\\plus{}ac^2\\plus{}bc^2 \\geq a^2c\\plus{}b^2c\\plus{}c^3\\plus{}2$\r\nNote that $ ab^2\\plus{}ac^2 \\geq 2$\r\n$ a^3\\plus{}b^3 \\geq a^2c\\plus{}b^2c$\r\n$ a^2b \\geq c^3$\r\nSo $ M \\geq 0$\r\n$ N \\geq 0$<=>$ a^2c\\plus{}b^2c\\plus{}c^3 \\geq 2$\r\nThis ineq true because $ a^2c\\plus{}b^2c \\geq 2$\r\nDone!", "Solution_6": "[quote=\"arqady\"]\n$ \\frac {abc(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}}\\geq 2abc + \\prod (b + c - a)$ is true already. :wink:[/quote]\r\nWe can rewrite this ineq as:\r\n$ \\frac {abc(2ab + 2bc + 2ca - a^2 - b^2 - c^2)}{a^{2} + b^{2} + c^{2}}\\geq \\prod (b + c - a)$ \r\nWe can suppose that $ a = x + y, b = y + z, c = z + x$ where $ x,y,z > 0$\r\nThen we have: \r\n${ \\frac {(x + y)(y + z)(z + x)(4xy + 4yz + 4zx)}{2x^{2} + 2y^{2} + 2z^{2} + 2xy + 2yz + 2zx}}\\geq 8xyz$ \r\nI.e.\r\n${ \\frac {(x + y)(y + z)(z + x)(xy + yz + zx)}{x^{2} + y^{2} + z^{2} + xy + yz + zx}}\\geq 4xyz$\r\nWe can rewrite this ineq as:\r\n$ \\sum_{sym}x^{3}y^{2} + \\sum_{sym}x^{3}yz + 2\\sum_{sym}x^{2}y^{2}z\\ge2\\sum_{sym}x^{3}yz + 2\\sum_{sym}x^{2}y^{2}z$\r\nBut last ineq is true by Muirhead's ineq\r\nDone! :)" } { "Tag": [ "inequalities", "geometry", "geometry unsolved" ], "Problem": "Given a pentagon $ABCDE$ has all sides equal $a$. Construct another pentagon $A'C'E'B'D'$ such that $DC=A'B',AB=C'B',ED=C'D',CB=D'E',AE=E'A'$. If sum of square of diagonals of the first pentagon is $s^{2}$. What's sum of square of sides of the second pentagon. Or find counterexample to show that we have not enough informations.\r\n\r\n[img]http://www.vcharkarn.com/vcafe/dekvit/uploaded_pics/CS22320x13.gif[/img]", "Solution_1": "We have not enough informations. I will just show how to construct a dynamic sketch of the situation: Draw a point E' in the plane, and two points A' and D' on the circle with center E' and radius $a$. Draw the circle with center A' and radius $a$, and choose a point B' on this circle. Let the circles with centers B' and D' and with radius $a$ intersect each other at a point C'. Then, the pentagon A'C'E'B'D' will have all of its diagonals equal to $a$. There are some troubles - sometimes, the circles with centers B' and D' and with radius $a$ don't intersect, and often the pentagon A'C'E'B'D' won't be convex, though it is clear that this doesn't really matter - but you can evade them all by moving around the points A', E', D' and, most importantly, B'. Now move B' along the circle with center A' and radius $a$, and you will see that the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' is variable. This means that we cannot compute it from $a$ and $s$.\r\n\r\nA more interesting question is whether we can give some bounds. E. g., is the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' always $\\geq 5\\left(\\frac{1-\\sqrt5}{2}\\right)^{2}a^{2}$ ?\r\n\r\n[b]EDIT:[/b] Yes, it is! It follows from the right part of the inequality http://www.mathlinks.ro/Forum/viewtopic.php?t=112482 .\r\n\r\n darij", "Solution_2": "[quote=\"darij grinberg\"]Draw a point E' in the plane, and two points A' and D' on the circle with center E' and radius s. Draw the circle with center A' and radius s, and choose a point B' on this circle. Let the circles with centers B' and D' and with radius s intersect each other at a point C'. Then, the pentagon A'C'E'B'D' will have all of its diagonals equal to s. [/quote]\r\n\r\nBut the problem told that the pentagon A'C'E'B'D' have all of its diagonals equal to $a$,is'n it?", "Solution_3": "[quote=\"mathpk\"]But the problem told that the pentagon A'C'E'B'D' have all of its diagonals equal to $a$,is'n it?[/quote]\r\n\r\nI had confused a with s. Now I have corrected my post. Anyway, the knowledge of s doesn't change anything, as we can still vary the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' without changing the pentagon ABCDE (so that both values a and s remain constant).\r\n\r\n Darij", "Solution_4": "How you see that the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' is variable?", "Solution_5": "[quote=\"mathpk\"]How you see that the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' is variable?[/quote]\r\n\r\nDynamic computer geometry, as it would be quite a pain to do this by hand.\r\n\r\n darij", "Solution_6": "I see... btw this problem is in my national competition. How to do this by hand?\r\n\r\nps. Thanks for helping me :) I had learned from you how to do geometry problems by using dynamic computer geometry. :D" } { "Tag": [ "function", "MIT", "college" ], "Problem": "Hi all,\r\nJust wanted to note that we created an online graphing calculator:\r\n [url]http://fooplot.com/[/url]\r\nwhich should work in most modern browsers (except Safari). Hope you find it useful - share it, let us know if you find it useful, see any bugs, or have any suggestions about what would be useful. It's primarily targeted to help the educational community.\r\n\r\nAlso, if you ever need a quick plot of a function, just type in the function right after the URL:\r\n [url]http://fooplot.com/exp(-x^2)[/url]", "Solution_1": "[quote=\"dheera\"]Hi all,\nJust wanted to note that we created an online graphing calculator:\n [url]http://fooplot.com/[/url]\nwhich should work in most modern browsers (except Safari). Hope you find it useful - share it, let us know if you find it useful, see any bugs, or have any suggestions about what would be useful. It's primarily targeted to help the educational community.\n\nAlso, if you ever need a quick plot of a function, just type in the function right after the URL:\n [url]http://fooplot.com/exp(-x^2)[/url][/quote]\r\n\r\ngrr I hate users who make an account just to advertise things.....burn him!", "Solution_2": "He is a new user, but he's an MIT friend of Joe Laurendi, (joml88), and this really is a good website if you were looking for a quick way to graph things (which I was a few days ago).", "Solution_3": "[quote=\"mysmartmouth\"]He is a new user, but he's an MIT friend of Joe Laurendi, (joml88), and this really is a good website if you were looking for a quick way to graph things (which I was a few days ago).[/quote]\r\n\r\nI know but at least some people would have the decency to put them in the right place or something....", "Solution_4": "[quote=\"7h3.D3m0n.117\"][quote=\"mysmartmouth\"]He is a new user, but he's an MIT friend of Joe Laurendi, (joml88), and this really is a good website if you were looking for a quick way to graph things (which I was a few days ago).[/quote]\n\nI know but at least some people would have the decency to put them in the right place or something....[/quote]\r\nMods, please send this topic to 'High School>>>>Other Problem Solving Topics Forum'. Thank you in advance (if this ever gets sent there :P)" } { "Tag": [ "geometry", "rectangle", "combinatorics unsolved", "combinatorics" ], "Problem": "In a multiple choice test there were 4 questions and 3 possible answers for each question. A group of students was tested and it turned out that for any three of them there was a question which the three students answered differently. What is the maximum number of students tested?", "Solution_1": "I put in a 3x4 rectangle the answer of 3 students\nI get:\n OOO\n OOO\n OOO \n OOO\nthen one line have to contain 3 different answer thus 3! possibilities\nin the other space can be any answer, thus 3^9 possibilities\n--> 3! * 3^9\n\nthis is the maximum possibilities of the question of 3 generic students that are binomial(x,3)\n--> binomial(x,3) < 3! * 3^9 --> the maximum x is x=27 \n\nthe answer is 27 students\n\n\nSorry for my bad math and my worst english\n\nz." } { "Tag": [ "AMC" ], "Problem": "The solution they give makes very little sense to me. Could someone explain it, especially how they determined that the points must be one of the 5 given ones.\r\n\r\nThank you all\r\nBTW I'm new to this site, so if I'm doing anything incorrectly, please let me know.\r\n\r\nJB", "Solution_1": "Hey John,\r\n\r\nOne of the great things you'll learn to love about AOPS is the math jams. Look at the transcript from this night's math jam. You can click the tab next to Forum." } { "Tag": [ "topology", "Ring Theory", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $ M$ compact Hausdorff topological space. For $ t\\in M$ we define $ \\chi_t: C(M)\\rightarrow \\mathbb{C}$ as evaluation at point $ t$: $ \\chi_t(f)\\equal{}f(t),f\\in C(M)$. Prove that $ t\\rightarrow \\chi_t$ is homeomorphism for $ M$ to $ \\sigma(C(M))$. If by that homeomorphism we identify $ t$ with $ \\chi_t$, prove that Gelfand transformation is an identity.\r\n\r\n[hide=\"what I have done so far\"] I proved that $ t\\rightarrow \\chi_t$ is continuous, and that $ Ker \\chi_t$ is maximal ideal in $ C(M)$ ( since it codimension equals 1 ). I don't know how to continue from that, it's told that we have to notice that every maximal ideal is of that form, to prove that, we have to assume otherwise and use the fact that $ M$ is compact.[/hide]\r\n\r\nAgain any hints are welcome!", "Solution_1": "if $ \\mathfrak{m} \\in \\sigma(C(M))$ is not of the form, for every $ t \\in M$ there is some $ f \\in \\mathfrak{m}$ such that $ f(t) \\neq 0$, i.e. $ f$ does not vanish on an open neighborhood of $ t$. now use compactness to construct a unit in $ \\mathfrak{m}$, which is impossible.\r\n \r\nthe openness of $ t \\to \\chi_t$ is directly verified or by using that this is a continuous bijection from a compact space into a hausdorff space. have a look at the gelfand transform, then it's clear that it becomes the identity here." } { "Tag": [ "induction", "geometry", "parallelogram", "analytic geometry", "combinatorial geometry", "combinatorics unsolved", "combinatorics" ], "Problem": "Suppose there are $997$ points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least $1991$ red points in the plane. Can you find a special case with exactly $1991$ red points?", "Solution_1": "I'm curious as to whether or not this proof works...\r\n\r\nWe will prove by induction that given $n \\ge 2$ points in a plane, there are at least $2n-3$ distinct midpoints.\r\n\r\nBase Case: $n = 2$, we clearly have one midpoint.\r\n\r\nInduction Step: Call one of the points on the convex hull of the $n+1$ points $P$. Then take the convex hull of the other $n$ points not including $P$. We know $P$ is outside of the convex hull. By our inductive hypothesis, these $n$ points have at least $2n-3$ distinct midpoints. But there must exist two (consecutive) vertices $A$ and $B$ of the convex hull such that the midpoints of $AP$ and $BP$ lie outside of the convex hull and cannot be the same as any of the other midpoints (which clearly have to lie inside the convex hull). Hence there are at least $2n-3+2 = 2(n+1)-3$ distinct midpoints, completing the induction. (For the case that all the points are collinear, just take $A$ and $B$ the closest points to $P$ and we create two new midpoints.)\r\n\r\nSo for $n = 997$, we have at least $2(997)-3 = 1991$ midpoints.\r\n\r\nAnd if we take the points $(0,0); (2,0); \\ldots; (1992,0)$, we have exactly $1991$ distinct midpoints, namely $(1,0); (2,0); \\ldots; (1991,0)$.", "Solution_2": "I have not yet got the complete solution. But I believe this is the first step.\r\n\r\nConsider a point A1, when it is joined with the rest of the 996 points, no two lines shall intersect. Therefore there shall be 996 red points. And I noticed that 1991=996 +995.\r\nSo, I am trying on these lines.", "Solution_3": "Let points $A_i$ and $A_j$ be such that segment $A_iA_j$ has the greatest length of all other segments that are connecting some two points. Then construct segments $A_iA_1$, $A_iA_2$ ... $A_iA_n$ and $A_jA_1$, $A_jA_2$ ... $A_jA_n$. Those segments have distinct midpoints, because if any segment $A_iA_x$ have the same midpoint with some $A_jA_y$ then quadriliteral $A_iA_jA_xA_y$ is parallelogram with diagonals $A_iA_x$ and $A_jA_y$ so $A_iA_x>A_iA_j$ or $A_jA_y>A_iA_j$. So we have at least $995\\cdot 2 +1=1997$ distinct midpoints (\"1\" is the midpoint of $A_iA_j$). \r\n\r\n\r\n[color=blue][b]Other solution:[/b][/color] \r\nSort points in the coordinate plane by their $y$ coordinates (points with same $y$ coordinate are sorted by $x$ coordinate). Then take this segments $A_1A_2$, $A_2A_3$... $A_{n-1}A_n$ and $A_1A_3$, $A_3A_5$ and $A_2A_4$, $A_4A_6$... \r\nIt isn't hard to show that all this segments have distinct midpoints and so on \r\n(I am not so sure about this second solution, because I solve this problem in that way a few months ago and I wrote this right from my head, so maybe something is missing, but I don't think so ... :) )\r\n\r\nSpecial case is when all points are on a line and $A_1A_2=A_2A_3=...=A_{n-1}A_n$\r\n\r\nbye", "Solution_4": "My attempt at a solution:\n\nAs before, plot the points on a coordinate plane. Now, choose the uppermost point, and if not unique the leftmost of these to be $P_1$. Similarly, choose the lowermost, rightmost point to be $P_2$. Consider all the segments with $P_1$ as a vertex. There are $996$. Similarly, there are $996$ with $P_2$, and only one is double counted, i.e. $P_1P_2$. Thus, there are atleast $1991$ unique red points, since all of these midpoints must be unique. (Proof is left as a simple exercise in inequalities..) Hence, we are done. \n\nNote: This also proves it for the more general $2n-3$, as stated above. \n\nFun fact: This problem was unknowingly solved while doing the famous \"windmill problem\".. :P", "Solution_5": "Impose a Cartesian coordinate system on the set of points such that they have all distinct $y$-coordinates; let these be $y_1, y_2, \\cdots, y_{997}$ in increasing order. Furthermore, let $A_i$ be the point with $y$-coordinate $y_i$.\n\nNow, consider the midpoints of $\\overline{A_iA_{i+1}}$ for $1 \\leq i \\leq 996$; there are $996$ such midpoints, and by the order none of them have the same $y$-coordinate, and thus they do not coincide. Furthermore, consider the midpoints of $\\overline{A_iA_{i+2}}$ for $1 \\leq i \\leq 995$. These cannot coincide with any of the former midpoints because the $y_i$ are all distinct. This yields $996+995=1991$ distinct midpoints, as desired.\n\nFor a construction, take all the points to be equally spaced along a line.", "Solution_6": "Let, the points, be $(a_1, b_1), (a_2, b_2),....., (a_{997},b_{997}).$Now, take the midpoints of the points, $(a_i, b_i), (a_{i+1}, b_{i+1}).$ There, are $996,$ of these, and these do not coincide. Now, take the points, $(a_i, b_i), (a_{i+2}, b_{i+2}),$ which holds, $995,$ midpoints. These, again can not coincide, hence, this is $996+995=1991,$ hence, proved." } { "Tag": [], "Problem": "Solve this ecuation :\r\n\r\n$ (3x\\minus{}m\\minus{}n\\minus{}p)(\\frac{1}{x\\minus{}m}\\plus{}\\frac{1}{x\\minus{}n}\\plus{}\\frac{1}{x\\minus{}p})\\equal{}1$\r\n\r\n$ m,n,p\\in \\mathbb{R}\\ ,\\ m\\neq n\\neq p\\neq m$ are constants .", "Solution_1": "hello, your equation is equivalent with $ \\frac{(m\\plus{}n\\minus{}2x)(m\\plus{}p\\minus{}2x)(n\\plus{}p\\minus{}2x)}{(m\\minus{}x)(n\\minus{}x)(p\\minus{}x)}\\equal{}0$, so we have the following solutions $ x_1\\equal{}\\frac{m\\plus{}n}{2}$, $ x_2\\equal{}\\frac{m\\plus{}p}{2}$, $ x_3\\equal{}\\frac{n\\plus{}p}{2}$.\r\nSonnhard." } { "Tag": [ "function", "topology", "calculus", "calculus computations" ], "Problem": "We know that continuous functions map compact sets to compact sets.\r\n\r\n1) Whats an example of a closed set $ S\\subset \\mathbb{R}$ and a continuous function $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that $ f(S)$ is not closed?\r\n\r\n2) Whats an example of an open set $ U\\subset \\mathbb{R}$ and a continuous function $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that $ f(U)$ is not open?", "Solution_1": "[quote=\"wizard\"]1) Whats an example of a closed set $ S\\subset \\mathbb{R}$ and a continuous function $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that $ f(S)$ is not closed?[/quote]\n\n\\[ \\left[ {1, \\plus{} \\infty } \\right)\\xrightarrow{{f\\left( x \\right): \\equal{} \\frac{1}\n{{\\left| x \\right| \\plus{} 1}}}}\\left( {0,\\frac{1}\n{2}} \\right].\n\\]\n\n[quote=\"wizard\"]2) Whats an example of an open set $ U\\subset \\mathbb{R}$ and a continuous function $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that $ f(U)$ is not open?[/quote]\r\n\\[ \\left( { \\minus{} 1,1} \\right)\\xrightarrow{{f\\left( x \\right): \\equal{} \\left| x \\right|}}\\left[ {0,1} \\right).\r\n\\]" } { "Tag": [ "superior algebra", "superior algebra theorems" ], "Problem": "Hi, I was wondering how Frey's Equation, the equation that describes FLT as an elliptic curve, y^2 = X(X-x^n)(X-y^n), is derived from the general form of elliptic curves, y^2 =x ^3 +ax + b.", "Solution_1": "See G. Frey, [i]Links between stable elliptic curves and certain diohantine equations[/i] Ann. Univ. Sarav. Ser. Math. (1986), 719-722." } { "Tag": [], "Problem": "This problem was proposed on KBAHT (maybe it's a problem in Russian MO) :\r\n From $a+b+c+d=S$ and $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=S$ , we infer : $\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}+\\frac{1}{1-d} =S$ . Find $S$ ?\r\n Can you please help me to find a solution of this problem ? Thanks .", "Solution_1": "[quote=\"jarod\"]This problem was proposed on KBAHT (maybe it's a problem in Russian MO) :\n From $a+b+c+d=S$ and $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=S$ , we infer : $\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}+\\frac{1}{1-d} =S$ . Find $S$ ?\n Can you please help me to find a solution of this problem ? Thanks .[/quote]\r\n\r\nIt'a a problem from the 4th step(fo 5 steps) of RMO 2005.\r\nSolution:\r\n\r\nFrom $a+b+c+d=S$ and $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=S$ , we infer : $\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}+\\frac{1}{1-d} =S$\r\nalso $\\frac{1}{1-\\frac{1}{a}}+\\frac{1}{1-\\frac{1}{b}}+\\frac{1}{1-\\frac{1}{c}}+\\frac{1}{1-\\frac{1}{d}} =S$\r\nBut $\\frac{1}{1-t}+\\frac{1}{1-\\frac{1}{t}}=1$\r\nSo $2S=4 \\Rightarrow S=2$", "Solution_2": "Examples of such numbers:\r\n$a=2+\\sqrt3$\r\n$b=2-\\sqrt3$\r\n$c=d=-1$" } { "Tag": [], "Problem": "A three-digit bumber $\\overline{abc}$ (in decimal representation) is such that\r\n\r\n(i) its hundreds digit is equal to the sum of the other two digits, and\r\n\r\n(ii)$b(c+1)=52-4a$.\r\n\r\nFind all such numbers.", "Solution_1": "$a=b+c$\r\n$b(c+1)=52-4a$\r\n\r\n$b(c+1)=52-4(b+c)$\r\n$bc+5b+4c=52$\r\n$(b+4)(c+5)=72$\r\nso the numbers:\r\n$844,853,981,927$ satisfy the conditions" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let $ \\{A_1,A_2,...\\}$ be a sequence of finite sets. Is the product $ P \\equal{} \\Pi_{i \\equal{} 1}^{\\infty}A_i$ countable?", "Solution_1": "Is the product $ \\{0,1\\}^{\\mathbb N}$ countable?" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ x,y,z>0$,then:\r\n\r\n$ 1\\minus{}\\frac{8xyz}{(y\\plus{}z)(z\\plus{}x)(x\\plus{}y)}\\geq \\frac{z\\minus{}y}{y\\plus{}z}\\plus{}\\frac{x\\minus{}z}{z\\plus{}x}\\plus{}\\frac{y\\minus{}x}{x\\plus{}y} \\geq \\frac{8xyz}{(y\\plus{}z)(z\\plus{}x)(x\\plus{}y)}\\minus{}1$;", "Solution_1": "$ x_1,x_2,x_3,x_4 > 0$,prove that:\r\n$ 2(1 \\minus{} \\frac{16x_1x_2x_3x_4}{(x_1 \\plus{} x_2)(x_2 \\plus{} x_3)(x_3 \\plus{} x_4)(x_4 \\plus{} x_1)}\\geq \\frac {x_1 \\minus{} x_2}{x_1 \\plus{} x_2} \\plus{} \\frac {x_2 \\minus{} x_3}{x_2 \\plus{} x_3} \\plus{} \\frac {x_3 \\minus{} x_4}{x_3 \\plus{} x_4} \\plus{} \\frac {x_4 \\minus{} x_1}{x_4 \\plus{} x_1}\\geq \\minus{} 2(1 \\minus{} \\frac{16x_1x_2x_3x_4}{(x_1 \\plus{} x_2)(x_2 \\plus{} x_3)(x_3 \\plus{} x_4)(x_4 \\plus{} x_1)}$", "Solution_2": "Let $ x_1,x_2,...,x_n > 0,x_{n \\plus{} 1} \\equal{} x_1,n \\geq 3$,prove that:\r\n\r\n$ (n \\minus{} 2)(1 \\minus{} \\frac {2^n\\prod_{i \\equal{} 1}^n{x_i}}{\\prod_{i \\equal{} 1}^n{(x_i \\plus{} x_{i \\plus{} 1})}})\\geq \\sum_{i \\equal{} 1}^n{\\frac {x_i \\minus{} x_{i \\plus{} 1}}{x_i \\plus{} x_{i \\plus{} 1}}} \\geq \\minus{} (n \\minus{} 2)(1 \\minus{} \\frac {2^n\\prod_{i \\equal{} 1}^n{x_i}}{\\prod_{i \\equal{} 1}^n{(x_i \\plus{} x_{i \\plus{} 1})}})$\r\n\r\n\r\nBQ", "Solution_3": "[size=150]I said,once:\nOur vision should be to wide and wide,\nFrom a small place,\nfond out fine and nice![/size]", "Solution_4": "[size=200]yes ,\nOur vision should be wide and wide,[/size]\r\nas Follows:\r\n\r\nLet $ x_1,x_2,...,x_n > 0,x_{n \\plus{} 1} \\equal{} x_1,n \\geq 3,t$is odd number ,prove that:\r\n\r\n$ (n \\minus{} 2)(1 \\minus{} \\frac {2^n\\prod_{i \\equal{} 1}^n{x_i}}{\\prod_{i \\equal{} 1}^n{(x_i \\plus{} x_{i \\plus{} 1})}})\\geq \\sum_{i \\equal{} 1}^n{\\frac {(x_i \\minus{} x_{i \\plus{} 1})^t}{(x_i \\plus{} x_{i \\plus{} 1})^t}} \\geq \\minus{} (n \\minus{} 2)(1 \\minus{} \\frac {2^n\\prod_{i \\equal{} 1}^n{x_i}}{\\prod_{i \\equal{} 1}^n{(x_i \\plus{} x_{i \\plus{} 1})}})$\r\n\r\n\r\nBQ", "Solution_5": "$ \\frac{2x\\minus{}y\\minus{}z}{2x\\plus{}z\\plus{}y}\\plus{}\\frac{2y\\minus{}z\\minus{}x}{2y\\plus{}x\\plus{}z}\\plus{}\\frac{2z\\minus{}x\\minus{}y}{2z\\plus{}y\\plus{}x} \\geq \\frac{ 8xyz}{(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)}\\minus{}1.$\r\n\r\n\r\n$ \\frac{2x\\minus{}y\\minus{}z}{2x\\plus{}z\\plus{}y}\\plus{}\\frac{2y\\minus{}z\\minus{}x}{2y\\plus{}x\\plus{}z}\\plus{}\\frac{2z\\minus{}x\\minus{}y}{2z\\plus{}y\\plus{}x} \\le \\minus{}\\frac{ 8xyz}{(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)}\\plus{}1.$", "Solution_6": "$ 2 - \\frac {32x_1x_2x_3x_4}{(x_1 + x_2)(x_2 + x_3)(x_3 + x_4)(x_4 + x_1)} \\geq \\frac {3x_1 - x_2 - x_3 - x_4}{3x_1 + x_2 + x_3 + x_4} + \\frac{3x_2 - x_3 - x_4 - x_1}{3x_2 + x_3 + x_4 + x_1} + \\frac {3x_3 - x_4 - x_1 - x_2}{3x_3 + x_4 + x_1 + x_2} + \\frac {3x_4 - x_1 - x_2 - x_3}{3x_4 + x_1 + x_2 + x_3}$\r\n\r\n\r\n$ -2 + \\frac {32x_1x_2x_3x_4}{(x_1 + x_2)(x_2 + x_3)(x_3 + x_4)(x_4 + x_1)} \\le \\frac {3x_1 - x_2 - x_3 - x_4}{3x_1 + x_2 + x_3 + x_4} + \\frac{3x_2 - x_3 - x_4 - x_1}{3x_2 + x_3 + x_4 + x_1} + \\frac {3x_3 - x_4 - x_1 - x_2}{3x_3 + x_4 + x_1 + x_2} + \\frac {3x_4 - x_1 - x_2 - x_3}{3x_4 + x_1 + x_2 + x_3}$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "this somehow relates to my other question, anyways,\r\n\r\nConsider a uniform distribution of mass on the solid ball of radius $ r$ about the origin. Show that the gravitational field at point $ \\mathbf{x}$ is the same as if the mass closer to the origin than $ \\mathbf{x}$ were all located at the origin and the mass farther from the origin than $ \\mathbf{x}$ (if any) were absent.", "Solution_1": "This is the Shell Theorem from physics. Look in any calculus-based physics text on mechanics and there should be a proof." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Show that for all integer k, there are infinitely many natural numbers n such that k*2^2^n+1 is composite.\r\n\r\nI know that it is an old problem from PEN but I can't catch the solution gave in the original.Can somebody post a more detail and complete solution?\r\nThank you very much!", "Solution_1": "If a solution is not clear, ask in the corresponding thread. So post your questions in the PEN subforum.", "Solution_2": "You are right ZetaX :) \r\nBut do you have any idea towards this problem?I think that it worth discussing here :)", "Solution_3": "I am curious about this result.Can somebody give me a proof?Or just a \"given k\" such as $ k\\equal{}1$?", "Solution_4": "we can always find a prime p such that 2^2^n+1=p-1.", "Solution_5": "[quote=\"hxy09\"]I think that it worth discussing here[/quote]\r\nNo. Just revive the PEN topic; spreading around discussion of one problem in multiple topics is one of the biggest pains in the ass about this forum." } { "Tag": [ "geometry", "perimeter", "geometry unsolved" ], "Problem": "A point P is given inside a triangle ABC. A line l through P not con-\r\ntaining A intersects AB and AC at Q and R, respectively. Find the line\r\nl such that the perimeter of \u25b3AQR is least possible.", "Solution_1": "[url]http://www.imocompendium.com/othercomp/Nd/NdMC1990.pdf[/url]\r\n\r\nTo construct a line $ l$ through $ P,$ such that the $ \\triangle AQR$ has a given perimeter $ p,$ draw a circle centered at $ A$ with radius $ s \\equal{} \\frac{p}{2}$ intersecting the rays $ (AB, (AC$ at $ S, T,$ then a circle $ (E)$ tangent to $ (AB, (AC$ at $ S, T$ and finally, a tangent from $ P$ to $ (E),$ intersecting $ (AB, (AC$ at $ Q, R.$ $ (E)$ is the A-excircle of the $ \\triangle AQR$ with the given perimeter $ p.$\r\n\r\nFor minimum perimeter $ p$, $ P$ has to be the tangency point of $ QR$ with $ (E).$ Draw a circle $ (E)$ tangent to $ (AB, (AC$ and passing through $ P$ and then draw a tangent of $ (E)$ at $ P$ intersecting $ (AB, (AC$ at $ Q, R.$ 2 circles exist, tangent to $ (AB, (AC$ and passing through $ P.$ Pick the one, which lies outside of the resulting $ \\triangle AQR.$", "Solution_2": "[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14460[/img]\r\n\r\n(point line line) Apolonius problem\r\n\r\n :thumbup: :clap: \r\n\r\n[url]http://garciacapitan.auna.com/bella/htm/apo-ppr.htm[/url]" } { "Tag": [ "probability", "function" ], "Problem": "A fair coin is tossed. Let X be the number of tosses for four tails. Find the probability function of X. What is the probability that the fourth tail is preceded by\r\n\r\na) exactly 3 heads?\r\nb) at least 3 heads?", "Solution_1": "a)\r\n[hide] Since the fourth tail has to be preceeded by $h$ number of heads, there are three tails with the $h$ number of heads preceeding the fourth. The number of ways to arrange that is then $\\frac {(h+3)!}{h!3!}$. The probability of getting any one arrangement is $\\frac {1}{2}^{h+3}$. Then the probability that $h$ number of heads preceeds the fourth toss is $(\\frac {1}{2})^{h+3} * \\frac {(h+3)!}{h!3!}$. So the probability that the fourth toss is preceeded by exactly three heads is $(\\frac {1}{2})^6 * \\frac {6!}{3!3!} = \\frac {5}{16}$.[/hide]", "Solution_2": "Sorry it posted twice.", "Solution_3": "[quote=\"allrighty\"]A fair coin is tossed. Let X be the number of tosses for four tails. Find the probability function of X. What is the probability that the fourth tail is preceded by\n\na) exactly 3 heads?\nb) at least 3 heads?[/quote]\r\n\r\nThe probability fn. of X can be determined as follows.\r\n\r\nIf a total of X tosses are reqd for 4 tails, there has to be exactly 3 tails in the first X-1 tosses, the remaining toss results being heads.\r\n\r\nSo P(X) = C(X-1,3)*(1/2)^X\r\n\r\na> \r\nFor case(a), X has to be 3 (heads)+ 3 (intermdeiate tails)+1( last tail) =7\r\n\r\nSo P(E) = C(7-1,3)*(1/2)^7 = C(6,3)/128\r\n\r\nb> Here let's calculate the probability of the complement event , i.e. the no. of preceding heads is at most 2.\r\n The probability of the complement is\r\n\r\nthen P(X=4) + P(X=5) +P(X=6) corresponding to the no. of heads being 0,1and 2 respectively\r\n\r\nThen P(E) = 1 - ( P(X=4) + P(X=5) +P(X=6))\r\n\r\nThe values P(X=i) can be calculated from the expression for P(X) substituting the values 4,5,6" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "There is nothing more annoying than studying math, so Peter is trying to minimize his study time for\r\npassing MAT100. The overall mark will be composed by 15% from the problem sets, 45% from the term\r\ntests and 40% from the final exam. Peter expects his mark P for the problem sets to be proportional to the\r\ntime x he spends on them (let\u2019s say, P = x), his mark for the term tests to be T = (2x)/3 + y, where y is the time\r\nhe spends studying specifically for the term tests and his mark F for the final to be F = x/2 + y/2 + z, where z\r\nis the time he studies specifically for the final. \r\n\r\nWhat is the minimum amount of time for achieving the pass mark of 50 points, and how should he break down this time?\r\n\r\n(The above model is not realistic, because in reality the marks P; T and F cannot go higher than 100\r\neach. Comment on how this would influence your answer.)", "Solution_1": "$ P(x,y,z) \\equal{} x$, $ T(x,y,z) \\equal{} \\frac {2}{3} x \\plus{}y$, $ F(x,y,z) \\equal{} \\frac {1}{2}x \\plus{} \\frac {1}{2}y \\plus{} z$\r\n\r\nwant $ S(x,y,z) \\equal{} .15P \\plus{} .45T \\plus{} .4F \\equal{} .65x \\plus{} .65y \\plus{} .4z \\equal{} 50$\r\n\r\nto be our restriction, and we want to minimize\r\n\r\n$ g(x,y,z) \\equal{} x \\plus{} y \\plus{} z$\r\n\r\nso.. it would just be lagrange multipliers?\r\n\r\nthe more interesting question is how you would implement the max score concept..\r\n\r\nmaybe something like $ 100(1 \\minus{} e^{ \\minus{} x})$ but this means youll never reach 100.. which isnt realistic..", "Solution_2": "lagrange multipliers dont work. is there another method to solving this ?" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Find all $ n\\in\\mathbb{N}$ such that polynomial\n\\[ P(x) = (x - 1)(x - 2) \\cdots (x -n)\\]\ncan be represented as $Q(R(x))$, for some polynomials $ Q(x)$ and $R(x)$ with degree greater than $1$.", "Solution_1": "[quote=\"Erken\"]Find all $ n\\in\\mathbb{N}$,such that polynomial\n$ P(x) \\equal{} (x \\minus{} 1)(x \\minus{} 2)\\dots (x \\minus{} n)$\ncan be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$.[/quote]\r\n\r\nIt seems I misunderstood something because this problem seems trivial :\r\n\r\nAny non constant polynomial $ P(x)$ can be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$ :\r\n\r\nTake $ R(x)\\equal{}x\\plus{}a$ and $ Q(x)\\equal{}P(x\\minus{}a)$", "Solution_2": "[quote=\"pco\"][quote=\"Erken\"]Find all $ n\\in\\mathbb{N}$,such that polynomial\n$ P(x) \\equal{} (x \\minus{} 1)(x \\minus{} 2)\\dots (x \\minus{} n)$\ncan be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$.[/quote]\n\nIt seems I misunderstood something because this problem seems trivial :\n\nAny non constant polynomial $ P(x)$ can be represented as $ Q(R(x))$,for some non-constant polynomials $ Q(x),R(x)$ :\n\nTake $ R(x) \\equal{} x \\plus{} a$ and $ Q(x) \\equal{} P(x \\minus{} a)$[/quote]\r\nSorry,i forgot to write,that \r\n$ deg(R),deg(Q)\\geq 2$.", "Solution_3": "[hide=\"hint\"]\nConsider two cases,when $ n$ is odd,and when it is an even number :wink: \n[/hide]", "Solution_4": "If $ n > 2$ is even, then\r\n$ P(x) \\equal{} \\prod_{k \\equal{} 1}^{n/2}(x \\minus{} k)(x \\minus{} (n \\plus{} 1 \\minus{} k)) \\\\\r\n\\hspace*{0.3in} \\equal{} \\prod_{k \\equal{} 1}^{n/2}((x \\minus{} 1)(x \\minus{} n) \\plus{} k(n \\plus{} 1 \\minus{} k) \\minus{} n) \\equal{} Q(R(x))$\r\nwith $ R(x) \\equal{} a(x \\minus{} 1)(x \\minus{} n) \\plus{} b\\in\\mathbb{R}[x]$ non-linear\r\nand $ Q(x) \\equal{} \\prod_{k \\equal{} 1}^{n/2}\\left({x \\minus{} b\\over a} \\plus{} k(n \\plus{} 1 \\minus{} k) \\minus{} n\\right)\\in\\mathbb{R}[x]$ non-linear\r\nfor any $ a,b\\in\\mathbb{R},a\\not \\equal{} 0$.\r\n\r\nI'll show, that these are the only $ Q(x),R(x)\\in\\mathbb{R}[x]$ non-linear with $ P(x) \\equal{} Q(R(x))$.\r\n\r\nSay $ q: \\equal{} \\deg Q\\ge 2$ and $ r: \\equal{} \\deg R\\ge 2$. Then $ n \\equal{} qr$.\r\nWe can replace $ R(x)$ with $ {1\\over a}R(x)$ and $ Q(x)$ with $ Q(ax)$ for any $ a\\in\\mathbb{R},a\\not \\equal{} 0$.\r\nSimilar we can replace $ R(x)$ with $ R(x) \\minus{} R(1)$ and $ Q(x)$ with $ Q(x \\plus{} R(1))$.\r\nSo we can assume $ R(x)$ has leading coefficient $ 1$ and $ R(1) \\equal{} 0$.\r\n\r\nFrom [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=733142&search_id=2134256501#733142]here[/url] we see, that the strictly monotonic sequences\r\n$ R(1 \\plus{} kq),R(2 \\plus{} kq),\\ldots,R((k \\plus{} 1)q)$ and \r\n$ R(1 \\plus{} (k \\plus{} 1)q),R(2 \\plus{} (k \\plus{} 1)q),\\ldots,R((k \\plus{} 2)q)$\r\nrun through all the roots of $ Q(x)$ in converse order for all $ k \\equal{} 0,1,\\ldots ,r \\minus{} 2$.\r\n\r\nThis gives\r\n$ (*)$ $ R(k) \\equal{} R(2q \\plus{} 1 \\minus{} k)$ for all $ k \\equal{} 1,2,\\ldots 2q$.\r\nand\r\n$ (**)$ $ R(k) \\equal{} R(k \\plus{} 2q)$ for all $ k \\equal{} 1,\\ldots ,n \\minus{} 2q$.\r\n\r\nFor $ r\\ge 3$ and $ (q,r)\\not \\equal{} (2,3)$ we have $ 1 < (q \\minus{} 1)(r \\minus{} 2)$ and $ r \\minus{} 1 < qr \\minus{} 2q \\equal{} n \\minus{} 2q$.\r\nThen by $ (**)$ $ R(x) \\minus{} R(x \\plus{} 2q)$ has $ n \\minus{} 2q$ different roots and degree $ \\le\\deg R \\minus{} 1 < n \\minus{} 2q$.\r\nSo $ R(x) \\equal{} R(x \\plus{} 2q)$ and $ R(1) \\equal{} 0$ gives infinitely many roots $ 1 \\plus{} 2qk$ of $ R(x)$ for $ k\\in\\mathbb{N}$,\r\nand therefore $ R(x)\\equiv 0$. Contradiction!\r\n\r\nFor $ (q,r) \\equal{} (2,3)$ we get $ 0 \\equal{} R(1) \\equal{} R(4) \\equal{} R(5)$ using $ (*)$ and $ (**)$.\r\nThis gives $ R(x) \\equal{} (x \\minus{} 1)(x \\minus{} 4)(x \\minus{} 5)$ and $ R(2) \\equal{} 6,R(3) \\equal{} 4$, which contradicts $ (*)$.\r\n\r\nSo $ r \\equal{} 2$ and $ n \\equal{} 2q > 2$ even and $ (*)$ gives $ 0 \\equal{} R(1) \\equal{} R(n)$.\r\nThen $ R(x) \\equal{} (x \\minus{} 1)(x \\minus{} n)$ and $ P(x) \\equal{} Q(R(x)) \\equal{} \\tilde{Q}(R(x))$ \r\nwith $ \\tilde{Q}(x) \\equal{} \\prod_{k \\equal{} 1}^{n/2}(x \\plus{} k(n \\plus{} 1 \\minus{} k) \\minus{} n)\\in\\mathbb{R}[x]$.\r\nThis gives $ Q(x) \\equal{} \\tilde{Q}(x)$, which completes the proof.\r\n($ Q(x) \\minus{} \\tilde{Q}(x)$ has infinitely many roots $ x\\equal{}R(y),y\\in\\mathbb{R}$)" } { "Tag": [ "inequalities", "calculus", "integration" ], "Problem": "Find all integer solutions to the system of equations:\r\n$x^{3}+y^{3}+z^{3}=3=x+y+z$", "Solution_1": "[b]Solution.[/b] For $x,y,z\\geq 0$ by Arithmetic-Cubic Mean Inequality, $\\sqrt[3]{\\frac{x^{3}+y^{3}+z^{3}}{3}}=1 \\geq \\frac{x+y+z}{3}=1$, where equality holds if and only if $x=y=z$. Since equality holds, $3x^{3}=3\\implies x=y=z=1$. As a result, the only solution is $(x,y,z)=(1,1,1)$ for this case. If we are also considering negative integers, using the well known fact that the only integral solutions to $x^{3}+y^{3}+z^{3}=3$, where at least one of $x,y,z$ is less than 0, then $(x,y,z)=(4,4,-5)$, including all its permutations. \r\n\r\nTherefore, all solutions are $(x,y,z,)=(1,1,1),(4,4,-5),(4,-5,4), (-5,4,4)$.$\\blacksquare$" } { "Tag": [ "calculus", "limit" ], "Problem": "I hope this doesn't qualify as Calculus (it shouldn't ...)\r\n\r\nFor the convergent sequence with $\\displaystyle a_1=1$ and $\\displaystyle a_{n+1}=\\sqrt{3+2a_n}$, determine $\\displaystyle \\lim_{n\\to \\infty}{a_n}$", "Solution_1": "[hide=\"Might not be right\"]I'm not too sure this is right, but as n gets bigger, there are larger and larger strings of $\\sqrt{3+2\\sqrt{3+2\\sqrt{3+2\\sqrt{3+ \\cdots 2\\sqrt{5}}}}}$ Let this equal x, then $x^2$ gets closer to $3+2x$ as the series of sqrts gets longer, solving $x^2=3+2x$ the two solutions are $x=-1, x=3$ and because it can't be negative, then therefore it's 3. This is the only way I can think of how to do such a problem.[/hide]", "Solution_2": "Before you employ that trick, you first must prove that the limit exists.\r\n\r\n[hide=\"Hint\"]Prove that the sequence is strictly increasing and bounded from above.[/hide]", "Solution_3": "it already says that it converges...\"for the convergent series\"\r\n\r\nwhen it converges, for large $n$, we can say $a_{n+1}=a_n$\r\n\r\n$x^2=3+2x \\implies x=-1,3$ and of course the negative is not the answer since we square roots only have a positive range, so $\\boxed{3}$", "Solution_4": "$a_{n+1}-3=\\sqrt{3+2a_n}-3=\\frac{2(a_n-3)}{\\sqrt{3+2a_n}+3},$ thus we have\r\n\r\n$|a_{n+1}-3|\\leq \\frac{2}{3}|a_n-3|,$ yielding $0\\leq |a_n-3|\\leq \\left(\\frac{2}{3}\\right)^{n-1}|a_1-3|.$\r\n\r\n$\\therefore \\lim_{n\\to\\infty} a_n=3.$" } { "Tag": [ "induction", "quadratics", "Diophantine equation", "number theory unsolved", "number theory" ], "Problem": "The sequence $ (x_n)_{n\\geq1}$ is defined by:\r\n$ x_1\\equal{}1,x_2\\equal{}2,$\r\n$ x_{n\\plus{}2}\\equal{}4x_{n\\plus{}1}\\minus{}x_n$ $ \\forall n\\geq1$\r\nProve that $ x_{2008}$ is the sum of $ 2$ perfect squares.", "Solution_1": "[quote=\"KDS\"]The sequence $ (x_n)_{n\\geq1}$ is defined by:\n$ x_1 \\equal{} 1,x_2 \\equal{} 2,$\n$ x_{n \\plus{} 2} \\equal{} 4x_{n \\plus{} 1} \\minus{} x_n$ $ \\forall n\\geq1$\nProve that $ x_{2008}$ is the sum of $ 2$ perfect squares.[/quote]\r\n\r\nLet $ u\\equal{}2\\minus{}\\sqrt 3$ and $ v\\equal{}\\frac 1u\\equal{}2\\plus{}\\sqrt 3$ the two roots of $ x^2\\minus{}4x\\plus{}1\\equal{}0$. We have $ x_n\\equal{}\\frac{u^{n\\minus{}1}\\plus{}v^{n\\minus{}1}}2$\r\n\r\nThen, $ x_{2p}\\minus{}1\\equal{}\\frac{u^{2p\\minus{}1}\\minus{}2\\plus{}v^{2p\\minus{}1}}2$ $ \\equal{}(\\frac{\\sqrt 3\\plus{}1}2u^p)^2\\minus{}1\\plus{}(\\frac{\\sqrt 3\\minus{}1}2v^p)^2$ $ \\equal{}(\\frac{\\sqrt 3\\plus{}1}2u^p\\minus{}\\frac{\\sqrt 3\\minus{}1}2v^p)^2$\r\n\r\nAnd so $ x_{2p}\\equal{}y_p^2\\plus{}1$ where the sequence $ y_p$ is $ y_n\\equal{}\\frac{\\sqrt 3\\minus{}1}2v^n\\minus{}\\frac{\\sqrt 3\\plus{}1}2u^n$\r\n\r\nAnd so the sequence $ y_n$ may be defined as :\r\n$ y_1\\equal{}1$\r\n$ y_2\\equal{}5$\r\n$ y_{n\\plus{}2}\\equal{}4y_{n\\plus{}1}\\minus{}y_n$\r\n\r\nSo $ y_n\\in\\mathbb Z$ and, since $ x_{2p}\\equal{}y_p^2\\plus{}1$, we get that any member of rank even of the sequence $ x_n$ is the sum of two perfect squares.\r\n\r\nHence the result.", "Solution_2": "in fact, $ x_{n}\\equal{}X_{n\\minus{}1},n \\geq 2$, for which $ X_{i}$ are the answers of the pell equation:$ X^2\\minus{}3Y^2\\equal{}1$(we can prove this by either induction or (what pco did) finding the formula of $ x_{n}$).the answers are easily obtained by:\r\n$ X_{n\\plus{}1}\\equal{}2X_{n}\\plus{}3Y_{n},Y_{n\\plus{}1}\\equal{}X_{n}\\plus{}2Y_{n}$.we can easily see that $ X_{2k\\minus{}1}$ is even. Since \r\n$ (X_{n}\\minus{}1)(X_{n}\\plus{}1)\\equal{}3Y_{2k\\minus{}1}^2,\\gcd(X_{n}\\minus{}1,X_{n}\\plus{}1)\\equal{}1 (\\mbox{for odd n}) \\implies X_{n}\\plus{}1\\equal{}3q^2,X_{n}\\minus{}1\\equal{}s^2,(s,q)\\equal{}1,sq\\equal{}Y_{2k\\minus{}1}, \\implies X_{n}\\equal{}s^2\\plus{}1$\r\nbut we had $ x_{2k}\\equal{}X_{2k\\minus{}1}$.So we r DONE!(note that $ 2$ is a quadratic non-residue modulo $ 3$ so the case $ X_{n}\\plus{}1\\equal{}s^2,X_{n}\\minus{}1\\equal{}3q^2$ is ommitted.)\r\nu can prove that if $ p$ is a prime and $ p \\equiv \\pm 3 \\mod {8}$ and the fondamental answer of the pell equation $ X^2\\minus{}pY^2\\equal{}1$ is $ (0,1)$ modulo $ 2$, then the rank of odd answers of this pell equation have the same property." } { "Tag": [ "ratio", "conics", "function", "geometry proposed", "geometry" ], "Problem": "Let A be be a point outside the circle C, and AB and AC be the two tangents from A to this circle C. Let L be an arbitrary tangent to C that cuts AB and AC in P and Q. A line through P parallel to AC cuts BC in R. Prove that while L varies, QR passes through a fixed point. :)", "Solution_1": "Suppose $T$ is intersection of $QR$ and $AB$ If the tangent from $T$ to circle (not $AB$ the other) is' not parallel to $AC$ by Brianchon theorem $PR$ , $AC$ and tangent from $T$ to circle are concurrent.\r\nBut $PR$ is parallel to $AC$ contradiction. So $T$ is constant.", "Solution_2": "Well done ;) . A very nice solution. \r\n\r\nSee my solution( its a little bit longer):\r\nI will prove that the pole of QR is on a fixed line. The pole of QR is the intersection of the polars of Q and R. Let L be tangent to C in X. So the polar of Q is XC . we name the line through P parallel to AC: d. So the polar of R is the line through A and the pole of d .The pole of d (which is Y) is on the diameter of C perpendicular to d( which is the diameter through C which we name CD) .it lies also on BX. So we have found Y=( CD, BX). Let Z be the intersection of CX and AY( so Z is the pole of QR). \r\nNow to finish: its obvious that Z is on BD which is constant ( because AY ,the polar of R, passes through Z so the polar of Z passes through R so Z is on BD). :D", "Solution_3": "Here's what I did: we must show that the application from line $AC$ to line $BC$ which maps $Q\\to R$ is a perspectivity. In order to do this we must show that it's a homographic (preserving the cross-ratio) transformation and it maps $C\\to C$. \r\n\r\nIt's easy to see that it's homographic because it's the composition of $Q\\to P$ (from line $AC$ to line $AB$; it's wel-known that this sort of transformation is homographic (even if $(C)$ is a conic, not only a circle)) and $P\\to R$ (from line $AB$ to line $BC$).\r\n\r\nWe must now show that $C\\to C$. In this case $Q=C\\Rightarrow L=AC\\Rightarrow P=A\\Rightarrow PR=AC\\Rightarrow R=AC\\cap BC=C$, and we're done.", "Solution_4": "grobber, i didn't understand your solution properly could u explain more?", "Solution_5": "Assume you have two lines $d_1,\\ d_2$ and a point $X$ (which doesn't belong to $d_1$ or $d_2$). Take a variable line $d$ through $X$ which cuts $d_i$ in $A_i$. The function $f:d_1\\to d_2,\\ f(A_1)=A_2$ is called perspectivity and it'sa a particular case of function from $d_1$ to $d_2$ which preserves the cross-ratio of four points. \r\n\r\nIf you have $g:d_1\\to d_2$ s.t. $g$ preserves the cross-ratio, it can be shown (I won't show it here; try searching the Net for stuff like \"homographic transformation\", \"perspectivity\" etc.) that this is a perspectivity iff $g(O)=O$, where $O=d_1\\cap d_2$. This is exactly what I did here.", "Solution_6": "oh! Grobber I know homographic transformation I had not understood the last part of your proof. :) :( :blush: ;)", "Solution_7": "Oh.. Sorry for lecturing you :D Is it Ok now?", "Solution_8": "last line is'nt clear to me properly :?", "Solution_9": "What I'm trying to say is that according to the theorem I meantioned in one of my previous messages on this topic, in order to show that the transformation $Q\\to R$ is a perspectivity from line $AC$ to line $BC$ (i.e. $QR$ passes through a fixed point), we must show that by this transformation the intersection point of the two lines $AC$ and $BC$ turns into itself. This means we want to prove that when $Q$ coincides with $C$, by constructing $R$, we also find it to coincide with $C$. That's why the last line begins with $Q=C$ and ends with $R=C$ (what we want to show is the implication $Q=C\\Rightarrow R=C$).", "Solution_10": "[quote=\"Omid Hatami\"]Suppose $T$ is intersection of $QR$ and $AB$ If the tangent from $T$ to circle (not $AB$ the other) is' not parallel to $AC$ by Brianchon theorem $PR$ , $AC$ and tangent from $T$ to circle are concurrent.\nBut $PR$ is parallel to $AC$ contradiction. So $T$ is constant.[/quote]\nI don't understand why we have $PR$ , $AC$ and tangent from $T$ to circle are concurrent by Brianchon theorem" } { "Tag": [ "geometry", "3D geometry", "sphere", "function" ], "Problem": "The volume of a sphere of radius r is V = (4/3)(pi)(r^3); the surface area S of this sphere is S = 4(pi)(r^2). \r\n\r\n(a) Express the volume V as a function of the surface area S.\r\n\r\n(b) If the surface area doubles, how does the volume change?", "Solution_1": "${V(r)=\\frac{4 \\pi r^{3}}{3}}$\r\n${S(r)=4 \\pi r^{2}}$\r\n\r\nIt' s easy to see\r\n\r\n${V'(r)=S(r)}$\r\n\r\nOr\r\n\r\n${V(r)=\\frac{S(r)^{3/2}}{6 \\sqrt{\\pi }}}$\r\n\r\nTo double the suface need\r\n\r\n${S\\left(\\sqrt{2}r\\right)=2 S(r)}$\r\n\r\nSo Volume\r\n\r\n${V\\left(\\sqrt{2}r\\right)=2 \\sqrt{2}V(r)}$", "Solution_2": "a good tip.\r\n\r\nif the side length of a polygon is multiplied by a factor of $x$, the area is multipled by a factor of $x^{2}$. \r\n\r\nsame with area-volume" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Let the sequence $(x_n)$ define by:\r\n\r\n$x_1=1$\r\n$x_{m+n}=\\frac{1}{2}.(x_{2m}+x_{2n}), \\forall m,n \\in \\mathbb{N}^*$\r\n\r\nFind $x_{2007}$.", "Solution_1": "We have $x_{n+2}=\\frac 12 (x_4+x_{2n}),x_{n+1}=\\frac 12 (x_2+x_{2n}) \\Longrightarrow x_{n+2}=x{n+1}+h,h=\\frac{x_4-x_2}{2}$.\r\nFor all h and $x_2$ sequence $x_{n+1}=x_2+h(n-1)$ is solution." } { "Tag": [ "articles", "LaTeX" ], "Problem": "1. How do you make a space between two paragraphs?\r\n\r\n2. Why is it that when I type \r\n\r\n$\\framebox{a) \\ 90^{\\circ}}$\r\n\r\nit doesn't work. I want to get it so that the box is completely around my answer a) 90 degrees", "Solution_1": "\\framebox {a) \\ 90^{\\circ}}", "Solution_2": "\\framebox{} doesn't work as well as \\boxed{}.\r\n\r\n[code]\\boxed{90^{\\circ}}[/code]\r\n\r\n$\\boxed{90^{\\circ}}$", "Solution_3": "how do you skip a line between paragraphs? right now, everything is just lumped together", "Solution_4": "You can insert extra space by using\r\n\\bigskip \\\\\r\nor\r\n\\medskip \\\\\r\nor, if you want to vary the space, by using \r\n\\vspace{2ex} \\\\\r\n\r\nYou can use any number before ex (1ex=height of an x in the current font) and can use other units such as pt, in, mm etc.\r\n\r\nBut please remember, the style you are using may look best with no extra line between paragraphs. If you pick up a book this is the way most of them are printed.", "Solution_5": "this is exactly the result I wanted.\r\n\r\nhowever, the \\medskip\\\\ apparently adds to an \"error\" whenever I compile. is there a reason for this?", "Solution_6": "If you let us know the error and a simple document in which you get that error we can try and sort out what is happening.", "Solution_7": "whenever I add a \\medskip\\\\ an error shows up\r\n\r\nfor example:\r\n\r\n\\documentclass{article}\r\n\\begin{document}\r\n\r\nThis document has an error.\r\n\r\n\\medskip\\\\\r\n\r\nI don't know why.\r\n\r\n\\end{document}", "Solution_8": "This works though:\r\n[code]\\documentclass{article} \n\\begin{document} \n\nThis document has an error. \n\\medskip\\\\ \n\nI don't know why. \n\n\\end{document}[/code]The problem was the line between [i]This document has an error[/i] and \\medskip\\\\ . You'd get the same error [i]There's no line here to end [/i] even with \\\\ because LaTeX thinks you want a new paragraph rather than a new line.", "Solution_9": "I made the change you suggested but now instead of \"error\" I have \"bad box\"", "Solution_10": "Whoops! Missed that as you can ignore bad boxes if the printout looks OK. But to get rid of that warning I should have got rid of the other blank line after \\medskip \\\\ :\r\n[code]\\documentclass{article} \n\\begin{document} \n\nThis document has an error. \n\\medskip \\\\ \nI don't know why. \n\n\\end{document}[/code]This should have no errors or warnings.", "Solution_11": "I use the command:\r\n\r\n\\vskip 2in\r\n\r\nYou can play with this to make it perfect. Examples:\r\n\r\n\\vskip .5in\r\n\\vskip 3cm", "Solution_12": "\\vskip is a TeX primitive so may need care in its use. It is usually best to stick to LaTeX commands which are less likely to have side effects. In this case, the equivalent LaTeX command is \\vspace{2in} or whatever.\r\n\r\nFor example, I found this on a newsgroup and it gives an interesting error not flagged up by LaTeX.\r\n[code]\\documentclass{article} \n\\begin{document} \nBeware TeX primitives.\n\\vskip2cm \nPlus, in some cases errors are hard to find. \n\\end{document}[/code]", "Solution_13": "That's true, but sometimes TeX primitives let you get around built-in limitations. Of course, those limitations are probably built in for a reason... I haven't experimented with this much, but does \\vskip2cm make a difference if you do \\vskip 2cm (with a space) instead?", "Solution_14": "I think TeX swallows the space so it doesn't matter. I agree that if a TeX primitive works then use it; I have used \\def where I should have used \\newcommand. Kopka & Daly put it this way\r\n[quote]The primitive TeX commands form the bedrock of any format, and anything defined with them will always do exactly what the programmer expected. However, the equivalent LaTeX command could actually do more as time goes on. The \\newcommand checks for name clashes with existing commands, for example. It might even be possible that a debugging device that keeps track of all redefinitions will be added later; any commands defined with \\def would be excluded from such a scheme. Even now in LaTeX2e, there is something like this to keep track of all files input with middle and high-level commands.[/quote]\r\nThey then go on to discuss the fragility of some low-level commands and say that while one should always try to use high-level commands there will be occasions where this is not possible.\r\nYou may find the discussion about \\vskip interesting in [url=http://groups.google.co.uk/group/comp.text.tex/browse_frm/thread/7f9388fd02753030]comp.text.tex[/url] where I found the example I gave earlier." } { "Tag": [ "inequalities", "Euler", "inequalities solved" ], "Problem": "prove that in any triangle the following inequality holds:\r\n\r\n(cyc sum(bc cosA))/(cyc sum (a sin A))>=2r.", "Solution_1": "I really love these kind of inequalities! :D:D\r\n\r\nsolution:\r\n\r\ntry to prove the following identity:\r\n(cyc sum(bc cosA))/(cyc sum (a sin A))=R\r\nand with Euler you get R>=2r.\r\n\r\ntschuess!", "Solution_2": "Proof of the identity which Lagrangia uses in the solution\r\n We know that bc cosA=(1/2)*(b*b+c*c-a*a)\r\n That means that \\sum bc cosA=(1/2)(a*a+b*b+c*c) and using the fact that a*a=4R*R*sinA*sinA So we have \\sum bc cosA=2R*R( \\sum sinA*sinA)\r\n We also know that a*sinA=2RsinA*sinA That means that \\sum a*sinA=2R( \\sum sinA*sinA)\r\n So \\sum bc cosA/ \\sum a*sinA=R which is from Euler Theorem>=2r \r\n \r\n I make that proof because I think that the solution of Lagrangia is too \"empty\" without it.\r\n\r\n Hope it is corect!", "Solution_3": "indeed it is kinda \"empty\".. but I gave him a hint! :D :D", "Solution_4": "well, the hint that Lagrangia gave me was enough! thanks for the proof RNicula!", "Solution_5": "I was looking for a more direct proof, without using that identity. I guess that this proof is pretty neat!\r\n\r\ntschuess!\r\n:D" } { "Tag": [ "trigonometry", "search", "geometry proposed", "geometry" ], "Problem": "The similar isosceles triangles $ \\Delta AC_1 B,\\Delta BA_1 C$ and $ \\Delta CB_1 A$ with bases $ AB,BC$ and $ AC$ respectively are constructed externally on the sides of non-isosceles triangle $ \\Delta ABC$. Prove that if $ A_1 B_1 \\equal{} B_1 C_1$, then $ \\angle BAC_1 \\equal{} 30^\\circ$.", "Solution_1": "[quote=\"Edward_Tur\"] [color=darkred]The similar isosceles triangles $ \\Delta AC_1 B,\\Delta BA_1 C$ and $ \\Delta CB_1 A$ with bases $ AB,BC$ and $ AC$ respectively are constructed externally on the sides of non-isosceles triangle $ \\Delta ABC$. Prove that if $ A_1 B_1 = B_1 C_1$, then $ \\angle BAC_1 = 30^\\circ$.[/color] [/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] Denote $ m(\\widehat {BCA_1}) = \\phi$ and $ \\frac {1}{2\\cos\\phi} = \\rho$ . Thus, $ \\|\\begin{array}{c} A_1B = A_1C = \\rho a \\\\\n\\ B_1C = B_1A = \\rho b \\\\\n\\ C_1A = C_1B = \\rho c\\end{array}$ \n\nand $ \\|\\begin{array}{c} m(\\widehat {A_1BC_1}) = B + 2\\phi \\\\\n\\ m(\\widehat {A_1CB_1}) = C + 2\\phi\\end{array}$ . Apply the [b]generalized Pythagoras' theorem[/b] :\n\n$ \\|\\begin{array}{ccc} \\triangle A_1CB_1\\ : \\ \\ A_1B_1^2 = \\rho^2[a^2 + b^2 - 2ab\\cos (C + 2\\phi )] \\\\\n \\\\\n\\triangle A_1BC_1\\ : \\ \\ A_1C_1^2 = \\rho^2[a^2 + c^2 - 2ac\\cos (B + 2\\phi )]\\end{array}\\ \\ \\Longrightarrow$\n\n$ \\|\\begin{array}{c} A_1B_1^2 = \\rho^2[a^2 + b^2 - 2ab(\\cos C\\cos 2\\phi - \\sin C\\sin 2\\phi)] \\\\\n \\\\\nA_1C_1^2 = \\rho^2[a^2 + c^2 - 2ac(\\cos B\\cos 2\\phi - \\sin B\\sin 2\\phi)]\\end{array}$ . Thus\n\n$ A_1B_1 = A_1C_1\\ \\ \\Longleftrightarrow\\ \\ b^2 - 2ab\\cos C\\cos 2\\phi = c^2 - 2ac\\cos B\\cos 2\\phi$ , i.e.\n\n$ \\cos 2\\phi = \\frac {b^2 - c^2}{2ab\\cos C - 2ac\\cos B}$ $ \\Longleftrightarrow$ $ \\cos 2\\phi = \\frac 12$ $ \\Longleftrightarrow$ $ \\boxed {\\ \\phi = 30^{\\circ}\\ }$ .\n\nI use the relations $ ab\\sin C = ac\\sin B$ and $ ab\\cos C - ac\\cos B = b^2 - c^2$ .[/color]\r\n\r\nSee and http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1642104637&t=45678" } { "Tag": [ "limit", "algebra unsolved", "algebra" ], "Problem": "$a_n>0 \\forall n\\ge 1$, $\\lim_{n \\rightarrow +\\infty} \\frac{\\sum_1^n a_i}{\\Pi_1^n(1+a_i)}=0$.Prove that : $\\sum_1^{+\\infty} a_n=+\\infty$", "Solution_1": "[hide]\nSuppose $\\sum_{n=1}^\\infty a_i = L$ where $L$ is a finite positive real.\n\nThen $\\frac{\\sum a_i}{\\prod (1+a_i)} \\ge \\frac{\\sum a_i}{\\left(\\frac{n+\\sum a_i}{n}\\right)^n}$ by AM-GM.\n\nBut $\\lim_{n \\to \\infty} \\left(\\frac{n+\\sum a_i}{n}\\right)^n = 1$ because $L$ is finite.\n\nSo $\\lim_{n \\to \\infty} \\frac{\\sum a_i}{\\prod (1+a_i)} \\ge \\lim_{n \\to \\infty} \\frac{\\sum a_i}{\\left(\\frac{n+\\sum a_i}{n}\\right)^n} = \\lim_{n \\to \\infty} \\sum a_i = L > 0$. Contradiction. Hence we must have\n\n$\\sum_{n=1}^\\infty a_i = \\infty$.\n[/hide]" } { "Tag": [ "calculus", "logarithms", "function", "inequalities", "complex analysis", "advanced fields", "advanced fields unsolved" ], "Problem": "Let p denote partitions and f fibonacci seqence. For each positive intger k, prove that f(n)>p(n+k) hold for all +ve integers n but finitely many values.\r\n\r\nBomb", "Solution_1": "Partitions grow as $\\exp(c\\sqrt{n})$, and Fibonacci sequence exponentially.", "Solution_2": "Of course, but the problem is, I guess, to prove this special case using combinatorics only.\r\nI think Erdos had a nice upper bound for $p(n)$ with a combinatorial proof.", "Solution_3": "What do you call \"combinatorial\"? I may prove that $\\ln p(n)=c\\sqrt{n}+o(\\sqrt{n})$, where $c=2\\sqrt{\\zeta(2)}$ without using generating functions and complex analysis at all, so any schoolboy, if he wants, may understand it. Maybe, remainder term may be improved, and we may even show that $np(n)e^{-c\\sqrt{n}}$ has a limit (known to be equal to $\\frac1{4\\sqrt{3}}$).", "Solution_4": "Fedor, that is extremely interesting. Can you explain the idea of this elementary method for $p(n)$ asymptotics?\r\n\r\nBy \"combinatorial\" proof I meant that inequalities are derived by constructing functions between finite sets. Erdos argument was (something like) $p(1+2+...+n) < 4^n$ by constructing an injection from partitions to subsets of a set of size $2n$.", "Solution_5": "The idea is to use an identity $np(n)=\\sum\\limits_{k=0}^{n-1} p(k)\\sigma(n-k)$, where $\\sigma(k)$ means the number of divisors. Then, use Abel transform:\r\n\r\n\\[\r\n\\sum_{k=0}^{n-1} p(k)\\sigma(n-k)=\\sum_{k=0}^{n} p(k)(F(n-k)-F(n-k-1))=\\sum_{k=0}^{n-1} (p(k)-p(k-1))F(n-k),\r\n\\]\r\n\r\nwhere $F(k)=\\sum_{i=0}^k \\sigma(k),\\,F(0)=0$. Note that $p(n)$ increases (it is obvious), so we may replace $F(n-k)$ to its lower and upper bound to get lower and upper estimates of $RHS$. It is easy to show that $F(k)=\\zeta(2)(1+2+\\dots+k)+o(k^2)$ (people say that $o(k^2)$ is at fact $O(\\ln\\ln k)$. But I do not know whether it is elementary). So, do this and you get estimates $np(n)\\ge c'\\sum_{k=0}^{n-1} p(k)(n-k)$ and $np(n)\\le c''\\sum_{k=0}^{n-1} p(k)(n-k)$ for aribitrary $c'{0}$ then $ A_j^{*}\\equal{}\\lambda_{j}A_{j}^{\\minus{}1}$ and if $ \\lambda_{j}\\equal{}0$ then $ A_{j}\\equal{}0$; therefore $ A_{j}A_{j}^*\\equal{}A_{j}^{*}A_j$. QED." } { "Tag": [ "videos", "calculus", "derivative", "function", "parameterization", "real analysis", "real analysis unsolved" ], "Problem": "I always get excited when a video game gives me the opportunity to do a little math. It usually doesn't help me much, but at least I can feel like I'm exercising my mind :)\r\n\r\nI came up with this equation and I'm completely stumped. I'm trying to solve it for [i]x[/i] in terms of [i]a[/i]:\r\n\r\nx=(50/39)*[1-(1-a)^(15/x+6)]+(5/3)*(1-a)^(15/x+6)\r\n\r\nI don't even know where to begin. I've had up to Calc III, so I feel like I should know how to solve this, but it's been a while. Anyone have any ideas?\r\n\r\nThanks.", "Solution_1": "Seems pretty transcendental to me.. We could approximate something for a given $ a$, but in situation like this I wouldn't say that anything useful can be done (correct me if I'm wrong, I might have overseen something simple). I'd like to know more about the context: what was the game?", "Solution_2": "Okay, I did a bit of simplifying (just multiplying through the parentheses) and came up with \r\n\r\nx=50/39+(5/13)(1-a)^(15/x+6).\r\n\r\nThe game is World of Warcraft, and here's the scenario. I'll try to explain it in term that can be understood by someone who doesn't know anything about the game.\r\n\r\nI have a kitty cat that likes to kill monsters. It does this by \"attacking\" them and \"biting\" them. A bite happens once every 2.5 seconds, and this rate never changes. An attack normally happens every 5/3 seconds, but sometimes it happens 30% faster (once every 50/39 seconds).\r\n\r\nBefore I explain how the attack speed changes, I'll throw out there that my ultimate goal is to find the average attack speed ([i]x[/i]).\r\n\r\nThe attack speed increase is in effect while my cat has a \"buff\", which, when applied, lasts for fifteen seconds. The buff is acquired by attacking or biting. Whenever she attacks or bites, she there is a chance ([i]a[/i]) that she will gain the buff for the next fifteen seconds. This happens whether or not she already has it. (If she already has the buff, its duration is refreshed to fifteen seconds; she cannot have more than one buff at a time).\r\n\r\nNow here's where the math comes in. First, I solved for the chance she'll have the buff at any given time:\r\n\r\nc=1-(1-a)^n, where n is the number of attacks or bites that have happened in the last fifteen seconds.\r\n\r\nn was just the number of attacks per fifteen seconds (15/x) plus the number of bites per fifteen seconds (15/2.5). Thus:\r\n\r\nn=15/x+6\r\n\r\nand the chance that she'll have the buff at any given time is:\r\n\r\nc=1-(1-a)^(15/x+6)\r\n\r\nNow, to find the average attack speed [i]x[/i], I had to do a time-weighted average of the two attack speeds. That would be (attack speed [b]with[/b] the buff)*(fraction of time spent [b]with[/b] the buff)+(attack speed [b]without[/b] the buff)*(fraction of time spent [b]without[/b] the buff):\r\n\r\nx=(50/39)c+(5/3)(1-c)\r\n\r\nSubstitute and simplify to get\r\n\r\nx=50/39+(5/13)(1-a)^(15/x+6).\r\n\r\nAnd that's where I am now :)", "Solution_3": "With an equation like this, there's no reason to expect an exact analytical solution. That leaves numerical methods, which can give a very close approximation to the solution. Here's Newton's method:\r\n\r\nStep 1: Rearrange the equation into the form $ f(x)\\equal{}0$. Find a formula for the derivative $ f'$.\r\nStep 2: Guess an initial value $ x_0$.\r\nStep 3: Set $ x_{n\\plus{}1}\\equal{}x_n\\minus{}\\frac{f(x_n)}{f'(x_n)}$. Repeat as $ n$ increases, until the numbers stop noticeably changing. The final $ x_n$ is a good approximation to the true root $ x$.\r\n\r\nIf you use a bad initial guess, it may not work. If this happens, try again with a new starting point. The idea behind this method is that we approximate $ f$ with a linear function and solve the resulting linear equation. It converges very rapidly if you start close enough, with $ |x_{n\\plus{}1}\\minus{}x|$ being proportional to $ |x_n\\minus{}x|^2$.\r\n\r\nYour example isn't quite in a usable form; we would need a numerical value for the parameter $ a$ in order to use this." } { "Tag": [ "integration", "real analysis", "real analysis unsolved" ], "Problem": "Find\r\n\r\n $ \\int( x^{x^{x}} \\plus{} x^{x} )dx$ :lol:", "Solution_1": "Actually, I beg to differ. This can't be evaluated in terms of elementary functions." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "calculus computations" ], "Problem": "$ \\int \\tan x \\ln(\\sec x) \\ dx$", "Solution_1": "[hide]For $ x \\in \\left]\\minus{} \\frac{\\pi}{2}, \\frac{\\pi}{2}\\right[$, we have\n\n$ P \\equal{} \\int (\\minus{} \\tan x) \\ln(\\cos x)\\,dx \\equal{} \\frac12 \\ln^2(\\cos x) \\plus{} C$.[/hide]", "Solution_2": "Let $ I$ be the proposed integral.\r\n[hide]if $ u = \\sec x \\Rightarrow I = \\Int {{du\\over u}.ln u$\nIntegrating by parts with $ dv = {{du}\\over u}$ and $ w = \\ln u$ we get to $ I = {{\\ln^2\\sec x}\\over 2} + C$ [/hide]" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Define a sequence $ X_{0}$, $ X_{1}$, ... of rational numbers by $ X_{0}=2$ and $ X_{n+1}=X_{n}-\\frac{1}{X_{n}}$ for $ n\\geq 0$. Is the sequence bounded?", "Solution_1": "It was first asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31823]here[/url]. Unfortunately, I still cannot add much to what I wrote then :(.", "Solution_2": "No convergence.\r\nThe recurrence is the formula for finding the zeros of the function $ f(x)=e^{\\frac{x^{2}}{2}}$ by the Newton-Raphson method.\r\nThis function has no zeros, so the series does not converge.", "Solution_3": "That much was clear enough from the beginning: the limit would satisfy $ x=x-\\frac{1}{x}$, i.e., $ \\frac{1}{x}=0$, which is impossible. The question is whether the sequence stays bounded or, if you prefer, whether $ 0$ is not a limit point of this sequence." } { "Tag": [ "search", "arithmetic sequence", "number theory", "relatively prime" ], "Problem": "The game is as this follows:\r\n\r\nI'll think on an integer number between 1 and 100000. You will ask me for characteristics of the number (just yes/no questions), and I will answer all the questions. If you make posts like:\r\n\r\nis it 1?\r\nis it 2?\r\nis it 3?\r\n...\r\n\r\nI will NOT answer. \r\nIf you guess the number, you esrn 12 points. I will deduce points if:\r\n\r\na) You make too many questions\r\nb) You post too many spam\r\nc) You cheat\r\nd) any fail at the search\r\ne) You late too many time.\r\n\r\nWE BEGIN RIGHT NOW!!!", "Solution_1": "epic fail....", "Solution_2": "Is it a prime?", "Solution_3": "Is it even?", "Solution_4": "Is it greater than or equal to 50,000?", "Solution_5": "Is it a primitive root mod 226?", "Solution_6": "Is it a prime? [b]NO[/b]\r\nIs it even? [b]NO[/b]\r\nIs it greater than or equal to 50,000? [b]NO[/b]\r\nIs it a primitive root mod 226? [b]What does it mean? :P [/b]", "Solution_7": "Is it less than 100?", "Solution_8": "Is it less than 100? [b]NO[/b]", "Solution_9": "Is it greater than or equal to 25,000?", "Solution_10": "Is it greater than or equal to 25,000? [b]NO[/b]", "Solution_11": "Is it greater than or equal to 12,500?", "Solution_12": "Is it greater than or equal to 12,500? [b]YES[/b]", "Solution_13": "[hide=\"Observation\"] The restriction that the answer be either yes or no means that we can (at best) guarantee to divide our search space in half with each question (which is equivalent to asking for another bit in the binary representation of the desired number), so for those of you wondering whether there is a better chain of questioning than the one that is currently being pursued, there isn't. [/hide]", "Solution_14": "Is it greater than or equal to 18,750?", "Solution_15": "is the number less than 26?", "Solution_16": "YES!!!The message is too small. Please make the message longer before submitting.", "Solution_17": "is the number less than 12?", "Solution_18": "YES jhgfiweurafgirewfuwerfgvbiuewrgtwhywr", "Solution_19": "is the number one of the digits in your username?", "Solution_20": "YESThe message is too small. Please make the message longer before submitting.", "Solution_21": "is it 1?\r\n\r\nblah blah blah long message", "Solution_22": "oh yes :winner_second:\r\n\r\nYou made too many questions by yourself so -3\r\n\r\ncf249: 9\r\ndynamo729: 9 \r\npanjia123: 9 \r\n\r\nnew number", "Solution_23": "Is it less than 100?\r\n\r\n(ps how much is too much? [because I wont do it again, I just didnt know sorry!])", "Solution_24": "NO.\r\n\r\nhehe, you wasted a question", "Solution_25": "and, too much is: I count the number of pages wasted. Each one is a point less. But you made too many questions yourself, so I penalized you.", "Solution_26": "Is it greater than 50000?", "Solution_27": "Is it greater than 50000? NO", "Solution_28": "Is it greater than 20,000?", "Solution_29": "Is it greater than 20,000? YES" } { "Tag": [ "induction", "algebra unsolved", "algebra" ], "Problem": "Let $r_1=2$ and $r_n = \\prod^{n-1}_{k=1} r_i + 1$, $n \\geq 2.$ Prove that among all sets of positive integers such that $\\sum^{n}_{k=1} \\frac{1}{a_i} < 1,$ the partial sequences $r_1,r_2, ... , r_n$ are the one that gets nearer to 1.", "Solution_1": "Proof:Lemma:\\[ 1-(\\frac{1}{r_1}+\\frac{1}{r_2}+\\ldots+\\frac{1}{r_n})=\\frac{1}{r_1r_2\\ldots r_n} \\]\r\nThe proof of Lemma:When $n=1$,it's easy to see it's correct.\r\nSuppose it's correct for $n-1(n\\geq2)$,\r\nthen \\[ 1-(\\frac{1}{r_1}+\\frac{1}{r_2}+\\ldots+\\frac{1}{r_n})=\\frac{1}{r_1r_2\\ldots r_{n-1}}-\\frac{1}{r_n}=\\frac{1}{r_n-1}-\\frac{1}{r_n}=\\frac{1}{(r_n-1)r_n}=\\frac{1}{r_1r_2\\ldots r_n} \\]\r\nThat means the lemma is correct.\r\n\r\nWe only need to show that\r\n$\\forall a_1\\leq a_2\\leq\\ldots\\leq a_n\\in\\mathbb{N},\\frac{1}{a_1}+\\frac{1}{a_2}+\\ldots+\\frac{1}{a_n}<1$,\r\nthen we must have \\[ \\frac{1}{a_1}+\\frac{1}{a_2}+\\ldots+\\frac{1}{a_n}\\leq\\frac{1}{r_1}+\\frac{1}{r_2}+\\ldots+\\frac{1}{r_n} \\]\r\nWe prove it by induction:\r\nWhen $n=1$,$\\frac{1}{a_1}<1$,$a_1\\geq2=r_1$.So $\\frac{1}{a_1}\\leq\\frac{1}{r_1}$.\r\nSuppose the conclusion is correct for all positive integer smaller than $n$,and there exist $a_1,a_2,\\ldots,a_n$ such that\r\n\\[ \\frac{1}{r_1}+\\frac{1}{r_2}+\\ldots+\\frac{1}{r_n}<\\frac{1}{a_1}+\\frac{1}{a_2}+\\ldots+\\frac{1}{a_n}<1 \\eqno(*) \\]\r\nBy Lemma\\[ 1-(\\frac{1}{r_1}+\\frac{1}{r_2}+\\ldots+\\frac{1}{r_n})=\\frac{1}{r_1r_2\\ldots r_n} \\]\r\nAnd it's obvious that \\[ 1-(\\frac{1}{a_1}+\\frac{1}{a_2}+\\ldots+\\frac{1}{a_n})\\leq\\frac{1}{a_1a_2\\ldots a_n} \\]\r\nBecause $\\frac{1}{a_1}+\\frac{1}{a_2}+\\ldots+\\frac{1}{a_n}=\\frac{x}{a_1a_2\\ldots a_n}<1,x\\in\\mathbb{Z}$\r\nSo $x\\sum_{i=1}^n\\frac{a_i}{r_i}\\geq n\\sqrt[n]{\\frac{a_1a_2\\ldots a_n}{r_1r_2\\ldots r_n}} \\]\r\nSo \\[ a_1a_2\\ldots a_n R$ because $ 250 > 13 (\\sqrt 8 + \\sqrt {13} + \\sqrt {104})\\ .$" } { "Tag": [ "analytic geometry", "resistors" ], "Problem": "Set up a coordinate plane and place 1 ohm resistors between ever pair of adjacent lattice points. What is the resistance between any 2 adjacent lattice points? (This is not a trick question!)", "Solution_1": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151232[/url]", "Solution_2": "Isn't the resistance simply $ 0.5$ Ohms??", "Solution_3": "yes, it's $ 0.5$ Ohms. I thought [b]dominicleejun[/b] was asking for the resistance between any two points :)", "Solution_4": "Adjacent, as far as I doesn't mean any two points :lol:. jk", "Solution_5": "what is the resistance between the points (0,0) and (x,y)?\n\nsorry for revive.." } { "Tag": [ "function", "algebra", "system of equations", "inequalities proposed", "inequalities" ], "Problem": "For $ a,b,c$ positive real numbers such that $ a\\plus{}b\\plus{}c\\equal{}1$, determine the minimum and maximum value of $ 2a^2\\plus{}2b^2\\plus{}2c^2\\plus{}9abc$.\r\n\r\nsorry if it's a silly problem for you :(", "Solution_1": "[quote=\"Raja Oktovin\"]For $ a,b,c$ positive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 1$, determine the minimum and maximum value of $ 2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} 9abc$.\n\nsorry if it's a silly problem for you :([/quote]\r\n$ A\\equal{}2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} 9abc\\equal{}(2a^2 \\plus{} 2b^2 \\plus{} 2c^2)(a\\plus{}b\\plus{}c)\\plus{}9abc\\ge (a\\plus{}b\\plus{}c)^3\\equal{}1$ its equivalent\r\n$ a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc\\ge a^2(b\\plus{}c)\\plus{}b^2(a\\plus{}c)\\plus{}c^2(a\\plus{}b)$ and its true by Schur. A=1 when a=b=c=1/3 And its obvious that $ A\\equal{}(2a^2 \\plus{} 2b^2 \\plus{} 2c^2)(a\\plus{}b\\plus{}c)\\plus{}9abc\\le 2(a\\plus{}b\\plus{}c)^3$ (just open brackets) A=2 a=1,b,c=0 :wink:", "Solution_2": "halo...\r\n\r\ni don't understand your solution.. sorry.. \r\n\r\nplease explain :( \r\n\r\n\r\nRaja", "Solution_3": "Do you want to mean that it lies within $ [1,2]$??", "Solution_4": "[quote=\"Raja Oktovin\"]For $ a,b,c$ positive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 1$, determine the minimum and maximum value of $ 2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} 9abc$.\n\nsorry if it's a silly problem for you :([/quote]\r\nTry to use function $ F(a,b,c) \\equal{} 2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} \\alpha (a \\plus{} b \\plus{} c \\minus{} 1)$.\r\n\r\nSo, in the extremum $ \\frac {df}{da} \\equal{} 0$, $ \\frac {df}{db} \\equal{} 0$, $ \\frac {df}{dc} \\equal{} 0$, and $ a \\plus{} b \\plus{} c \\equal{} 1$.\r\n\r\nAfter solving this system of equations (with unknown $ a,b,c,\\alpha$), you will have that $ a \\equal{} b \\equal{} c \\equal{} 1/3$ - only one extremum-point.\r\n\r\nSo, $ F(1/3,1/3,1/3) \\equal{} 1$.\r\n\r\nNow, if one of $ a,b,c$ is 0 (for example, $ c$), then you will have $ a \\plus{} b \\equal{} 1$, and $ F(a,b,0) \\equal{} 2a^2 \\plus{} 2b^2 \\equal{} 2(a^2 \\plus{} (1 \\minus{} a)^2) \\equal{} 2(2a^2 \\minus{} 2a \\plus{} 1)$ - the minimum value of $ F(a,b,0)$ if and only if $ a \\equal{} b \\equal{} 1/2$.\r\n\r\nSo, $ F(1/2,1/2,0) \\equal{} 1$.\r\n\r\nIf both $ b \\equal{} c \\equal{} 0$, then $ a \\equal{} 1$, and $ F(1,0,0) \\equal{} 2$.\r\n\r\nSo, the minimum value is 1, and the maximum is 2.", "Solution_5": "[quote=\"Raja Oktovin\"]halo...\n\ni don't understand your solution.. sorry.. \n\nplease explain :( \n\n\nRaja[/quote] Whats you don't understand?" } { "Tag": [ "calculus" ], "Problem": "A businessman has some money in his pocket.For the first two days he spends (1/3)+100YTL of it.And for the last two days spends (1/5)+100 of available money.After 4 days there is 140YTL in his pocket.How much money did he have at start?", "Solution_1": "What is the YTL? Is that some type currency?", "Solution_2": "sorry it's turkish liras :) it isn't important. :)", "Solution_3": "[quote=\"Shahzada\"]A businessman has some money in his pocket.For the first two days he spends (1/3)+100YTL of it.And for the last two days spends (1/5)+100 of available money.After 4 days there is 140YTL in his pocket.How much money did he have at start?[/quote]\r\n[hide=\"Work Backwards\"] \nSay that $x-(\\frac{x}{5}+100)=140$. Solving, we find x to be 300. Again, say that $y-(\\frac{y}{5}+100)=300$. Thus, we have that y=500. Now, say that $a-(\\frac{a}{3}+100)=500$. Thus, a=900. Now, we have that $b-(\\frac{b}{3}+100)=900$. Thus, $b=\\boxed{1500}$, which is our answer. [/hide]", "Solution_4": "Yes it's true.This is another simple way.\r\n[hide][img]http://img128.imageshack.us/img128/8992/dfgoc4.gif[/img][/hide]", "Solution_5": "that not needs more than on simple equation \r\nbut its beneficial for mental calculus[/hide]" } { "Tag": [], "Problem": "OK. The second revolution has started!\r\n\r\nRules: \r\n\r\nThere are 10 ppl, 1 king, 1 queen, 2 knights (one known to public and one unknown), 1 renegade, and 5 citizens in each country, bringing the total to 20 ppl.\r\n\r\nKing tries to kill the queen and 3 citizens without losing any knights or kill all citizens and queen to win the game. \r\n\r\nQueen must kill King firsthand to win. \r\n\r\nRenegades try to have lowest number of points when the kingdom falls.\r\n\r\nCitizens have same objective as queen. \r\n\r\nKnights can cooperate (have same goal) as the king, or kill queen, other knight, and the king. \r\n\r\nKing starts w/ 20 pts, known knight 20 pts, and others 10 pts. \r\n\r\nAll type of people except renegade can only win if the kingdom wins.\r\n\r\nWhen queen or knight (2nd objective for knight) completes their objective before kingdom falls, the person becomes the king.\r\n\r\nKingdom falls when the king is dead; but obviously it doesn't fall if the other kingdom has already fallen. \r\n\r\nEveryone will be private except King and the known knight. \r\n\r\nDaily, king will receive 3 pts, known knight 2 pts, and 1 pt for rest. \r\n\r\nA person may attack once per day, removing same number of points from both sides and removal can only be up to 2 pts, and attack 1 person only. \r\n\r\nKing can execute someone by giving up 15 pts every 3 days. \r\n\r\nAlso, king may not be attacked unless both knights are dead. Also, anyone can give points (as much as wanted, once per day) to the king, but only king can give points to knights. \r\n\r\nI will PM who's who if all 10 ppl are signed up; and king and knight will be announced.\r\n\r\nMob Lynch: during the whole game, a Rest (those who are not known) has 1 vote to mob lynch someone. Whenever majority is met, the person will be lynched (cannot be known knight or the king). Majority does not include abstained votes. \r\n\r\nThere are 9 events (one of them happens twice, so total 10) that the player may choose in order. I have decided them, but I may change them so I am not posting yet. Every day, 1st person to post what actions to take (doesn't matter with the kingdom, same effect will go on to same kingdom regardless of the poster. The poster also may choose to not to use any of events.) i will have events posted later.\r\n\r\nCurrent signups:\r\n\r\n1.(^_^)\r\n2.packman2812\r\n3.chenhsi\r\n\r\n17 or 7 spots still available (if i can't get 20 people, i will do this by having only one kingdom).", "Solution_1": "I'll play.", "Solution_2": "i will join", "Solution_3": "I'll join. I think others after me should join with PM to make this topic not clog up. (Hehe, hypocrisy!)", "Solution_4": "Updated current signups:\r\n\r\n1.(^_^) \r\n2.packman2812 \r\n3.chenhsi \r\n4.lotsofmath\r\n5.budi713\r\n6.Brut3Forc3\r\n7.mnmath\r\n\r\nComeon people, it'll be more fun when we have 20 people!", "Solution_5": "I could join, but right now, I'm vacationing in Boston, so I might not be able to get internet access everyday. I hope everyone forgives me for that. \r\n\r\n/in", "Solution_6": "Joining. Blah Blah Blah", "Solution_7": "Updated signups:\r\n\r\n1.(^_^)\r\n2.packman2812\r\n3.chenhsi\r\n4.lotsofmath\r\n5.budi713\r\n6.Brut3Forc3\r\n7.mnmath\r\n8.142857\r\n9.alexhhmun\r\n10.BOGTRO\r\n\r\ndo you want to start or get more people?", "Solution_8": "I cant, sorry.", "Solution_9": "more please.", "Solution_10": "Looks like I missed it.\r\n\r\nIf you guys need more people, I'm here!", "Solution_11": "please wait. There are more than 20 active people in this forum. Continue waiting.", "Solution_12": "I'll join", "Solution_13": "me want to join :)\r\n\r\nUpdated signups: \r\n\r\n1.(^_^) \r\n2.packman2812 \r\n3.chenhsi \r\n4.lotsofmath \r\n5.budi713 \r\n6.Brut3Forc3 \r\n7.mnmath \r\n8.142857 \r\n9.alexhhmun \r\n10.BOGTRO\r\n11.142857 \r\n12.iLord\r\n13.Garyzx\r\n14.Crazychinesemaniac\r\n\r\nUnless 6 more people join anyone below BOGTRO will not play I guess, which includes me, so come on people...I wanna play!!!", "Solution_14": "6 more please.", "Solution_15": "Why not? I have time. I'll sign up.", "Solution_16": "i think iLord wants to sign up...", "Solution_17": "Yeah, I want to be in.", "Solution_18": "C'mon people, we need signups!!! The game has to start quickly, or many of us will have to leave!", "Solution_19": "Yeah...... I have to leave in 10 days or so and I won't be back for long......\r\n\r\nI PMed Revolution I players, only dynamo and iLord responded.\r\n\r\nMaybe we can cut down some:\r\n\r\n1 king, 1 knight, 1 queen, and 4 citizens each country.\r\n\r\nIf we don't get any more people by this Friday, I will start like this.\r\n\r\nLet's hope some people join. (especially since MOSP is over, i think).\r\n\r\n1.(^_^)\r\n2.packman2812\r\n3.chenhsi\r\n4.lotsofmath\r\n5.budi713\r\n6.Brut3Forc3\r\n7.mnmath\r\n8.142857\r\n9.alexhhmun\r\n10.BOGTRO\r\n11.142857\r\n12.iLord\r\n13.Garyzx\r\n14.Crazychinesemaniac \r\n\r\nPM people you can reach so we can get more people.\r\n\r\nAnyways, I am posting 10 events scheduled and the first person each day may choose which (or not choose at all)\r\n\r\n1. Everyone must give up 1 point. If this is not done by the end of the day, the person loses 5 points.\r\n2. A person from each kingdom must give 2 points to respective king.IFINDBTEOTD (if this is not done by the end of the day) king can execute someone without 15 points.\r\n3. Everyone except king receives 1 point. Knight receives 2 extra points.\r\n4. People may PM me a number by subtracting 1 point. The person to PM the largest number loses that much, and 2nd largest receives that many. You can only bid up to how much you have.\r\n5. People may PM me a number by subtracting 1 point. The person to PM the lowest unique positive integer receives 5 points.\r\n6. Knights of both kingdom must give up points how many each king wants to.\r\n7. King may give up 10 points to revive a dead knight (knight starting with 10 points) or give up 15 points to bribe a citizen into a knight (point become the number of points left of that citizen)\r\n8. Declare War; a Rest (unknowns) with lowest points on both sides die. ( if multiple, at most 2 will die. Choosing it will be by getting last people on the list. This may not be used for first 3 days) X 2\r\n9. Swap a Rest with highest points from each kingdom (if multiple, all will swap.)", "Solution_20": "eh I'll join.", "Solution_21": "I think we should wait a bit longer...", "Solution_22": "The problem is, that i have to go (so does budi, alexhmun, (^_^) and 1=2 at least) will leave for AMP pretty soon (or July 13 :P )\r\n\r\nSo I have no option unless I want to mod it during school, which will be quite painful.\r\n\r\nI have to hurry up with this.\r\n\r\nPlease put this on your sigs if you want:\r\n\r\n[code]Start [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212218]Revolution: The War![/url]\n :help:[/code]", "Solution_23": "[quote=\"dannyhamtx\"]The problem is, that i have to go (so does budi, alexhmun, (^_^) and 1=2 at least) will leave for AMP pretty soon (or July 13 :P )\n\nSo I have no option unless I want to mod it during school, which will be quite painful.\n\nI have to hurry up with this.\n\nPlease put this on your sigs if you want:\n\n[code]Start [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212218]Revolution: The War![/url]\n :help:[/code][/quote]\r\n\r\nYou misspelled my name... *sniffles* \r\n\r\nOh, and maybe we change that into \r\n[code]Start [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212218]Revolution 2: The War![/url]", "Solution_24": "This game should be freezed now. I hope mods will not consider this \"dead.\"\r\n\r\nIt MAY start again when we come back from AMP which is August 3rd.\r\n\r\nSee you till then and i hope more ppl still sign up anyways.", "Solution_25": "I sign up\r\n\r\n :D", "Solution_26": "1.(^_^)\r\n2.packman2812\r\n3.chenhsi\r\n4.lotsofmath\r\n5.budi713\r\n6.Brut3Forc3\r\n7.mnmath\r\n8.142857\r\n9.alexhhmun\r\n10.BOGTRO\r\n11.142857\r\n12.iLord\r\n13.Garyzx\r\n14.Crazychinesemaniac\r\n15.mathlete9999\r\n16.!!!!!\r\n\r\nWe just need 4 more people now\r\n\r\nI hope the new people join now *sigh*", "Solution_27": "Finally, I almost forgot about this game :D", "Solution_28": "Is this considered dead? just curious. Sorry for posting if it is :D .", "Solution_29": "[quote=\"dannyhamtx\"]1.(^_^)\n2.pac[color=red]k[/color]man2812\n3.chenhsi\n4.lotsofmath\n5.budi713\n6.Brut3Forc3\n7.mnmath\n8.142857\n9.alexhhmun\n10.BOGTRO\n11.142857\n12.iLord\n13.Garyzx\n14.Crazychinesemaniac\n15.mathlete9999\n16.!!!!!\n\nWe just need 4 more people now\n\nI hope the new people join now *sigh*[/quote]\r\n\r\npackman $ \\neq$ pacman :mad:\r\n\r\nAnd yeah, this is possibly dead." } { "Tag": [ "algebra", "polynomial", "inequalities", "vector", "function", "algebra unsolved" ], "Problem": "Find all polynomials $ P(x)$ of degree $ \\le n$ with nonnegative real coefficients\r\n\r\nsuch that the inequality $ P(x)P(1/x) \\le (P(1))^2$ holds for all positive real\r\n\r\nnumbers x.", "Solution_1": "By Cauchy-Schwarz, \\[ P(x)P(x^{\\minus{}1})\\equal{}\\left(\\sum x^ka_k\\right)\\left(\\sum x^{\\minus{}k}a_k\\right)\\geq\\left(\\sum a_k\\right)^2\\equal{}P(1)^2\\](valid only for positive real $ x$).\r\nTherefore, $ P(x)P(x^{\\minus{}1})\\equal{}P(1)^2$, and equality holds above. The equality case for the above inequality is when the two \"vectors\" are parallel... i.e. each \"component\" $ x^ka_k\\equal{}N\\cdot x^{\\minus{}k}a_k$, for some constant $ N$. When $ k\\equal{}0$, $ N$ drops out to be $ 1$. We conclude that $ x^k\\in\\pm1$ for all $ k$. Since $ x$ is an arbitrary positive real, $ k$ can't take on the value one - that is, \\[ \\boxed{P(x)\\text{ is a constant function}}\\]", "Solution_2": "Let $ P(x)$ be such a polynomial, and let its degree be $ k$. Since all of the coefficients of $ P(x)$ are nonnegative, $ P(x)P\\left(\\frac{1}{x}\\right)\\ge 0$ for all $ x\\ge 0$. We consider the behavior of $ P(x)P\\left(\\frac{1}{x}\\right)$ as $ x\\to\\infty$. Clearly, $ P(x)$ goes to infinity as $ x^k$. It follows, since $ P(x)P\\left(\\frac{1}{x}\\right)$ is bounded, that $ P\\left(\\frac{1}{x}\\right)$ goes to zero as fast or faster than $ x^{\\minus{}k}$, or equivalently, $ P(x)$ has a zero of multiplicity not less than $ k$ at zero. But is $ P(x)$ is a polynomial of degree $ k$, this implies that it has exactly one zero of multiplicity $ k$ at zero. The only polynomials satisfying this requirement are those of the form $ Cx^k$. It is easy to see that, if $ C\\ge 0$ and $ 0\\le k\\le n$, then $ Cx^k$ all the required conditions.", "Solution_3": "nice , thank you" } { "Tag": [ "limit", "combinatorics open", "combinatorics" ], "Problem": "Does there exist an real number $ c < \\frac{1}{2}$ and a positive integer $ m$ such that for every positive integer $ n$ and family $ \\mathcal F$ of subsets of $ \\{1, \\ldots, n\\}$, we have: If $ |A \\cap B \\cap C| \\geq m$ for all $ A, B, C \\in \\mathcal F$, then $ |F| \\leq c \\cdot 2^n$.", "Solution_1": "That seems to be a nice question! Before proceeding, how about the case of two-set-intersection, i.e. replacing the $ |A \\cap B|$ with $ A \\cap B \\cap C|$ in the condition? Is there such a $ c$ in this case? This simpler version seems to be interesting too ...", "Solution_2": "The similar problem with $ A \\cap B$ is not that difficult: There is no such $ c$. For a proof, consider an arbitrary positive integer $ m$. For every $ n \\in \\mathbb N$, define $ \\mathcal F_n \\equal{} \\{A \\subset \\{1, 2, \\ldots, n\\} : |A| \\geq \\frac{n\\plus{}m}{2}\\}$. We have $ A,B \\in \\mathcal F \\Rightarrow |A \\cap B| \\geq m$ and $ \\lim_{n \\rightarrow \\infty} \\frac{|\\mathcal F_n|}{2^n} \\equal{} \\frac{1}{2}$." } { "Tag": [], "Problem": "A wine dealer, who carries his wine in identical barrels, travels through\r\n\r\nthree cities. Each city charges a different tax which is collected upon \r\n\r\nentrance into the city. The first city takes $\\frac{1}{3}$ of his wine \r\n\r\nas a tax, the second city takes $\\frac{1}{4}$ of his wine, and\r\n\r\n the third city takes $\\frac{1}{5}$ of his wine. \r\n\r\nAfter paying all these taxes he sold all of his remaining 72 barrels \r\n\r\nof wine. How many barrels of wine did he have originally?", "Solution_1": "[hide]$72\\cdot\\frac{5}{4}\\cdot\\frac{4}{3}\\cdot\\frac{3}{2}=\\boxed{180}$[/hide]", "Solution_2": "[hide=\"Solution\"]If he had $x$ barrels, then:\n\n1. After the first city he was left with ${2\\over 3}x$ barrels;\n\n2. After the second city he was left with ${3\\over 4}\\cdot{2\\over 3}x={1\\over 2}x$ barrels;\n\n3. After the third city he was left with ${4\\over 5}\\cdot{1\\over 2}x={2\\over 5}x$ barrels.\n\n${2\\over 5}x=72\\iff x=180$ barrels.[/hide]", "Solution_3": "[hide]He ended up with 72 barrels after 1/5 of them were taken, so $72=4/5*x=90.$ $90=3/4*y=120$, and $120=2/3*z=180$\n[/hide]", "Solution_4": "[/hide]4/5*3/4*2/3=2/5. This represents the fraction of wine that was taken from him. 72 is 2/5ths of [b]180[/b][hide][/hide]", "Solution_5": "[quote=\"moldlee\"]A wine dealer, who carries his wine in identical barrels, travels through\n\nthree cities. Each city charges a different tax which is collected upon \n\nentrance into the city. The first city takes $\\frac{1}{3}$ of his wine \n\nas a tax, the second city takes $\\frac{1}{4}$ of his wine, and\n\n the third city takes $\\frac{1}{5}$ of his wine. \n\nAfter paying all these taxes he sold all of his remaining 72 barrels \n\nof wine. How many barrels of wine did he have originally?[/quote]\r\n$\\frac{4}{5}\\cdot{3}{4}\\cdot{2}{3}\\cdot 72=\\frac{2}{5}\\cdot 72=\\boxed{180}$" } { "Tag": [ "limit", "function", "calculus", "derivative", "geometry", "3D geometry", "calculus computations" ], "Problem": "My book states the following:\r\n\r\nIf the alternating series \r\n\r\n$\\sum_{n=1}^{\\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + ... (b_n > 0)$\r\nsatisfies\r\n(i) $b_{n+1} \\leq b_n$ for all $n$\r\n\r\n(ii) $\\lim_{n \\rightarrow \\infty} b_n = 0$\r\n\r\nthen the series is convergent.\r\n\r\n\r\n\r\nThat much I understand....I read the proof and it makes sense to me.\r\n\r\n\r\n\r\nOne of the examples says to test the following series for convergence or divergence.\r\n$\\sum_{n=1}^{\\infty} (-1)^{n+1} \\frac{n^2}{n^3 + 1}$\r\n\r\nIt says to find out if $b_n = \\frac{n^2}{n^3 + 1}$ is decreasing we take the function $f(x) = \\frac{x^2}{x^3 + 1}$\r\nAnd since the derivative is $f'(x) = \\frac{x(2-x^3)}{{x^3 +1}^2}$ is decreasing on the interval (cube root 2, $\\infty$) which means that $b_{n+1} < b_n$ when $n \\geq 2$\r\n\r\nIt goes on to show the limit as n approaches infinity is zero, so the given series converges....\r\n\r\n\r\n\r\nBut in the definition of the test, it says it should be $b_{n+1} < b_n$ for all $n$, yet this example considers only $n$ greater or equal to 2. Doesn't it really matter if $b_n$ is ultimately decreasing, or is it for all $n$?\r\n\r\nIt seems that this example conflicts with the definition....", "Solution_1": "If you remove a finite number of terms from a series, the nature of the series remains the same. If the series you get from removing the first $2$ terms converges then the initial series converges as well.", "Solution_2": "Okay, thanks\r\n\r\nThat's what I thought because a definite number of terms would converge, so it shouldn't matter to add that to a convergent series..." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Given the function $ f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that $ f(x^3 \\plus{} y^3) \\equal{} (x\\plus{}y)(f(x)^2 \\minus{} f(x)f(y) \\plus{} f(y)^2)$, $ \\forall x,y \\in \\mathbb{R}$. Prove that $ f(2010x)\\equal{}2010f(x)$.", "Solution_1": "[quote=\"rajelaz\"]There's function $ f(x) \\mathbb{R} \\rightarrow \\mathbb{R}$ such that $ f(x^3 \\plus{} y^3) \\equal{} (x \\plus{} y)(f(x)^2 \\minus{} f(x)f(y) \\plus{} f(y)^2)$ , $ \\forall x,y \\in \\mathbb{R}$ .. prove that $ f(2010x) \\equal{} 2010f(x)$[/quote]\r\nLet $ P(x,y)$ be the assertion $ f(x^3\\plus{}y^3)\\equal{}(x\\plus{}y)(f(x)^2\\minus{}f(x)f(y)\\plus{}f(y)^2)$\r\nLet $ \\mathbb S\\equal{}\\{x\\in\\mathbb R$ such that $ f(xy)\\equal{}xf(y)$ $ \\forall y\\in\\mathbb R\\}$\r\n\r\n$ 1\\in\\mathbb S$\r\n$ P(0,0)$ $ \\implies$ $ f(0)\\equal{}0$ and $ 0\\in\\mathbb S$\r\n$ P(\\sqrt[3]x,0)$ $ \\implies$ $ f(x)\\equal{}\\sqrt[3]xf(\\sqrt[3]x)^2$ and so $ f(x)$ and $ x$ have same signs.\r\nIf $ p,q\\in\\mathbb S$, then $ P(p\\sqrt[3]x,q\\sqrt[3]x)$ $ \\implies$ $ p^3\\plus{}q^3\\in\\mathbb S$ (1)\r\n\r\nIf $ p\\in\\mathbb S$, then :\r\n$ P(x,0)$ $ \\implies$ $ f(x^3)\\equal{}xf(x)^2$\r\n$ P(\\sqrt[3]px,0)$ $ \\implies$ $ pf(x^3)\\equal{}f(px^3)\\equal{}\\sqrt[3]pxf(\\sqrt[3]px)^2$\r\nComparing these two lines, we get, for $ x\\neq 0$, and using the fact that $ f(x)$ and $ x$ have same signs : $ f(\\sqrt[3]px)\\equal{}\\sqrt[3]pf(x)$ and so $ \\sqrt[3]p\\in\\mathbb S$ (2)\r\n\r\nNow :\r\n$ p\\in\\mathbb S$ $ \\implies$ $ \\sqrt[3]p\\in\\mathbb S$ (using (2))\r\n$ \\sqrt[3]p\\in\\mathbb S$ and $ 1\\in\\mathbb S$ $ \\implies$ $ p\\plus{}1\\in\\mathbb S$ (using (1))\r\n\r\nSo $ p\\in\\mathbb S$ $ \\implies$ $ p\\plus{}1\\in\\mathbb S$, and, since $ 1\\in\\mathbb S$, $ \\mathbb N\\subseteq\\mathbb S$ and so $ 2010\\in\\mathbb S$\r\nQ.E.D." } { "Tag": [ "LaTeX" ], "Problem": "\"Most fractions look better in \\displaystyle (remember, you don't need the \\displaystyle declaration if you are in \\[...\\ or \\$\\$...\\$\\$ mode.)\"\r\nit doesn't work! \\[(\\frac{2-x}{2})^{2}=2x\\] but\r\n$ \\left(\\frac{2-x}{2}\\right)^{2}=2x$\r\nwhy??", "Solution_1": "Parentheses will never get bigger or smaller on their own (as far as I know) -- you always have to use \\left and \\right if you want them bigger.", "Solution_2": "To answer the question: on the forum we have displaystyle on automatically." } { "Tag": [ "geometry", "3D geometry", "analytic geometry", "greatest common divisor" ], "Problem": "A 150X324X375 rectangular solid is made by gluing together 1X1X1 cubes. An internal diagonal of this solid passes through the interiors of how many of the 1X1X1 cubes?", "Solution_1": "[hide=\"Hint\"]\nPIE. :wink: [/hide]", "Solution_2": "This is just like the chessboard and a line problem, just in three dimensions, so to get the answer wouldn't you just add the dimentions of the figure?", "Solution_3": "[quote=\"Kamil Witek\"]This is just like the chessboard and a line problem, just in three dimensions, so to get the answer wouldn't you just add the dimentions of the figure?[/quote]\r\n[hide=\"I am pretty sure the answer is:\"]\n\\[150+324+375-\\left[ (150,324)+(324,375)+(375,150) \\right]+(150,324,375) = \\boxed{768}\\] where $(a,b)$ is the greatest common factor of $a$ and $b$. We count the number of times the line must pass through a lattice plane (integer constant), then subtract the number of times that it goes through an edge of two cubes (i.e. two of three coordinates are integers); but we overcount, so by the Principle of Inclusion-Exclusion we have to add back the number of times the diagonal goes through a vertex of a cube. [/hide]" } { "Tag": [ "limit", "induction", "real analysis", "real analysis unsolved" ], "Problem": "Let $I=(0,1)$, and let $C$ be a countable subset of $I$. Also let $(a_{n})_{n\\ge 1}$ be a sequence of positive reals with $\\sum a_{n}\\le 1$. \r\n\r\nProve that we can place disjoint compact intervals $I_{n}, n\\ge 1$ inside $I$ such that the length of $I_{n}$ is $a_{n}$ for all $n\\ge 1$ and every point in $C$ belongs to the interior of some $I_{n}$.", "Solution_1": "We consider $(x_{n})$ a sequence s.t. every number in $C$ appears exactly once in the sequence.\r\nFirstly, we consider the case $\\sum_{n\\geq 1}a_{n}=a\\in(0,1)$\r\n$I_{n}: =[c_{n},d_{n}], \\forall n\\geq 1$\r\nLet $(b_{n})_{n\\geq 1}$ be a sequence of positive reals s.t. $\\sum_{n\\geq 1}b_{n}=1-a$\r\nWe take $I_{1}=[b_{1},b_{1}+a_{1}]$ and $J_{n}=\\bigcup_{k=1}^{n}I_{k}$. \r\n\r\nLet $M=\\{c_{n}\\mid n\\in \\mathbb{N}\\}$\r\nWe assume we have fixed the intervals $I_{1},...,I_{k-1}$\r\nLet $t$ be the smallest integer for which $x_{t}\\notin J_{k-1}$\r\n\r\n1)If $x_{t}< \\max M$ and $a_{k}|b|\nBy (1), we can make the substitution: a=2c, b=2d\nWhere cd=-24\nd cannot equal to 2 (or else it violates (4))\n|c|,|d| cannot equal to 4 (or else it violates (6))\nRemember that c is negative.\nThe only integer pairs left (c,d) are:\n(-1,24),(-2,12),(-3,8),(-4,6),(-8,3),(-24,1)\nWe can now remove those that are factors of each other:\n(-3,8),(-4,6),(-8,3)\nAlso remember from (2). The only (c,d) pair that satisfies all criteria is (-8,3)\nwhich leads us that the two numbers are -16 and 6.\n\nQED[/hide]" } { "Tag": [ "parameterization" ], "Problem": "prove that the equation $ x^2\\plus{}y^2\\plus{}1\\equal{}z^2$ has infinitely many solution", "Solution_1": "[quote=\"aadil\"]prove that the equation $ x^2 \\plus{} y^2 \\plus{} 1 \\equal{} z^2$ has infinitely many solution[/quote]\r\n\r\n$ (2k^2)^2\\plus{}(2k)^2\\plus{}1\\equal{}(2k^2\\plus{}1)^2$\r\n\r\nSo $ (x,y,z)\\equal{}(2k^2,2k,2k^2\\plus{}1)$ is solution for all $ k$, done.", "Solution_2": "[hide=\"a much less intelligent approach...\"]Since $ y^2 \\plus{} 1 \\equal{} z^2$ has infinitely many solutions, \n\n\n$ (0,k,\\sqrt {k^2 \\plus{} 1})$ works for $ x^2 \\plus{} y^2 \\plus{} 1 \\equal{} z^2$.[/hide]\r\n\r\nEDIT: Sorry, too careless :blush:", "Solution_3": "I suppose he meant infinitely many solutions in integers, which invalidates chrischris's proof but stephencheng's proof remains correct.\r\n\r\n(Otherwise, you can just set $ z\\equal{}\\sqrt{x^2\\plus{}y^2\\plus{}1}$ and then [i]any[/i] ordered pair $ (x,y)$ works.)", "Solution_4": "[quote=\"Yongyi781\"]I suppose he meant infinitely many solutions in integers, which invalidates chrischris's proof but stephencheng's proof remains correct.\n\n(Otherwise, you can just set $ z \\equal{} \\sqrt {x^2 \\plus{} y^2 \\plus{} 1}$ and then [i]any[/i] ordered pair $ (x,y)$ works.)[/quote]\r\n\r\n0 is a integer. You meant naturals?", "Solution_5": "$ (0,0,1)$ is the only solution that contains 0's. :maybe:", "Solution_6": "[hide=\"Discussion\"] One way to solve this problem is to write it as $ x^2 \\plus{} 1 \\equal{} z^2 \\minus{} y^2 \\equal{} (z \\minus{} y)(z \\plus{} y)$. Then it's reasonable to set $ z \\minus{} y \\equal{} 1, z \\plus{} y \\equal{} x^2 \\plus{} 1$ for $ x \\equal{} 2k$ even, which gives stephencheng's parameterization.\n\nYou can use similar techniques to deduce the parameterization of the [url=http://en.wikipedia.org/wiki/Pythagorean_quadruple]Pythagorean quadruples[/url], and then it remains merely to specialize. [/hide]" } { "Tag": [ "function", "algebra", "polynomial", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $a,b\\in\\mathbb{R}, a0$ there exists a polynomial $P$ such that\r\n\\[\\int_{a}^{b}|f(x)-P(x)|<\\epsilon. \\]", "Solution_1": "does it work with $f(x)=\\frac{1}{x}$ on $]0;1[$ ?", "Solution_2": "alekk has a point. You'd better add a condition on $f$ beyond merely being continuous on the open interval. You need some kind of control of behavior at the endpoints.\r\n\r\nNow, it turns out that a sufficient condition would be for $f$ to be absolutely integrable on $(a,b)$ - that is, that \r\n\r\n$\\int_{a}^{b}|f(x)|\\,dx<\\infty.$\r\n\r\nBut once you put in this condition, you'll find that you don't really need for $f$ to be continuous - the continuity doesn't even help." } { "Tag": [ "arithmetic sequence", "AMC" ], "Problem": "When the mean, median, and mode of the list\r\n10; 2; 5; 2; 4; 2; x\r\nare arranged in increasing order, they form a non-constant arithmetic progres-\r\nsion. What is the sum of all possible real value of x ?\r\n\r\n(A) 3 (B) 6 (C) 9 (D) 17 (E) 20\r\n\r\nI was really confused by non-constant arithmetic progression :!:", "Solution_1": "It was actually #23\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=438929#438929" } { "Tag": [ "geometry", "circumcircle", "search", "perpendicular bisector" ], "Problem": "I need help with these 5 geometry problems:\r\n\r\n1. A semicircle \u0393 is drawn on one side of a straight line l. C and D are points on \u0393. The tangents to \u0393 at C and D meet l at B and A respectively, with the center of the semicircle between them. Let E be the point of intersection of AC and BD, and F the point on l such that EF is perpendicular to l. Prove that EF bisects angle CFD.\r\n\r\n2. (1995 IMO) Let A, B, C, and D be distinct points on a line, in that order. The circles with diameters AC and BD intersect at X and Y. O is an arbitrary point on the line XY but not on AD. CO intersects the circle with diameter AC again at M, and BO intersects the other circle again at N. Prove that the lines AM, DN, and XY are concurrent.\r\n\r\n3. (2007 IMO) In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area.\r\n\r\n4. In the plane we are given two circles intersecting at X and Y. Prove that there exist four points P, Q, R, S with the following property: For every circle touching the two given circles at A and B, and meeting the lines XY at C and D, each of the lines AC, AD, BC, BD passes through one of these points.\r\n\r\n5. (2000 IMO) Two circles G1 and G2 intersect at M and N. Let AB be the line tangent to these circles at A and B, respectively, such that M lies closer to AB than N. Let CD be the line parallel to AB and passing through M, with C on G1 and D on G2. Lines AC and BD meet at E; lines AN and CD meet at P; lines BN and CD meet at Q. Show that EP = EQ.\r\n\r\nThanks!\r\n\r\n :) :) :)", "Solution_1": "How about searching some of these next time? [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=365179#365179]2[/url],[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=894655#894655]3[/url],[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1021401687&t=301]4[/url],[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354110#354110]5[/url]", "Solution_2": "Still, can anyone solve 1, its driving me crazy :help: :bomb: :wallbash:", "Solution_3": "[quote=\"SHSMCMXCIV\"]1. A semicircle \u0393 is drawn on one side of a straight line l. C and D are points on \u0393. The tangents to \u0393 at C and D meet l at B and A respectively, with the center of the semicircle between them. Let E be the point of intersection of AC and BD, and F the point on l such that EF is perpendicular to l. Prove that EF bisects angle CFD.\n[/quote]\r\nLook here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=352892#p352892" } { "Tag": [ "algebra", "polynomial", "floor function", "ceiling function", "logarithms", "search", "number theory unsolved" ], "Problem": "Does there exist a positive integer $ n$ such that $ n!$ starts with $ 2005$?", "Solution_1": "Yes. the sequence {n!}, n=1,2,3,... is what one calls a left-normal sequence, i.e. it can start with any string of digits in any base.\r\n\r\nOther left normal sequences: \r\n{p_n}, the nth prime\r\n{F_n}, the nth Fibonacci number\r\n{P(n)}, where P is a nonconstant polynomial in Z[x]\r\n\r\nSuhaimi", "Solution_2": "I have never heard anything like that before. :blush: \r\n\r\nCan you prove it? And do you have any reference about left-normal sequence?", "Solution_3": "The first four decimal digits of $ n!$ can be written as\r\n\\[ d \\equal{} \\left\\lfloor \\frac {n!}{10^{\\left\\lceil\\log_{10} n!\\right\\rceil \\minus{} 4}}\\right\\rfloor\r\n\\]\r\nReplace $ n!$ by Stirling's approximation (which agrees to more than four digits for sufficiently large $ n$), and then use the fact that $ \\lbrace an\\minus{}\\lfloor an \\rfloor \\rbrace_{n\\in\\mathbb{N}}$ is dense in $ (0,1)$ for any irrational $ a$. \r\n\r\nThe first few values are $ 4968!\\approx 2.0059\\cdot 10^{16207}$, $ 6230!\\approx 2.0056\\cdot 10^{20936}$, $ 6527!\\approx 2.0050\\cdot 10^{22066}$. There are infinitely many such values, and the argument works not just for $ 2005$ but any integer.", "Solution_4": "I get this problem from a competition in Singapore for secondary school student (about 15 years old). I think there must be a simpler solution, correct?", "Solution_5": "Have a look here for a more general statement\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=237215850&t=23417\r\n\r\nPierre." } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all positive intergers n, satisfying the following condition:\r\nfor all pairs of the positive intergers p, q, if p+q is a square number , then p^n + q^n is also a square.\r\n\r\nI suppose the only solution to the problem is n=1, but I haven't got the proof, I'll appreciate that if anyone could help me.", "Solution_1": "$ 1 \\plus{} (k^2 \\minus{} 1)$ is a square for all $ k$, but $ 1^n \\plus{} (k^2 \\minus{} 1)^n$ can't be a square for sufficiently large $ k$; it's the standard bounding trick. Alternately, the Mason-Stothers theorem might work, since if the above is a square for all $ k$ it must be the square of a polynomial.", "Solution_2": "I've tried that idea, but I still can't get the exact proof. Could you explain it more clearly?", "Solution_3": "Okay, the bounding trick is a little annoying to work out for all $ n$, so I'll just use the Mason-Stothers theorem. It's a well-known result (posted on the board many times) that if a polynomial such as $ P(k) \\equal{} 1 \\plus{} (k^2 \\minus{} 1)^n$ is square for all positive integers $ k$ then it must in fact be the square of a polynomial $ Q(k)$, hence\r\n\r\n$ 1 \\plus{} (k^2 \\minus{} 1)^n \\equal{} Q(k)^2$\r\n\r\nas a polynomial identity. By the [url=http://mathworld.wolfram.com/MasonsTheorem.html]Mason-Stothers theorem[/url], $ (k^2 \\minus{} 1)^n Q(k)^2$ must have at least $ 2n \\plus{} 1$ distinct roots. But since $ Q(k)$ has degree $ n$, this polynomial can have at most $ n \\plus{} 2$ distinct roots. It follows that $ n \\le 1$." } { "Tag": [ "function", "integration", "calculus", "derivative", "calculus computations" ], "Problem": "Can anyone tell me how to solve a ordinary differential equation?\r\n\r\nI know this is kind of easy, but it's rather common in calculus.\r\n\r\n :lol: :blush:", "Solution_1": "You provided no examples. There are assorted techniques. You must [b]classify[/b] your problem.\r\n\r\nFirst classifying question: What is the order of the equation? First order? Or second or higher order? (Note that there's some room for creativity here: $y''+k(y')^{2}=g$ is ostensibly a second order equation but we can view it as a first order equation for the function $y'.$)\r\n\r\nIf it is first order: now ask whether it is linear. That is, can it be written as $y'+p(x)y=q(x)?$ If so, then put it into that standard form and use the multiplying factor technique. (Multiply both sides by $\\mu(x)=e^{\\int p(x)\\,dx}$ and then recognize the left hand side as the derivative of a product.)\r\n\r\nIf it is not linear, then ask if it is separable. That is, can it be factored as $y'=f(x)g(y).$ If so, then employ the technique of separation of variables.\r\n\r\nEquations that are both linear and separable pose no problem; both methods work (or at least they \"work\" if we can do the integrals involved).\r\n\r\nMany first order differential equations are neither linear nor separable. There are a number of other possible narrower classifications, each bringing its own specialized technique. We will not be able to usefully classify all of them.\r\n\r\nNow, what if the equation is second or higher order? We start by asking if it's linear - hoping the answer is \"yes\" because there are few available techniques for nonlinear equations. If linear we ask if it is constant coefficient, and we ask whether it is homogenous or nonhomogeneous. Our answers to each of these classifying questions bring various techniques into play." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "geometric transformation", "reflection", "AMC" ], "Problem": "Can someone restate the problem for the tetrahedron(i dont have my booklet) because i got 18, and i just want to know how i got the extra 3\r\n\r\nheres what i have\r\n4 faces same color(like BBBB) 3 choices\r\n3 same(like BBBW), 1 different 6 choices\r\n2 same 2 same (so like BBWW) 3 choices\r\n2 same 2 different (so like BBWR)-is this suppose to be 3 choices, because i get BBRW RRWB WWBR, but cant you also have BBWR or does it say the tetrahedrons can be reflections of eachother?", "Solution_1": "Yeah for 3 colors, there are 3 possibilities.\r\n\r\nThink of it this way. If you have BBWR. Lay the tetrahedron so that the edge between the two B faces is on the ground horizontally. Then the face on the left top is W and the right top is R. Buti f you spin it 180 degrees about the y axis, R is on the left and W is on the left, which is BBRW, so BBWR=BBRW", "Solution_2": "dang.. if i ddint think about it i wouldve guessed it right.. i guessed A first because i have 1 other A up to that point in the test.." } { "Tag": [ "IMO Shortlist", "inequalities unsolved", "inequalities" ], "Problem": "for real numbers $\\(x_1\\,\\(x_2\\,...\\(x_n\\ $prove that $\\frac{x_1}{1+x_1^2}\\ + \\frac{x_2}{1+x_1^2+x_2^2}+...+\\frac{x_n}{1+x_1^2+x_2^2+...+x_n^2}\\ < \\sqrt{n}\\ $", "Solution_1": "It's IMO Shortlist 2001 algebra problem 3. See http://www.mathlinks.ro/Forum/viewtopic.php?t=17449 , http://www.kalva.demon.co.uk/short/soln/sh01a3.html , or http://www.mathlinks.ro/Forum/viewtopic.php?t=15624 .\r\n\r\n darij" } { "Tag": [ "probability" ], "Problem": "On the number line, point $ A$ is located at 0, point $ B$ is located at 4, and point $ C$ is located at 6.\n[asy]defaultpen(1);\ndraw((0,0)--(6,0));\nfor (int i=0;i<7;++i){\n draw((i,-.1)--(i,.1));\n label(string(i),(i,-.1),(0,-1));\n}\nlabel(\"\\(A\\)\",(0,0),(0,1));\nlabel(\"\\(B\\)\",(4,0),(0,1));\nlabel(\"\\(C\\)\",(6,0),(0,1));[/asy]\nA dart randomly lands somewhere on the number line between $ A$ and $ C$. What is the probability that it lands closer to $ B$ than it does to $ A$ and $ C$?", "Solution_1": "The distance between the midpoint of A and B, and the midpoint of B and C yields where the dart would be closer to B than the other two.\r\n\r\nThis is from 2 to 5, or 3 units. There is a total of 6 units where the dart could land, so the probability is $ \\frac{3}{6} \\equal{} \\frac{1}{2}$" } { "Tag": [ "analytic geometry", "function", "Asymptote" ], "Problem": "Say I want to label the coordinates of a point, but it's coordinates are irrational. Is there a way for me to make the coordinates like this?\r\n[code]\ndraw((\\sqrt{2},\\sqrt{3})--(21-3\\sqrt{5},\\sqrt{67}));\n[/code]\r\nOr do I have to just round the value using a calculator?", "Solution_1": "I think sqrt function is built in Asymptote.\r\n\r\n[asy]unitsize(1cm); draw(D(sqrt(2),sqrt(3))--D(23-3*sqrt(5),sqrt(67)));[/asy]", "Solution_2": "Yes, Asymptote is a full programming language with built-in functions like sqrt(), exp(), log(), sin(), etc. Just write\r\n\r\ndraw((sqrt(2),sqrt(3))--(21-3*sqrt(5),sqrt(67))); \r\n\r\nand, voila:\r\n[asy]draw((sqrt(2),sqrt(3))--(21-3*sqrt(5),sqrt(67)));[/asy]\r\n\r\n@Valentin: It is D((2,3)), not D(2,3) (first parentheses for the function, second parentheses for the pair :). I almost thought you discovered a bug in cse5 when I saw your horizontal line)", "Solution_3": "Eh ok thanks guys.", "Solution_4": "I suspect you do not know the best news yet: you do not need to compute the coordinates of intersection points yourself. The intersectionpoint() (or IP() in cse5) command does the job for you. I'm writing this because I strongly suspect that your crazy coordinates were obtained by intersecting a circle and a line or something like that :)", "Solution_5": "No, they were actually made up, but thanks anyways.\r\nAnyways, do you just do something like:\r\n[code]\nIP(((0,0)--(1,1)),((1,0)--(0,1)));\n[/code]\r\nOr what?", "Solution_6": "Yes, though you do not need that many parentheses: IP((0,0)--(1,1),(1,0)--(0,1)); is good enough. So, here is an example with a circle and a line:\r\n[code]\npath P=D(CR((0,0),3.6)), Q=D((4,0)--(0,5));\nD(\"A\",IP(P,Q),W); D(\"B\",IP(P,Q,1));\n[/code]\r\ngives \r\n[asy]path P=D(CR((0,0),3.6)), Q=D((4,0)--(0,5));\nD(\"A\",IP(P,Q),W); D(\"B\",IP(P,Q,1));[/asy]", "Solution_7": "Theoretically you can just click on the image and get the code :D", "Solution_8": "[quote=\"fedja\"][code]\npath P=D(CR((0,0),3.6)), Q=D((4,0)--(0,5));\nD(\"A\",IP(P,Q),W); D(\"B\",IP(P,Q,1));\n[/code][/quote]\r\n\r\nHow is it determined which intersectionpoint is labeled \"A\"?", "Solution_9": "[quote=\"undefined117\"]How is it determined which intersectionpoint is labeled \"A\"?[/quote]\r\nNotice that there is an extra argument in the second [b]IP()[/b] command. That third argument represents the index of the point that you want. For example: If two paths intersect $ 3$ times, it can be $ 0$, $ 1$, or $ 2$. Also, since $ 0$ is the default, it is not necessary in the first [b]IP()[/b] command." } { "Tag": [ "combinatorics open", "combinatorics" ], "Problem": "Hi,\r\n\r\nLet P(n) be the number of ways to express a positive integer n in the sum of some positive integers (including itself).\r\nThen \r\nP(1)=1 (1)\r\nP(2)=2 (1+1,2)\r\nP(3)=3 (1+1+1, 1+2, 3)\r\nP(4)=5 (1+1+1+1, 1+1+2, 1+3, 2+2, 4)\r\nP(5)=7 (1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3, 5)\r\n.......\r\n\r\nP(n)=?\r\n\r\nIs it possible to generalize a formula for P(n) or the relation between P(n+1) and P(n)", "Solution_1": "http://mathworld.wolfram.com/PartitionFunctionP.html" } { "Tag": [ "search", "geometry", "incenter", "circumcircle", "geometric transformation", "reflection", "homothety" ], "Problem": "Let AB be a diameter of circle center at O. Pick an arbitrary point P on segment AO and suppose the circle center at P with radius AP intersect AB again at C. Now, let D be one of the two points on the arc of bigger circle(center at O) that DC=DB. Let Q be the center of the circle tangent to DC and the two other circles(i.e. those center at O and P). Prove QC is perpendicular to AB.", "Solution_1": "[quote=\"greentreeroad\"]Let AB be a diameter of circle center at O. Pick an arbitrary point P on segment AO and suppose the circle center at P with radius AP intersect AB again at C. Now, let D be one of the two points on the arc of bigger circle(center at O) that DC=DB. Let Q be the center of the circle tangent to DC and the two other circles(i.e. those center at O and P). Prove QC is perpendicular to AB.[/quote]\r\n\r\nPosted before, but I am lazy to search for it. I just remember that the proofs were complicated (including mine).\r\n\r\nCD cuts the circle (P) with diameter AC again at E. F is the foot of perpendicular from E to AC. Take a circle (Q) centered on the perpendicular to AC at C, tangent to the ray (CD at U and (externally) tangent to (P); there is only one. Parallel to AC through U cuts CQ at V. The right $ \\triangle CUQ \\sim \\triangle AEC$ with their U-, E-altitudes UV, EF are similar with coefficient $ k \\equal{} \\frac {CQ}{AC} \\equal{} \\frac {QU}{CE} \\equal{} \\frac {CV}{AF}.$ Since (P), (Q) with radii PC, QU are externally tangent,\r\n\r\n$ (PC \\plus{} QU)^2 \\equal{} PC^2 \\plus{} CQ^2,\\ AC \\cdot QU \\plus{} QU^2 \\equal{} CQ^2.$\r\n\r\nUsing $ CE^2 \\equal{} AC \\cdot FC$ and substituting for $ CQ^2 \\equal{} QU^2 \\cdot \\frac {AC^2}{CE^2} \\equal{} QU^2 \\cdot \\frac {AC}{FC}$ and calculating radius QU of (Q),\r\n\r\n$ AC \\cdot QU \\plus{} QU^2 \\equal{} QU^2 \\cdot \\frac {AC}{FC},\\ QU \\equal{} \\frac {AC \\cdot FC}{AF} \\equal{} \\frac {CE^2}{AF}.$\r\n\r\nIt follows that CV = CE. Let the perpendicular lines UV, EF intersect at Q' and let (Q') be a circle with center Q' and radius Q'V, tangent to CQ at V. Since $ \\frac {FQ'}{AF} \\equal{} \\frac {CV}{AF} \\equal{} \\frac {CQ}{AC} \\equal{} k,$ the points Q', Q, A are collinear. From similar right $ \\triangle EFC \\sim \\triangle CUQ,$\r\n\r\n$ q \\equal{} \\frac {Q'V}{QU} \\equal{} \\frac {FC}{QU} \\equal{} \\frac {CE}{CQ} \\equal{} \\frac {CV}{CQ} \\equal{} \\frac {FQ'}{CQ} \\equal{} \\frac {AQ'}{AQ},$\r\n\r\nso that A is the external similarity center of the circles (Q'), (Q). Inversion in a circle (A) with center A and radius AC takes the circle (P) with diameter AC into the common tangent CQ of (A), (P). The inversion center A is the external similarity center of (Q'), (Q) and (Q') is tangent at V to the inversion image CQ of (P). Consequently, this inversion takes (Q) into (Q'). Let $ W \\in (Q')$ be diametrically opposite to V and let tangent to (Q') at W, parallel to CQ, cut the line ECD at K. As EFQ' is midparallel of CQ, KW, it follows that E is the midpoint of CK. As $ AE \\perp CK,$ it follows that $ K \\in (A).$ Let WK cut the circle (A) again at L. The line KL is inversion image of a circle (O) passing through the inversion center A, centered on the ray (AC, and internally tangent to the circle (Q); there is only one. Let B be the other intersection of (O) with the ray (AC. Since AC = AK = AL, C is incenter of the isosceles $ \\triangle BKL$ with circumcircle (O), the line KCD bisects the angle $ \\angle LKB$ and DL = DB = DC. QED.", "Solution_2": "Here is the link which yetti was talking about:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=536144173&t=18266", "Solution_3": "I did not remember this link, I was talking about a different one: [url]http://www.mathlinks.ro/viewtopic.php?t=38681[/url].\r\n\r\nBased on the above solution, i.e., on C being incenter of the isosceles $ \\triangle BKL,$ we can generalize:\r\n\r\nLet $ \\triangle ABC$ be scalene (non-isosceles) with incenter I. The angle bisectors AI, BI, CI cut the triangle circumcircle (O) again at X, Y, Z. (P) is a circle with diameter IX. (Q) is incircle of the curvilinear triangle BIZ, tangent to the rays IB, IZ and internally tangent to (O). (R) is incircle of the curviliear triangle CIY, tangent to the rays IC, IY and internally to (O). Prove that the circles (Q), (R) are both externally tangent to (P).", "Solution_4": "Here is a very simple solution of the generalization, which can be applied without change to the special Sangaku case of an isosceles $ \\triangle ABC$\r\n(see [url]http://www.cut-the-knot.org/Curriculum/Geometry/CirclesAndRegularTriangle.shtml[/url]).\r\n\r\nLet $ (O)$ be the circumcircle and $ I$ the incenter of a $ \\triangle ABC.$ Perpendicular to $ OI$ at $ I$ cuts $ (O)$ at $ M, N.$ The circumcircle $ (O)$ cuts a circle $ \\mathcal J$ with center $ I$ and radius $ IM \\equal{} IN$ in diametrically opposite points. Consequently, inversion in $ \\mathcal J$ with negative power $ \\minus{} IM^2$ (ordinary inversion followed by reflection in the inversion center) takes the circumcircle $ (O)$ to itself and $ X$ to $ A$. The circle $ (P)$ with diameter $ IX$ passing through the inversion center goes to a line perpendicular to $ IX$ at $ A,$ or the external bisector of the angle $ \\angle CAB.$ Lines $ BIY, CIZ$ passing through the inversion center also go into themselves. Let $ I_b, I_c$ be the B-, C-excenters of the $ \\triangle ABC.$ Inversion image $ (Q')$ of the circle $ (Q)$ is tangent to $ II_b \\equiv BI,$ to $ II_c \\equiv IZ,$ and externally tangent to the 9-point circle $ (O)$ of the $ \\triangle II_bI_c.$ Therefore, $ (Q')$ is its excircle in the angle $ \\angle I_bI_cI,$ also tangent to the external bisector $ I_cAI_b$ of the angle $ \\angle CAB,$ the inversion image of the circle $ (P).$ Similarly, inversion image $ (R')$ of $ (R)$ is excircle of the $ \\triangle II_bI_c$ in the angle $ \\angle II_bI_c,$ also tangent to $ I_bI_c.$ As a result, the original circles $ (Q), (R)$ are tangent to the original circle $ (P)$ with diameter $ IX,$ obviously externally.", "Solution_5": "Great generalization yetti! this is my solution.\r\nLet $ M,N$ be midpoint of $ IB,IC$ respectively. Since $ X$ is circumcenter of triangle $ IBC$, $ \\angle IMX \\equal{} \\angle INX \\equal{} 90$ meaning that $ (P)$ pass $ M,N$.So $ (P)$ is circumcircle of triangle $ IMN$. we know that $ (Q)$ tangent to $ IB,IZ$ which are side of triangle $ IMN$. Hence, to prove $ (Q)$ tangent to circumcircle of triangle $ IMN$, we need to show $ (Q)$ is $ I$-exmixtilinear circle of triangle $ IMN$.( I don' know what is exatly name of this). Let $ (Q)$ touch $ IB,IC$ at $ P_1,P_2$. From well know fact about mixtilinear circle, It is suffice to show $ P_1P_2$ pass $ I_N$ where $ I_N$ is $ N$-excenter of triangle $ IMN$ (In fact, what I mean well known fact about mixtilinear circle is that if A-mixtilinear circle touch $ AB,AC$ at $ P_1,P_2$ respectively, then $ P_1P_2$ pass incenter of $ ABC$. Extroversion must be incenter->excenter) . It is known that $ P_1P_2$ pass incenter of triangle $ BYZ,BCZ$, and we call incenter of traingle $ BCZ$ by $ S$. So consequantly, what we need to show is $ \\angle SI_NI \\equal{} 90$. In triangle $ BCZ$, since $ ZB \\equal{} ZI$, we can show $ S$ lie on circumcircle of $ IBC$.So $ S$ is intersection of circumcircle of triangle $ IBC$ with bisector of $ \\angle ICB$. Triangle $ IBC$ and $ IMN$ are homothety with center $ I$ and homothety ratio $ \\frac {1}{2}$. Let midpoint of $ IS$ be $ S'$ which is intersection of circumcirlce of triangle $ IMN$ with bisector of $ \\angle INM$. From well known fact, $ S'$ is circumcenter of triangle $ II_NM$ and together with fact that $ S'$ is midpoint of $ IS$, $ IS$ is diameter of circumcircle of triangle $ II_NM$. Hecne $ \\angle SI_NI \\equal{} 90$ and we are done.", "Solution_6": "we can generalize:\r\n\r\nLet quadrilateral $ ABCD$ inscribe in circle $ (O)$. Let diagonal $ AC,BD$ intersect at $ E$. Let incircle of triangle $ ABE$ and $ CDE$ touch $ AC,BD$ at $ X,Y,Z,W$ respectively. $ (P)$ is incircle of the curvilinear triangle $ AEB$, tangent to the rays $ EA$, $ EB$ and internally tangent to $ (O)$. $ (Q)$ is incircle of the curvilinear triangle $ CED$, tangent to the rays $ EC$, $ ED$ and internally tangent to $ (O)$. Prove that circumcircle of triangle $ EXW$ and $ EYZ$ both externally tangent to $ (P),(Q)$.", "Solution_7": "Well known lemma: P is arbitrary point on the segment BC of a $ \\triangle ABC$ with incircle (I) touchng BC at D. (J), (K) are incircles of $ \\triangle ABP, \\triangle ACP.$ Their common internal tangent onther than AP goes through D and the angle $ \\angle JDK$ is therefore right. \r\n\r\nMinor variation: P is arbitrary point on the ray (BC beyond C. (J) is incircle of $ \\triangle ABP$ as before, (K) is the A-excircle of $ \\triangle ACP.$ The common external tangent of (J), (K) other than AP goes through D and the angle $ \\angle JDK$ is therefore right. [hide=\"Proof is the same.\"] (J), (K) touch BC at E, F.\n\n$ ED \\equal{} PC \\plus{} CD \\minus{} PE \\equal{} PC \\plus{} \\frac{_1}{^2}(BC \\plus{} CA \\minus{} AB) \\minus{} \\frac{_1}{^2}(BC \\plus{} CP \\plus{} PA \\minus{} AB) \\equal{}$\n$ \\equal{} \\frac{_1}{^2}(AC \\plus{} CP \\minus{} PA) \\equal{} PF$ $ \\Longrightarrow$\n\n$ PD \\equal{} PE \\plus{} ED \\equal{} PE \\plus{} PF$ equals to the common external tangent length of (J), (K). The 2 common external tangents of (J), (K) cut segment length on the comon internal tangent BC equal to their common external tangent length PD, therefore their common external tangent other than AP goes through D. [/hide]\r\n\r\nProblem: XYZW is isosceles tapezoid with diagonal intersection E $ \\Longrightarrow$ $ \\odot(EXW) \\cong \\odot(EYZ)$ are symmetrical WRT the bisector of the angle $ \\angle AEB \\equal{} \\angle DEC,$ same as (P), (Q) $ \\Longrightarrow$ it is sufficient to show tangency of (Q) and $ \\odot(EYZ).$ $ \\triangle ABE \\sim \\triangle DCE$ $ \\Longrightarrow$\r\n\r\n$ \\frac{EY}{EZ} \\equal{} \\frac{BE \\plus{} EA \\minus{} AB}{DE \\plus{} EC \\minus{} DC} \\equal{} \\frac{EB}{EC}$ $ \\Longrightarrow$ $ YZ \\parallel BC.$\r\n\r\nLet (I) be the incircle of $ \\triangle CDE$ (tangent to CE, DE at Z, W), (J) the incircle of $ \\triangle BCD,$ (K) the incircle of $ \\triangle BCE,$ and (L) the C-excircle of $ \\triangle BCE.$ Let F be foot of perpendicular from J to the angle bisector EL of the angle $ \\angle AEB \\equal{} \\angle DEC.$ KELB is cyclic with a circumcircle (N) because of the right angles at B, E. JFLB is cyclic with a circumcircle (M) because of right angles at B, F. By the lemma, the angle $ \\angle JWL$ is also right and $ W \\in (M).$ ZW (parallel to EK) cuts the angle bisector CKL at S. Then\r\n\r\n$ \\angle WSL \\equal{} \\angle EKL \\equal{} \\angle EBL \\equal{} \\angle WBL$ $ \\Longrightarrow$ $ S \\in (M).$\r\n\r\n$ SW \\parallel JF$ $ \\Longrightarrow$ JFWS is isosceles trapezoid and $ \\angle FWS \\equal{} \\angle WSJ \\equal{} \\angle WBJ.$ The $ \\triangle FWZ$ is isosceles with FW = FZ,\r\n\r\n$ \\angle ZFW \\equal{} 180^\\circ \\minus{} 2 \\angle FWS \\equal{} 180^\\circ \\minus{} 2 \\angle WBJ \\equal{} 180^\\circ \\minus{} \\angle WBC \\equal{} 180^\\circ \\minus{} \\angle WYZ$ $ \\Longrightarrow$\r\n\r\nYZFW is cyclic and YF bisects the angle $ \\angle EYZ \\equal{} \\angle WYZ.$ Since EF also bisects the angle $ \\angle WEZ,$ F is the Y-excenter of the $ \\triangle EYZ.$ Let JF cut CE, DE at U, V, then F is the midpoint of UV. By Thebault theorem, U, V are tangency points with CE, DE of the incircle (Q) of curvilinear $ \\triangle DEC$ internally tangent to the circumcircle (O) of the $ \\triangle BCD.$ Since the Y-excenter F of the $ \\triangle EYZ$ is the midpoint of UV, (Q) is also the mixtilinear Y-excircle of the $ \\triangle EYZ,$ externally tangent to its circumcircle. QED.", "Solution_8": "Dear Yetti, Leonhard Euler and Mathlinkers,\r\nfinally, you can see a completely synthetical proof of this nice san Gaku on my site :\r\nhttp://perso.orange.fr/jl.ayme vol. 4 La fameuse san Gaku de lma prefecture de Gumma (1803)\r\nSincerely\r\nJean-Louis" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "[i] Let $ a,b,c,d$ be positive real numbers such that :\n\n$ a\\plus{}b\\plus{}c\\plus{}d \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2$\n\n Prove that :\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{}d^3 \\plus{} a \\plus{}b \\plus{} c \\plus{} d \\leq 8$ [/i]", "Solution_1": "The inequality doesn't hold for all positive reals.\r\n\r\nIsn't that for $ a,\\ b,\\ c,\\ d\\geq 2,\\ a\\plus{}b\\plus{}c\\plus{}d\\equal{}a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2\\Longrightarrow a^3\\plus{}b^3\\plus{}c^3\\plus{}d^3\\plus{}a\\plus{}b\\plus{}c\\plus{}d\\geq 8$? :wink:", "Solution_2": "[quote=\"kunny\"]The inequality doesn't hold for all positive reals.\n\nIsn't that for $ a,\\ b,\\ c,\\ d\\geq 2,\\ a \\plus{} b \\plus{} c \\plus{} d \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2\\Longrightarrow a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} d^3 \\plus{} a \\plus{} b \\plus{} c \\plus{} d\\geq 8$? :wink:[/quote]\r\n\r\nFor $ a,b,c,d \\ge 2$ the condition cant be true my friend :wink:", "Solution_3": "You are right. :blush: I take a rest, see you.", "Solution_4": "[quote=\"nguyenvuthanhha\"][i] Let $ a,b,c,d$ be positive real numbers such that :\n\n$ a \\plus{} b \\plus{} c \\plus{} d \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2$\n\n Prove that :\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} d^3 \\plus{} a \\plus{} b \\plus{} c \\plus{} d \\leq 8$ [/i][/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=58853\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=59758", "Solution_5": "[quote=\"Dimitris X\"][quote=\"kunny\"]The inequality doesn't hold for all positive reals.\n\nIsn't that for $ a,\\ b,\\ c,\\ d\\geq 2,\\ a \\plus{} b \\plus{} c \\plus{} d \\equal{} a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2\\Longrightarrow a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} d^3 \\plus{} a \\plus{} b \\plus{} c \\plus{} d\\geq 8$? :wink:[/quote]\n\nFor $ a,b,c,d \\ge 2$ the condition cant be true my friend :wink:[/quote]\r\nthe numbers $ a,b,c,d$ can not be all greater then $ 2$ at least one of them is greater than $ 1$ and an other one is smaller than $ 1$", "Solution_6": "I do know it. You should see the former post when you post as rule. :wink:" } { "Tag": [ "abstract algebra", "linear algebra", "matrix", "algebra", "polynomial", "complex numbers", "linear algebra unsolved" ], "Problem": "let A be a n x n complex matrix. let K denote the space of all n x n complex matrices and define an operator T on K by T(B)=AB-BA. show that the rank of this operator is atmost n^2-n. \r\n\r\ni tried looking at the kernel. the kernel consists of all matrices which commute with A, now we know that scalar multpiles of I commute with any general matrix........am i on the right lines? if not, give me some hints", "Solution_1": "only scalar multiples of unity commute with every matrix, not all diagonal matrices", "Solution_2": "yeah, sorry stupid mistake. i've edited it now", "Solution_3": "[quote=\"kouboy\"]let A be a n x n complex matrix. let K denote the space of all n x n complex matrices and define an operator T on K by T(B)=AB-BA. show that the rank of this operator is atmost n^2-n. \n\ni tried looking at the kernel. the kernel consists of all matrices which commute with A, now we know that scalar multpiles of I commute with any general matrix........am i on the right lines? if not, give me some hints[/quote]\r\nAs you noticed, the kernel of this operator consists of all the matrices $B$ which commutes with $A$. Now, there is an interesting result hich you may try to prove(otherwise ask): the dimension of the space of solutions of the equation $AX=XA$ in $M_n(\\mathbb{C})$ is $\\sum_{\\lambda} \\sum_{i,j} min(a_i(\\lambda),a_j(\\lambda))$, where $\\lambda$ goes over the eigenvalues of $A$ and $a_i(\\lambda)$ is the size of a Jordan block corresponding to $\\lambda$.\r\nSo, since $\\sum_{\\lambda} \\sum_{i,j} min(a_i(\\lambda),a_j(\\lambda)) \\geq \\sum_{\\lambda} \\sum_i a_i(\\lambda)=n$, you get that $dimKer T \\geq n$, but $dimKer T +dimIm T =n^2$, so you get the desired result.", "Solution_4": "well i dont no about jordan forms yet though i dare say i'll be knowing it in another couple of days. is there any other approach ? elementary i mean. anyway thanks for replying eugene. once i read jordan forms , i'll come back to this prob and try to prove your claim. thanks again", "Solution_5": "Interesting! But is it necessary that you work over the complex numbers?\r\n\r\nI guess it suffices that your characteristic polynomial of A splits in your field?\r\n\r\nAnd what if not?use splitting fields??", "Solution_6": "[quote=\"fredbel6\"]Interesting! But is it necessary that you work over the complex numbers?\n\nI guess it suffices that your characteristic polynomial of A splits in your field?\n\nAnd what if not?use splitting fields??[/quote]\r\n\r\nActually, the proof i know uses exactly the form of Jordan blocks, so we need to have the characteristic polynomial split over the field. But, who knows, maybe you or someone else will manage to prove this for example using Rational Canonical Form which exists over any field. It would be very nice." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "incenter", "sphere", "radical axis", "power of a point" ], "Problem": "For a tetrahedron SABC,the incircle of \u25b3ABC has the incenter I and touches BC,CA,AB at points E,F,D respectively.let A',B',C' be points on SA,SB,SC s.t. AA'=AD,BB'=BE,CC'=CF and let S' be the point diametrically opposite to S on the circumsphere of SABC.assume that the line SI is an altitude of SABC.show that S'A'=S'B'=S'C'.\r\nPlease :help: :ewpu:", "Solution_1": "Let $ \\mathcal{T}_a, \\mathcal{T}_b, \\mathcal{T}_c$ be the spheres with centers $ A,B,C$ passing through $ (D,F,A'),$ $(D,E,B'),$ $(E,FC').$ $I$ is the radical center of the circles centered at $ A,B,C$ passing through $(F,D),$ $(D,E),(E,F).$ Therefore, the perpendicular line to the face $ ABC$ through $ I$ is the radical axis of $ \\mathcal{T}_a, \\mathcal{T}_b, \\mathcal{T}_c$ $\\Longrightarrow$ powers of $ S$ to $ \\mathcal{T}_a, \\mathcal{T}_b, \\mathcal{T}_c$ equal $ k^2.$\n\n$ SB^2 \\minus{} (BB')^2 \\equal{} SC^2 \\minus{} (CC')^2 \\equal{} SA^2 \\minus{} (AA')^2 \\equal{} k^2$\n\nPower of $ B$ with respect to the circumsphere $ (O,R)$ is $ R^2 \\minus{} (B'O)^2 \\equal{} BB' \\cdot B'S$\n\nNote that in the triangle $ \\triangle B'SS',$ the segment $ B'O$ is the median issuing from $B'.$ Thus\n\n$ (B'O)^2 \\equal{} \\frac {_1}{^2}(SB')^2 \\plus{} \\frac {_1}{^2}(S'B')^2 \\minus{} \\frac {_1}{^4}(2R)^2$\n\nSubstituting $ (BO')^2$ from the previous expression yields\n\n$ (S'B')^2 \\equal{} 4R^2 \\minus{} (SB')^2 \\minus{} 2BB' \\cdot B'S$\n\n$ (S'B')^2 \\equal{} 4R^2 \\minus{} SB^2 \\minus{} (BB')^2 \\plus{} 2BS \\cdot BB' \\minus{} 2BB' \\cdot B'S$\n\n$ (S'B')^2 \\equal{} 4R^2 \\minus{} SB^2 \\minus{} (BB')^2 \\plus{} 2BB'(BS \\minus{} B'S)$\n\n$ (S'B')^2 \\equal{} 4R^2 \\minus{} SB^2 + (BB')^2$\n\n$ (S'B')^2 \\equal{} 4R^2 \\minus{} k^2$\n\nSince this latter expression is independent of the chosen vertex, we conclude that\n\n$S'A' \\equal{} S'B' \\equal{} S'C' \\equal{} \\sqrt {4R^2 \\minus{} k^2}$" } { "Tag": [ "complex analysis", "integration", "logarithms", "complex numbers", "complex analysis unsolved" ], "Problem": "Suppose F is holomorphic in the unit disc, and\r\n\r\n$ \\sup_{0 \\le r < 1}$ $ \\frac {1}{2\\pi}$ $ \\int_{ \\minus{} \\pi}^{\\pi}$ $ \\log^ \\plus{} |F(re^{i\\theta})|$ $ d\\theta$ < $ \\infty$ ,\r\n\r\nwhere $ \\log^ \\plus{} u$=log $ u$ if $ u \\ge 1$, and $ \\log^ \\plus{} u \\equal{} 0$ if $ u < 1$.\r\n\r\nThen $ lim_{r \\to 1} F(re^{i \\theta})$ exists for almost every $ \\theta$.", "Solution_1": "Are you not bracketing the exponent on purpose?\r\n\r\nDo you want $ F(re^{i\\theta})$ in the two places that it occurs?", "Solution_2": "Consider the case in which $ F$ has no zeros in the open unit disc. (We can use Blaschke products to arrive at this from the general case.) Then $ F$ has a holomorphic logarithm on the unit disc, and the given condition tells us something about the real part of that logarithm. \r\n\r\nCan you proceed from there?", "Solution_3": "Thanks.\r\nNow, I see that $ F$ has a holomorphic logarithm on the unit disc.\r\nBut I don't know what 'the given condition tells us something about the real part of that logarithm' means." } { "Tag": [ "polynomial", "algebra", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "in the proof of eisenstein criterion i see that P a prime ideal of R is\r\nconsidered.\r\nwhat does it mean that $a_{0}\\in P^{2}$ ?\r\nand that $f(x)=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ is monic and is from R[x] and has all coeffs in P\r\nis reduced modulo P and the resulting polynomial $\\widehat{f(x)}=x^{n}$.\r\nwhere have all the coeffs gone ?\r\nwhy is it that 1 wich is coeff of $x^{n}$ remains after reduction ?\r\ni know what reduction is but i really dont get whats going on with all\r\nthe coeffs disappearing because of it this time.", "Solution_1": "what i'm actually really intrigued about is the following counter-example\r\nif i have $f(x)=x^{5}+2x^{4}+4x^{3}+1$\r\nnow if i reduce it modulo $\\mathbb{Z}_{5}$(5 is prime) i get\r\n$\\widehat{f(x)}=x^{5}+\\hat{2}x^{4}+\\hat{4}x^{3}+\\hat{1}$ but\r\nfrom what i know those coefficients don't actually disappear but are\r\njust reduced modulo $\\mathbb{Z}_{5}$\r\nmoreover if the coefficients are already from $Z_{5}$(supposing),\r\nthen after reduction they remain the same(in my oppinion).\r\nwhere am i wrong ?", "Solution_2": "in the mean time i understood what an element that is in $P^{2}$\r\nlooks like($a \\in P^{2}=(p^{2})$ => $p^{2}| a$).\r\ni would like to share that with others as well :lol:", "Solution_3": "Sorry, but what book is this?\r\ndummit foote (is that the name? Author?)", "Solution_4": "this is the book\r\nhttp://www.emba.uvm.edu/~foote/book_cover.gif", "Solution_5": "Nevermind. It is here:\r\nhttp://www.amazon.com/Abstract-Algebra-David-S-Dummit/dp/0471433349\r\n\r\nedit: sorry didn't see your post\r\n\r\n\r\nIs it a good book?", "Solution_6": "you would have to ask someone with more experience like ZetaX for example.\r\nbut in my oppinion its a good book,it keeps stuff going in a practical sense.\r\nit has concrete examples and avoids getting into formalisations and\r\nthat kind of things,wich is good,but for my curriculum it isnt sufficient.\r\ni would think that if i would be able to read it all and to read all of serge lang\r\ni would be able to solve harder problems also.\r\nbut to answer your question short : for begginers(like me) i think its a very good book", "Solution_7": "ooooh,now i follow the f i chose as an example earlier does not\r\nrespect the hypothesis of eisenstein criterion.\r\nall coeffs have to be divisible by p hence they are all 0 in P\r\nnow i get it." } { "Tag": [ "integration", "geometry", "parameterization", "calculus", "rotation", "real analysis", "real analysis theorems" ], "Problem": "okay, i need some clarification on some formulae.\r\n\r\nto compute the volume about the $ x$ axis: $ \\pi\\int_a^b f^2.$\r\n\r\nto compute the volume about the $ y$ axis: $ 2\\pi\\int_a^b xf(x)\\,dx.$\r\n\r\nto compute the volume about the $ x$ axis when given $ x \\equal{} f(y),$ $ 2\\pi\\int_a^byf(y)\\,dy.$\r\n\r\nto compute the volume about the $ y$ axis when given $ x \\equal{} f(y),$ $ \\pi\\int_a^bf(y)^2\\,dy.$\r\n\r\nto compute the area of surface of revolution about the $ x$ axis $ 2\\pi\\int_a^bf(x)\\sqrt {1 \\plus{} (f'(x))^2}\\,dx.$\r\n\r\nwhat's the formula to compute the area of surface of revolution about the $ y$ axis?", "Solution_1": "[quote=\"Null\"]what's the formula to compute the area of surface of revolution about the $ y$ axis?[/quote]\r\nThe surface area element is $ dA\\equal{}2\\pi x\\,ds.$\r\n\r\nSince $ ds$ can be rendered in various ways depending on whether you're integrating with respect to $ dx,$ $ dy,$ or $ dt$ where $ t$ is a parameter, it's probably best to start out with this simplified form; most of your other formulas should best be written this way as well.\r\n\r\n$ ds^2\\equal{}dx^2\\plus{}dy^2,$ so\r\n\r\n$ ds\\equal{}\\sqrt{1\\plus{}\\left(\\frac{dy}{dx}\\right)^2}\\,dx$ or $ ds\\equal{}\\sqrt{\\left(\\frac{dx}{dy}\\right)^2\\plus{}1}\\,dy$, or $ ds\\equal{}\\sqrt{\\left(\\frac{dy}{dt}\\right)^2\\plus{}\\left(\\frac{dy}{dt}\\right)^2}\\,dt.$", "Solution_2": "i see...\r\n\r\nmmm, so i want to compute the area of a surface of revolution given the curve $ y\\equal{}x^2$ for $ 0\\le x\\le3$ about the $ y$ axis, what's the integral?", "Solution_3": "Probably easiest to do that $ dx,$ so $ ds\\equal{}\\sqrt{1\\plus{}4x^2}\\,dx$ and the integral is\r\n\r\n$ \\int_0^32\\pi x\\sqrt{1\\plus{}4x^2}\\,dx$\r\n\r\nWhich (rarity) can actually be computed in closed form.\r\n\r\nThat's pretty close to one of the problems on the test I gave this morning; I had it as $ y\\equal{}\\sqrt{4x},$ $ 0\\le x\\le 4,$ rotated about the $ x$ axis - but it's the same sort of shape, a parabaloid of revolution.", "Solution_4": "thanks so much Kent for your time.\r\n\r\nlast question, are the above formulae i wrote before the correct ones? i mean, i don't want to mess it up by using a wrong formula.", "Solution_5": "A suggestion: rather than remembering the formulas, it's much more productive to remember the [i]pictures[/i] associated with the formulas. Understanding how to produce a picture and then use the picture in order to write down the appropriate integral is a skill that might be of some use to you in your life (e.g., if you continue taking mathematics or engineering courses, it helps to have some visual intuition for three-dimensional objects); having the formulas memorized is unlikely to be of much use after you're done with this section of your course. That said, your formulas look right to me.", "Solution_6": "[quote=\"JBL\"]A suggestion: rather than remembering the formulas, it's much more productive to remember the [i]pictures[/i] associated with the formulas.[/quote]\r\nThat's a very close paraphrase of something I've told my current class maybe a dozen times, since it's worth repeating." } { "Tag": [ "ratio", "AMC", "AIME", "binomial coefficients" ], "Problem": "The binomial coefficients $\\binom{n}{m}$ can be arranged in rows (with the $n$th row $\\binom{n}{0}$, $\\binom{n}{1}$, ... $\\binom{n}{n}$) to form Pascal's triangle. In which row are there three consecutive entries in the ratio $3 : 4 : 5$?\r\n\r\nEDIT: why are problems #2, #3, and #4 on AIME 1992 so easy? #1 is much harder... :P", "Solution_1": "Unless I missed something, you're right, this is very easy for an AIME problem.\r\n\r\n[hide=\"solution\"]Suppose $\\binom{n}{m-1},\\binom{n}m,\\binom{n}{m+1}$ have the ratio $3: 4: 5$. We can easily use the definition $\\binom{r}{k}=\\frac{r!}{k!(r-k)!}$ to get the equations $\\frac{m}{n-m+1}=\\frac34$ and $\\frac{m+1}{n-m}=\\frac45$. These become a pair of linear equations in two variables, which is easy to solve. This gives $m=27$ and $n=62$.[/hide]", "Solution_2": "That's also what I got. :)", "Solution_3": "#1 isn't really hard, its just worded badly on Kalva. :P" } { "Tag": [ "ceiling function", "induction", "combinatorics unsolved", "combinatorics" ], "Problem": "Given 2005 distinct numbers $a_1,\\,a_2,\\dots,a_{2005}$. By one question, we may take three different indices $1\\le i4$; for $n=4$ we need $3$ questions, $2$ don't suffice). We ask questions $\\{i,i+1,i+2\\}$ for odd $i$ between $1$ and $n-3$, and after that we ask the question $\\{1,3,n\\}$. For odd $n$ we do something very similar.", "Solution_3": "I do not understand teh first part, grobber. What if we ask for some triple wich intersect with first $m$ numbers after few questions concerning next group? For example: 123, then 234, then 567, then 514?", "Solution_4": "But the order of the questions does not matter, so we ask them in the order we desire. If after we have asked some questions, there is another question regarding one of the numbers we have asked before, then we ask that one. In this case, we simply reorder the questions so that $514$ is the third one.", "Solution_5": "It is not clear. aybe, our strategy uses previous answers before asking further questions. Say, we ask 514 only if 2>1>3, else we ask 523.", "Solution_6": "You ask a question which includes three numbers. After this, if there still are some questions which include on of the numbers used, choose one and ask it. After that, look once more for questions in our list which include one of the numbers already used, and if there are such questions, choose one and ask it. What exactly isn't clear? :? If there are more questions using numbers already used, just choose one of them. We do this until we no longer can, i.e. until there are no questions using the numbers previously used.", "Solution_7": "Or, I am stupid, I forgot, which problem are you talking about (I thought about 9.4.)", "Solution_8": "[quote=\"grobber\"]I think it's $1003$, actually, and, for general $n$ instead of $2005$, I think it's $\\left\\lceil\\frac n2\\right\\rceil$. [/quote]\r\n\r\nIf we ask instead of triplets for k-tuples, the answer is $\\left\\lceil\\frac {n}{k-1}\\right\\rceil$ ?", "Solution_9": "[quote=\"grobber\"]\nIf we do this until it's no longer possible, it means that each question we ask adds at most two more indices, so when we have to stop we will have determined all the numbers with the respective indices and we will have used at least $\\frac{m}{2}$ questions, where $m$ is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other $n-m$ indices, the ones we haven't touched so far, and the procedure is repeated.\n[/quote]\r\n\r\nBut when we begin each time, we use three indices. So we might be able to determine three indices by that question...", "Solution_10": "Hmmm I think grobber's solution works (basically induction, right?) and is probably the most intuitive. However, here's another way I will roughly outline. Suppose less than n/2 questions are used. Then more than half the elements have been asked about only once (obviously every element is asked about at least once). Then there exists some question such that two of the elements it concerns were only asked about once. But then we can not distinguish between these two, so we need at least (roughly) n/2 questions.", "Solution_11": "[quote=\"probability1.01\"]Then there exists some question such that two of the elements it concerns were only asked about once. [/quote]\r\n\r\nHmm.. how does this come?", "Solution_12": "[quote=\"dragonfire\"][quote=\"probability1.01\"]Then there exists some question such that two of the elements it concerns were only asked about once. [/quote]\n\nHmm.. how does this come?[/quote]\r\n\r\nThere are more than n/2 elements with the property of [asked about only once]. Each of these belongs to one of fewer than n/2 questions. Of course, there is some casework for even and odd n, but I think you can figure it out.", "Solution_13": "Here is an amusing solution that is different from the above.\n\n[hide = Solution]\n\nFor showing that $1003$ questions indeed suffice, do the same thing as [b] grobber [/b] did above. Now, we will show that $1003$ questions are indeed necessary. Suppose, for contradiction, that we were somehow able to find out all the numbers with the $1002$ questions $(x_1, y_1, z_1), (x_2, y_2, z_2), \\cdots, (x_{1002}, y_{1002}, z_{1002}),$ where the $x_i$'s, $y_i$'s, and $z_i$'s are all in $[1, 2005].$ Call a positive integer $1 \\le n \\le 2005$ [b] sketchy [/b] if it only appears one time among the $x_i$'s, $y_i$'s, and $z_i$'s. In other words, in the multiset of all $x_i$'s, $y_i$'s, and $z_i$'s (of size $3 \\cdot 2005 = 6015$, call this multiset $M$), $n$ only appears once. Else, call $n$ [b] rigorous [/b]. Note that any rigorous $n$ must appear at least twice, since otherwise we would never know what $a_n$ is. Now, let's define the function $f(n)$ for $1 \\le n \\le 2005$ by letting it equal $\\frac1a$, where $a$ is the number of times that $n$ appears in $M$. Observe that the sum $\\sum_{k=1}^{1002} (f(x_k) + f(y_k) + f(z_k))$ is exactly equal to $2005$. Now, observe that $f(x_k) + f(y_k) + f(z_k) \\le 2$ for all $1 \\le k \\le 1002.$ Indeed, if it was greater than $2$, then we must have that at least two of $f(x_k), f(y_k), f(z_k)$ are $1$'s, WLOG $f(x_k) = f(y_k) = 1.$ However, this means that the values of $a_{x_k}$ and $a_{y_k}$ are indistinguishable, impossible. Therefore, as $f(x_k) + f(y_k) + f(z_k) \\le 2$, we have that $2005 = \\sum_{k=1}^{1002} (f(x_k) + f(y_k) + f(z_k)) \\le 2 \\cdot 1002 = 2004,$ contradiction. Hence, our assumption that we finished in $1002$ questions was incorrect, and the answer is $\\boxed{1003}.$ \n\n[/hide]" } { "Tag": [ "probability", "AMC", "AIME", "AIME II", "number theory", "relatively prime" ], "Problem": "Source: AIME II 2001 \r\n11. Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 mathches, the probabilities that Club Truncator will win, lose, or tie are each $\\frac{1}{3}$. The probability that Club Truncator will finish the season with more wins than losses is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.", "Solution_1": "Isn't this an application of the multinomial theorem? Of course, we don't necessarily have to use it here.", "Solution_2": "Clearly the probability that the team wins more than looses is the same as the probability if the team looses more than wins. Call that $P$ and add those together. Then add in the times where the same wins and losses, call this $Q$. This covers all possibilities so: \\[2P+Q=1\\] We can easily calculate $Q$ because that is just the probability of having let P(win,loss,tie) denote the probability that occurs. Then:\r\n\r\n$P(0,0,6)=\\frac{6!}{0!0!6!}*\\frac{1}{3^{6}}$\r\n\r\n$P(1,1,4)=\\frac{6!}{1!1!4!}*\\frac{1}{3^{6}}$\r\n\r\n$P(2,2,2)=\\frac{6!}{2!2!2!}*\\frac{1}{3^{6}}$\r\n\r\n$P(3,3,0)=\\frac{6!}{3!3!0!}*\\frac{1}{3^{6}}$\r\n\r\nAdd these to get \\[Q=\\frac{1}{3^{6}}*(1+30+90+20)=\\frac{47}{243}\\] Then we can compute $P$: \\[P=\\frac{1-Q}{2}=\\frac{98}{243}\\] It follows that $m+n=98+243=\\boxed{341}$" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "let f and g two functions continuous on [a, b] let us suppose that for any x of [a, b] we have f (x) sup g (x). show that there exists a strictly positive such f (x) sup= g (x) +a", "Solution_1": "Sorry diane, I don't understand you :maybe: \r\nWhat does \"f(x) sup g(x)\" and \"f(x) sup = g(x) + a\" mean?\r\nIs \"sup\" the supremum?", "Solution_2": "f(x)>g(x) ,f(x)>=g(x)+a", "Solution_3": "$f,g$ continuous $\\Rightarrow$ $f-g$ continuous on $[a,b]$. Hence $f-g$ achieves its minimum on $[a,b]$, $c$. Because $\\forall x \\in [a,b]: (f-g)(x) > 0$, we have $c>0$. \r\nThen $f(x) = g(x)+(f-g)(x) \\geq g(x)+c > g(x)+\\frac{c}{2}( \\forall x \\in [a,b])$" } { "Tag": [ "geometry", "perimeter" ], "Problem": "In the figure, the sides $AB$ and $AC$ of the equilateral triangle $ABC$ have been extended, and the circle tangent to these extended sides and to $BC$, as shown, has radius 4. What is the perimeter of $\\triangle ABC$?\r\n\r\n[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=416[/img]\r\n\r\na) $4 \\sqrt 3$\r\nb) $6 \\sqrt 3$\r\nc) $8 \\sqrt 3$\r\nd) 12\r\ne) 14", "Solution_1": "[hide]Connect the centre of the ex-circle and $A$. Then $h_{BC}=4$, hence: $C=8\\sqrt{3}$. [/hide]", "Solution_2": "[hide]\nlabel the center of the circle $O$ and the perpendicular dropped onto the extension of $AB$, $D$. $OAD$ is a 30-60-90 triangle. Let the point of tangency between $\\triangle ABC$ and circle with center $O$ be labeled $P$. $PAB$ is also a 30-60-90 triangle.\n\nso $AO = 8$ and $AP = 4$ thus $AB = \\frac{8}{\\sqrt{3}}$ and the perimeter = $8\\sqrt{3}$ c)[/hide]", "Solution_3": "[quote=\"agolsme\"][hide]\n$OAD$ is a 30-60-90 triangle.[/hide][/quote]\r\nWhy is that? :huh:", "Solution_4": "Because it's a perpendicular so one of its angles is 90 and one of its angles is an angle of the triangle $ABC$ which is equilateral so it's 60. The last one must be 30. :D", "Solution_5": "Oh, it's equilateral. I didn't catch that, looking at the drawing. :roll: (Yes, I understand that it's hard to make good drawings. I have trouble with it, too. :P)", "Solution_6": "[quote=\"JesusFreak197\"]Oh, it's equilateral. I didn't catch that, looking at the drawing. :roll: (Yes, I understand that it's hard to make good drawings. I have trouble with it, too. :P)[/quote]\r\nIf you really need help in drawing pictures, you can try a find a Geometer's Sketchpad on the internet.", "Solution_7": "LOL. My drawing is perfectly fine. :rotfl: \r\n\r\nPlus, if Jesusfreak read the question [b]carefully[/b], it should've been clear.", "Solution_8": "[quote=\"shobber\"]If you really need help in drawing pictures, you can try a find a Geometer's Sketchpad on the internet.[/quote]\r\n\r\nWell, I have never yet drawn a picture on the computer, and I don't think Geometer's Sketchpad helps with normal drawings... does it? Anyway, I generally do pretty well, but a lot of times I'll get proportions way off and stuff. :P", "Solution_9": "[quote=\"JesusFreak197\"][quote=\"shobber\"]If you really need help in drawing pictures, you can try a find a Geometer's Sketchpad on the internet.[/quote]\n\nWell, I have never yet drawn a picture on the computer, and I don't think Geometer's Sketchpad helps with normal drawings... does it? Anyway, I generally do pretty well, but a lot of times I'll get proportions way off and stuff. :P[/quote]\r\n\r\nGeometer's Sketchpad is really good, but it isn't free :( MetaPost is probably the best though.", "Solution_10": "[quote=\"shobber\"][hide]Connect the centre of the ex-circle and $A$. Then $h_{BC}=4$, hence: $C=8\\sqrt{3}$. [/hide][/quote]\r\n\r\nWhat is $h_{BC}$?" } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "[color=purple] Let an acute triangle $ ABC$ in plane which has three exraii $ r_{a},r_{b},r_{c}$ and three medians $ m_{a},m_{b},m_{c}$. With any $ k\\ge0$ prove that \n $ \\frac {r_{b}^k \\plus{} r_{c}^k}{m_{b}^k \\plus{} m_{c}^k} \\plus{} \\frac {r_{c}^k \\plus{} r_{a}^k}{m_{c}^k \\plus{} m_{a}^k} \\plus{} \\frac {r_{a}^k \\plus{} r_{b}^k}{m_{a}^k \\plus{} m_{b}^k}\\ge 3$\n [/color]", "Solution_1": "[quote=\"evarist\"][color=purple] Let an acute triangle $ ABC$ in plane which has three exraii $ r_{a},r_{b},r_{c}$ and three medians $ m_{a},m_{b},m_{c}$. With any $ k\\ge1$ prove that \n $ \\frac {r_{b}^k \\plus{} r_{c}^k}{m_{b}^k \\plus{} m_{c}^k} \\plus{} \\frac {r_{c}^k \\plus{} r_{a}^k}{m_{c}^k \\plus{} m_{a}^k} \\plus{} \\frac {r_{a}^k \\plus{} r_{b}^k}{m_{a}^k \\plus{} m_{b}^k}\\ge 3$\n [/color][/quote]" } { "Tag": [], "Problem": "Sa se rezolve ecuatia : [x]=x{x}", "Solution_1": "Da !\r\nEste o ecuatie draguta cu multe solutii" } { "Tag": [], "Problem": "Solve in $\\mathbb{R}$:\r\n\\[ \\begin{displaymath} \\left\\{ \\begin{array}{11} x^2+3y=9 \\\\ y^4+4(2x-3)y^2-48(x+y)+155=0 \\end{array} \\right \\end{displaymath} \\]", "Solution_1": "$\\begin{displaymath} \\left\\{ \\begin{array}{11} x^2+3y=9 ---(1) \\\\ y^4+4(2x-3)y^2-48(x+y)+155=0 ---(2)\\end{array} \\right \\end{displaymath}$\r\n*******************************************************************\r\n[hide]From (1), $y=\\frac{9-x^2}3$.\nSubstitute into (2) and simply:\n\n$\\frac{1}{81}x^8-\\frac 4 9 x^6+\\frac 8 9 x^5+\\frac{14}3x^4-16x^3+4x^2+24x-16 = 0$\n\n$\\frac 1{81}(x^2-6x+12)(x^2+6x+6)(x^4-18x^2+36x-18)=0$\n\n....... :!:\n[/hide]" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find the least positive integer $ n$ with the following property: if $ n$ distinct sums of the form $ x_p\\plus{}x_q\\plus{}x_r,1\\le p k$ contradiction!\r\n$ r > 0$, $ a_{0}\\leq b_{0}$ $ \\implies$\r\n$ a_{0}.b_{0} \\plus{} b_{0}^{2} > a_{0}^{2} \\plus{} b_{0}^{2} > ka_0.b_0$ $ \\implies$\r\n$ ka_0.b_0 \\minus{} b_0^{2} < a_{0}.b_{0}$ $ \\implies$ $ r < a_0$\r\ncontradiction!", "Solution_3": "[quote=\"poonprang\"]2.FInd all integer solution to the equation\n\n(x^2-y^2)^2=1+16y [/quote]\r\n\r\nObviously $ y\\geq 0$ and $ 1\\plus{}16y$ is an odd square. So we can write $ 1\\plus{}16y\\equal{}(2n\\plus{}1)^2$ for some $ n\\geq 0$\r\n\r\n$ \\implies$ $ y\\equal{}\\frac{n^2\\plus{}n}{4}$ and $ x^2\\minus{}y^2\\equal{}\\pm(2n\\plus{}1)$ $ \\implies$ $ x^2\\equal{}y^2\\pm(2n\\plus{}1)$\r\n\r\n$ \\implies$ $ 2n\\plus{}1\\geq 2y\\minus{}1$ and so $ y\\leq n\\plus{}1$ and so $ \\frac{n^2\\plus{}n}{4}\\leq n\\plus{}1$ and so $ 0\\leq n\\leq 5$. Then :\r\n\r\n$ n\\equal{}0$ $ \\implies$ $ y\\equal{}0$ $ \\implies$ $ (x^2)^2\\equal{}1$ and two solutions $ (\\minus{}1,0)$ and $ (1,0)$\r\n\r\n$ n\\equal{}1$ $ \\implies$ $ y\\equal{}\\frac 12$ so no solution\r\n\r\n$ n\\equal{}2$ $ \\implies$ $ y\\equal{}\\frac 32$ so no solution\r\n\r\n$ n\\equal{}3$ $ \\implies$ $ y\\equal{}3$ $ \\implies$ $ (x^2\\minus{}9)^2\\equal{}49$ and two solutions $ (\\minus{}4,3)$ and $ (4,3)$\r\n\r\n$ n\\equal{}4$ $ \\implies$ $ y\\equal{}5$ $ \\implies$ $ (x^2\\minus{}25)^2\\equal{}81$ and two solutions $ (\\minus{}4,5)$ and $ (4,5)$\r\n\r\n$ n\\equal{}5$ $ \\implies$ $ y\\equal{}\\frac {15}2$ so no solution\r\n\r\nHence the solutions : $ (\\minus{}1,0),(1,0),(\\minus{}4,3),(4,3),(\\minus{}4,5)$ and $ (4,5)$", "Solution_4": "$ 2)(x^{2}\\minus{}y^{2})^{2}\\equal{}16y\\plus{}1$ $ \\implies$\r\n$ (x\\minus{}y)^{2}.(x\\plus{}y)^{2}\\equal{}16y\\plus{}1$ \r\n$ (x\\plus{}y)^{2}\\leq 16y\\plus{}1$ if $ x\\geq 5$ $ \\implies$\r\n$ y^{2}\\plus{}10y\\plus{}25\\leq 16y\\plus{}1$ $ \\implies$ $ y^{2}\\minus{}6y\\plus{}24\\leq0$ Contradiction!\r\n$ x\\leq 4$ we know if x is a solution -x is a solution.\r\n$ x\\geq 0$\r\n$ i)x\\equal{}0$ $ \\implies$ $ y^{4}\\equal{}16y\\plus{}1$ $ \\implies$ $ y|1 ,y\\equal{}1,\\minus{}1$ $ \\implies$ we get any solution!\r\n$ ii)x\\equal{}1$ $ \\implies$ $ y\\plus{}1|16y\\plus{}1$ $ y\\plus{}1|15$\r\n$ y\\plus{}1\\equal{}1,y\\plus{}1\\equal{}3,y\\plus{}1\\equal{}5,y\\plus{}1\\equal{}15$ $ \\implies$\r\n$ y\\equal{}0,y\\equal{}2,y\\equal{}4,y\\equal{}14$ $ \\implies$ $ (1,0)$ is a solution.\r\n$ iii)x\\equal{}2$ $ \\implies$ $ y\\plus{}4|16y\\plus{}1$ $ \\implies$ $ y\\plus{}4|63$\r\n$ y\\plus{}4\\equal{}7,y\\plus{}4\\equal{}9,y\\plus{}4\\equal{}21,y\\plus{}4\\equal{}63$ and $ y^{2}\\minus{}16y\\minus{}5\\leq 0$ \r\n$ y\\leq 16$ $ \\implies$ $ y\\equal{}3,y\\equal{}5$ $ \\implies$ we get any solution!\r\n$ iv)x\\equal{}3$ $ \\implies$ $ y^{2}\\plus{}6y\\plus{}9\\leq 16y\\plus{}1$ $ \\implies$\r\n$ y^{2}\\minus{}10y\\plus{}9\\leq 0$ $ \\implies$ $ 1\\leq y \\leq 9$ \r\n$ y\\plus{}9|16y\\plus{}1$ $ \\implies$ $ y\\plus{}9|143$ $ y\\plus{}9\\equal{}13 ,y\\plus{}9\\equal{}11$\r\n$ \\implies$ $ y\\equal{}4,y\\equal{}2$ $ \\implies$ we get any solution!\r\n$ v) x\\equal{}4$ $ \\implies$ $ y^{2}\\plus{}8y\\plus{}16\\leq 16y\\plus{}1$ $ \\implies$\r\n$ y^{2}\\minus{}8y\\plus{}15\\leq0$ $ \\implies$ $ (y\\minus{}5)(y\\minus{}3)\\leq0$ $ \\implies$ $ y\\equal{}3,y\\equal{}4,y\\equal{}5$\r\n$ \\implies$ $ (4,5),(4,3)$ are solution.\r\nall solution $ \\{(\\minus{}1,0),(1,0),(4,3),(4,5),(\\minus{}4,3),(\\minus{}4,5)\\}$ :wink:", "Solution_5": "[quote=\"poonprang\"]3.Let a b c d e positive integer such that \n\n abcde=a+b+c+d+e Find the maximum possible value of max{a,b,c,d,e}\n[/quote]\r\n\r\nWlog say $ a\\geq b\\geq c\\geq d\\geq e\\geq 1$\r\n\r\nIf $ d\\geq 2$, then $ abcde\\geq 8a$ while $ a \\plus{} b \\plus{} c \\plus{} d \\plus{} e\\leq 5a$. So $ d \\equal{} e \\equal{} 1$ and the equation is $ abc \\equal{} a \\plus{} b \\plus{} c \\plus{} 2$\r\nIf $ c\\geq 3$, $ abc\\geq 9a$ while $ a \\plus{} b \\plus{} c \\plus{} 2\\leq 3a \\plus{} 2\\leq 5a$ so either $ c \\equal{} 2$, either $ c \\equal{} 1$\r\nIf $ c \\equal{} 2$, the equation is $ 2ab \\equal{} a \\plus{} b \\plus{} 4$ and $ a \\equal{} b \\equal{} 2$\r\nIf $ c \\equal{} 1$, the equation is $ ab \\equal{} a \\plus{} b \\plus{} 3$ and so $ a(b \\minus{} 1) \\equal{} b \\plus{} 3$ and so $ b \\minus{} 1|b \\plus{} 3$ and so $ b\\in\\{2,3,5\\}$\r\n\r\n$ b \\equal{} 2$ $ \\implies$ $ a \\equal{} 5$\r\n$ b \\equal{} 3$ $ \\implies$ $ a \\equal{} 3$\r\n$ b \\equal{} 5$ $ \\implies$ $ a \\equal{} 2$, impossible since $ a\\geq b$\r\n\r\nSo $ abcde \\equal{} a \\plus{} b \\plus{} c \\plus{} d \\plus{} e$ has just 3 ordered solutions in positive integers : $ (2,2,2,1,1),(5,2,1,1,1)$ and $ (3,3,1,1,1)$\r\n\r\nHence the answer : $ 5$", "Solution_6": "$ 1 \\equal{} \\frac {1}{bcde} \\plus{} \\frac {1}{acde} \\plus{} \\frac {1}{abde} \\plus{} \\frac {1}{abce} \\plus{} \\frac {1}{abcd}$\r\n$ a\\geq b\\geq c \\geq d\\geq e$ $ \\implies$ \r\n$ e\\geq 2$ $ \\implies$ $ \\frac {1}{bcde} \\plus{} \\frac {1}{acde} \\plus{} \\frac {1}{abde} \\plus{} \\frac {1}{abce} \\plus{} \\frac {1}{abcd} < 1$\r\n$ e \\equal{} 1$ $ \\implies$ $ d\\geq 2$ $ \\implies$\r\n$ \\frac {1}{bcde} \\plus{} \\frac {1}{acde} \\plus{} \\frac {1}{abde} \\plus{} \\frac {1}{abce} \\plus{} \\frac {1}{abcd} < 1$\r\n$ d \\equal{} 1$ $ \\implies$\r\n$ abc \\equal{} a \\plus{} b \\plus{} c \\plus{} 2$ $ \\implies$\r\n$ 1 \\equal{} \\frac {1}{bc} \\plus{} \\frac {1}{ac} \\plus{} \\frac {1}{ab} \\plus{} \\frac {2}{abc}$ $ \\implies$\r\n$ c\\geq 3$ $ \\frac {1}{bc} \\plus{} \\frac {1}{ac} \\plus{} \\frac {1}{ab} \\plus{} \\frac {2} {abc} < 1$\r\n$ c \\equal{} 1$ V $ c \\equal{} 2$\r\n$ i)c \\equal{} 1,ab \\equal{} a \\plus{} b \\plus{} 3$ $ \\implies$ $ a(b \\minus{} 1) \\equal{} b \\plus{} 3$ \r\n$ \\implies$ $ b \\minus{} 1|4$ $ b \\minus{} 1 \\equal{} 1,b \\minus{} 1 \\equal{} 2,b \\minus{} 1 \\equal{} 4$ $ \\implies$\r\n$ b \\equal{} 2,b \\equal{} 3,b \\equal{} 5$ $ \\implies$ $ a \\equal{} 5,a \\equal{} 3,a \\equal{} 2$ \r\n$ a\\geq b$ $ \\implies$ $ (5,2),(3,3,)$\r\n$ ii)c \\equal{} 2,2ab \\equal{} a \\plus{} b \\plus{} 3$ $ \\implies$ $ a(2b \\minus{} 1) \\equal{} b \\plus{} 3$\r\n$ 2b \\minus{} 1\\leq b \\plus{} 3$ $ \\implies$ $ b\\leq 4$ \r\n$ (b \\equal{} 1,a \\equal{} 4),(b \\equal{} 4,a \\equal{} 1)$ $ \\implies$ $ (4,1)$\r\n :wink:", "Solution_7": "[quote=\"cnyd\"] $ ii)c \\equal{} 2,2ab \\equal{} a \\plus{} b \\plus{} 3$ [/quote]\r\n\r\nNo, $ 2ab\\equal{}a\\plus{}b\\plus{}4$ \r\n :wink:", "Solution_8": "sorry.\r\n$ ii)2ab \\equal{} a \\plus{} b \\plus{} 4$ $ \\implies$\r\n$ a(2b \\minus{} 1) \\equal{} (b \\plus{} 4)$ $ \\implies$\r\n$ 2b \\minus{} 1\\leq b \\plus{} 4$ $ \\implies$\r\n$ b\\leq 5$ $ \\implies$ $ (b\\geq2)$\r\n$ b \\equal{} 2,a \\equal{} 2$\r\nall solution\r\n$ \\{(2,2,2,1,1),(5,2,1,1,1),(3,3,1,1,1)\\}$ :wink:" } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Prove that in any triangle $ ABC$ there are the inequalities \r\n\r\n$ 1\\blacktriangleright\\ \\ \\boxed {\\ \\frac {1}{2R^2}\\ \\le\\ \\frac {1}{b^2 \\plus{} c^2} \\plus{} \\frac {1}{c^2 \\plus{} a^2} \\plus{} \\frac {1}{a^2 \\plus{} b^2}\\ \\le\\ \\frac {1}{4Rr}\\ }$ .\r\n\r\n$ 2\\blacktriangleright\\ \\ \\frac 12 \\plus{} \\frac rR\\ \\le \\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}\\ \\le\\ 1$ .\r\n\r\n[b]Remark.[/b] You can use the first inequality from [url=http://www.mathlinks.ro/viewtopic.php?t=293296][color=red][b]here[/b][/color][/url].", "Solution_1": "hello, for your first inequality, setting $ R \\equal{} \\frac {abc}{4A}$ with $ A\\equal{}\\sqrt {s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}$ and $ s \\equal{} \\frac {a \\plus{} b \\plus{} c}{2}$ your inequality is equivalent to\r\n$ \\frac {8s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}{a^2b^2c^2}\\le \\frac {1}{b^2 \\plus{} c^2} \\plus{} \\frac {1}{c^2 \\plus{} a^2} \\plus{} \\frac {1}{a^2 \\plus{} b^2}$ and after some algebraic manipulations we get\r\n$ a^2b^2(a^2 \\minus{} b^2)^2(a^2 \\plus{} b^2) \\plus{} a^2c^2(a^2 \\minus{} c^2)^2(a^2 \\plus{} c^2) \\plus{} b^2c^2(b^2 \\minus{} c^2)^2(b^2 \\plus{} c^2)\\geq0$ which is true.\r\nSonnhard.", "Solution_2": "hello, for the second one we have the right-hande side as $ ab \\plus{} bc \\plus{} ca \\le a^2 \\plus{} b^2 \\plus{} c^2$, this is well-known. The left-hand side is equivalent with\r\n$ \\frac {1}{2} \\plus{} \\frac {4A^2}{s \\cdot abc}\\le\\frac {ab \\plus{} bc \\plus{} ca}{a^2 \\plus{} b^2 \\plus{} c^2}$. Let $ c$ the minimum of $ a,b$, then we get after same algebra that the inequation above is equivalent to\r\n$ (a \\minus{} b)^2\\left(a^2(a \\plus{} b \\minus{} c) \\plus{} ab(b \\minus{} c) \\plus{} b^2(b \\minus{} c)\\right) \\plus{} c^3(a \\minus{} c)(b \\minus{} c)\\geq0$.\r\nSonnhard.", "Solution_3": "hello, to the right-hande side of your first inequality:\r\nwe have $ \\frac{1}{4Rr}\\equal{}\\frac{s}{abc}$ with $ R\\equal{}\\frac{abc}{4A}$ and $ r\\equal{}\\frac{A}{s}$. So we have to prove\r\n$ \\frac{1}{a^2\\plus{}b^2}\\plus{}\\frac{1}{a^2\\plus{}c^2}\\plus{}\\frac{1}{b^2\\plus{}c^2}\\le \\frac{s}{abc}$ with $ s\\equal{}\\frac{a\\plus{}b\\plus{}c}{2}$. By the AM-GM inequality we get\r\n$ \\frac{1}{b^2\\plus{}c^2}\\le \\frac{1}{2bc}$\r\n$ \\frac{1}{a^2\\plus{}c^2}\\le \\frac{1}{2ac}$\r\n$ \\frac{1}{b^2\\plus{}a^2}\\le \\frac{1}{2ba}$\r\nadding the right-hande sides above we get $ \\frac{1}{2}\\left(\\frac{a\\plus{}b\\plus{}c}{abc}\\right)$\r\nSonnhard." } { "Tag": [ "inequalities", "function", "inequalities proposed" ], "Problem": "Let $t_{1},t_{2},\\cdots,t_{n}$ be positive reals such that $\\boxed{t_{1}t_{2}\\cdots t_{n}=1}$. Prove that there exists an integer $1\\leq k\\leq n$ such that $t_{k}(t_{k+1}+1)\\geq 2$\r\n :)", "Solution_1": "assume the contrary we get that:\r\n\r\n$t_{1}(t_{2}+1)<2$\r\n.\r\n.\r\n.\r\n$t_{n-1}(t_{n}+1)<2$\r\n$t_{n}(t_{1}+1)<2$\r\n\r\nmultiplying the above $n$ inequalities we get that:\r\n\r\n$t_{1}t_{2}...t_{n}(t_{1}+1)...(t_{n}+1)<2^{n}$\r\n\r\nso $(t_{1}+1)...(t_{n}+1)<2^{n}$\r\n\r\nnow by $AM-GM$ we have that $t_{i}+1\\geq 2\\sqrt{t_{i}}$ applying this for the last inequality we get that:\r\n\r\n$2^{n}\\sqrt{t_{1}...t_{n}}\\leq (t_{1}+1)...(t_{n}+1)<2^{n}$\r\n\r\nso we get that:\r\n\r\n$2^{n}\\leq (t_{1}+1)...(t_{n}+1)<2^{n}$\r\n\r\nso we get that $2^{n}<2^{n}$ which is a contradiction...\r\n\r\nso there exists $k$ s.t $t_{k}(t_{k+1}+1)\\geq 2$", "Solution_2": "Umm, Babak, just a quick note... The contradiction of $\\ge$ is actually $<$, not $\\le$.", "Solution_3": "yeah sorry it was a little typo,I edited my post... \r\ntnx :)", "Solution_4": "[quote=\"BaBaK Ghalebi\"]\n$t_{1}(t_{2}+1)<2$\n.\n.\n.\n$t_{n-1}(t_{n}+1)<2$\n\nmultiplying the above $n$ inequalities we get that:\n\n$t_{1}t_{2}...t_{n}(t_{1}+1)...(t_{n}+1)<2^{n}$\n\n[/quote]\r\n\r\nmultiplying the above inequalities you'll get that:\r\n\r\n$t_{1}t_{2}...t_{n-1}(t_{2}+1)...(t_{n}+1)<2^{n}$\r\n\r\nactually the correct thing is $1\\leq k\\leq n$ where we'll define $a_{n+1}=a_{1}$ , well i must have said this in the problem...anyway you're solution is correct and we'll have:\r\n\r\n$t_{1}(t_{2}+1)<2$\r\n.\r\n.\r\n.\r\n$t_{n-1}(t_{n}+1)<2$\r\n$t_{n}(t_{1}+1)<2$\r\n\r\nNow by multiplying these inequalities we'll have:\r\n\r\n$t_{1}t_{2}...t_{n}(t_{1}+1)...(t_{n}+1)<2^{n}$\r\n\r\nand ...", "Solution_5": "Then, you can generalize like this:\r\n\r\nLet $f$ be a function defined on integers which takes up positive real values, with period $n$ such that $f(2007)f(2008)f(2009)\\cdots f(n+2006)=1$. Show that for every integer $l$ there exists infinitely many integers $k$ such that $f(k)(f(k+l)+1)\\ge 2$.\r\n\r\nThe proof is quite similar. Am I right? What do you think about the generalization?", "Solution_6": "yeah I have edited my post...\r\nand about the generalization,well if for every $l$ there exists such $k$ then by the period of $f$ we`ll have infinitely many such $k$s...\r\nthere might be better generalizations which gives another condition on $f$ (different from periodic) and asks for infinetly many $k$s...", "Solution_7": "Yeah... I agree, there probably is a better generalization, but this is all I can think of right now... It is essentially the same problem as the one sinajackson posted, but it seems kind of more difficult when you see it for the first time because it's more abstract.", "Solution_8": "yeah at the first look it seems pretty difficult specially the fact that $f$ is a function not a polynomial..." } { "Tag": [], "Problem": "", "Solution_1": "Uh yes. That has happened to me before. What is so significant about it?", "Solution_2": "What would be interesting if it was a ranking game, and both people got it correct.", "Solution_3": "Holy.. Yeah that would be interesting. And 1-1! :lol:", "Solution_4": "What if that happens on the last question of a countdown?", "Solution_5": "It probably would result in not a draw, but as if both won at the same time, so both players earn rating points.", "Solution_6": "Just because they only show 3 digits doesn't mean they entered at the *exact* same time. So probably the one who entered it a nanosecond before.", "Solution_7": "in a countdown, it would glitch, as always." } { "Tag": [ "trigonometry", "algebra", "function", "domain", "rotation", "analytic geometry", "geometry" ], "Problem": "[size=134][b]Problem 6[/b] (13 points)[/size]\r\n[size=134][b]Casmir Effect[/b][/size]\r\n\r\n[b]Part-A[/b] What is the pressure of ideal gases when the temperature goes to absolute zero? (2 points)\r\n\r\n[b]Part-B[/b] Casmir proposed in the 1950\u2019s that vacuum is actually filled with virtual electromagnetic waves, and the energies stored in the vacuum produces an observable force between two parallel metallic plates separated by distance $d$. The electromagnetic energy stored between the plates is given in Quantum Mechanics by $E=\\frac12 \\sum_{nR)$ from the sphere center. The image charge $q'$ should be inside the sphere (outside the space outside the sphere) and on the line joining the sphere center and $q$, as shown in Figure-B. The combined contribution of $q$ and $q'$ is to make the potential everywhere on the sphere surface zero. Find the position and the value of $q'$. (4 points)\r\n\r\n[b]Part-B[/b]\r\nAs shown in Figure-C, a problem often encountered in atomic force microscopy is to determine the force of the sample on the probe tip, which is a conductor sphere of radius $R$ held at electric potential $V$ at distance $h_{0}$ from a large conducting sample surface at zero potential. In the following we will apply the 'image charge' method step by step to find the force.\r\n\r\n(a) We start by putting a point charge $q_{0}$ inside the sphere such that the sphere surface is an equal-potential one with voltage $V$, while ignoring the effect of the conductor plate. Determine $q_{0}$ in terms of $V$ and $R$. (1 point)\r\n\r\n(b) Determine the value and position ($h_{1}$) of the image charge of $q_{0}$ and call it $q_{1}$, such that the combined contribution of them is to make the sample surface a zero-potential plane. (2 points)\r\n\r\n(c) The presence of $q_{1}$ now makes the conductor sphere surface no longer equal-potential. Put another point charge $q_{2}$ inside the sphere such that the combined effect of $q_{0}, q_{1}$ and $q_{2}$ is to make the potential on the sphere surface equal to $V$ again. Determine $q_{2}$ and its position $h_{2}$. (2 points)\r\n\r\n(d) Repeat (b) to determine the image charge $q_{3}$ of $q_{2}$, and repeat (c) to find the image charge $q_{4}$ of $q_{3}$. Derive the general expression between $h_{2n}$ and $h_{2(n+1)}$, $q_{2n}$ and $q_{2(n+1)}$, and $q_{2n+1}$ and $q_{2(n+1)+1}$, $n=0,1,2,\\cdots$ (3 points)\r\n\r\n(e) Derive the total force of the sample on the sphere [b][i][u]in the form of a summation over an infinite series[/u][/i][/b]. (2 points)\r\n\r\n(f) Suppose the force in (e) is $1.1\\times10^{-12}\\,\\text N$ when $V=V_{0}$, $R=1.0 \\times10^{-8}\\,\\text m$, and $h_{0}=5.0\\times10^{-8}\\,\\text m$, find the force when $V= 2V_{0}, R = 1.0\\,\\text m$, and $h_{0}= 5.0\\,\\text m$. (4 points)\r\n\r\n(g) Given $R/h_{0}= 1/51$, to determine the force in (e) up to ~1% accuracy, what are the terms that should be kept? (2 points)\r\n\r\n(h) Estimate the order of magnitude of the relative error caused by the terms neglected in (g). (2 points)" } { "Tag": [ "geometry", "perimeter" ], "Problem": "The areas of two squares differ by 100 square units, and the perimeters of the two squares differ by 10 units. What is the perimeter, in units, of the smaller square?", "Solution_1": "$ a^2\\minus{}b^2 \\equal{} 100$\r\n\r\n$ 4a\\minus{}4b\\equal{}10$\r\n$ a\\minus{}b\\equal{}\\frac{5}{2}$\r\n\r\n$ (a\\plus{}b)\\frac{5}{2} \\equal{} 100$\r\n$ a\\plus{}b\\equal{} 40$\r\n$ a\\minus{}b\\equal{}\\frac{5}{2}$\r\n_________________(+)__\r\n\r\n$ 2a \\equal{} \\frac{85}{2}$\r\n$ a\\equal{}\\frac{85}{4}$\r\n$ 4b\\equal{} \\boxed{75}$" } { "Tag": [ "function", "calculus", "derivative", "parameterization", "real analysis", "real analysis unsolved" ], "Problem": "Hey I had a quick question on differentials\r\n\r\nWhen you differentiate\r\n\r\n$p_{i}= \\sqrt{(x_{i}-x_{u})^{2}+(y_{i}-y_{u})^{2}+(z_{i}-z_{u})^{2}}$\r\n\r\ndon't you get:\r\n\r\n$\\Delta p_{i}=-\\Big( \\frac{ (x_{i}-x_{u}) \\Delta x_{u}+(y_{i}-y_{u}) \\Delta y_{u}+(z_{i}-z_{u}) \\Delta z_{u}}{p_{i}-b_{u}}\\Big)+\\Delta b_{u}$\r\n\r\nand not\r\n\r\n$\\Delta p_{i}= \\frac{ (x_{i}-x_{u}) \\Delta x_{u}+(y_{i}-y_{u}) \\Delta y_{u}+(z_{i}-z_{u}) \\Delta z_{u}}{p_{i}-b_{u}}+\\Delta b_{u}$\r\n\r\nwhich is what my book says?", "Solution_1": "The notation is unclear. $p$ is a function of up to 6 variables. Which ones are variables for differentiation and which are treated as fixed parameters? Also $b_{u}$ is undefined and $\\Delta$ usually means something unrelated to this problem." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Prove that $ ab^5 \\plus{} bc^5 \\plus{} ca^5\\ge abc(a^2b \\plus{} b^2c \\plus{} c^2a)$, for $ a,b,c > 0$.", "Solution_1": "Please delete this post.", "Solution_2": "Hint: Try using AM GM with Power mean inequality.", "Solution_3": "$ \\frac {8ab^5 \\plus{} 2bc^5 \\plus{} 11ca^5}{21} \\ge a^3b^2c$ by AM-GM. Add this cyclically. Can we use Muirhead, I thought it had to be symmetric in all variables?", "Solution_4": "dgreenb801 i have tried muirhead on this one the case comes of to be somthing like this...\r\n\r\n$ a^5c\\plus{}a^5b\\plus{}b^5a\\plus{}b^5c\\plus{}c^5a\\plus{}c^b\\ge\\;a^3b^2c\\plus{}a^3bc^2\\plus{}ab^3c^2\\plus{}a^2b^3c\\plus{}ab^2c^3\\plus{}a^2bc^3$\r\n\r\nnow we get a sum of two similar inequalities..that we require to prove\r\n\r\ni think this can be AM GM (but seriously i don't know how...i'm still trying)", "Solution_5": "again: see here\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=103761 :) :)" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "If $ a_{0}\\equal{}a_{1}\\equal{}1$ and $ a_{n\\plus{}1}\\equal{}2a_{n}\\minus{}a_{n\\minus{}1}$ then $ a_n$ is a perfect square for all $ n$ and also $ a_{n}\\equal{}b_{n}^2$ where $ b_{n}$ is the general term of the sequence given by $ b_{0}\\equal{}b_{1}\\equal{}1$ and $ b_{n\\plus{}1}\\equal{}kb_{n}\\minus{}b_{n\\minus{}1}$ for all $ n\\geq1$.", "Solution_1": "[quote=\"manjil\"]If $ a_{0} \\equal{} a_{1} \\equal{} 1$ and $ a_{n \\plus{} 1} \\equal{} 2a_{n} \\minus{} a_{n \\minus{} 1}$ then $ a_n$ is a perfect square for all $ n$ and also $ a_{n} \\equal{} b_{n}^2$ where $ b_{n}$ is the general term of the sequence given by $ b_{0} \\equal{} b_{1} \\equal{} 1$ and $ b_{n \\plus{} 1} \\equal{} kb_{n} \\minus{} b_{n \\minus{} 1}$ for all $ n\\geq1$.[/quote]\r\n\r\n$ a_n\\equal{}1\\equal{}1^2$ $ \\forall n$\r\n$ b_n\\equal{}1$ $ \\forall n$ and $ b_{n \\plus{} 1} \\equal{} 2b_{n} \\minus{} b_{n \\minus{} 1}$ for all $ n\\geq1$" } { "Tag": [ "summer program", "MathPath" ], "Problem": "Sherwin buys three cans of beer and two math textbooks. Each math textbook costs four and a half times as much as a bottle of beer. If he spent all of his money on beer instead of beer and math, how many six-packs of beer could he have bought, assuming the price of a can of beer stays the same?", "Solution_1": "[quote=\"Iron Fist\"]Sherwin buys three cans of beer and two math textbooks. Each math textbook costs four and a half times as much as a bottle of beer. If he spent all of his money on beer instead of beer and math, how many six-packs of beer could he have bought, assuming the price of a can of beer stays the same?[/quote] [hide=\"Solution\"] Let's say the beer costs $x$ dollars, which means the math textbooks each cost $4.5x$. In total, Sherwin spends $3x+2\\times4.5x=3x+9x=12x$. This means he can buy 12 beers or $\\boxed{2}$ six-packs of beer. :D [/hide]", "Solution_2": "im truly amazed how did u come up wit that equation... i really feel like everyone here is naturally good at math, im struggling so hard to try to understand", "Solution_3": "Lol, it comes from practice... and being forced to do math when you're young...", "Solution_4": "[hide]Let b be the price of beer.\nHe spends a total of $2(4\\frac{1}{2})b+3b=9b+3b=12b$\nso he could of bought 2 six packs.[/hide]", "Solution_5": "[hide]Say the price of a beer is $\\text{s}$.\n\nSo, the amount of money I was supposed to spend on both mathbooks and beer is $4.5x*2+3x=9x+3x=12x$\n\nSo with the amount of money to buy 12 beers, I can get $2$ six packs of beer! :D \n\nBut with my chinese bargaining skills, I could have probably gotten 24 beers for the price of 12. ;) [/hide]", "Solution_6": "Yeah, that is my answer.\r\n\r\nSherwin, are you okay?", "Solution_7": "[hide=\"answer\"]4[/hide]", "Solution_8": "[quote=\"daermon\"][hide=\"answer\"]4[/hide][/quote]\r\nThat's wrong. :?", "Solution_9": "[quote=\"lotrgreengrapes7926\"][quote=\"daermon\"][hide=\"answer\"]4[/hide][/quote]\nThat's wrong. :?[/quote]\r\n\r\nGood job lotrgreengrapes7926, you are correct for saying that is incorrect. ;)", "Solution_10": "The answer is 2, 2 books cost the same amount as 9 beer cans, so there are 12 beer cans in all, so there are 2 packs.\r\n\r\nSherwin, I'm sending you to church every time you get drunk.", "Solution_11": "oops. you guys are right.\r\n\r\ndrunken_math, are you the same Sherwin that came in second at the MathPath ping pong tournament?", "Solution_12": "[hide]Call the beer price $b$. That means the textbook money is $\\frac{9}{2}b$. $3b+2\\left(\\frac{9}{2}\\right)=12b$. $\\frac{12}{6}=2$, so he can buy $\\boxed{2}$ 6-packs of beer.[/hide]", "Solution_13": "(4.5*2+3)/6=2", "Solution_14": "[hide]he buys beer and math textbooks?!?!? :| [/hide]\n[hide]assume that a beer's price=b and textbook's price is m...We know that m=4,5b...we also know that he has 3b +2m=3b+2 *(4,5b)=3b+9b=12b 12b/6b=[color=red]2[/color][/hide]", "Solution_15": "[quote=\"Iron Fist\"]Sherwin buys three cans of beer and two math textbooks. Each math textbook costs four and a half times as much as a bottle of beer. If he spent all of his money on beer instead of beer and math, how many six-packs of beer could he have bought, assuming the price of a can of beer stays the same?[/quote]\r\n\r\n2 six packs" } { "Tag": [ "real analysis", "real analysis unsolved" ], "Problem": "Let F be an ordered field in which every strictly monotone increasing sequence bounded above converges. Prove that F is complete.", "Solution_1": "For any Cauchy sequence, consider its limit inferior." } { "Tag": [ "search", "LaTeX" ], "Problem": "This was the 1st problem in the South African talent search round 5.\r\n\r\nProve that there exist\r\na) 5 points in the plane so that among all the triangles with vertices among these points, there are 8 right-angled triangles.\r\nb) 64 points in the plane so that there are at least 2005 right-angled triangles with vertices among these points.", "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58598]www.artofproblemsolving.com/Forum/viewtopic.php?t=58598[/url]", "Solution_2": "Oh, sorry. Well South Africa quite often uses problems from other olympiads, I should have known it would be somewhere", "Solution_3": "Don't worry about it! I was just posting the link as a reference.", "Solution_4": "Oh good. Well thanks :)", "Solution_5": "I think this works for part a: (0,0),(a,0),(0,a),(a,a),(a/2,a/2). By the way how do you get that fancy mth expository font that everyone writes in? Copy it from Word Equation Editor?", "Solution_6": "You mean $\\text {this }$ kind of writing? It's called $\\text{\\LaTeX} \\text { and can be used to typeset mathematics like this: } \\sqrt{5+\\frac{x}{\\sqrt{x+\\sqrt{2}}}}.$\r\n\r\nClick here to begin how to learn it: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php\r\n\r\n:D :D :D" } { "Tag": [ "algebra", "polynomial", "calculus", "integration", "function", "domain", "induction" ], "Problem": "Let $ A\\left[x\\right]$ be the ring of polynomials with coefficients in a commutative ring $ A$. Let $ f\\equal{} a_0 \\plus{} a_1x \\plus{} a_2x^2 \\plus{} .... \\plus{} a_nx^n$ be an element in $ A\\left[x\\right]$.\r\n\r\n[b]1.[/b] Show that $ f$ is a unit iff $ a_0$ is a unit in $ A$ and $ a_1$, $ a_2$, ..., $ a_n$ are nilpotent.\r\n\r\n[b]2.[/b] Show that $ f$ is a zero divisor iff there exists $ a\\neq 0$ in $ A$ such that $ a\\cdot f\\equal{}0$.", "Solution_1": "1. Let $g=b_0 + b_1 x + ... + b_m x^m$ be an inverse polynomial of $f=a_0 + a_1 x + \\dotsc + a_n x^n$. Then $b_0 a_0 = 1$, so that $a_0$ is a unit in $A$. The trick is now to consider the nilradical of $A$, which consists of the nilpotent elements of $A$, but which is also the intersection of all prime ideals of $A$. So let's take a prime ideal $P$ of $A$. When $\\overline{f} , \\overline{g}$ are the corresponding polynomials over $A/P$, we have $\\overline{f} \\overline{g} = 1$, i.e. $\\overline{f}$ is a unit in $(A/P)[x]$. But since $A/P$ is an integral domain, this means that $\\overline{f}$ is constant (this is well-known, and can be proved by working with the degree of a polynomial). Thus, all coefficients $\\overline{a_i}$ with $i>0$ have to be zero in $A/P$, i.e. we have $a_i \\in P$. Because $P$ was arbitrary, $a_i$ is nilpotent.\r\n\r\nNow let $a_0$ be a unit in $A$ and $a_i$ nilpotent for all $i>0$. We have to find an inverse polynomial $g = b_0 + b_1 x + \\dotsc$ of $f$. We'll do that \"backwards\": The condition $f g = 1$ means $a_0 b_0 = 1$ and $a_0 b_k + \\dotsc + a_k b_0 = 0$ for all $k>0$. Well, we find an appropriate $b_0$ because $a_0$ is a unit in $A$. The recursive formula $b_k = -b_0 (a_1 b_{k-1} + \\dotsc + a_k b_0)$ determines all $b_i$, s.t. $f$ is a unit in $A[[x]]$. But it is easy to see that the $b_i$ vanish once $i$ is greater than $n=deg(f)$ and the nilpotency indices of the $a_i$. So $f$ is a unit in $A[x]$.\r\n\r\n2. If $a*f = 0$ for some $0 \\neq a \\in A$, then $f$ is of course a zero divisor in $A[x]$. Now let $f$ be a zero divisor in $A[x]$, say $fg = 0$ for some $0 \\neq g \\in A[x]$. Write $f = a_0 + a_1 x + \\dotsc + a_n x^n$. First assume $g a_i=0$ for all $i$. There's a coefficient $a \\neq 0$ in $g$. It suffices $a f_i= 0$ for all $i$, hence $af = 0$, and we're done. Else there's a maximal $i \\leq n$ s.t. $g a_i \\neq 0$. Then $0 = gf = g a_{i+1} x^{i+1} + \\dotsc + g a_n x^n$. The potentially hightest monomial in this polynomial is $b a_i x^{m+i}$, when $b x^m$ is the highest monomial in $g$. Thus, $b a_i = 0$, and the hightest monomial in $a_i g$ is smaller than the one in $g$. But every monomial in $a_i g$ occours - apart from the coefficient - in $g$. So we can make an induction and get an $g$ s.t. $g a_i = 0$ for all $i$. But this implies, as we have seen, the existence of some $0 \\neq a \\in A$ s.t. $a f = 0$.", "Solution_2": "When you say P von A, do you mean A/P (A mod P)? :lol: thanks for your help", "Solution_3": "I meant \"P of A\", see the edit. A/P is the factor ring A modulo P ...", "Solution_4": "How do we know that the nilradical is the intersection of all prime ideals in A?", "Solution_5": "Sorry, but why does A/P being a domain imply that f must be constant. :)", "Solution_6": "You've posted the first question in another topic. For the second:\r\n\r\nWhen $R$ is a domain, then also $R[x]$, and the degree of polynomials satisfies $\\deg(f*g)=\\deg(f)+\\deg(g)$. So, $f*g=1$ implies $\\deg(f) = 0$, s.t. $f$ is constant.", "Solution_7": "[quote=\"-oo-\"] The recursive formula $ b_k \\equal{} \\minus{} b_0 (a_1 b_{k \\minus{} 1} \\plus{} \\dotsc \\plus{} a_k b_0)$ determines all $ b_i$, s.t. $ f$ is a unit in $ A[[x]]$. But it is easy to see that the $ b_i$ vanish once $ i$ is greater than $ n \\equal{} deg(f)$ and the nilpotency indices of the $ a_i$. So $ f$ is a unit in $ A[x]$.\n[/quote]\n\nCould u clarify this statement please? why we go to study $ f$ in $ A[[x]]$?how we know $ f$ will be unit of choosing $ b_i$? \n\nThank you $ \\minus{}oo\\minus{}$\n\n[color=red][[b]Moderator edit:[/b] The statement by -oo- quoted above is wrong. In fact, the $b_i$ don't necessarily vanish once $i$ is greater than $ n \\equal{} deg(f)$ and the nilpotency indices of the $ a_i$. Instead, they necessarily vanish once $i$ is greater than the sum of the nilpotency indices of the $ a_i$. This is not immediately trivial from -oo-'s argument, however. Correct solutions of problem 1 has been given by -oo- in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=89417 . Problem 2 has been solved at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=425496 .][/color]" } { "Tag": [ "quadratics", "geometry", "rectangle", "perimeter", "algebra", "Pythagorean Theorem" ], "Problem": "Can you help me with these two questions please.\r\n\r\n\r\nThe sides of a right angled triangle are 2x cm, (x+3)cm, and the hypotenuse is (2x + 2.4)cm. Use Pythagorus' theorem to form a quadratic equation in x and then solve it to find x. \r\n\r\n\r\nThe area of a rectangle with length (x + 4.6)cm and width (x - 2.1)cm is 134.63cm squared. a) Form a quadratic equation and solve it to find x to two decimal places. b) What is the rectangle's perimeter to one decimal place? \r\n\r\n\r\nCharikaar", "Solution_1": "Is this homework?\r\n\r\n-sigh.", "Solution_2": "NO.\r\n\r\nthese are two questions on AQA past papers that i can't solve.\r\n\r\nI know little about pythagraous\r\n\r\na2+b2=c2\r\n\r\nbut can't use it to solve the above problems so please help.\r\n\r\n\r\nCharikaar", "Solution_3": "Okay, so you know the Pythagorean Theorem. That's a start. How do you think you should apply it to the first question?", "Solution_4": "[hide=\"Hint for problem 1\"]Call the expressions for the 2 legs a and b, and call the hypotenuse c.\nThrough the pythagorean theorem, we know that [tex]a^2+b^2=c^2[/tex] where a and b are the legs and c is the hypotenuse.\nWe can \"plug in\" the given lengths of the sides to the pythagorean theorem to get [tex](2x)^2+(x+3)^2=(2x+2.4)^2[/tex][/hide]\r\n\r\nP.S: No offense, but I think this problem belongs in the \"Getting Started\" section of the High School Forum, or mabye even in the upper levels of the Middle School Forum.", "Solution_5": "thaks for the hint.\r\n\r\ni think i can do it now...........let me have a it a go.\r\ni think i will have to use (a+b)^2=a^2+b^2+2ab\r\n\r\ncharikaar" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ J$ be a finitely generated ideal such that $ J\\equal{}J^2$. Prove that $ J$ is principal and generated by an idempotent.\r\n\r\nThis looks like a straightforward question, but I do not know how to prove this is principal or how this is generated by an idempotent. How should I go about showing this?", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=181620\r\n\r\n darij" } { "Tag": [ "function", "vector", "algebra", "linear equation", "calculus", "calculus computations" ], "Problem": "On a recent test, I had a problem of the form\r\n$ ay'' \\plus{} by' \\plus{} cy^2 \\equal{} 0$\r\nThe last term was obviously supposed to be $ cy$, but I was just curious whether the problem as printed was actually solvable. I believe it isn't, but I have no idea how to go about proving something like that.", "Solution_1": "You need to distinguish between the essentially algebraic question \"does there exist a technique for writing solutions to this in closed form\" and the analytic question \"does this differential equation have solutions.\" I'm usually more interested in the second question than the first.\r\n\r\nAs for the first question: I don't know of any such techniques, but I won't swear that that's the final answer.\r\n\r\nAs for the second question: the hypotheses of the existence-uniqueness theorem are met locally at every point. (I am assuming that $ a\\ne 0.$) That is, given any point $ (t_0,y_0,y_0')\\in\\mathbb{R}^3$ there exists a $ \\delta>0$ and a unique twice-differentiable function $ \\phi$ on $ (t_0\\minus{}\\delta,t_0\\plus{}\\delta)$ such that $ a\\phi''(t)\\plus{}b\\phi'(t)\\plus{}c(\\phi(t))^2\\equal{}0,\\phi(t_0)\\equal{}y_0,$ and $ \\phi'(t_0)\\equal{}y_0'.$\r\n\r\nOf course, the fact that this is a non-linear equation takes away several of our tools. The set of solutions is not a vector space. Independence is not a meaningful word. The Wronskian doesn't do much good. And we can't assume that solutions exist globally. (That is, we might need the $ \\delta$ in what I said above.)" } { "Tag": [ "inequalities" ], "Problem": "I wanted to see if anyone can help me out with this problem:\r\n \\sqrt a 2 + 8abc + \\sqrt b 2 +8abc + \\sqrt c 2+8abc \\leq 3(a+b+c) but you may find this puzzle more interesting:", "Solution_1": "[quote=\"Cauchy_Schwartz\"]I wanted to see if anyone can help me out with this problem:\n \\sqrt a 2 + 8abc + \\sqrt b 2 +8abc + \\sqrt c 2+8abc \\leq 3(a+b+c)[/quote]\n\nShould this be solved using your nickname? :D\n(BTW, maybe it should be \\sqrt a 2 + 8bc, not 8abc ? Else, it's not really homogeneous...)\n\n[quote=\"Cauchy_Schwartz\"]but you may find this puzzle more interesting:[/quote]\r\n\r\nWell, I feel this should be moved to the \"Ultimative Unsolved Challenges\" section...\r\n\r\n Darij", "Solution_2": "Thanks for the observation, you are right it is 8bc, 8ca, 8ab, i did a mistake when i wrote it. I see that my problem wasn't a real mystery, I should try composing something more difficult.", "Solution_3": "Concerning the inequality, I have posted [url=http://mathlinks.ro/viewtopic.php?p=15777]a proof in the inequalities subforum[/url].\r\n\r\n Darij", "Solution_4": "yes i found the puzzle very interesting and i bet it is a horse!", "Solution_5": "Kangaroo maybe :D" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "For infinitely many points on the plane it is known that distance between any two of them is a whole number. Prove that they all are on the same line.", "Solution_1": "You should read this book as a reference:\r\n[hide=\"Challenging Mathematical Problems with Elementary solutions\"]http://books.google.com/books?id=nowbdLTlr7YC&dq=problems+with+elementary+solutions+yaglom&printsec=frontcover&source=web&ots=JAgYNY3shM&sig=t5i_8A2yGbSXZDQ6tuwUTgMCHsM#PPA5,M1[/hide]\r\nit is problem 109b there :wink:" } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Does there exist a power of 2 such that permuting its digits gives another power of 2?\r\n\r\n[i]Iran 1999[/i]", "Solution_1": "Can this power have 0 in it?\r\nFor example is 12 a permutation of 102?\r\nI ask this because the problem would be easier if the 2 powers had the same number of digits.", "Solution_2": "They have the same number of digits I think :)", "Solution_3": "I think these two fact can help\r\n\r\n1) For every n, n - p(n) divide by 9, where p(n) number, receiving from n by permutating its digits.\r\n2) 1/ 9 < n/p(n) < 9.", "Solution_4": "Sorry, zeros can be placed first Mihai :)", "Solution_5": "The problem is very simple. Suppose that 2^n is formed by a permutation of the digits of 2^m and that m>n. In particular, 2^m and 2^n have the same sum of digits, which means they are congruent modulo 9. From here we find that m-n>=6. Thus, 2^m>=64*2^n and so 2^m has more digits than 2^n, false.", "Solution_6": "[quote=\"harazi\"]The problem is very simple. Suppose that 2^n is formed by a permutation of the digits of 2^m and that m>n. In particular, 2^m and 2^n have the same sum of digits, which means they are congruent modulo 9. From here we find that m-n>=6. Thus, 2^m>=64*2^n and so 2^m has more digits than 2^n, false.[/quote]\n\nTake a look at what Arne said, harazi :\n[quote=\"Arne\"]Sorry, zeros can be placed first Mihai :) [/quote].\r\n\r\nNow you see that $2^m$ and $2^n$ needn't have the same number of digits.", "Solution_7": "Yes, I heard that long time before. But I would really like to see a solution of this problem.", "Solution_8": "it was the first or fourth problem of iran second round. and im sure the number of digits are equale.\nand in this case it doesnt hve any solution" } { "Tag": [ "function", "AMC", "USA(J)MO", "USAMO", "calculus", "derivative", "Putnam" ], "Problem": "The number of odd divisors of $n$ equals the number of arithmetic progressions (in steps of one) which sum to $n$.\r\n\r\nFor example, 15 has 4 odd divisors (1,3,5,15) and 15 is the sum of four arithmetic progressions (15, 1+2+3+4+5, 7+8, 4+5+6).\r\n\r\nThis can be solved with a simple bijection, but it can also be solved with generating functions.\r\n\r\nLet $A(n)$ be the number of odd divisors of $n$. Let $B(n)$ be the number of ways to express $n$ as an arithmetic progression. We have,\r\n\\[A(n)=[x^n]\\sum_{i=1}^{\\infty} \\sum_{k=1}^{\\infty} x^{k(2i-1)}\\\\\r\n=[x^n] \\frac{x}{1-x}+\\frac{x^3}{1-x^3}+\\frac{x^5}{1-x^5}+\\ldots\\]\r\n(or you could sum over $i$ then $k$)\r\n\\[B(n)=[x^n]\\sum_{k=1}^{\\infty} \\frac{x^{\\frac{k(k+1)}{2}}}{1-x^k}\\\\\r\n=[x^n] \\frac{x}{1-x}+\\frac{x^3}{1-x^2}+\\frac{x^6}{1-x^3}+\\ldots\\]\r\nHence we need to show,\r\n\\[\\frac{x}{1-x} + \\frac{x^3}{1-x^3} + \\frac{x^5}{1-x^5} + \\ldots=\\frac{x}{1-x} + \\frac{x^3}{1-x^2} + \\frac{x^6}{1-x^3} + \\ldots\\]\r\nI'm not sure how to do this and I can't find this solution anywhere on the internet. Does anyone have any ideas?", "Solution_1": "I didn't even understand hoe you het $ A(n) $ and $B(n) $ . Can you plz explain. I'm kinda new in this field!! :(", "Solution_2": "[quote=\"Rushil\"]I didn't even understand hoe you het $A(n)$ and $B(n)$ . Can you plz explain. I'm kinda new in this field!! :([/quote]\r\n\r\nSure. $[x^n]$ denotes the coeffecient of $x^n$. For example, $[x^2]2x^3+5x^2+3x=5$.\r\n\r\nTo understand the generating function for $A(n)$, note that $(2i-1)$ will hit every odd number. IF there exsists a $k$ such that $k*(2i-1)=n$, then $(2i-1)$ is an odd divisor of $n$ AND the term $x^{k(2i-1)}$ adds 1 to $[x^n]$.\r\n\r\nTo understand the generating function for $B(n)$, note that the first arithmetic progression of legnth $k$ is $1+2+\\ldots+k=\\frac{k(k+1)}{2}$. Add 1 to each term to get the next arithmetic progression of length $k$. $(1+1)+(2+1)+\\ldots+(k+1)=1+2+\\ldots+k+k=\\frac{k(k+1)}{2}+k$. It should be clear that if $\\frac{k(k+1)}{2}+ik=n$ for some $i$, that $n$ can be expressed as an arithmetic progression of length $k$ AND the term $x^{\\frac{k(k+1)}{2}+ik}$ will contribute 1 to $[x^n]$. Thus,\r\n\\[ B(n)=[x^n]\\sum_{k=1}^{\\infty} \\sum_{i=0}^{\\infty} x^{\\frac{k(k+1)}{2}+ik}=[x^n]\\sum_{k=1}^{\\infty} \\frac{x^{\\frac{k(k+1)}{2}}}{1-x^k} \\]\r\n\r\n\r\nGenerating functions are a great tool. I will use the following examples (which might be hard to follow) of contest problems to make my point.\r\n\r\n\r\n[b]Problem 5 of the 1986 USAMO:[/b] [i]If $p$ is a partition, $f(p)=$ number of 1's in $p$, and $g(p)=$ number of distinct integer sizes in $p$. Show $\\sum f(p)=\\sum g(p)$, where the sums are taken over all partitions of $n$.\n\nFor example, the partitions of 3 are 3, 2+1, 1+1+1. $f(3)=0$, $f(2+1)=1$, $f(1+1+1)=3$ and $g(3)=1$, $g(2+1)=2$, $g(1+1+1)=1$.\n\nIn other words, if you take a random partition of $n$, the expected number of 1's equals the expected number of distinct integer sizes.[/i]\r\n\r\nHopefully it is clear that,\r\n\r\n\\begin{eqnarray*} \\sum f(p)&=& [x^n](x+2x^2+3x^3+\\ldots)(1+x^2+x^4+\\ldots)(1+x^3+x^6+\\ldots)\\cdots\\\\ \\ &=&[x^n]\\frac{x}{(1-x)^2 (1-x^2)(1-x^3)\\cdots} \\end{eqnarray*}\r\nNow let,\r\n\r\n\\begin{eqnarray*} G(x,y)&=&(1+xy+x^2y+\\ldots)(1+x^2y+x^4y+\\ldots)(1+x^3y+x^6y+\\ldots)\\cdots\\\\ \\ &=&\\left(1+\\frac{xy}{1-x}\\right)\\left(1+\\frac{x^2y}{1-x^2}\\right)\\left(1+\\frac{x^3y}{1-x^3}\\right)\\cdots \\end{eqnarray*}\r\n\r\nHopefully it is clear that,\r\n\r\n\\begin{eqnarray*} \\sum g(p)&=&[x^n]\\left.\\frac{\\partial G(x,y)}{\\partial y}\\right]_{y=1}\\\\ \\ &=& [x^n]\\left.G(x,y)\\sum_{k=1}^{\\infty} \\frac{x^k}{1-x^k+x^ky}\\right]_{y=1}\\\\ \\ &=&[x^n]\\frac{1}{(1-x)(1-x^2)(1-x^3)\\cdots} \\frac{x}{1-x}\\\\ \\ &=&\\sum f(p) \\end{eqnarray*}\r\n\r\n\r\nThe derivative was evaluated using the simple formula for differentiation of a product,\r\n\\[ \\frac{\\partial}{\\partial x} f_1(x)f_2(x)\\cdots f_n(x)=f_1(x)f_2(x)\\cdots f_n(x)\\sum_{k=1}^{n}\\frac{f_k'(x)}{f_k(x)} \\]\r\n\r\n\r\n[b]Problem B4 of the 1991 Putnam:[/b] [i]$p$ is a prime greater than 2. Prove[/i]\r\n\\[ \\sum_{n=0}^{p} {p\\choose n} {{p+n}\\choose n} \\equiv 2^p+1\\mod p^2 \\]\r\n\r\n\\begin{eqnarray*} \\sum_{n=0}^{p} {p\\choose n} {{p+n}\\choose n}&=&[x^p]\\sum_{n=0}^{p} {{p\\choose n}(1+x)^{p+n}}\\\\ \\ &=&[x^p](1+x)^p(1+(1+x))^p\\\\ \\ &=&\\sum_{n=0}^{p} {{p\\choose n}{p\\choose p-n}2^n}\\\\ \\ &\\equiv& 2^p+1\\mod p^2 \\end{eqnarray*}\r\n\r\n\r\n[b]Problem A1 of the 2003 Putnam:[/b] [i]Let $c(n)$ denote the number of partitions of $n$ such that no pair of integers in a partition differs by more than 1. For example, the valid partitions of 4 are 4, 2+2, 2+1+1, 1+1+1+1. So $c(4)=4$. Compute $c(n)$.[/i]\r\n\r\n\\begin{eqnarray*} c(n)&=&[x^n]\\sum_{k=1}^{\\infty} {\\frac{x^k}{1-x^k}\\frac{1}{1-x^{k+1}}}\\\\ \\ &=&[x^n]\\frac{1}{(1-x)^2}\\sum_{k=1}^{\\infty} \\frac{x^k}{(1+x+\\ldots+x^{k-1})(1+x+\\ldots+x^k)}\\\\ \\ &=&[x^n]\\frac{1}{(1-x)^2} \\lim_{N\\rightarrow\\infty} \\sum_{k=1}^{N} (\\frac{1}{1+x+\\ldots+x^{k-1}}-\\frac{1}{1+x+\\ldots+x^k})\\\\ \\ &=&[x^n]\\frac{1}{(1-x)^2} \\lim_{N\\rightarrow\\infty} (1-\\frac{1-x}{1-x^{N+1}})\\\\ \\ &=&[x^n]\\frac{x}{(1-x)^2}\\\\ \\ &=& n \\end{eqnarray*}\r\n\r\n\r\n[b]Problem A6 of the 1990 Putnam:[/b] [i]Call an ordered pair $(A,B)$ of subsets of $[n]$ \"admissible\" if $a>|B|$ for each $a\\in A$ and $b>|A|$ for each $b\\in B$. How many admissible ordered pairs of subsets of $[10]$ are there?[/i]\r\n\r\nLet $c(n)$ denote the number of admissible ordered pairs of subsets of $[n]$. Hopefully it is clear that,\r\n\\[ c(n)=\\sum_{a+b\\leq n} {{n-b\\choose a}{n-a\\choose b}} \\]\r\nThus,\r\n\r\n\\begin{eqnarray*} c(n)&=&[x^n]\\sum_{k=0}^{\\infty}x^k\\sum_{a=0}^{\\infty}{ {a+k\\choose a}x^a}\\sum_{b=0}^{\\infty} {{b+k\\choose b}x^b}\\\\ \\ &=&[x^n]\\sum_{k=0}^{\\infty}\\left(\\frac{1}{(1-x)^{k+1}}\\right)^2\\\\ \\ &=&[x^n]\\frac{1}{(1-x)^2}\\sum_{k=0}^{\\infty}\\left(\\frac{x}{(1-x)^2}\\right)^k\\\\ \\ &=&[x^n]\\frac{1}{1-3x+x^2}\\\\ \\ &=&[x^n]\\frac{1}{(1-x)^2-x}\\\\ \\ &=&[x^{2n}]\\frac{1}{(1-x^2)^2-x^2}\\\\ \\ &=&[x^{2n}]\\frac{1}{(1-x-x^2)(1+x-x^2)}\\\\ \\ &=&[x^{2n}]\\frac{1}{2}\\left(\\frac{1+x}{1-x-x^2}+\\frac{1+(-x)}{1-(-x)-(-x)^2}\\right)\\\\ \\ &=&[x^{2n}]\\frac{1+x}{1-x-x^2}\\\\ \\ &=&F_{2n}+F_{2n+1}\\\\ \\ &=& F_{2n+2} \\end{eqnarray*}\r\n\r\nSo $c(10)=F_{22}=17711$.\r\n\r\nNote that we used the known generating function for Fibanacci Numbers, $F_n=[x^n]\\frac{x}{1-x-x^2}$.\r\n\r\n\r\n[b]Problem B1 of the 1996 Putnam:[/b] [i]A subset of $[n]$ is selfish if it contains its own cardinality as an element, and the other elements are greater than the cardinality. Let $b(n)$ denote the number of selfish subsets of $[n]$. Compute $b(n)$.[/i]\r\n\r\n\\begin{eqnarray*} b(n)&=&[x^n]\\sum_{m=0}^{\\infty}x^m\\sum_{k=0}^{\\infty}{x^k(x+x^2+x^3+\\ldots)^{k-1}}\\\\ \\ &=&[x^n]\\frac{x}{1-x-x^2}\\\\ \\ &=&F_n \\end{eqnarray*}\r\n\r\n\r\n[b]Problem A5 of the 1956 Putnam:[/b] [i]How many binary strings of length $n$ with $k$ 1's are there such that no 1's are next to each other?[/i]\r\n\r\nLet $S$ denote the number of binary strings.\r\n\r\n\\begin{eqnarray*} S&=&[x^n](x+x^2+x^3+\\ldots)(x^2+x^3+x^4+\\ldots)^{k-1}(1+x+x^2+\\ldots)\\\\ \\ &=&[x^n]\\frac{x^{2k-1}}{(1-x)^{k+1}}\\\\ \\ &=&[x^{n-2k+1}]\\frac{1}{(1-x)^{k+1}}\\\\ \\ &=&{n-k+1\\choose k} \\end{eqnarray*}\r\n\r\n\r\n[b]Unknown Source:[/b] [i]An increasing sequence of nonnegative integers $a_0,a_1,a_2,\\ldots$ has the following property: every nonnegative integer $n$ can be expressed uniquely in the form $n=a_i+2a_j$ where $i$ and $j$ are not necessarily distinct. Determine $a_{2004}$.[/i]\r\n\r\nLet,\r\n\\[ F(x)=\\sum_{i=0}^{\\infty} x^{a_i} \\]\r\nThen the unique representation property gives,\r\n\\[ F(x)F(x^2)=\\sum_{i=0}^{\\infty}\\sum_{j=0}^{\\infty} x^{a_i+2a_j}=\\sum_{n=0}^{\\infty} x^n=\\frac{1}{1-x} \\]\r\nThus,\r\n\\[ F(x^2)F(x^4)=\\frac{1}{1-x^2}=\\frac{F(x)F(x^2)}{1+x} \\]\r\nThus,\r\n\\[ F(x)=(1+x)F(x^4) \\]\r\nBy iteration,\r\n\\[ F(x)=(1+x)(1+x^4)(1+x^{16})(1+x^{64})\\cdots \\]\r\nSo, $$ is the sequence of nonnegative integers that are sums of distinct powers of 4.\r\n\r\nHopefully it is clear that the representation of $a_i$ in base 4 [i]looks[/i] like the binary representation of $i$. Since $2004=11111010100_{(2)}$ it follows that $a_{2004}=11111010100_{(4)}=1396816$.\r\n\r\n\r\n\r\nIf you like this stuff, you should look up the Catalan Reccurence which arises in a number of problems. The idea behind solving the Catalan Reccurence with generating functions can be used in other problems. For example, the Vandermonde identity,\r\n\\[ \\sum_{k=0}^{n}{{a\\choose k}{b\\choose n-k}}=[x^n]\\sum_{k=0}^{a} {{a\\choose k}x^k}\\sum_{m=0}^{b} {{b\\choose m}x^m}=[x^n](1+x)^{a+b}={a+b\\choose n} \\]\r\nThe problem of Derrangements is also something to check out. The method is nice. It uses an exponential generating function (to make the generating function converge), and then shows the generating function satisfies a differential equation.\r\n\r\nAnd, the Roots of Unity Filter is a powerful tool. It allows you create one function from another by extracting every $k$'th power of $x$ for any natural number $k$. I have seen a couple of contest problems where the Roots of Unity Filter could be used.", "Solution_3": "I don't understand how Negative_3 got \\begin{eqnarray*} \\sum f(p)&=& [x^n](x+2x^2+3x^3+\\ldots)(1+x^2+x^4+\\ldots)(1+x^3+x^6+\\ldots)\\cdots\\\\ \\ &=&[x^n]\\frac{x}{(1-x)^2 (1-x^2)(1-x^3)\\cdots} \\end{eqnarray*} Why is $(1+x^2+x^4+\\cdots)(1+x^3+x^6+\\cdots)\\cdots$ in the factorization on the RHS, and what does it mean combinatorially?\n\nP.S.: Correct me if I am wrong but I believe that $(x+2x^2+3x^3+\\cdots)$ is on the RHS to signify whether there are $1,2,3,\\cdots$ ones in a partition $p$ of $n$. Also, sorry for bumping a really old thread." } { "Tag": [ "geometry", "inradius" ], "Problem": "Please download the doc file:\r\n\r\n[url]http://www.mp95.com/ChinaNationalMiddleSchoolMathTournament2009_session2.doc[/url]", "Solution_1": "I am simply unable to open the file!", "Solution_2": "Not everybody has Microsoft Word so here it is as a picture", "Solution_3": "This is a middle school competitions, and has proofs, wow.", "Solution_4": "[quote=\"HiDN428\"]This is a middle school competitions, and has proofs, wow.[/quote]\r\n\r\nIf without proofs, how can the students be selected? Actually, students focus on proofs\r\nrather than calculations.", "Solution_5": "mn=3.57\nya need to find AD,CD,BD. which are 1.8,15.676,16/5 resp.\nthen ya calculate the inradius of each triangle. ya add them and get 3.57 ( the two incircles will touch CD at a particular point, through which MN goes.\nobviously MN is perp. to CD." } { "Tag": [ "counting", "distinguishability", "probability", "binomial coefficients", "AMC" ], "Problem": "How many non-empty subsets $ S$ of $ \\{1, 2, 3, \\ldots, 15\\}$ have the following two properties?\r\n\r\n(1) No two consecutive integers belong to $ S$.\r\n(2) If $ S$ contains $ k$ elements, then $ S$ contains no number less than $ k$.\r\n\r\n$ \\textbf{(A) } 277\\qquad \\textbf{(B) } 311\\qquad \\textbf{(C) } 376\\qquad \\textbf{(D) } 377\\qquad \\textbf{(E) } 405$", "Solution_1": "i had no other choice but to brute force it out!\r\nwasn't that bad though", "Solution_2": "If A is the smallest number in the subset...\r\n\r\nk = 1:\r\n\r\nFor all A, there's 1 possibility, 15 possible A, 15 subsets.\r\n\r\nk = 2:\r\n\r\nA = 2, then we take the sum of A = 4,5,6,...,15 for k = 1.\r\nA = 3, then we take the sum of A = 5,6,7,...,15 for k = 1.\r\n\r\nKeep going until you finish k = 5.", "Solution_3": "Well for $k$ element-subset we much choose from $15-(k-1)$ elements, now the consecutive part is a bit tricky. \r\n\r\nWe make a 1-1 correspondence. consider the set of sequence ddddddddd, 15-2(k-1) d's, we can choose anyway k d we like, so so far \r\n(15-2(k-1)Ck, now add k-1 \"|\" each between 2 consecutive d's chosen. Each chosen d represent a number from k to 15, the pth d corresponds to number k+p-1 (counting the dividing \"|\"). So yeah $(15-2(k-1))Ck$ is the number of subsets for $k$ and sum that up to 5.", "Solution_4": "For this, it is 15 for k=1 which is trivial then its 1 summation, 2 sums, 3sums, and 4 sums.\r\n\r\n[hide]\nk=1, 15 (A zero-sum - $\\displaystyle\\sum_{x=1}^{15}1$)\nk=2, $\\displaystyle\\sum_{x=1}^{12}x$\nk=3 $\\displaystyle\\sum_{y=1}^{9}{\\displaystyle\\sum_{x=1}^{y}x}$\nk=4$\\displaystyle\\sum_{z=1}^{6}{\\displaystyle\\sum_{y=1}^{z}{\\displaystyle\\sum_{x=1}^{y}x}}$\nk=5$\\displaystyle\\sum_{a=1}^{3}{\\displaystyle\\sum_{z=1}^{a}{\\displaystyle\\sum_{y=1}^{z}{\\displaystyle\\sum_{x=1}^{y}x}}}$\n\nUse TI-89, add em up and you get $\\boxed{405}$[/hide]\r\n\r\nsee the pattern?", "Solution_5": "[quote=\"qweretyq\"]For this, it is 15 for k=1 which is trivial then its 1 summation, 2 sums, 3sums, and 4 sums.\n\n[hide]\nk=1, 15 (A zero-sum - $\\displaystyle\\sum_{x=1}^{15}1$)\nk=2, $\\displaystyle\\sum_{x=1}^{12}x$\nk=3 $\\displaystyle\\sum_{y=1}^{9}{\\displaystyle\\sum_{x=1}^{y}x}$\nk=4$\\displaystyle\\sum_{z=1}^{6}{\\displaystyle\\sum_{y=1}^{z}{\\displaystyle\\sum_{x=1}^{y}x}}$\nk=5$\\displaystyle\\sum_{a=1}^{3}{\\displaystyle\\sum_{z=1}^{a}{\\displaystyle\\sum_{y=1}^{z}{\\displaystyle\\sum_{x=1}^{y}x}}}$\n\nUse TI-89, add em up and you get $\\boxed{405}$[/hide]\n\nsee the pattern?[/quote]\r\n\r\nSolution without TI89:\r\n\r\nCan easily calculate table of values:\r\n\r\n$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $\\boxed{15}$\r\n$1$ $3$ $6$ $10$ $15$ $21$ $28$ $36$ $45$ $55$ $66$ $\\boxed{78}$\r\n$1$ $4$ $10$ $20$ $35$ $56$ $84$ $120$ $\\boxed{165}$\r\n$1$ $5$ $15$ $35$ $70$ $\\boxed{126}$\r\n$1$ $6$ $\\boxed{21}$\r\n\r\nwhere each term can be found by adding the term above it, and the term to the left of the one above it. (Then add, answer E = 405)", "Solution_6": "[hide=\"Answer\"]Each single-element subset works: $15$.\n\nFor two-element subsets, we cannot use $1$, and cannot use consecutive integers, so we essentially have $\\binom{13}{2}=78$.\n\nFor three-element subsets, we cannot use $1$ or $2$, and we cannot use consecutive integers.\n\nWe continue this and notice that we are essentially just removing two numbers each time: the minimum, and one more consecutive integer. We take this to choosing $5$ integers from $7$, so our sum is $\\binom{15}{1}+\\binom{13}{2}+\\binom{11}{3}+\\binom{9}{4}+\\binom{7}{5}=405\\Rightarrow \\boxed{E}$. Notice that we cannot do another one, because we would have $\\binom{5}{6}$, which is impossible.[/hide]\r\n\r\nThis problem only took me 4 minutes. :o", "Solution_7": "[hide=\"a recursive solution that for some reason produces the wrong answer\"]\n\nLet $ A_n$ denote the set of subsets satisfying the two conditions for the set $ \\{1, 2, \\ldots , n \\}$. We partition $ A_n$ into two categories: those containing $ n$ and those not. In the former case, each subset is in $ A_{n\\minus{}1}$. In the latter category, we map the subset to a subset of $ A_{n\\minus{}3}$ using the following bijection: remove $ n$ and reduce each remaining number by $ 1$. \n\\[ \\{ a_1, a_2, \\ldots , a_k, n \\} \\rightarrow \\{ a_1\\minus{}1, a_2\\minus{}1, \\ldots , a_k\\minus{}1 \\}\n\\]\nSince the least element and the number of elements are both reduced by $ 1$, the conditions are still satisfied. Since $ a_k \\leq n\\minus{}2$, $ a_k\\minus{}1 \\leq n\\minus{}3$, so we get a subset of $ A_{n\\minus{}3}$. We get the following recursive relation: $ a_n \\equal{} a_{n\\minus{}3} \\plus{} a_{n\\minus{}1}$. Starting with $ a_1 \\equal{}1 , a_2 \\equal{} 2, a_3 \\equal{} 4$, i get $ a_{15}\\equal{}230$, which isn't right. :( \n[/hide]", "Solution_8": "[quote=\"Phelpedo\"][hide=\"a recursive solution that for some reason produces the wrong answer\"]\n\nLet $ A_n$ denote the set of subsets satisfying the two conditions for the set $ \\{1, 2, \\ldots , n \\}$. We partition $ A_n$ into two categories: those containing $ n$ and those not. In the former case, each subset is in $ A_{n \\minus{} 1}$. In the latter category, we map the subset to a subset of $ A_{n \\minus{} 3}$ using the following bijection: remove $ n$ and reduce each remaining number by $ 1$.\n\\[ \\{ a_1, a_2, \\ldots , a_k, n \\} \\rightarrow \\{ a_1 \\minus{} 1, a_2 \\minus{} 1, \\ldots , a_k \\minus{} 1 \\}\n\\]\nSince the least element and the number of elements are both reduced by $ 1$, the conditions are still satisfied. Since $ a_k \\leq n \\minus{} 2$, $ a_k \\minus{} 1 \\leq n \\minus{} 3$, so we get a subset of $ A_{n \\minus{} 3}$. We get the following recursive relation: $ a_n \\equal{} a_{n \\minus{} 3} \\plus{} a_{n \\minus{} 1}$. Starting with $ a_1 \\equal{} 1 , a_2 \\equal{} 2, a_3 \\equal{} 4$, i get $ a_{15} \\equal{} 230$, which isn't right. :( \n[/hide][/quote]\r\n\r\nFirst, that would be $ a_{14} \\equal{} 230$. $ a_{15} \\equal{} 387$ is still incorrect though, but it's encouragingly closer.\r\n\r\nWatch the element $ \\{n\\}$ and you'll see what goes wrong. Adjusting the recurrence accordingly (and $ a_3 \\equal{} 3$, yes?) yields the correct answer.\r\n\r\nInteresting solution.", "Solution_9": "please dont post on topics over a year old", "Solution_10": "Why? It shows another way to solve the problem.", "Solution_11": "This is the topic linked to in the Contests section of this site. I was always under the impression that such topics are always open for additional discussion, for the purpose of questions/additional solutions.", "Solution_12": "Revival of [i]any[/i] old topic is not only okay, but positively good, if you wish to contribute something genuinely new and worthwhile, for example, a new solution, a flaw in an old one.", "Solution_13": "[quote=\"JesusFreak197\"][hide=\"Answer\"] $ \\binom{15}{1} \\plus{} \\binom{13}{2} \\plus{} \\binom{11}{3} \\plus{} \\binom{9}{4} \\plus{} \\binom{7}{5} \\equal{} 405\\Rightarrow \\boxed{E}$. Notice that we cannot do another one, because we would have $ \\binom{5}{6}$, which is impossible.[/hide]\n[/quote]\r\nI don't understand your solution. $ \\binom{13}{2}$ <- you could be choosing two consecutive integers among the 13 integers. for example, you could be choosing 7,8. in other words how can you guareentee you don't have two consecutive numbers in your subset?", "Solution_14": "To justify $ {13 \\choose 2}$, let's make the selected numbers an ordered pair $ (a,b)$, with both $ a,b \\in \\{1,2,\\ldots,13\\}$ and $ a < b$. We will say that this is equivalent to having a 2 element subset containing $ a\\plus{}1$ and $ b\\plus{}2$. Because $ a \\geq 1$, $ b\\plus{}2 \\geq a\\plus{}1 \\geq 2$, so condition 2 is satisfied. And because $ b > a$, $ b \\geq a \\plus{} 1$. This means there is at least one integer between $ b\\plus{}2$ and $ a\\plus{}1$, so condition 1 is satisfied. Similar reasoning with ordered 3-tuples, 4-tuples, etc. will justify the other binomial coefficients in that sum.", "Solution_15": "how would someone do the method mentioned above by beta involving the dddddd's and a divider with a -1 correspondence? [/quote]", "Solution_16": "I got it right but it took me like 10 minutes to get it. I just counted the ways. I could've used my time on other easier problems.", "Solution_17": "[quote=\"turtlecloud\"]how would someone do the method mentioned above by beta involving the dddddd's and a divider with a -1 correspondence? [/quote]\n\nThat method is known as stars and bars (or sticks and stones or balls and urns, whatever). It basically gives the number of ways to distribute $k$ indistinguishable objects into $n$ bins and is commonly used in combinatorics. The formula is $\\dbinom{n+k-1}{k}$.", "Solution_18": "Yeah introduction to counting and probability by Dave Patrick is really nice.", "Solution_19": "Can someone please completely explain the recursive solution?", "Solution_20": "[quote=MellowMelon]To justify $ {13 \\choose 2}$, let's make the selected numbers an ordered pair $ (a,b)$, with both $ a,b \\in \\{1,2,\\ldots,13\\}$ and $ a < b$. We will say that this is equivalent to having a 2 element subset containing $ a\\plus{}1$ and $ b\\plus{}2$. Because $ a \\geq 1$, $ b\\plus{}2 \\geq a\\plus{}1 \\geq 2$, so condition 2 is satisfied. And because $ b > a$, $ b \\geq a \\plus{} 1$. This means there is at least one integer between $ b\\plus{}2$ and $ a\\plus{}1$, so condition 1 is satisfied. Similar reasoning with ordered 3-tuples, 4-tuples, etc. will justify the other binomial coefficients in that sum.[/quote]\n\nWait does he mean $a-2, b-1$? And also can someone illustrate this with $\\binom{11}{3}$?", "Solution_21": "The largest possible element is $n-k+1$. Since this process is easily reversible, we have a bijection. You should have: ${15 \\choose 1} + {14 \\choose 2} + {13 \\choose 3} + ... + {9 \\choose 7} + {8 \\choose 8}$\n\nThe second condition states that no element in our original configurationcan be less than $k$, which translates to subtracting $k$ from each binomial coefficient. Now we have, after we cancel all the terms ${n \\choose k}$ where $n < k$, ${15 \\choose 1} + {13 \\choose 2} + {11 \\choose 3} + {9 \\choose 4} + {7 \\choose 5}= 15 + 78 + 165 + 126 + 21 = \\boxed{405}. The answer is $E$.", "Solution_22": "Can you explain the bijection a bit more? Thanks :)", "Solution_23": "one of the easiest #25s on the 12 I have ever seen.\n[hide=Solution]\nObviously, we can use engineer's induction. For the $1$ case, there are $\\binom{15}{1}=15$ subsets. For the $2$ case, there are just $\\binom{13}{2}$ ways. For the $3$ case, the are $\\binom{11}{3}$ ways by the hockey-stick identity. In fact, the hockey-stick identity saves all the computation. Summing over all the cases, ending when $m2. We could tweak it to work in any field with more than three elements (in which it shows that $A=0$), but we need more when the field is $\\mathbb{Z}_2$.\r\n\r\nIn particular, we need an invertible matrix $B$ such that $B-I$ is also invertible. Let $p(x)=x^n+b_{n-1}x^{n+1}+\\cdots+b_1x+b_0$ be such that $p(1)$ and $p(0)$ are nonzero. We can do this by choosing $p$ irreducible, or by merely choosing $b_{n-1}=-1,b_0=1$ and $b_i=0$ for each other $i$. Now the matrix $B=\\begin{bmatrix}0&0&0&\\cdots&0&-b_0\\\\ 1&0&0&\\cdots&0&-b_1\\\\0&1&0&\\cdots&0&-b_2\\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots&\\vdots\\\\ 0&0&0&\\cdots&0&-b_{n-2}\\\\0&0&0&\\cdots&1&-b_{n-1}\\end{bmatrix}$ has determinant $(-1)^np(0)$, and $B-I$ has determinant $(-1)^np(1)$. In the example above, $p(0)=p(1)=1$.\r\nNow $A(B-I)=0$. Since $B-I$ is invertible, $A=0$.\r\n\r\nThe above works in any (finite) field.", "Solution_5": "Another solution, i would like to share it here, hope it is ok.\r\n\r\nIf $A_1,...,A_m$ are all the matrices X with $det(X) \\neq 0$ they are forming a group $M = U(M_n(Z_p))$.\r\nSuppose A is one of this $A_i$.\r\n$A^2 = (A_1 + ... + A_m)*A_1 + ... + (A_1 + ... + A_m)*A_m = (A_1 + ... + A_m) + ... + (A_1 + ... + A_m) = m*A$.\r\nI used $(A_1 + ... + A_m)*A_i = (A_1 + ... + A_m)$, in some other order maybe, because M is group.\r\nBut A is invertible, so $A = mI_n$. [X is invertible in $M_n(Z_p)$ if and only if $det(X) \\neq 0$, because p is prime].\r\nOn the other hand p | m, hence $A = O_n$ which has determinant 0, contradiction.\r\n\r\nFor p | m, we consider the relation $m = card(U(M_n(Z_p))) = (p^n - 1)(p^n - p)...(p^n - p^{n-1})$ [classical one].\r\nThis can be proved like this: an element X in $U(M_n(Z_p))$ can be build by choosing the first column, then the second, and so on.\r\nWe need to take care that $det(X) \\neq 0$.\r\nThe first column can be choose from $Z_p$ in $p^n - 1$ ways [all $p^n$ possible cases without one case, all elements are 0].\r\nThe second column can be choose from $Z_p$ in $p^n - p$ ways [all $p^n$ possible cases minus the cardinal of the generated space of the first already filled column].\r\nThe third column can be choose from $Z_p$ in $p^n - p^2$ ways [all $p^n$ possible cases minus the cardinal of the generated space of the first and second already filled columns].\r\nAnd so on through the last column. :)" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c>\\minus{}1$. Prove that\r\n\r\n$ \\sqrt{\\frac{a\\plus{}2}{a\\plus{}b\\plus{}5}}\\plus{}\\sqrt{\\frac{b\\plus{}2}{b\\plus{}c\\plus{}5}}\\plus{}\\sqrt{\\frac{c\\plus{}2}{c\\plus{}a\\plus{}5}}\\leq\\frac{3\\sqrt{2}}{2}$", "Solution_1": "It's obviously trues because: $ \\sum\\ \\sqrt{\\frac{x}{x\\plus{}y}} \\leq\\ \\frac{3}{\\sqrt{2}}$ :maybe: \r\nWith $ x\\equal{}a\\plus{}1$...", "Solution_2": "[quote=\"nguoivn\"]It's obviously trues because: $ \\sum\\ \\sqrt {\\frac {x}{x \\plus{} y}} \\leq\\ \\frac {3}{\\sqrt {2}}$ :maybe: \nWith $ x \\equal{} a \\plus{} 1$...[/quote]\r\n\r\nThank you very much.", "Solution_3": "[quote=\"nguoivn\"]It's obviously trues because: $ \\sum\\ \\sqrt {\\frac {x}{x \\plus{} y}} \\leq\\ \\frac {3}{\\sqrt {2}}$ :maybe: \nWith $ x \\equal{} a \\plus{} 1$...[/quote]\nI think with $ x\\equal{}a\\plus{}1;y\\equal{}b\\plus{}1;z\\equal{}c\\plus{}1$, this ineq \n\n$ \\Leftrightarrow \\sum \\sqrt{\\frac{x\\plus{}1}{x\\plus{}y\\plus{}3}}\\le \\frac{3}{\\sqrt{2}}$\n\n[quote=\"Ligouras\"]Let $ a,b,c > \\minus{} 1$. Prove that\n\n$ \\sqrt {\\frac {a \\plus{} 2}{a \\plus{} b \\plus{} 5}} \\plus{} \\sqrt {\\frac {b \\plus{} 2}{b \\plus{} c \\plus{} 5}} \\plus{} \\sqrt {\\frac {c \\plus{} 2}{c \\plus{} a \\plus{} 5}}\\leq\\frac {3\\sqrt {2}}{2}$[/quote]", "Solution_4": "MAY be nguoivn means $ x\\equal{}a\\plus{}2$... :wink:", "Solution_5": "[quote=\"anas\"]MAY be nguoivn means $ x \\equal{} a \\plus{} 2$... :wink:[/quote]\r\n\r\nEven with $ x\\equal{}a\\plus{}2...$, we have this ineq\r\n\r\n$ \\Leftrightarrow \\sum \\sqrt{\\frac{x}{x\\plus{}y\\plus{}1}}\\le \\frac{3}{\\sqrt{2}}$", "Solution_6": "[quote=\"gigaman\"][quote=\"anas\"]MAY be nguoivn means $ x \\equal{} a \\plus{} 2$... :wink:[/quote]\n\nEven with $ x \\equal{} a \\plus{} 2...$, we have this ineq\n\n$ \\Leftrightarrow \\sum \\sqrt {\\frac {x}{x \\plus{} y \\plus{} 1}}\\le \\frac {3}{\\sqrt {2}}$[/quote]\r\n\r\nbut : $ \\frac{x}{x\\plus{}y\\plus{}1} \\leq \\frac{x}{x\\plus{}y}$ :D", "Solution_7": "[quote=\"anas\"][quote=\"gigaman\"][quote=\"anas\"]MAY be nguoivn means $ x \\equal{} a \\plus{} 2$... :wink:[/quote]\n\nEven with $ x \\equal{} a \\plus{} 2...$, we have this ineq\n\n$ \\Leftrightarrow \\sum \\sqrt {\\frac {x}{x \\plus{} y \\plus{} 1}}\\le \\frac {3}{\\sqrt {2}}$[/quote]\n\nbut : $ \\frac {x}{x \\plus{} y \\plus{} 1} \\leq \\frac {x}{x \\plus{} y}$ :D[/quote]\n\nI know, I just say if\n\n[quote=\"gigaman\"][quote=\"anas\"]MAY be nguoivn means $ x \\equal{} a \\plus{} 2$... :wink:[/quote]\n\nEven with $ x \\equal{} a \\plus{} 2...$, we have this ineq\n\n$ \\Leftrightarrow \\sum \\sqrt {\\frac {x}{x \\plus{} y \\plus{} 1}}\\le \\frac {3}{\\sqrt {2}}$[/quote]", "Solution_8": "[quote=\"gigaman\"]\nI think with $ x \\equal{} a \\plus{} 1;y \\equal{} b \\plus{} 1;z \\equal{} c \\plus{} 1$, this ineq \n\n$ \\Leftrightarrow \\sum \\sqrt {\\frac {x \\plus{} 1}{x \\plus{} y \\plus{} 3}}\\le \\frac {3}{\\sqrt {2}}$\n\n[/quote]\r\nBut $ \\sum \\sqrt {\\frac {x \\plus{} 1}{x \\plus{} y \\plus{} 3}}\\le \\sum \\sqrt {\\frac {x \\plus{} 1}{x \\plus{} y \\plus{} 2}}$ (obviously trues) :wink:" } { "Tag": [ "MATHCOUNTS" ], "Problem": "The front and back covers of a book are each 3 mm thick, and a stack of 25 sheets of paper is 4 mm thick. Al\u2019s book is 38 mm thick, and all of the pages are numbered front and back starting with page\r\n1. What is the number of the last page facing the back cover?", "Solution_1": "$ 3$ mm for each of the front and back. So there's $ 32$ mm for the pages, dividing by $ 4$ yields $ 8 \\times 25\\equal{}200$ sheets, for $ \\boxed{400}$ pages.", "Solution_2": "Actually the answer to the question is the [i]number of the last page facing the back cover[/i]. (Using LaTeX's work.) After finding that there are 400 pages, you can know that the number of the last page is 400. This is because the backside of the first sheet is numbered 2, the backside of the second sheet is numbered 4, and so on." } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Up to my knowledge, a semi-convex function ($f(\\frac{x+y}{2})\\ge \\frac{f(x)+f(y)}{2}$) need not to be convex. Can you give me a counterexample.", "Solution_1": "What you wrote means \"semi-concave\".", "Solution_2": "The question was answered in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=293926#p293926]http://www.mathlinks.ro/Forum/viewtopic.php?t=46490 post #10[/url]. The answer in brief: Take a non-continuous solution of the Cauchy functional equation.\r\n\r\n Darij", "Solution_3": "Thank you. I didn't think about that functions they have a lot of interesting properties also. For example their graphs are dense in $\\mathbb{R}^2$ ;)" } { "Tag": [ "function", "modular arithmetic", "quadratics", "number theory", "prime numbers", "algebra unsolved", "algebra" ], "Problem": "For odd prime numbers $ p$, find all functions $ f: \\mathbb{Z}\\rightarrow\\mathbb{Z}$ satisfying following conditions:\r\n\r\n(1) If $ m\\equiv n\\mod p$ , $ f(m)\\equal{}f(n)$.\r\n(2) For all integers $ m$ and $ n$, $ f(mn)\\equal{}f(m)f(n)$.", "Solution_1": "[quote=\"FantasyLover\"]For odd prime numbers $ p$, find all functions $ f: \\mathbb{Z}\\rightarrow\\mathbb{Z}$ satisfying following conditions:\n\n(1) If $ m\\equiv n\\mod p$ , $ f(m) \\equal{} f(n)$.\n(2) For all integers $ m$ and $ n$, $ f(mn) \\equal{} f(m)f(n)$.[/quote]\r\n\r\n$ f(n) \\equal{} 1$ $ \\forall n$ is a trivial solution.\r\n$ f(n) \\equal{} 0$ $ \\forall n$ is a trivial solution.\r\n$ f(n) \\equal{} 1$ $ \\forall n\\neq 0\\pmod p$ and $ f(n)\\equal{}0$ $ \\forall n\\equal{}0\\pmod p$ is a trivial solution.\r\n\r\nIf $ f(u)\\equal{}0$ for some $ u\\neq 0\\pmod p$, then let $ y\\equal{}\\frac{v}{u}\\pmod p$. then $ v\\equal{}uy\\pmod p$ and so $ f(v)\\equal{}f(u)f(y)\\equal{}0$ $ \\forall v$\r\n\r\nConsider now we have $ m$ such that $ f(m)\\notin\\{0,1\\}$ \r\n$ f(0) \\equal{} f(m)f(0)$ $ \\implies$ $ f(0) \\equal{} 0$.\r\n$ f(m*1) \\equal{} f(m)f(1)$ $ \\implies$ $ f(1) \\equal{} 1$\r\nLet $ x\\neq 0\\pmod p$. Exists $ y$ such that $ xy \\equal{} 1\\pmod p$ so $ f(xy) \\equal{} f(1)$ so $ f(x)f(y) \\equal{} 1$ and so $ f(x)\\in\\{ \\minus{} 1,1\\}$ $ \\forall x\\neq0\\pmod p$ (since $ f(x)\\in\\mathbb Z$ and $ f(x)|1$). So $ f(m) \\equal{} \\minus{} 1$\r\n\r\nLet now $ u\\neq 0\\pmod p$ a quadratic residue mod p. we have $ u \\equal{} v^2\\pmod p$ and so $ f(u) \\equal{} (f(v)^2 \\equal{} 1$\r\nSo $ m$ is not a quadratic residue mod p.\r\nLet $ w$ a non quadratic residue mod p. $ mw$ is a quadratic residue and so $ f(mw) \\equal{} 1$. But $ f(mw) \\equal{} f(m)f(w) \\equal{} \\minus{} f(w)$ and so $ f(w) \\equal{} \\minus{} 1$\r\n\r\nSo $ f(x)$ is the Legendre symbol $ \\left(\\frac {x}{p}\\right)$\r\n\r\nAnd all the solutions are \r\n\r\n$ f(x) \\equal{} 0$\r\n\r\n$ f(x) \\equal{} 1$\r\n\r\n$ f(x) \\equal{} 0$ $ \\forall x\\equal{}0\\pmod p$ and $ f(x)\\equal{}1$ $ \\forall x\\neq 0\\pmod p$\r\n\r\n$ f(x) \\equal{} \\left(\\frac {x}{p}\\right)$" } { "Tag": [ "calculus", "integration", "number theory", "prime factorization" ], "Problem": "Find all n from 1-50 such that the fraction $n!/((a+1)!(n-a)!)$ is an integer for all integers a such that $0< a < n$", "Solution_1": "[hide]First, we can rewrite the expression as $\\frac{{n+1\\choose a+1}}{n+1}$.\n\n(defining new variables for lemma)\n-------------------------------\nWe know that ${m \\choose n}$ must be divisible by m if m is prime, and 0 Quit\r\nHit STAT\r\nHit the right arrow cursor so that the STAT Menu shows CALC in bold.\r\n\r\nYou should now have a list of items.\r\nScroll down until you see CubicReg (item #6). This stands for cubic regression.\r\n\r\nHit Enter. This will bring you to the home screen and paste the function CubicReg on it.\r\nThis function needs arguments, namely, the lists to run the regression on.\r\n\r\nWe want to graph L1(time) as X values, and L2(number of ppl) as Y values.\r\nHit 2ND --> L1\r\nHit the comma key\r\nHit 2ND --> L2\r\nHit the comma key\r\nHit VARS\r\nHit Right Cursor\r\nHit enter over Y-VARS\r\nHit 1 for FUNCTION\r\nHit 1 to store the regression in function Y1.\r\n\r\nNow hit enter.\r\nLet the calculator run for a bit, it shouldn't take long.\r\n\r\nYou now have a function stored in the Y1 Space.\r\nHit the grey Y= button at the top of the calc.\r\nCheck to make sure you have a function.\r\nNow, hit the window key.\r\nSet these values:\r\n\r\nXmin = -1\r\nXmax = 6\r\nXscl = 1\r\nYmin = 1400\r\nYmax = 2200\r\nYscl = 100\r\nXres=1 (in general ALWAYS leave this as 1)\r\n\r\nWe're almost done.\r\nHit 2ND --> Stat Plot\r\nMouse over until 1: is bolded.\r\nHit Enter\r\nSet the Xlist to L1, and the Ylist to L2\r\nSet the mark to the middle one. (or whatever you want)\r\nSet the type to the the upper right most type (scatter plot).\r\nFinally, turn the scatter plot on by mousing over the ON option and hitting enter.\r\n\r\nHere's a summary of what you just did:\r\nInputted data into to lists.\r\nRan a cubic regression to find a trend line.\r\nSaved the trend line to the Y1 function.\r\nMade a scatter plot with the data you inputted.\r\n\r\nWe can now hit graph.\r\nThis shows you the data points, and the line we made.\r\n\r\nThat should help with the rest of the problem." } { "Tag": [], "Problem": "For all numbers [i]p[/i] and [i]q[/i], where [i]p[/i] is not equal to 4. let [i]p[/i]//[i]q[/i] be defined as [i]pq[/i]/[i]p[/i]-4. For what value of [i]p[/i] does [i]p[/i]//7 = 21\r\n\r\nThis isn't a multiple choice question, but a grid-in. I tried to just plug in numbers, but that is tedious and takes some time. Is there a formula for this?", "Solution_1": "[hide]7p / (p-4) = 21\np = 3p - 12\n2p = 12\np = 6[/hide]", "Solution_2": "For all numbers p and q, where p is not equal to 4. let p//q be defined as pq/p-4. For what value of p does p//7 = 21 \r\n\r\nThis isn't a multiple choice question, but a grid-in. I tried to just plug in numbers, but that is tedious and takes some time. Is there a formula for this?\r\n\r\n\r\nthere isn't a formula because it's already defined in the problem.\r\nso in $p$//$7$ remember that 7 is the $q$ so we just plug in 7 to the defined equation. so\r\n\r\n$\\frac{7p}{p-4}= 21$\r\n\r\nim pretty sure you know how to find p from here. \r\ndon't plug in numbers.\r\n\r\na high school way of solving the fraction would be\r\n$21(p-4)=7p$\r\n$21p-84=7p$\r\n$-84=-14p$\r\n$p=6$" } { "Tag": [], "Problem": "How many nonnegative solutions are there to $a+b+c+d=17$, provided that $d \\leq 12$? \r\n\r\nWondering if there's a solution not involving case analysis.", "Solution_1": "I'm pretty sure you can do it without case analysis (at least, not very many cases), by partitioning.\r\n\r\nIf we did not have the requirement that $d\\leq12$, there would be $\\binom{18}{3}$ solutions. To eliminate the cases where $d>12$, we consider the first partition to be $d$; we see that when $d=12+k$, with $00$ and $ n$ is an integer greater than 1\r\nProve that\r\n\r\n$ \\frac {(a\\plus{}b\\plus{}c)^2}{a^2\\plus{}b^2\\plus{}c^2} \\leq \\sqrt[n] {\\frac {2a}{a\\plus{}b}} \\plus{}\\sqrt[n] {\\frac {2b}{b\\plus{}c}} \\plus{}\\sqrt[n] {\\frac {2c}{c\\plus{}a}} \\leq 3$\r\n\r\n\r\nPlease help me. :( \r\nI have posted this in IE open Question, but nobody help me.", "Solution_1": "For right inequality: \r\n\r\nif we define $ x \\equal{} \\frac {2a}{a \\plus{} b} ,y \\equal{} \\frac {2b}{b \\plus{} c} ,z \\equal{} \\frac {2c}{c \\plus{} a}$\r\n\r\nby power mean inequality we have\r\n\\[ (\\frac {x^{\\frac {1}{n}} \\plus{} y^{\\frac {1}{n}} \\plus{} z^{\\frac {1}{n}}}{3})^n\\le (\\frac {x^{\\frac {1}{2}} \\plus{} y^{\\frac {1}{2}} \\plus{} z^{\\frac {1}{2}}}{3})^2\r\n\\]\r\nso it's enough to prove that\r\n\\[ A \\equal{} \\sqrt {\\frac {2a}{a \\plus{} b}} \\plus{} \\sqrt {\\frac {2b}{b \\plus{} c}} \\plus{} \\sqrt {\\frac {2c}{c \\plus{} a}} \\leq 3\r\n\\]\r\nby cawchy we know\r\n\\[ (A)^2\\le (2(a \\plus{} b \\plus{} c))(\\sum\\frac {2a}{(a \\plus{} b)(a \\plus{} c)})\r\n\\]\r\nso we must prove that\r\n\\[ 4((a \\plus{} b \\plus{} c)(\\sum ab \\plus{} ac))\\le9(\\sum a^2b \\plus{} a^2c \\plus{} 2abc)\r\n\\]\r\nor\r\n\\[ 6abc\\le \\sum a^2b \\plus{} a^2c\r\n\\]\r\nthat is AM-GM", "Solution_2": "What's about the left side?\r\nMaybe it 's harder", "Solution_3": "The left inequality is wrong if we have a big $ n$", "Solution_4": "Why? Can you show me one example? :huh: :huh:", "Solution_5": "[quote=\"Dread Phantom\"]Why? Can you show me one example? :huh: :huh:[/quote]\r\n\r\n \\[ n\\rightarrow\\infty , a\\equal{}2,b\\equal{}1,c\\equal{}1\\]", "Solution_6": "Emm , but this example is wrong.\r\nI have tried to find several example but I can't find one." } { "Tag": [ "LaTeX", "Asymptote", "AoPSwiki", "articles", "integration", "trigonometry", "parameterization" ], "Problem": "How can I have latex text wrap about an asymptote image? Let's say I want text on the left side, and an image to the text's right side.", "Solution_1": "You would wrap the text around in the same way as you would any image. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=wrapfig]wrapfig package[/url] will do this. The documentation is in .../tex/latex/wrapfig/wrapfig.sty at the end.\r\nAdapting the AoPSWiki example at [[Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]] to put the text next to the yellow smiley:\r\n[code]\\documentclass{article}\n\\usepackage[pdftex]{graphicx}\n\\usepackage{asymptote}\n\\usepackage{wrapfig}\n\\begin{document}\nHello. \nI like to make pics with Asymptote like this one. This line carries on over the image:\\\\\n\\begin{wrapfigure}{r}{1in}\n \\begin{asy}\n include graph;\n size(1inch);\n filldraw(circle((0,0),1),yellow,black);\n fill(circle((-.3,.4),.1),black);\n fill(circle((.3,.4),.1),black);\n draw(arc((0,0),.5,-140,-40));\n\\end{asy}\n\\end{wrapfigure}\n\nIt makes me happy, since I can still type my normal LaTeX stuff around it: \n\\(\\int_0^{\\pi}{\\sin{x}}\\,dx=2\\) \n\nMore text that goes along the line but will be wrapped when it reaches the image\\\\\nMore text that goes along the line but will be wrapped when it reaches the image\\\\\nMore text that goes along the line but will be wrapped when it reaches the image\\\\\nMore text that goes along the line but will be wrapped when it reaches the image or carry on underneath it.\n\\end{document}[/code]\r\nDo read the wrapfig documentation as it has a lot more parameters you can adjust.", "Solution_2": "How about if I want do this inside a table cell? (Like the the middle column of the table i was describing in the other thread)", "Solution_3": "Just put the asy code inside the cell as if it were text.", "Solution_4": "For some reason the formatting got all messed up.", "Solution_5": "Possibly for that particular cell you want to have\r\n\\begin{minipage}\r\nyour_cell_text&image_here\r\n\\end{minipage}\r\n? Minipages are often useful." } { "Tag": [ "linear algebra", "matrix", "inequalities", "inequalities unsolved" ], "Problem": "For positive integers $ n,k$ such that $ 1\\leq k\\leq2n$ prove that:\r\n\r\n\r\n$ \\begin{pmatrix}2n+1 \\\\ k-1 \\end{pmatrix}+\\begin{pmatrix}2n+1 \\\\ k+1 \\end{pmatrix}\\geq2\\frac{n+1}{n+2}\\begin{pmatrix}2n+1 \\\\ k \\end{pmatrix}$", "Solution_1": "Adding in both members $ 2\\left(\\begin{array}{c}2n+1 k\\end{array}\\right)$ and using the recurrance formula for combinations, the inequality becomes\r\n$ \\left(\\begin{array}{c}2n+3 k+1\\end{array}\\right)\\geq 2\\left(\\begin{array}2n+1 k\\end{array}\\right)\\cdot\\left(\\frac{n+1}{n+2}+1\\right)\\Leftrightarrow \\frac{n+1}{(k+1)(2n-k+2)}\\geq\\frac{1}{n+2}\\Leftrightarrow$\r\n$ \\Leftrightarrow (n-k)(n-k+1)\\geq 0$\r\nIf $ k \\angle B > \\angle C$. Let $O$ be the midpoint of the circumscribed circle, and $I$ the midpoint of the inscribed circle, and call $D$ the midpoint of $BC$.\r\nLet $M$ be the midpoint of the arc $BC$, not containing $A$. Let $F$ be a point on $ID$, such that $D$ is the midpoint of $IF$. Let $Q$ be the intersection of $MF$ with the circumscribed circle.\r\n\r\n(a) Show that $Q$ lies on the arc $AC$, not containing $B$.\r\n\r\n(b) Show that $AQ-BQ+CQ =0$.", "Solution_1": "Hmmm seems to have been pretty tricky! Part a is pretty easy, so I'll just do part b (assuming part a).\r\n\r\nLet P be the point on BQ such that IP || MQ. We first show that AIPB is cyclic--this is true since $\\angle IPQ = \\angle PQM = \\angle BAI$. Then let N be the midpoint of arc AB. It is well-known that NA = NB = NI, so we also have NA = NP. Since $\\angle NQA = \\angle NQP$, we also then have QA = QP. Now pick the point R such that RBQC is a parallelogram. By virtue of the fact that QM bisects $\\angle BQC$, we have BP = BR = QC. Hence QB = QP + BP = QA + QC.", "Solution_2": "[quote=\"probability1.01\"]Now pick the point R such that RBQC is a parallelogram. By virtue of the fact that QM bisects $\\angle BQC$, we have BP = BR = QC. Hence QB = QP + BP = QA + QC.[/quote]\r\nMaybe i have a blind spot but didnt get this part :blush: (particulary where do you use that D is the middle of IF). Will you please clarify?", "Solution_3": "I forgot to note that R must lie on IP due to all the symmetry with respect to D.", "Solution_4": "NIce!!\r\n\r\nI don't think you needed N though: a_{32}$ then we put $ a_{29}$ in the subset with $ a_{32}$, if not, we put $ a_{29}$ with $ a_{31}$ and $ a_{30}$. In every step, we put $ a_k$ in the subset with the smaller sum. If the two subsets are equal, we can choose, it's the same thing. When we finish (we put $ a_1$) the two subsets are equal...[/hide]\r\nIf I have a good memory, this is the strategy, and it look it works, but I don't really know how to prove it, and I totally forgot the problem after doing this... I post this because no one post nothing, then maybe someone gets inspired by my strategy :p\r\n(And as I said in posts before, excuse my poor english...)" } { "Tag": [ "search", "ratio" ], "Problem": "A crew of 80 men can do a job in 24 days. If the contractor increases the work force by 1/2 and productivity is unchanged, how many days will be saved by adding the additional workers?\r\n\r\nAnswer:\r\n[hide]8.0[/hide]", "Solution_1": "[hide=\"Solution\"]The $ 80$ men each contribute $ 24$ days of work. Therefore, $ 80\\times 24 \\equal{} 1920$ days of work are put into the project. An increase in the workforce by $ 1/2$ yields $ 120$ men. If Therefore, since productivity is unchanged, each man works $ 1920/120 \\equal{} 16$ days, so the job will be done in $ 16$ days now. $ 24 \\minus{} 16 \\equal{} \\boxed{8}$ days are saved.[/hide]", "Solution_2": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302912url]Search Please[/url]", "Solution_3": "[hide=\"A slightly more straightforward solution\"]\nWe have the ratio men:days as 80:24.\nWe want there to be 1.5 more men, right? So we multiply the mens side by 1.5. Since we have more men to work on the same amount of job, the days are going to decrease, right? So we divide 24 days by 1.5 since the two are proportional. Now if we have 80x1.5=120 men, we need 24/1.5=16 days to work on the project. So 24-16=[b]8 [/b] days were saved [/hide]", "Solution_4": "If the contractor increases the work force by 1/2 this mean the men will become 80 + 40 =120\r\n\r\n(80) men can do a job in (24 ) \r\n\r\n(120) men can do a job in (x)\r\n\r\nthen 120x =1920 \r\n\r\ntherefore, x=16\r\n\r\nthe contractor will be saved 24-16=8 days\r\n\r\nsalahcool :gathering:" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "How to prove this??\r\n\r\nlim sup (an+bn)= 0, \\exists N^{(s)}_\\epsilon$ s.t. $ n\\ge N_\\epsilon \\to s_n < s\\plus{}\\epsilon$\r\n$ (ii): \\forall \\epsilon >0, \\forall N, \\exists N'>N$ s.t. $ |s_{N'} \\minus{} s |< \\epsilon$\r\n\r\n$ (i) \\equal{} \\lim \\sup s_n \\le s,(ii) \\equal{} \\lim \\sup s_n \\ge s$\r\n\r\nfix $ \\epsilon > 0$, then $ \\exists N\\equal{}\\max (N^{(a)}_{\\frac{\\epsilon}{2}}, N^{(b)}_{\\frac{\\epsilon}{2}})$\r\n\r\n$ n \\ge N \\to a_n \\plus{}b_n < a\\plus{} b \\plus{} \\epsilon \\to \\lim \\sup (a_n \\plus{} b_n ) \\le a \\plus{} b$\r\n\r\nif we could prove (ii) for general sequences, we would actually have $ \\lim \\sup (a_n \\plus{}b_n ) \\equal{} a \\plus{} b$\r\n\r\nbut we know this isnt true, for instance, $ a_n \\equal{} (\\minus{}1)^n, b_n \\equal{} \\frac{1}{2} (\\minus{}1)^{n\\plus{}1}$\r\n\r\nthen $ a_n \\plus{}b_n \\equal{} \\minus{}b_n$ which will give us our contradiction" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Help my old brother to get this nonlinear autonomous third system or is called The Lorenz Equation. \r\n$\\frac{dx}{dt}=\\sigma(-x+y)$\r\n$\\frac{dy}{dt}=rx-y-xz$\r\n$\\frac{dz}{dt}=-bz+xy$\r\nwhere $\\sigma$ , $r$ , and $b$ is constant.", "Solution_1": "What do you mean by \"get\"?\r\n\r\nThis is a nonlinear autonomous system. Finding closed-form solutions for $x,y,z$ in terms of $t$ would be painful and complicated at best, with a reasonable likelyhood of being impossible.\r\n\r\nIs perhaps the question to find and characterize (as stable, unstable, or whatever) all the stationary points of the autonomous system?" } { "Tag": [ "articles", "college" ], "Problem": "I heard the University of California branches and CSULB require a year of visual arts. Is that true? Though I am an American citizen, I live in India. Unfortunately, my schools never had that facility (visual arts). Does that make me ineligible? \r\nAlso, I want to know what the universities in California look for while giving scholarships. Actually, my family is really poor(well, at least when compared to people in USA) and cannot afford the fees unless I get plenty of scholarships or federal loans. But then, I do not know whether I am eligible for federal loans since I haven't lived in USA for a year. Though my SAT score is quite good\r\n(2250-Math 800, Reading 760, Writing 690) my GPA is around 2.8.\r\nThe only reason I am applying to universities in California is that I will be treated as an in-state student after my first year. I do not have any outstanding achievements except having made it to the IMO selection camp once (didn't do well at the camp). Well, I could attach two articles to my application (I can only wish they were difficult enough to be called research papers).\r\nAny information in this matter will be really appreciated and, to be honest, is badly needed. Also let me know which universities I should apply to.", "Solution_1": "Regarding your first question (the only one I have enough experience to answer), there is a requirement for one year of visual and performing arts. As an out of state student, I did not know about this requirement and only took a half year of music theory. When I tried to submit my application, it warned me that I was missing this requirement and said to explain in the Additional Information section. I wrote that my school only required a half a year of visual and performing arts and that I played the piano outside of school and was accepted to Berkeley and UCLA.\r\n\r\nThey wouldn't penalize you for something you didn't have access to, so explain that in the Additional Information section. Also if you have extracurricular experience with visual and performing arts, it could be worth mentioning there also." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Prove that if $\\overline{abc}$ is prime, then $b^{2}-4ac$ is not a perfect square.", "Solution_1": "By Cohn's theorem, $ax^{2}+bx+c$ is irreducible in $Q[X]$ and so it cannot have rational roots. Thus conclusion.", "Solution_2": "[quote=\"harazi\"]By Cohn's theorem, $ax^{2}+bx+c$ is irreducible in $Q[X]$ and so it cannot have rational roots. Thus conclusion.[/quote]\r\nWhat is Cohn's Theorem?", "Solution_3": "Try [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32826]this topic[/url]." } { "Tag": [], "Problem": "Anyone here have any programs for a TI-83+ in TI-BASIC that they could show me?", "Solution_1": "http://www.ticalc.org", "Solution_2": "how do you put the programs on the calcultor. :?", "Solution_3": "Do you have any kind of link cable? If not, then buy one. ;)", "Solution_4": "yeah, i have a link cable. What do i do with it?", "Solution_5": "Which kind is it? They usually come with instructions...", "Solution_6": "You can also read in the manual how to create basic programs. The language is very easy to learn and you will be able to make your own programs.\r\n\r\nWhat kinds of programs are you looking for? Games? Math? ...", "Solution_7": "i already have created 12 programs on my own, but i was wondering how to get Super Mario, or Snake.", "Solution_8": "you can download them online if you have a link cable and link it to the computer\r\nor, you can get it off a friend.", "Solution_9": "where do i link it in to the computer?", "Solution_10": "Look at the instructions! It depends on what type of cable you have.", "Solution_11": "The instructions didn't tell me anything!!!", "Solution_12": "Well it might be a good idea to start by telling us what type of cable you have..... :roll:" } { "Tag": [ "linear algebra", "matrix", "induction", "function" ], "Problem": "Problem Let $ A \\in M_n{\\mathbb{(C)}}$ , Prove that $ \\det(A\\overline{A} \\plus{} I_n) \\geq 0$ Here $ \\overline{A}$ is conjugate complex matrix of $ A$\r\n Proof not complete but easy understand .\r\n [b]By induction [/b]\r\n With $ n \\equal{} 1$ it is true .\r\n Suppose that it is correct to $ n \\geq 1$, now take $ A \\in M_{n \\plus{} 1}{\\mathbb{(C)}}$\r\ncase 1) If $ rank(A) \\equal{} r < n \\plus{} 1$ then the exist $ P,Q \\in GL_{n \\plus{} 1}{\\mathbb{(C)}}$ \r\nsuch that $ A \\equal{} P\\begin{bmatrix}I_r & O \\\\\r\nO & O\\end{bmatrix}Q$\r\n Put $ Q\\overline{P} \\equal{} \\begin{bmatrix}A_1 & A_2 \\\\\r\nA_3 & A_4\\end{bmatrix}$\r\n Here $ A_1 \\in M_r{\\mathbb{(C)}},A_4 \\in M_{n \\plus{} 1 \\minus{} r}{\\mathbb{(C)}}$,$ A_2 \\in M_{rx(n \\plus{} 1 \\minus{} r)}{\\mathbb{(C)}}$,$ A_3 \\in M_{(n \\plus{} 1 \\minus{} r)xr}{\\mathbb{(C)}}$\r\n Then $ A\\overline{A} \\equal{} P\\begin{bmatrix}I_r & O \\\\\r\nO & O\\end{bmatrix}Q\\overline{P}\\begin{bmatrix}I_r & O \\\\\r\nO & O\\end{bmatrix}\\overline{Q}$=$ P\\begin{bmatrix}I_r & O \\\\\r\nO & O\\end{bmatrix}\\begin{bmatrix}A_1 & A_2 \\\\\r\nA_3 & A_4\\end{bmatrix}\\begin{bmatrix}I_r & O \\\\\r\nO & O\\end{bmatrix}\\overline{Q}$=$ P\\begin{bmatrix}A_1 & O \\\\\r\nO & O\\end{bmatrix}\\overline{Q}$\r\n Use $ \\det(XY \\plus{} I) \\equal{} \\det(YX \\plus{} I)$ we have :\r\n $ \\det(A\\overline{A} \\plus{} I_{n \\plus{} 1}) \\equal{} \\det(\\overline{Q}P\\begin{bmatrix}A_1 & O \\\\\r\nO & O\\end{bmatrix} \\plus{} I_{n \\plus{} 1})$=$ \\det(\\begin{bmatrix}\\overline{A_1} & \\overline{A_2} \\\\\r\n\\overline{A_3} & \\overline{A_4}\\end{bmatrix}\\begin{bmatrix}A_1 & O \\\\\r\nO & O\\end{bmatrix} \\plus{} I_{n \\plus{} 1})$=$ \\det(\\begin{bmatrix}\\overline{A_1}A_1 & O \\\\\r\n\\overline{A_3}A_1 & O\\end{bmatrix} \\plus{} I_{n \\plus{} 1}) \\equal{} \\det(\\overline{A_1}A_1 \\plus{} I_r) \\geq 0$ (by induction $ r < (n \\plus{} 1)$) \r\ncase 2) [b]Now please make[/b] with $ rank(A) \\equal{} n \\plus{} 1$ (It is hard ) but It correct to $ k \\equal{} 1,2,..,n$ and $ rank(A) < n \\plus{} 1$ (I am so bad :blush: )\r\nPs : If you have a proof $ \\forall A \\in GL_n{\\mathbb{(C)}}$ that $ \\det(A\\overline{A} \\plus{} I_n) \\geq 0$ \r\n then It is so easy to prove that If $ rank(A) < n$ then $ \\det(A\\overline{A} \\plus{} I_n) \\geq 0$\r\n[hide=\" Answer with\"] $ rank(A) \\equal{} n \\plus{} 1$ If you have a solution please post it :clap2: [/hide]", "Solution_1": "You can see that If $ \\overline{P(\\lambda)} \\equal{} P(\\overline{\\lambda}) ,\\forall \\lambda \\in \\mathbb{C}$ , $ P \\in \\mathbb{K}[X]$\r\nthen $ P(x) \\in \\mathbb{R}[X]$\r\n Applied we have $ P(\\lambda) \\equal{} \\det(A\\overline{A} \\plus{} \\lambda I_n) \\in \\mathbb{R}[\\lambda]$ or $ A\\overline{A}$ have eigenvalues real $ \\lambda_1,...,\\lambda_k$ and conjugate complex $ r_1,\\overline{r_1},...,r_m,\\overline{r_m}$\r\nNow we consider $ f(A\\overline{A})$ such that $ f(\\lambda_i) \\plus{} 1 \\geq 0,1 \\leq i \\leq k$, and $ \\overline{f(r_j) \\plus{} 1} \\equal{} f(\\overline{r_j}) \\plus{} 1,1 \\leq j \\leq m$, $ k \\plus{} 2m \\equal{} n$ \r\nApplied for $ f(A\\overline{A}) \\equal{} e^{A\\overline{A}}$,$ f(A\\overline{A}) \\equal{} sin(A\\overline{A})$,$ f(A\\overline{A}) \\equal{} cos(A\\overline{A})$,$ f(A\\overline{A}) \\equal{} Ch(A\\overline{A})$ We have :\r\n $ \\det(e^{A\\overline{A}} \\plus{} I_n) \\geq 0$,$ \\det(sin{A\\overline{A}} \\plus{} I_n) \\geq 0$ ,$ \\det(cos{A\\overline{A}} \\plus{} I_n) \\geq 0$\r\n$ \\det(ch{A\\overline{A}}\\pm I_n) \\geq 0$ \r\n Note :Can have many function $ f$ above $ f \\equal{} 1$ are case $ \\det(A\\overline{A} \\plus{} I_{n})\\geq 0$ But $ f$ don't such that above\r\nadd For $ A_1,A_2,..,A_m \\in M_n{\\mathbb{(C)}}$ then $ \\det(e^{\\sum_{k \\equal{} 1}^{m}{A_k\\overline{A_k}}}) \\geq 0$\r\nIf $ A_iA_j \\equal{} A_jA_i 1 \\leq i,j \\leq m$ then $ f(A_1\\otimes{\\overline{A_1}},...,A_m\\otimes{\\overline{A_m}})$\r\n Please for more information about function $ f$ Thanks you very much ! :rotfl:\r\nQuestion : Let $ A,B \\in M_n{\\mathbb{(C)}}$ such that : $ AB \\equal{} BA$ then $ \\det(A\\overline{A} \\plus{} B\\overline{B} \\plus{} \\lambda I_n) \\in \\mathbb{R}[\\lambda]$ ?" } { "Tag": [ "\\/closed" ], "Problem": "When is the next AMC 12 class will be offered after the June session? Thanks!", "Solution_1": "Probably starting in October 2007, ending in late January or early February 2008 just before the AMC test." } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let $ ABCD$ be a convex quadrilateral such that $ \\angle DAB \\equal{}\\angle ABC \\equal{}\\angle BCD$. Let $ H$ and $ O$ denote the orthocenter and circumcenter of\r\nthe triangle $ ABC$. Prove that $ H,O,D$ are collinear.", "Solution_1": "I love this combination \"Easy\" + \"Unsolved Problems\"...\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=25859\r\n\r\n darij" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$\\ a,b,c\\ghe0$ .\r\n$\\frac{a}{a+\\sqrt{(a+b)(a+c)}}+\\frac{b}{b+\\sqrt{(b+c)(b+a)}}+\\frac{c}{c+\\sqrt{(c+a)(c+b)}}\\leq1$\r\n Good luck :D", "Solution_1": "we have : Ineq's BCS $(a+b)(b+c) \\geq (\\sqrt{bc}+\\sqrt{ba})^{2}$\r\n hence :$b+(a+b)(b+c) \\geq (\\sqrt{bc}+\\sqrt{ba})^{2}+b = \\sqrt{b}.( \\sqrt{a}+\\sqrt{b}+\\sqrt{c}).$\r\n from this, we prove pro very easy" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let x, y, z be positive real numbers,\r\nxyz=1\r\nProve:\r\n2/[(x + 1)^2 + y^2 + 1] + 2/[(y + 1)^2 + z^2 + 1] + 2/[(z + 1)^2 + x^2 + 1 ] <=1\r\n\r\nPls help. Can't do it for months. omg", "Solution_1": "Well, I don't have a complete solution so I'll post what I predict how it's gonna work :wink: \r\nFirst notice that $ (x\\plus{}1)^{2}\\plus{}y^{2}\\plus{}1\\equal{}x^{2}\\plus{}y^{2}\\plus{}2x\\plus{}2\\geq 2xy\\plus{}2x\\plus{}2\\equal{}2(xy\\plus{}y\\plus{}1)$\r\nthus:\r\n$ \\sum_{cyc}\\frac{2}{(x\\plus{}1)^{2}\\plus{}y^{2}\\plus{}1}\\leq \\sum_{cyc}\\frac{1}{xy\\plus{}y\\plus{}1}$\r\nWhat is missing in my pseudo-solution is the next part, being specific proving that:\r\n$ \\sum_{cyc}\\frac{1}{xy\\plus{}y\\plus{}1}\\leq \\sum_{cyc}\\frac{1}{xy\\plus{}z\\plus{}1}$\r\n(to be honest I'm not quite sure if it is really true :wink: )\r\nBut if that's true than our solution is finished, because $ \\forall_{a\\in R_{\\plus{}}}a\\plus{}\\frac{1}{a}\\geq 2$\r\nSo\r\n$ \\sum_{cyc}\\frac{1}{xy\\plus{}z\\plus{}1}\\equal{}\\sum_{cyc}\\frac{1}{\\frac{1}{z}\\plus{}z\\plus{}1}\\leq \\frac{1}{3}\\plus{}\\frac{1}{3}\\plus{}\\frac{1}{3}\\equal{}1$\r\nIt'd be great if someone would tell me if it is a good way to approach that problem :wink:", "Solution_2": "[quote=\"polskimisiek\"]Well, I don't have a complete solution so I'll post what I predict how it's gonna work :wink: \nFirst notice that $ (x \\plus{} 1)^{2} \\plus{} y^{2} \\plus{} 1 \\equal{} x^{2} \\plus{} y^{2} \\plus{} 2x \\plus{} 2\\geq 2xy \\plus{} 2x \\plus{} 2 \\equal{} 2(xy \\plus{} y \\plus{} 1)$\nthus:\n$ \\sum_{cyc}\\frac {2}{(x \\plus{} 1)^{2} \\plus{} y^{2} \\plus{} 1}\\leq \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1}$\nWhat is missing in my pseudo-solution is the next part, being specific proving that:\n$ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1}\\leq \\sum_{cyc}\\frac {1}{xy \\plus{} z \\plus{} 1}$\n(to be honest I'm not quite sure if it is really true :wink: )\nBut if that's true than our solution is finished, because $ \\forall_{a\\in R_{ \\plus{} }}a \\plus{} \\frac {1}{a}\\geq 2$\nSo\n$ \\sum_{cyc}\\frac {1}{xy \\plus{} z \\plus{} 1} \\equal{} \\sum_{cyc}\\frac {1}{\\frac {1}{z} \\plus{} z \\plus{} 1}\\leq \\frac {1}{3} \\plus{} \\frac {1}{3} \\plus{} \\frac {1}{3} \\equal{} 1$\nIt'd be great if someone would tell me if it is a good way to approach that problem :wink:[/quote]\r\n\r\nyou don't have to prove that part....if you look more carefully at $ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1}$, you will see that \r\n$ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1}\\equal{}1$ holds...cheers :wink:", "Solution_3": "[quote=\"behemont\"]if you look more carefully at $ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1}$, you will see that \n$ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1} \\equal{} 1$ holds...cheers :wink:[/quote]\r\nUff...Why is it 1?", "Solution_4": "[quote=\"Wolfbrother\"][quote=\"behemont\"]if you look more carefully at $ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1}$, you will see that \n$ \\sum_{cyc}\\frac {1}{xy \\plus{} y \\plus{} 1} \\equal{} 1$ holds...cheers :wink:[/quote]\nUff...Why is it 1?[/quote]\r\n$ \\frac {1}{xy \\plus{} y \\plus{} 1}\\plus{}\\frac{1}{yz\\plus{}z\\plus{}1}\\plus{}\\frac{1}{xz\\plus{}x\\plus{}1}\\equal{}$\r\n$ \\equal{}\\frac {1}{xy \\plus{} y \\plus{} 1}\\plus{}\\frac{xy}{y\\plus{}1\\plus{}xy}\\plus{}\\frac{y}{1\\plus{}xy\\plus{}y}\\equal{}1.$ :wink:", "Solution_5": "wow ty all", "Solution_6": "Hehehe, yup right. It turned out that I complicated everything myself :wink: Looking at it now it seems quite obvious that my \"part\" would even fail. :wink: Anyway also thanks." } { "Tag": [ "logarithms", "AMC", "AIME", "algebra", "system of equations", "AIME I", "binomial theorem" ], "Problem": "The system of equations\r\n\\begin{eqnarray*}\\log_{10}(2000xy) - (\\log_{10}x)(\\log_{10}y) & = & 4 \\\\\r\n\\log_{10}(2yz) - (\\log_{10}y)(\\log_{10}z) & = & 1 \\\\\r\n\\log_{10}(zx) - (\\log_{10}z)(\\log_{10}x) & = & 0 \\\\\r\n\\end{eqnarray*}\r\nhas two solutions $ (x_{1},y_{1},z_{1})$ and $ (x_{2},y_{2},z_{2}).$ Find $ y_{1} + y_{2}.$", "Solution_1": "k i get this but for some reason my values are complex :huh: \r\n[hide=\"Partial\"]\nfirst let:\n$a=\\log_{10}x,b=\\log_{10}y,c=\\log_{10}z$\nNotice:\n$\\log_{10}2000=\\log_{10}1000+\\log_{10}2=3+\\log_{1}0{2}$\n\nSo, Equation One:\n$3+\\log2+a+b-ab=4$\n$\\log2+a+b-ab=1$\n\n$\\log2+b+c-bc=log2+a+b-ab$\n$(c-a)-(cb+ab)=0$\n$(c-a)-b(c-a)=0$\n$(1-b)(c-a)=0$\n$\\therefore{b=1\\text{ or }a=c}$\n\nif $b=1$ then substituting it back in reveals no solutions..so then $a=c$\n\nSo sub $a=c$ into Equation three and you eventually get..\n$\\log2+2a-a^{2}=-3$\n$\\Rightarrow{a^{2}-2a-(3+\\log2)=0}$\n$\\Rightarrow{a^{2}-2a-\\log2000=0}$\n$a=\\frac{2\\pm\\sqrt{4+4\\log2000}}{2}=1\\pm\\sqrt{\\log20000}$\n\nWe want to find the two possible b's..so\nsub $a=1+\\sqrt{\\log20000}$ into 1\n\n$\\log2+(1+\\sqrt{4+\\log2})+b-(1+\\sqrt{4+\\log2})b=1$\n$\\Rightarrow{b_{1}=\\frac{\\log2+\\sqrt{4+\\log2}}{\\sqrt{4+\\log2}}}$\n\nsub $a=1-\\sqrt{4+\\log2}$ into 1\n\n$\\log2+(1-\\sqrt{4+\\log2})+b-(1-\\sqrt{4+\\log2})b=1$\n$\\Rightarrow{b_{2}=\\frac{-\\log2+\\sqrt{4+\\log2}}{\\sqrt{4+\\log2}}}$\n\n$y_{1}=10^{b_{1}}=10^{\\frac{\\log2+\\sqrt{4+\\log2}}{\\sqrt{4+\\log2}}}$\n\n$y_{2}=10^{b_{2}}=10^{\\frac{-\\log2+\\sqrt{4+\\log2}}{\\sqrt{4+\\log2}}}$\n\n$y_{1}+y_{2}=10^{\\frac{\\log2+\\sqrt{4+\\log2}}{\\sqrt{4+\\log2}}}+10^{\\frac{-\\log2+\\sqrt{4+\\log2}}{\\sqrt{4+\\log2}}}$\n\nif you can simplify this even more please do, I think there is a much better approach to this question\n[/hide]\r\nEDIT: yes i did have a sign error..", "Solution_2": "[quote=\"SM4RT\"]k i get this but for some reason my values are complex :huh: \n[hide=\"Partial\"]\nfirst let:\n$a=\\log_{10}x,b=\\log_{10}y,c=\\log_{10}z$\nNotice:\n$\\log_{10}2000=\\log_{10}1000+\\log_{10}2=3+\\log_{1}0{2}$\n\nSo, Equation One:\n$3+\\log2+a+b-ab=4$\n$\\log2+a+b-ab=1$\n\n$\\log2+b+c-bc=log2+a+b-ab$\n$(c-a)-(cb+ab)=0$\n$(c-a)-b(c-a)=0$\n$(1-b)(c-a)=0$\n$\\therefore{b=1\\text{ or }a=c}$\n\nif $b=1$ then substituting it back in reveals no solutions..so then $a=c$\n\nSo sub $a=c$ into Equation three and you eventually get..\n$\\log2+2a-a^{2}=3$\n$\\Rightarrow{a^{2}-2a+(3-\\log2)=0}$\n$\\Rightarrow{a^{2}-2a+\\log500=0}$\nhowever the discriminant is$4-4\\log500<{0}$\n\nWhy am i getting this??[/hide][/quote]\n\n[hide=\"SM4RT, maybe because of this?\"]Wait, $\\log(2000zx)-(\\log_{10}z)(\\log_{10}x) = 0$, using your substitutions, should simplify as follows:\n\n$\\log2+\\log(1000)+\\log_{10}z+\\log_{10}x-(\\log_{10}z)(\\log_{10}x) = 0$\n\n$\\log2+3+a+c-ac = 0$\n\nI think you switched the sign on the $3$.\n\nTo continue, Using your a = c result,\n\n$(\\log2+3)+2a-a^{2}= 0 a^{2}-2a-(\\log2+3) = 0$\n\nThe discriminant would then be $4+4\\log(2000)$, perfectly valid, but very messy. As a result, I'm stuck... Could someone check my work and expand upon this please?[/hide]", "Solution_3": "[quote=\"4everwise\"]The system of equations\n\\begin{eqnarray*}\\log_{10}(2000xy)-(\\log_{10}x)(\\log_{10}y) &=& 4\\\\ \\log_{10}(2yz)-(\\log_{10}y)(\\log_{10}z) &=& 1\\\\ \\log_{10}(2000zx)-(\\log_{10}z)(\\log_{10}x) &=& 0\\\\ \\end{eqnarray*}\nhas two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2}).$ Find $y_{1}+y_{2}.$[/quote]\r\n\r\nThis problem is wrong. I kept getting imaginary values for $\\log_{x}$ and $\\log_{z}$ until I checked and realized that the third equation should be $\\log_{10}(zx)-(\\log_{10}z)(\\log_{10}x) = 0$", "Solution_4": "[hide=\"Solution\"]\n[b]Solution.[/b]\n\\begin{eqnarray*}\\log_{10}(2000xy)-(\\log_{10}x)(\\log_{10}y) &=& 4\\\\ \\log_{10}(2yz)-(\\log_{10}y)(\\log_{10}z) &=& 1\\\\ \\log_{10}(2000zx)-(\\log_{10}z)(\\log_{10}x) &=& 0\\\\ \\end{eqnarray*}\nThe above system is equivalent to\n\\begin{eqnarray*}(\\log x-1)(\\log y-1) &=& \\log 2\\\\ (\\log y-1)(\\log z-1) &=& \\log 2\\\\ (\\log x-1)(\\log z-1) &=& 4+\\log 2\\\\ \\end{eqnarray*}\nLet $\\log x-1=u,\\log y-1=v,\\log z-1=w$. Hence,\n\\begin{eqnarray}uv &=& \\log 2\\\\ vw &=& \\log 2\\\\ uw &=& 4+\\log 2\\end{eqnarray}\nThus, from (1), $u=\\frac{\\log 2}{v}$. Using (3), $\\frac{w\\log 2}{v}=4+\\log 2$. Using (2), $w=\\pm\\sqrt{4+\\log 2}$. Then it follows that\n\\[(u,v,w)=\\left(\\pm\\sqrt{4+\\log 2},\\pm\\frac{\\log 2}{\\sqrt{4+\\log 2}},\\pm\\sqrt{4+\\log 2}\\right). \\]\nThis implies that\n\\[(x,y,z)=\\left(10^{\\pm\\sqrt{4+\\log 2}+1},10^{\\pm\\frac{\\log 2}{\\sqrt{4+\\log 2}}+1},10^{\\pm\\sqrt{4+\\log 2}+1}\\right) \\]\nAnd the conclusion follows.$\\blacksquare$[/hide]", "Solution_5": "Please hide your solutions, boxedexe.\r\n\r\n[hide=\"how are there multiple solutions?\"]\n\\begin{eqnarray*}log_{10}(\\frac{2000xy}{x+y}) &=& 4 \\\\ log_{10}(\\frac{2yz}{y+z}) &=& 1 \\\\ log_{10}(\\frac{zx}{x+z}) &=& 0 \\\\ \\frac{xy}{x+y}&=& 5 \\\\ \\frac{yz}{y+z}&=& 5 \\\\ \\frac{xz}{x+z}&=& 1 \\\\ \\frac{1}{x^{-1}+y^{-1}}&=& 5 \\\\ \\frac{1}{z^{-1}+y^{-1}}&=& 5 \\\\ \\frac{1}{x^{-1}+z^{-1}}&=& 1 \\\\ a&=&x^{-1}\\\\ b&=&y^{-1}\\\\ c&=&z^{-1}\\\\ a+b&=&\\frac{1}{5}\\\\ b+c&=&\\frac{1}{5}\\\\ a+b&=&b+c \\\\ a &=& c \\\\ x &=& z \\\\ a+a&=& 1 \\\\ a=c&=&\\frac{1}{2}\\\\ x=z&=&2\\\\ b&=&-\\frac{3}{10}\\\\ y &=&-\\frac{10}{3}\\end{eqnarray*}[/hide]", "Solution_6": "[quote=\"worthawholebean\"]Please hide your solutions, boxedexe.\n\n[hide=\"how are there multiple solutions?\"]\n\\begin{eqnarray*}log_{10}(\\frac{2000xy}{x+y}) &=& 4 \\\\ log_{10}(\\frac{2yz}{y+z}) &=& 1 \\\\ log_{10}(\\frac{zx}{x+z}) &=& 0 \\\\ \\frac{xy}{x+y}&=& 5 \\\\ \\frac{yz}{y+z}&=& 5 \\\\ \\frac{xz}{x+z}&=& 1 \\\\ \\frac{1}{x^{-1}+y^{-1}}&=& 5 \\\\ \\frac{1}{z^{-1}+y^{-1}}&=& 5 \\\\ \\frac{1}{x^{-1}+z^{-1}}&=& 1 \\\\ a&=&x^{-1}\\\\ b&=&y^{-1}\\\\ c&=&z^{-1}\\\\ a+b&=&\\frac{1}{5}\\\\ b+c&=&\\frac{1}{5}\\\\ a+b&=&b+c \\\\ a &=& c \\\\ x &=& z \\\\ a+a&=& 1 \\\\ a=c&=&\\frac{1}{2}\\\\ x=z&=&2\\\\ b&=&-\\frac{3}{10}\\\\ y &=&-\\frac{10}{3}\\end{eqnarray*}[/hide][/quote]\r\n$\\log x\\log y\\neq \\log (x+y)$ :wink:", "Solution_7": "[quote=\"boxedexe\"][quote=\"worthawholebean\"]Please hide your solutions, boxedexe.\n\n[hide=\"how are there multiple solutions?\"]\n\\begin{eqnarray*}log_{10}(\\frac{2000xy}{x+y}) &=& 4 \\\\ log_{10}(\\frac{2yz}{y+z}) &=& 1 \\\\ log_{10}(\\frac{zx}{x+z}) &=& 0 \\\\ \\frac{xy}{x+y}&=& 5 \\\\ \\frac{yz}{y+z}&=& 5 \\\\ \\frac{xz}{x+z}&=& 1 \\\\ \\frac{1}{x^{-1}+y^{-1}}&=& 5 \\\\ \\frac{1}{z^{-1}+y^{-1}}&=& 5 \\\\ \\frac{1}{x^{-1}+z^{-1}}&=& 1 \\\\ a&=&x^{-1}\\\\ b&=&y^{-1}\\\\ c&=&z^{-1}\\\\ a+b&=&\\frac{1}{5}\\\\ b+c&=&\\frac{1}{5}\\\\ a+b&=&b+c \\\\ a &=& c \\\\ x &=& z \\\\ a+a&=& 1 \\\\ a=c&=&\\frac{1}{2}\\\\ x=z&=&2\\\\ b&=&-\\frac{3}{10}\\\\ y &=&-\\frac{10}{3}\\end{eqnarray*}[/hide][/quote]\n$\\log x\\log y\\neq \\log (x+y)$ :wink:[/quote]\r\nWhoops.", "Solution_8": "I was doing over this problem..and I cant seem to find aninteger solutions for y1 + y2..\r\n\r\nIs this problem written wrong???", "Solution_9": "[quote=\"usaha\"]I was doing over this problem..and I cant seem to find aninteger solutions for y1 + y2..\n\nIs this problem written wrong???[/quote]\r\nIt never said it had to be an integer.", "Solution_10": "[quote=\"usaha\"]I was doing over this problem..and I cant seem to find aninteger solutions for y1 + y2..\n\nIs this problem written wrong???[/quote]\r\n\r\nIf you're thinking about the 2000 AIME, the problem on there had the last equation be \r\n\r\n$\\log zx-(\\log z)(\\log x)=0$", "Solution_11": "wow..that wouldnve made it a lot easier, the y1 and y2 are 5 adn 20 so 25...\r\n\r\nyou get the same thing, but you find that log x=log z and also log x=0 or 2.. plug them in and you get log y...", "Solution_12": "[quote=\"usaha\"]wow..that wouldnve made it a lot easier, the y1 and y2 are 5 adn 20 so 25...\n\nyou get the same thing, but you find that log x=log z and also log x=0 or 2.. plug them in and you get log y...[/quote]\r\n\r\nIs there an official solution for 2000's problems?", "Solution_13": "10^(1-log2)+10^(1+log2)=25", "Solution_14": "[hide=\"Full Solution to 2000 AIME I #9\"]Let $ \\log(\\omega) \\equal{} \\log_{10}(\\omega)$, and by this shorthand let $ a \\equal{} \\log(x)$, $ b \\equal{} \\log(y)$ and $ c \\equal{} \\log(z)$. Because $ \\log(pq) \\equal{} \\log(p) \\plus{} \\log(q)$, the system becomes\n\n$ \\log(2) \\plus{} 3 \\plus{} a \\plus{} b \\minus{} ab \\equal{} 4\\Rightarrow ab \\minus{} a \\minus{} b \\plus{} 1 \\equal{} \\minus{} \\log(2)\\Rightarrow (a \\minus{} 1)(b \\minus{} 1) \\equal{} \\log(2)$\n$ \\log(2) \\plus{} b \\plus{} c \\minus{} bc \\equal{} 1\\Rightarrow bc \\minus{} b \\minus{} c \\plus{} 1 \\equal{} \\log(2)\\Rightarrow (b \\minus{} 1)(c \\minus{} 1) \\equal{} \\log(2)$\n$ a \\plus{} c \\minus{} ac \\equal{} 0\\Rightarrow ac \\minus{} a \\minus{} c \\plus{} 1 \\equal{} 1\\Rightarrow (a \\minus{} 1)(c \\minus{} 1) \\equal{} 1$\n\nNow we substitute $ \\alpha \\equal{} a \\minus{} 1$, $ \\beta \\equal{} b \\minus{} 1$ and $ \\gamma \\equal{} c \\minus{} 1$ to get\n\n$ \\alpha\\beta \\equal{} \\log(2)$ (1)\n$ \\beta\\gamma \\equal{} \\log(2)$ (2)\n$ \\alpha\\gamma \\equal{} 1$ (3)\n\nWe take (1) divided by (2) and multiplied by (3):\n\n$ \\alpha^2 \\equal{} 1$,\n\nwhich gives us $ \\alpha \\equal{} \\pm 1$. Then $ \\beta \\equal{} \\pm\\log(2)$, so $ b \\equal{} 1\\pm\\log(2)$. Then the two solutions for $ b$ are\n\n$ b \\equal{} 1 \\plus{} \\log(2) \\equal{} \\log(10) \\plus{} \\log(2) \\equal{} \\log(20)$ and\n$ b \\equal{} 1 \\minus{} \\log(2) \\equal{} \\log(10) \\minus{} \\log(2) \\equal{} \\log(5)$.\n\nThus the two solutions for $ y$ are $ y_1 \\equal{} 20$ and $ y_2 \\equal{} 5$, so our answer is\n\n$ y_1 \\plus{} y_2 \\equal{} \\fbox{025}$.[/hide]", "Solution_15": "Have they made a hard logs question yet?\n\n[hide]\n\\[\\log_{10}(2000xy)-(\\log_{10}x)(\\log_{10}y) = 4. \\] \nTherefore,\n\\[\\log_{10}(2xy)-(\\log_{10}x)(\\log_{10}y) = 1. \\]\nBy comparing the first two equations, it becomes clear that\n\\[x=z \\].\nSolving the third equation gives\n\\[x=z=1,100 \\]\nPlugging this back into the equations gives\n\\[y=5,20 \\]\n\\[20+5 = 025. \\]\n\n[/hide]", "Solution_16": "[hide]First equation gives $3\\log_{10}2+\\log_{10}(x)+\\log_{10}(y)-\\log_{10}x\\log_{10}y=4$.\n\nWait first off let's let $\\log_{10}x=a, \\log_{10}y=b, \\log_{10}z=c$.\n\nTherefore $3\\log_{10}2+a+b-ab=4$ or therefore $3\\log_{10}2-(a-1)(b-1)=3$. This is the same as $(a-1)(b-1)=3\\log_{10}2-3$.\n\nThe second equation gives $\\log_{10}2+y+z-yz=1$. Therefore, $\\log_{10}2-(y-1)(z-1)=0$ or $\\log_{10}2=(b-1)(c-1)$. \n\nLastly, the third equation gives $a+c=ac$ or therefore $(a-1)(c-1)=1$. \n\nMultiply all the equations to give us $((a-1)(b-1)(c-1))^2=(3\\log_{10}2-3)(\\log_{10}2)$. \n\nWe care most about the variable $b$, so we go from teh equation $(a-1)(c-1)=1$ to give us $(b-1)^2=(3\\log_{10}2-3)(\\log_{10}2)$. First off, simplify this, to give us $\\log_{10}(2000)-\\log_{10}(1000)=\\log_{10}2$. Therefore $b-1=\\pm \\log_{10}2$. Therefore, substitute $b=\\log_{10}y$ to give us $\\log_{10}y-1=\\log_{10}2, -\\log_{10}2$.\n\nThe first gives us $\\log_{10}y=\\log_{10}(20)$ and second gives us $\\log_{10}y=\\log_{10}(5)$ so their sum is $\\boxed{25}$. [/hide]", "Solution_17": "[hide=\"fast solution\"]\nlet a=log x\nb=log y\nc= log z\n\nlog 2000xy-logxlogy=3+log 2+a+b-ab=4: (a-1)(b-1)=log2\ndo similar stuff to get (b-1)(c-1)=log 2 and (a-1)(c-1)=1\nThus b-1=plus or minus log 2 while both the others are plus or minus log 1\nthus y1+y2=10^(1+log2)+10^(1-log2)=10*10^(log2)+10*10^(-log2)=20+5=\n\n025\n\n[/hide]", "Solution_18": "[hide=\"bro\"]\nLet us denote the equations as 1,2,3 in descending order respectively.\nSubtracting $2$ from $1$ yields,\n$\\log 1000 + \\log 2 + \\log x + \\log y - \\log x\\log y - \\log 2 - \\log y - \\log z + \\log y\\log z = 3$\nWe get this from the log property $ \\log xy = \\log x + \\log y$\nFrom here it is easy to see,\n$ \\log x - \\log z + \\log y\\log z - \\log x\\log y = (1 - \\log y) (\\log x - \\log z) = 0$\n$\\rightarrow$ $x=z$ or $y=10$\nsubstituting $x=z$ we get $y= 20$ or $y=5$ and by checking if $y=10$ works, we find $y=5,20$ are the only solutions so,\n$y_1 + y_2 = 25$\nQED\n\n\n\n\n[/hide]", "Solution_19": "@Binomial-theorem: How is $ \\log_{10} (2000) = 3 \\log_{10} (2) $. I am getting $ 3 + \\log_{10} (2) $. But, with what you did, everything turns out beautifully. I did exactly what you did but I used $ 3 + \\log_{10} (2) $ and got a bunch of nice but non-working dead ends. Hm.", "Solution_20": "[hide=\"Can someone tell me where my reasoning fails, if x, y, and z are positive real numbers?\"]\n\nLet $a=\\log_{10}\\left(x\\right)$, $b=\\log_{10}\\left(y\\right)$, and $c=\\log_{10}\\left(z\\right)$. The last equation gives us\n\n$a+c-a\\cdot c=0$\n\nThus either $a+c=0$ or $\\dfrac{a^3+c^3}{a+c}=\\dfrac{\\left(a+c-a\\cdot c\\right)\\cdot\\left(a+c\\right)}{a+c}=0$. Let's prove $a+c=0$ with a proof by contradiction. If $a+c\\ne0$, then $\\dfrac{a^3+c^3}{a+c}=0$, so $a^3=-c^3$, so $a=-c$, a contradiction.\n\nThus, with that and $a^2+c^2-a\\cdot c=0$, $a=c=0$. Which doesn't look good.[/hide]", "Solution_21": "$a^3+c^3=(a+c)(a^2-ac+c^2)$, not $(a+c)(a+c-ac)$.", "Solution_22": "Ah, sorry, messed up a little, \\Edits\n\nThough that doesn't destroy my \"proof.\"\n\nEDIT: Oh. It does. Nvmd, the LHS isn't $a^2+c^2-a\\cdot c$.", "Solution_23": "[quote=SM4RT]k i get this but for some reason my values are complex :huh: \n[hide=\"Partial\"]\nfirst let:\n$a=\\log_{10}x,b=\\log_{10}y,c=\\log_{10}z$\nNotice:\n$\\log_{10}2000=\\log_{10}1000+\\log_{10}2=3+\\log_{1}0{2}$\n\nSo, Equation One:\n$3+\\log2+a+b-ab=4$\n$\\log2+a+b-ab=1$\n\n$\\log2+b+c-bc=log2+a+b-ab$\n$(c-a)-(cb+ab)=0$\n$(c-a)-b(c-a)=0$\n$(1-b)(c-a)=0$\n$\\therefore{b=1\\text{ or }a=c}$\n\nif $b=1$ then substituting it back in reveals no solutions..so then $a=c$\n\n[/quote]\n\nBy substituting a = c into the third equation, a = logx = 0 or 2. Plugging these values for equation 1, logy = log 5 for a = 0 and logy = log20 for a = 2. Didn't feel like it was too hard", "Solution_24": "I just tried this and only got one solution, because I got to $(\\log y)^2-\\log y^2=(\\log 5)^2-\\log 5^2.$ I looked at a solution and it showed that the other value is $20$ but used a different method. How do I find $y=20$ from here?\n\nEDIT: Oh I see! We complete the squares: $(b-1)^2=(\\log 5-1)^2\\Longleftrightarrow b=1\\pm(\\log 5-1)$ and thus the other solution is $b=-\\log5+\\log100=\\log20$.", "Solution_25": "In [url=https://youtu.be/sOyLnGJjVvc?t=372]this video[/url], at 6:12 he converts $\\log_{10} y_1=1-\\log 2$ to $\\frac{10}{10^{\\log_2}}=y_1$. He also converted $\\log_{10} y_2=1+\\log 2$ into $y_2=(10)(2)$. How exactly did he do this? I don't really understand which properties he used.", "Solution_26": "bump$ $", "Solution_27": "p5: Denote $\\log_{10}{x} = a, \\log_{10}{y} = b, \\log_{10}{z} = c$. Then $3 + \\log_{10}{2} + a + b - ab = 4, \\log_{10}{2} + b + c - bc = 1, a + c - ac = 0$. Simplifying gives $\\log_{10}{2} + a + b - ab = 1, \\log_{10}{2} + b + c - bc = 1, a + c - ac = 0$. Subtracting the first two equations gives $a - c - ab + bc = (a-c) - b(a-c) = (1-b)(a-c) = 0$. If $b = 1$, then the second equation gives $4+\\log_{10}{2} = 4$, which is obviously not possible. If $a = c$, then the third equation gives that either $a = c = 0$ or $a = c = 2$. In the first case, we get $b = 1 - \\log_{10}{2} = \\log_{10}{5} \\implies y = 5$. In the second case, we get $b = 1 + \\log_{10}{2} = \\log_{10}{5} \\implies y = 20$. The problems tells us that there are two solutions, so we don't have to worry about extraneous solutions. The answer is $5+20=\\boxed{25}$.", "Solution_28": "[quote=RJ5303707]In [url=https://youtu.be/sOyLnGJjVvc?t=372]this video[/url], at 6:12 he converts $\\log_{10} y_1=1-\\log 2$ to $\\frac{10}{10^{\\log_2}}=y_1$. He also converted $\\log_{10} y_2=1+\\log 2$ into $y_2=(10)(2)$. How exactly did he do this? I don't really understand which properties he used.[/quote]\n\nI don't know why he would do that for the first one (my youtube is blocked so I can't check if there's a typo) but here's what you do to solve it; write $1-\\log 2 = \\log 10 - \\log 2 = \\log \\frac{10}{2} = \\log 5$ and for the second one write $1+\\log 2 = \\log 10 + \\log 2 = \\log 10 \\cdot 2 = \\log 20$.", "Solution_29": "[b]Solution:[/b] Let $a = \\log_{10}x, b = \\log_{10}y, c=\\log_{10}z$. Then, by simple log manipulations we have that $\\log(2) + 3 + a + b - ab =4, \\log(2) + b + c - bc = 1, a + c - ac = 0$. Then, subtracting the first two equations, we have that $a - c - ab + bc = 0 \\Rightarrow (1-b)(a-c) = 0$. Clearly, $b=1$ fails. It follows that $a=c$. Using our third equation, we now see the only possible values are $(2, y_1, 2), (0, y_2, 0)$. One can use the system of equations once again to calculate these values as $10^{1 + \\log(2)} + 10^{1 - \\log(2)} = 20 + 5 = \\boxed{025}$." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let\r\n\r\n$ v_{m}=\\frac{\\sum_{1\\le{k_{1}}<{k_{2}}< ... <{k_{m}}\\le{n}}{x_{k_{1}}x_{k_{2}}...x_{k_{m}}}}{\\binom{n}{m}},(x_{1},\\cdots,x_{n})\\in{R_{+}^{n}},$\r\n\r\n\r\n\r\nprove that\r\n\r\n\\[ ({k-2})(v_2v_{k-2}-v_{k})\\ge({{k-1})(v_{1}v_{k-1}-v_{k}),k\\ge{3}.}\\]", "Solution_1": "Correction:\r\n\\[ ({k - 1})(v_{2}v_{k - 2} - v_{k})\\ge{2(k - 2)(v_{1}v_{k - 1} - v_{k}),k\\ge{3}.}\\]" } { "Tag": [ "calculus", "integration", "logarithms", "real analysis", "real analysis solved" ], "Problem": "Compute $\\int_{0}^{+\\infty}ln(thx)dx$ where\r\n$thx = \\frac{shx}{chx}$", "Solution_1": "American textbooks would use the notation tanh(x), sinh(x), and cosh(x) rather than thx, shx, or chx. \n\ntanh.\n\n\n\nThe answer is [hide]-(pi^2)/8.[/hide] Intermediate steps tomorrow.", "Solution_2": "The intermediate steps, as promised.\r\n\r\nLet $u =$ tanh$(x)$. Then $du = $1 - tanh$^2(x) dx = (1-u^2)dx.$ We rewrite this as $\\displaystyle dx = \\frac{du}{1-u^2}.$ As $x$ goes from 1 to $\\infty$, then $u$ goes from 0 to 1.\r\n\r\nMake that substitution, and the integral becomes:\r\n\r\n$\\displaystyle \\int_0^1 \\frac{\\ln u}{1-u^2}du = \\int_0^1 (\\sum_{k=0}^{\\infty}u^{2k}\\ln u) du = \\sum_{k=0}^{\\infty}\\int_0^1 (u^{2k}\\ln u) du = -\\sum_{k=0}^{\\infty}\\frac1{(2k+1)^2}.$\r\n\r\nThis sum is the negative of 3/4 of the sum $\\displaystyle \\sum_{n=1}^{\\infty}\\frac1{n^2}$, which is well known to be $\\displaystyle \\frac{\\pi^2}6.$ Therefore, the answer to the problem is $\\displaystyle -\\frac{\\pi^2}8.$\r\n\r\nThere are approaches to calculating $\\zeta (2)$, such as using the Fourier series of the integral of the square wave, that first calculate the sum of reciprocal odd squares - the sum we have here.\r\n\r\nThere are several possible justification for the interchange of sum and integral. The Monotone Convergence Theorem certainly suffices.", "Solution_3": "Nice proof Kent\r\nthx \r\n\r\nMoubinool OMARJEE" } { "Tag": [ "algebra", "polynomial", "algebra proposed" ], "Problem": "Let $ p > 2$ be a prime number. Find the least positive number $ a$ which can be represented as\r\n\\[ a \\equal{} (X \\minus{} 1)f(X) \\plus{} (X^{p \\minus{} 1} \\plus{} X^{p \\minus{} 2} \\plus{} \\cdots \\plus{} X \\plus{} 1)g(X),\r\n\\]\r\nwhere $ f(X)$ and $ g(X)$ are integer polynomials.\r\n\r\n[i]Mircea Becheanu[/i].", "Solution_1": "The remainder of $x^{p-1}+\\ldots+x+1$ when divided by $x-1$ is $p$, so $a=p$ works. It's also the smallest one, since plugging $1$ into $(x-1)f(x)+(x^{p-1}+\\ldots+x+1)g(x)=a$ gives $p|a$.\r\n\r\nWhy do we need $p$ to be prime? :? The remainder when we divide $\\frac{x^n-1}{x-1}$ to $x-1$ is always $n$ (since $x^{n-1}+\\ldots+x+1=(x-1)(x^{n-2}+2x^{n-3}+\\ldots+(n-2)x+(n-1))+n$), and the exact same argument as above shows that $n|a$ in the general case.. Sorry if I'm missing something obvious.", "Solution_2": "Are we to assume the degree of $f(x),g(x)$ are $\\geq 1$?", "Solution_3": "[quote=\"Fermat's Little Turtle\"]Are we to assume the degree of $f(x),g(x)$ are $\\geq 1$?[/quote]\nIt doesn't matter." } { "Tag": [ "geometry", "circumcircle", "trigonometry", "ratio" ], "Problem": "let ABC be a triangle with(A=90).and let (C) be its circumcircle .we draw AM wich is bisector of the diameter BC.\r\n\r\nwe draw CE ,wich is the bisector of the segment AM,and let it intersect (C) is point P.\r\n\r\nwe draw the altitude of triangle ,from point A.and let it intersect (C) in point Q.\r\n\r\nprove that : \r\n\r\n PQ bisects the segment AB", "Solution_1": "Let the median of $ AM $ meets $ AM $ at $ E $ and $ AB $ at $ F $.\r\n\r\n$ PQ $ bisect $ AB $ if and only if $ \\triangle{APQ}=\\triangle{BPQ} $\r\n\r\nSince $ \\angle{PAQ}+\\angle{PBQ}=180^o $, so we have: $ \\frac{\\triangle{APQ}}{\\triangle{BPQ}}=\\frac{AP \\cdot AQ}{BP \\cdot BQ} $\r\n\r\nSince $ M $ and $ E $ are mid-points, so use Menelaus theorem on triangle $ ABM $ and line $ CEF $:\r\n\\[\\frac{AF\\cdot BC \\cdot EA}{FB \\cdot MC \\cdot ME}=1\\]\r\n\\[\\Longleftrightarrow \\frac{AF}{FB}=\\frac12\\]\r\nBecause $ \\frac{AF}{FB}=\\frac{AP}{BP}\\frac{\\sin{\\angle{APC}}}{\\sin{\\angle{BPC}}}=\\frac{AP}{BP}\\sin{\\angle{ABC}} $\r\n\r\nHence: $ \\frac{AP}{BP}=\\frac{1}{2\\sin{\\angle{ABC}}} $\r\nPut it back to the original area ratio equation, we will have:\r\n\\[\\frac{\\triangle{APQ}}{\\triangle{BPQ}}=\\frac{1}{2\\sin{\\angle{ABC}}}\\frac{AQ}{BQ}=\\frac{AD}{\\sin{\\angle{ABC}}}\\cdot \\frac{1}{AB}=1\\]\r\n\r\nTherefore, $ PQ $ bisect segment $ AB $.", "Solution_2": ":yup: :first: very very good.ure a good solver ;)", "Solution_3": "angle ABC= angle CPQ(guess why?).\r\n\r\nangleBAM= angleABH=angleMBA. and it means that the quadrilaterall \r\n\r\nPDEA is cyclic.(D is the intersection of PQ and AB) and (E is the mid point of MA).\r\n\r\nif PDEA is cyclic we will have : angle PAD=angle PED.\r\n\r\nbut:angle PAD=anglePCB. and it means that angle PED=anglePCB.\r\n\r\nthen ED and and BC will be paralell. the problem is solved by looking at the two simillar triangle AED and AMB,we\r\n\r\ncan find that : AD=BD" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ C(O)$ . Denote $ E\\in AC\\cap BD$ .\r\n\r\nSuppose $ EB=3\\cdot ED$ and define $ \\{\\begin{array}{c}\r\nM\\in (BC)\\ ,\\ MB=MC\\\\\\\\\r\nN\\in (BE)\\ ,\\ NE=ED\\\\\\\\\\\r\nP\\in MN\\cap AB\\end{array}$ . Prove that $ EP\\perp EO$ .", "Solution_1": "We draw the line through the point $ M$ and parallel to $ AC,$ which intersects the line segments $ BD,\\ AB,$ at points $ K,\\ F$ respectively and so, we have $ BK \\equal{} KE$ and $ BF \\equal{} FA.$ \r\n\r\nBecause of now, the points $ K,\\ N,$ are the midpoints of the segments $ BE,\\ BD$ and because of $ BE \\equal{} 3\\cdot EB,$ we conclude that ( easy to prove ) $ BK \\equal{} 3\\cdot KN$ $ ,(1)$\r\n\r\nIt is easy to show that the points $ B,\\ M,\\ O,\\ N,\\ F$ are concyclic, with diameter of their circumcircle $ (O'),$ the segment $ BO.$\r\n\r\nThe point $ P,$ as the intersection point of the line segments $ BF,\\ MN,$ lies on the polar of $ K,$ with respect to the circle $ (O')$ and let $ S$ be the orthogonal projection of $ P,$ on the line segment $ O'K.$\r\n\r\nIt is clear that $ PS\\perp OE,$ because of $ O'K\\parallel OE,$ where $ O'$ is the center of $ (O').$\r\n\r\nWe denote the point $ E'\\equiv PS\\cap BD$ and it is enough to prove that $ E'\\equiv E.$\r\n\r\nBecause of the point $ E',$ is the intersection point of the line segment $ BN$ passing through the point $ K,$ from the polar of $ K$ with respect to the circle $ (O'),$ we conclude that the points $ E',\\ N,\\ K,\\ B,$ are in harmonic conjugation.\r\n\r\nSo, we have $ \\frac {E'B}{E'N} \\equal{} \\frac {KB}{KN}$ $ ,(2)$\r\n\r\nFrom $ (1),$ $ (2)$ $ \\Longrightarrow$ $ \\frac {E'B}{E'N} \\equal{} 3 \\equal{} \\frac {EB}{ED} \\equal{} \\frac {EB}{EN}$ $ ,(3)$\r\n\r\nFrom $ (3)$ $ \\Longrightarrow$ $ E'\\equiv E$ and the proof is completed.\r\n\r\n$ \\bullet$ This proof is dedicated to [b][size=100]Anastasios Vafeidis[/size][/b]. \r\n\r\nKostas Vittas.", "Solution_2": "[quote=\"Virgil Nicula\"] [color=darkred]Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w = C(O)$ . Denote $ E\\in AC\\cap BD$ .\n\nSuppose $ EB = 3\\cdot ED$ and define $ \\{\\begin{array}{c} M\\in (BC)\\ ,\\ MB = MC \\\\\n \\\\\nN\\in (BE)\\ ,\\ NE = ED \\\\\n \\\\\n\\ P\\in MN\\cap AB\\end{array}$ . Prove that $ EP\\perp EO$ .[/color] [/quote]\n\n[color=darkblue][b][u]Proof.[/u][/b] Denote $ \\{\\begin{array}{c} R\\in MN\\cap BC \\\\\n\\ S\\in PE\\cap DC\\end{array}$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \\overline {MNR}$ and the triangle $ BEC$ :\n\n$ \\frac {RE}{RC}\\cdot\\frac {MC}{MB}\\cdot\\frac {NB}{NE} = 1$ $ \\implies$ $ RC = 2\\cdot RE$ $ \\implies$ $ ER = EC$ . Thus, $ \\{\\begin{array}{c} ER = EC \\\\\n\\ EN = ED\\end{array}\\|$ $ \\implies$ $ \\overline {MNPR}\\parallel \\overline {DSC}$ $ \\implies$ $ EP = ES$ .\n\nApply the [b]property of butterfly[/b] to the interior point $ E\\in BD\\cap AC$ of the circle $ w$ : $ \\boxed {\\ EP = ES\\ \\Longleftrightarrow\\ EP\\perp EO\\ }$ .[/color]\n\n[quote=\"Virgil Nicula\"] [color=darkred][b][u]An easy extension.[/u][/b] Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w = C(O)$ . Define the points :\n\n$ E\\in AC\\cap BD$ ; $ \\{\\begin{array}{c}\nN\\in (BE)\\\\\\\nEN=ED\\end{array}$ ; $ \\{\\begin{array}{c}\nR\\in AC\\\\\\\nA\\in (RC)\\\\\\\nER=EC\\end{array}$ ; $ P\\in NR\\cap AB$ . Prove that $ EP\\perp EO$ .[/color] [/quote]", "Solution_3": "Dear Mathlinkers,\nthis problem comes also from\nDuman A. N., Problem 1669, Mathematics Magazine vol. 76, 2 (2003) 151\nand can be solved with the butterfly theorem\nsee\nhttp://perso.orange.fr/jl.ayme vol. 7 A new metamorphosis of the butterfly theorem p. 70\nSincerely\nJean-Louis\nSincerely\nJean-Louis", "Solution_4": "To avoid using butterfly theorem: let $PE$ cut the circumcircle at $K, H$ and cut $DC$ at $G$. It is obvious that $PE = EG$, if we can prove that $KP = GH$ then we will be done.\nUsing sinus rule we have $GC/sinE_{1} = EG/sinC_{1}$ and $AP/sinE_{1} = PE/sinA_{1}$, hence $GC/AP = sinA_{1}/sinC_{1}$. Similarly $BP/DG = sin D_{1}/sinB_{1}$. Thus $GC/AP = BP/DG$ or $GC.GD = PA.PB$ or $GH.GK = PK. PH$, it follows that $GH = KP$ (q.e.d)" } { "Tag": [ "analytic geometry", "FTW", "email", "ARML", "probability", "expected value" ], "Problem": "What scores do people think they got? What did people get as answers?\r\n\r\nI got 134, that is, 22 right, 2 wrong, 1 omit. I got 15 and 24 wrong, and I omitted 23.\r\n\r\nI don't remember all the answers, but here are a few, checked with others at my school:\r\n\r\n#25 is 26\r\n#24 is D\r\n#23 is 272\r\n#21 is E\r\n\r\nthe giant problem is 30\r\n#15 is 13\r\n\r\n\r\nAnd a certain freshman at my school got a very good score...he'll tell you more later...", "Solution_1": "aaaaaaaaaaaaaa", "Solution_2": "although i got more problems correct this year vs. last year,\r\ni thought it was harder\r\n\r\n(also consider that ive gotten better at math)", "Solution_3": "I thought last year's was a little harder.\r\nThere was that tricky expected value one and also the weird mass points problem.\r\nThis year's was pretty straightforward; only 23 required a lot of work.\r\n\r\nI will not make any claims about my score just in case.", "Solution_4": "Hint: He got a 150.\r\n\r\nI did not take it becuase my school doesn't do it this year (for some odd reason...)", "Solution_5": "Yongyi, why is your school so amazing but so stupid at the same time?\r\nThey let you take calc in 8th grade but not MAML...", "Solution_6": "I have no idea :(\r\n\r\n:noo: :noo:\r\n\r\nOn Tuesday, I knew MAML started then and I asked my coach about it, and he said there's no math competition today (meaning Tuesday).", "Solution_7": "I got an 88, same as last year, partially because I thought east was a different direction........ seemingly its along the positive x axis if you r at the origin. who knew?", "Solution_8": "I got 144, switched the people on #19. Oh well.", "Solution_9": "grr...made 2 careless mistakes\r\n\r\n\r\n132 this year... :( \r\n\r\ni thought the jack didn't move when the giant was chasing him...\r\n\r\ndidn't we all learn that when a big animal is chasing you, you stand still? (jurassic park any1?)", "Solution_10": "Yea...\r\nThat would make so much sense\r\nAn AMC level question in which there is only one thing moving...", "Solution_11": "i think i made the cutoff", "Solution_12": "cutoff last yr was 102", "Solution_13": "[quote=\"alkjash\"]Yea...\nThat would make so much sense\nAn AMC level question in which there is only one thing moving...[/quote]\r\n\r\nwell..mmo is so ez that i assumed that there was only one thing that was moving... :maybe: \r\n\r\nlike seriously, the first 10 questions are like free points", "Solution_14": "[quote=\"Pineappleperson\"]like seriously, the first 10 questions are like free points[/quote]\r\n\r\nlol it took me like 20 minutes to get through like the first 5 questions...then I happened to look at the clock and I realized I needed to work faster to get through the questions\r\n\r\n132: 21 right, 1 wrong, 3 omit\r\n\r\nI got 11 wrong (Somehow I managed to say: $ \\frac {5}{3} \\pi \\left(\\frac {r}{2}\\right)^2 \\equal{} \\frac {5}{4} \\pi r^2 \\equal{} \\cdots \\equal{} 54\\pi$. Because obviously $ \\frac {5}{3} \\pi \\left(\\frac {r}{2}\\right)^2 \\equal{} \\frac {5}{4} \\pi r^2$. :wallbash_red: ). I omitted 23 (too lazy to coordinate bash/trig bash it), 24 (thought the calculations were really long...only after the contest did I read the word \"integer\"), and 25 (is the answer 26 or 50? in other words, could \"coincide\" include the cases are in opposite directions but on the same line?)", "Solution_15": "[quote=\"CatalystOfNostalgia\"]Is the meeting being re-postponed because of snow?[/quote]\r\nIt just was.\r\n\r\nI'm 99% sure the finalist cutoff was settled by email (I participated in the discussion but it's hard to \"vote\" by email, and not everyone contributed to the conversation). The finalist cutoff was needed so L2 site numbers could be determined and the sites could be contacted last week. The semifinalist cutoff is probably set, just by tradition.\r\n\r\nI just sent an email to the board proposing we release the results at noon today using the agreed-upon finalist and traditional semifinalist cutoffs. If no one objects, I'll send an email here when the information is online. If the information isn't here by, say, 1 PM, send me a PM and I'll tell you your score and rank.", "Solution_16": "Why doesn't the MAML board just use a set formula, like setting the cutoff to be the highest possible score that will make at least 100 people be finalists?", "Solution_17": "[quote=\"Phelpedo\"]Why doesn't the MAML board just use a set formula, like setting the cutoff to be the highest possible score that will make at least 100 people be finalists?[/quote]\r\nBasically we do. We discuss the L1 cutoffs at the meeting mainly for the formality. (The meeting has to exist anyway, to discuss details of the state championship math meet, the L2 Olympiad, ARML funding, Math Day, etc.). In case there is something like a 400 way tie for 99th place, of course, a formula like you mention would break down. e.g., this year the choice was between 96 finalists and 120 finalists, and we chose the latter.\r\n\r\nI'll have the results up on the WOCOMAL website shortly.", "Solution_18": "Here are the finalists and semifinalists: http://www.wocomal.org/statistics/olympiad/200809/200809o_level1.html\r\n\r\nAs always, the names you see are the names that the Scantron picked up, so if there were miscodings (or incomplete codings) of names there will be errors. Let me know any errors that you see, and I'll fix them. Thanks.", "Solution_19": "Do only the finalists take the next round?", "Solution_20": "Mr. Yanco, would it be possible to also put up an unofficial list that includes Phillips Academy participants?", "Solution_21": "[quote=\"Phelpedo\"]would it be possible to also put up an unofficial list that includes Phillips Academy participants?[/quote] All of the PA participants who scored high enough to be a semifinalist or better are listed at the bottom of the page (as is the DA postgraduate and the Newton eighth grader). Were you asking for them to be sorted into the list?", "Solution_22": "Thanks for putting this up!\r\n\r\nLexington name corrections:\r\n\"Elkin Ro\" should be Welkin Uttaro. Welkin has failed to bubble his name correctly on pretty much all of his AMC tests...\r\n\"Kristine Im\" should be Kristine Kim. She is a junior.\r\nHa-Young as in Ha-Young Kang is hyphenated. Not a big deal.\r\n\"JaeYoon Lee\" should be Jaeyoon Lee, without a capital Y.\r\n\"Ju Zheng\" should be Julia Zheng (but in this case feel free to leave it as Ju purely for comedic value).", "Solution_23": "[quote=\"rjyanco\"][quote=\"Phelpedo\"]would it be possible to also put up an unofficial list that includes Phillips Academy participants?[/quote] All of the PA participants who scored high enough to be a semifinalist or better are listed at the bottom of the page (as is the DA postgraduate and the Newton eighth grader). Were you asking for them to be sorted into the list?[/quote]\r\n\r\nAh, I didn't look at the bottom. It's great as is, thanks.", "Solution_24": "I just fixed all of the name errors I was told about (five Lexington, one AB). Let me know if you see anything else.", "Solution_25": "This has been corrected on the main MAML site, but Sherrie Ng of Sharon should be Sherrie Wang. Thanks.\r\n\r\nThanks also for keeping us all up to date on just about everything.\r\n\r\nMM", "Solution_26": "Thanks for pointing that out -- the official results are now online at http://www.maml.org. (Yesterday the complete results, of all ~2500 students, were available there behind a very weak password. I didn't want to direct you there until that got straightened out.)", "Solution_27": "Just curious: how come 8th graders aren't allowed to advanced?", "Solution_28": "[quote=\"rjyanco\"]Thanks for pointing that out -- the official results are now online at http://www.maml.org. (Yesterday the complete results, of all ~2500 students, were available there behind a very weak password. I didn't want to direct you there until that got straightened out.)[/quote]\r\n\r\nI was wondering what that site was going to be used for; I guess now we know.\r\n\r\nJust to let you know, some (if not all) of the mistakes that were corrected on the wocomal site were not corrected on the maml site. For instance, \"Im Kristine\" of Lexington High School should be \"Kim Kristine\", and she's in 11th grade; \"Ro Elkin\" is really \"Uttaro Welkin\"; and so on.", "Solution_29": "[quote=\"calc rulz\"]Just curious: how come 8th graders aren't allowed to advanced?[/quote]\r\nThe obvious answer is that a majority of those on the MAML board voted that way. I believe the majority argument was that talented middle school students would get their chance to advance when they reached high school, so they shouldn't displace current (albeit less talented) high school students." } { "Tag": [ "inequalities", "calculus", "derivative", "Cauchy Inequality", "three variable inequality" ], "Problem": "Let $a,b,c$ be positive real numbers. Prove that\r\n\\[\\underset{cyc}{\\sum}\\frac{a}{b}\\ge \\underset{cyc}{\\sum}\\sqrt{\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}\\]\r\n:)", "Solution_1": "$\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\geq \\sqrt{\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}}$\r\n$\\frac{a^{2}}{b^{2}}+\\frac{b^{2}}{c^{2}}+\\frac{c^{2}}{a^{2}}+2(\\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}) \\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}+\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}+2(\\sqrt{\\frac{b^{2}+a^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{a^{2}+c^{2}}{a^{2}+b^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{c^{2}+a^{2}}})$\r\nLet's prove that $\\frac{a^{2}}{b^{2}}+2\\frac{b}{a}\\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\r\n$S=\\frac{a^{4}+2b^{3}a}{a^{2}b^{2}}-\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\r\n$S= \\frac{a^{4}c^{2}+2b^{5}a+2b^{3}ac^{2}-a^{2}b^{2}c^{2}}{a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\r\nBy AM-GM inequality $a^{4}c^{2}+2b^{3}ac^{2}=a^{4}c^{2}+b^{3}ac^{2}+b^{3}ac^{2}\\geq 3a^{2}b^{2}c^{2}$ \r\n$S \\geq \\frac{2b^{5}a+2a^{2}b^{2}c^{2}}{ a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\r\n$\\frac{b^{3}+ac^{2}}{ a(b^{2}+c^{2})}\\geq \\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\r\n$(b^{3}+ac^{2})^{2}(a^{2}+c^{2}) \\geq (b^{2}+c^{2})^{3}a^{2}$\r\n$a^{4}c^{4}+2ab^{3}c^{4}+2a^{3}c^{2}b^{3}\\geq 3a^{2}b^{4}c^{2}+3a^{2}b^{2}c^{4}$\r\n$\\Longrightarrow (a-b)^{2}(c^{4}(a+b)^{2}+c^{2}ab(a^{2}+b^{2})) \\geq 0$\r\nSo inequality proved.", "Solution_2": "I posted some time ago but there was noone interested in: http://www.mathlinks.ro/Forum/viewtopic.php?t=119604", "Solution_3": "[quote=\"cefer\"]$\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\geq \\sqrt{\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}}$\n$\\frac{a^{2}}{b^{2}}+\\frac{b^{2}}{c^{2}}+\\frac{c^{2}}{a^{2}}+2(\\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}) \\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}+\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}+2(\\sqrt{\\frac{b^{2}+a^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{a^{2}+c^{2}}{a^{2}+b^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{c^{2}+a^{2}}})$\nLet's prove that $\\frac{a^{2}}{b^{2}}+2\\frac{b}{a}\\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$S=\\frac{a^{4}+2b^{3}a}{a^{2}b^{2}}-\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$S= \\frac{a^{4}c^{2}+2b^{5}a+2b^{3}ac^{2}-a^{2}b^{2}c^{2}}{a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\nBy AM-GM inequality $a^{4}c^{2}+2b^{3}ac^{2}=a^{4}c^{2}+b^{3}ac^{2}+b^{3}ac^{2}\\geq 3a^{2}b^{2}c^{2}$ \n$S \\geq \\frac{2b^{5}a+2a^{2}b^{2}c^{2}}{ a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$\\frac{b^{3}+ac^{2}}{ a(b^{2}+c^{2})}\\geq \\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$(b^{3}+ac^{2})^{2}(a^{2}+c^{2}) \\geq (b^{2}+c^{2})^{3}a^{2}$\n$a^{4}c^{4}+2ab^{3}c^{4}+2a^{3}c^{2}b^{3}\\geq 3a^{2}b^{4}c^{2}+3a^{2}b^{2}c^{4}$\n$\\Longrightarrow (a-b)^{2}(c^{4}(a+b)^{2}+c^{2}ab(a^{2}+b^{2})) \\geq 0$\nSo inequality proved.[/quote]\r\ni haven't checked it yet, i'm too lazy :(. But i think my solution is very impressive because i used contradiction. :)", "Solution_4": "[quote=\"toanhocmuonmau\"][quote=\"cefer\"]$\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\geq \\sqrt{\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}}$\n$\\frac{a^{2}}{b^{2}}+\\frac{b^{2}}{c^{2}}+\\frac{c^{2}}{a^{2}}+2(\\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}) \\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}+\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}+2(\\sqrt{\\frac{b^{2}+a^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{a^{2}+c^{2}}{a^{2}+b^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{c^{2}+a^{2}}})$\nLet's prove that $\\frac{a^{2}}{b^{2}}+2\\frac{b}{a}\\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$S=\\frac{a^{4}+2b^{3}a}{a^{2}b^{2}}-\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$S= \\frac{a^{4}c^{2}+2b^{5}a+2b^{3}ac^{2}-a^{2}b^{2}c^{2}}{a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\nBy AM-GM inequality $a^{4}c^{2}+2b^{3}ac^{2}=a^{4}c^{2}+b^{3}ac^{2}+b^{3}ac^{2}\\geq 3a^{2}b^{2}c^{2}$ \n$S \\geq \\frac{2b^{5}a+2a^{2}b^{2}c^{2}}{ a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$\\frac{b^{3}+ac^{2}}{ a(b^{2}+c^{2})}\\geq \\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$(b^{3}+ac^{2})^{2}(a^{2}+c^{2}) \\geq (b^{2}+c^{2})^{3}a^{2}$\n$a^{4}c^{4}+2ab^{3}c^{4}+2a^{3}c^{2}b^{3}\\geq 3a^{2}b^{4}c^{2}+3a^{2}b^{2}c^{4}$\n$\\Longrightarrow (a-b)^{2}(c^{4}(a+b)^{2}+c^{2}ab(a^{2}+b^{2})) \\geq 0$\nSo inequality proved.[/quote]\ni haven't checked it yet, i'm too lazy :(. But i think my solution is very impressive because i used contradiction. :)[/quote]\r\nit is true", "Solution_5": "[quote=\"toanhocmuonmau\"]Let $a,b,c$ be positive real numbers. Prove that\n\\[\\underset{cyc}{\\sum}\\frac{a}{b}\\ge \\underset{cyc}{\\sum}\\sqrt{\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}\\]\n:)[/quote]\r\nActually, the following inequality holds for any non-zero real $x$:\r\n\\[\\underset{cyc}{\\sum}\\frac{a}{b}\\ge \\underset{cyc}{\\sum}{(\\frac{a^{x}+c^{x}}{b^{x}+c^{x}}})^\\frac 1{x}\\]", "Solution_6": "[quote=\"Vasc\"]\nActually, the following inequality holds for any non-zero real $x$:\n\\[\\underset{cyc}{\\sum}\\frac{a}{b}\\ge \\underset{cyc}{\\sum}{(\\frac{a^{x}+c^{x}}{b^{x}+c^{x}}})^\\frac 1{x}\\]\n[/quote]\r\nHow two prove it?", "Solution_7": "I am guessing that the only way is derivatives\r\n\r\nVasc, does\\[\\sum_{cyc}\\frac{a_{1}}{a_{2}}\\geq \\sum_{cyc}\\left(\\frac{a_{1}^{x}+a_{3}^{x}}{a_{2}^{x}+a_{3}^{x}}\\right)^{\\frac{1}{x}}\\]hold?", "Solution_8": "[quote=\"me@home\"]I am guessing that the only way is derivatives\n\nVasc, does\n\\[\\sum_{cyc}\\frac{a_{1}}{a_{2}}\\geq \\sum_{cyc}\\left(\\frac{a_{1}^{x}+a_{3}^{x}}{a_{2}^{x}+a_{3}^{x}}\\right)^{\\frac{1}{x}}\\]\nhold?[/quote]\r\nWithout derivatives.\r\nI think it is not true for $n>3$.", "Solution_9": "To prove $\\underset{cyc}{\\sum}\\frac{a}{b}\\ge \\underset{cyc}{\\sum}{(\\frac{a^{x}+c^{x}}{b^{x}+c^{x}}})^\\frac 1{x}$, we can use the known statement:\r\n\r\nLet $a,b,c$ and $x,y,z$ be positive numbers such that\r\n$abc=xyz$, $max\\{a,b,c\\}\\geq max \\{x,y,z\\}$ and $min\\{a,b,c\\}\\leq min \\{x,y,z\\}$.\r\nThen $a+b+c \\geq x+y+z$. :wink:", "Solution_10": "Dear toanhocmuonmau I am very curious to see your elegant solution as I cannot see it. \r\n\r\nThank you very much.", "Solution_11": "[quote=\"cefer\"]$\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\geq \\sqrt{\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}}$\n$\\frac{a^{2}}{b^{2}}+\\frac{b^{2}}{c^{2}}+\\frac{c^{2}}{a^{2}}+2(\\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b}) \\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\\frac{b^{2}+a^{2}}{c^{2}+a^{2}}+\\frac{c^{2}+b^{2}}{a^{2}+b^{2}}+2(\\sqrt{\\frac{b^{2}+a^{2}}{b^{2}+c^{2}}}+\\sqrt{\\frac{a^{2}+c^{2}}{a^{2}+b^{2}}}+\\sqrt{\\frac{c^{2}+b^{2}}{c^{2}+a^{2}}})$\nLet's prove that $\\frac{a^{2}}{b^{2}}+2\\frac{b}{a}\\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$S=\\frac{a^{4}+2b^{3}a}{a^{2}b^{2}}-\\frac{a^{2}+c^{2}}{b^{2}+c^{2}}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$S= \\frac{a^{4}c^{2}+2b^{5}a+2b^{3}ac^{2}-a^{2}b^{2}c^{2}}{a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\nBy AM-GM inequality $a^{4}c^{2}+2b^{3}ac^{2}=a^{4}c^{2}+b^{3}ac^{2}+b^{3}ac^{2}\\geq 3a^{2}b^{2}c^{2}$ \n$S \\geq \\frac{2b^{5}a+2a^{2}b^{2}c^{2}}{ a^{2}b^{2}(b^{2}+c^{2})}\\geq 2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$\\frac{b^{3}+ac^{2}}{ a(b^{2}+c^{2})}\\geq \\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n$(b^{3}+ac^{2})^{2}(a^{2}+c^{2}) \\geq (b^{2}+c^{2})^{3}a^{2}$\n$a^{4}c^{4}+2ab^{3}c^{4}+2a^{3}c^{2}b^{3}\\geq 3a^{2}b^{4}c^{2}+3a^{2}b^{2}c^{4}$\n$\\Longrightarrow (a-b)^{2}(c^{4}(a+b)^{2}+c^{2}ab(a^{2}+b^{2})) \\geq 0$\nSo inequality proved.[/quote]\r\nExcellent solution. How can you think about the squaring?", "Solution_12": "[quote=\"cefer\"]Let's prove that $\\frac{a^{2}}{b^{2}}+2\\frac{b}{a}\\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n[/quote]\r\nwonderful step :D , could you plz tell me how did you find out this ? :D", "Solution_13": "[quote=\"HTA\"][quote=\"cefer\"]Let's prove that $\\frac{a^{2}}{b^{2}}+2\\frac{b}{a}\\geq \\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+2\\sqrt{\\frac{b^{2}+c^{2}}{c^{2}+a^{2}}}$\n[/quote]\nwonderful step :D , could you plz tell me how did you find out this ? :D[/quote]\r\n\r\nAs you see I solved this problem in December, so I dont remember why I did this step,but may be I did it for a symmetry.", "Solution_14": "[quote=\"toanhocmuonmau\"]Let $x,y,z$ be positive real numbers. Prove that $\\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x}\\geq \\sqrt{\\frac{x^{2}+z^{2}}{y^{2}+z^{2}}}+\\sqrt{\\frac{y^{2}+x^{2}}{z^{2}+x^{2}}}+\\sqrt{\\frac{z^{2}+y^{2}}{x^{2}+y^{2}}}$.[/quote]\r\nUsing the Cauchy inequality,\r\n\r\n$\\sqrt{\\sum_{cyc}\\frac{x^{2}+z^{2}}{z}}\\sqrt{\\sum_{cyc}\\frac{z}{y^{2}+z^{2}}}\\geq\\sum_{cyc}{\\sqrt{\\frac{x^{2}+z^{2}}{y^{2}+z^{2}}}}$,\r\n\r\nwe have to prove\r\n\r\n$\\left(\\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x}\\right)^{2}\\geq\\left(\\frac{x^{2}+z^{2}}{z}+\\frac{y^{2}+x^{2}}{x}+\\frac{z^{2}+y^{2}}{y}\\right)\\left(\\frac{z}{y^{2}+z^{2}}+\\frac{x}{z^{2}+x^{2}}+\\frac{y}{x^{2}+y^{2}}\\right)$.\r\n[hide=\"Proof.\"]$\\left(\\sum_{cyc}\\frac{x}{y}\\right)^{2}-\\left(\\sum_{cyc}\\frac{x^{2}+z^{2}}{z}\\right)\\left(\\sum_{cyc}\\frac{z}{y^{2}+z^{2}}\\right)\\equiv\\frac{F(x,y,z)}{x^{2}y^{2}z^{2}(y^{2}+z^{2})(z^{2}+x^{2})(x^{2}+y^{2})}$.\n\n$F(x,y,z)=F(x,x+s,y+t)$\n\n$=32(s^{2}-st+t^{2})x^{10}+4(22s^{3}+s^{2}t+13st^{2}+22t^{3})x^{9}$\n\n$+4(28s^{4}+25s^{3}t+48s^{2}t^{2}+61st^{3}+28t^{4})x^{8}$\n\n$+2(41s^{5}+68s^{4}t+172s^{3}t^{2}+246s^{2}t^{3}+175st^{4}+41t^{5})x^{7}$\n\n$+(36s^{6}+86s^{5}t+315s^{4}t^{2}+582s^{3}t^{3}+599s^{2}t^{4}+272st^{5}+36t^{6})x^{6}$\n\n$+(9s^{7}+28s^{6}t+162s^{5}t^{2}+399s^{4}t^{3}+572s^{3}t^{4}+429s^{2}t^{5}+125st^{6}+9t^{7})x^{5}$\n\n$+(s^{8}+4s^{7}t+46s^{6}t^{2}+163s^{5}t^{3}+307s^{4}t^{4}+356s^{3}t^{5}+187s^{2}t^{6}+33st^{7}+t^{8})x^{4}$\n\n$+2st^{2}(3s^{6}+19s^{5}t+48s^{4}t^{2}+77s^{3}t^{3}+69s^{2}t^{4}+24st^{5}+2t^{6})x^{3}$\n\n$+s^{2}t^{3}(4s^{5}+17s^{4}t+36s^{3}t^{2}+47s^{2}t^{3}+32st^{4}+6t^{5})x^{2}$\n\n$+s^{3}t^{4}(s^{4}+5s^{3}t+7s^{2}t^{2}+8st^{3}+4t^{4})x+s^{4}t^{6}(s^{2}+t^{2})\\geq0$,\n\nwhich is clearly true for $x=\\min\\{x,y,z\\}$.[/hide]", "Solution_15": "Dear can_hang2007, can you please post your impressive solution by contradiction method?\r\n\r\nThank you very much.\r\n\r\nI am very curious to see it.", "Solution_16": "[quote=\"Vasc\"]To prove $ \\underset{cyc}{\\sum}\\frac {a}{b}\\ge \\underset{cyc}{\\sum}{(\\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\\frac 1{x}$, we can use the known statement:\n\nLet $ a,b,c$ and $ x,y,z$ be positive numbers such that\n$ abc = xyz$, $ max\\{a,b,c\\}\\geq max \\{x,y,z\\}$ and $ min\\{a,b,c\\}\\leq min \\{x,y,z\\}$.\nThen $ a + b + c \\geq x + y + z$. :wink:[/quote]\r\n\r\n\r\nWLOG $ a\\ge b\\ge c$ and $ x\\ge y\\ge z$ then $ a\\ge x$ , $ c\\le z$ and $ ab\\ge xy$ \r\n\r\nFrom Abel formula and AM-GM inequality we have\r\n\r\n${ a + b + c - x - y - z = (x - y)(\\frac {a}{x} - 1) + (y - z)(\\frac {a}{x} + \\frac {b}{y} - 2) + z(\\frac {a}{x} + \\frac {b}{y} + \\frac {c}{z} - 3) \\ge(x - y)(\\frac {a}{x} - 1) + (y - z)(2\\sqrt {\\frac {ab}{xy}} - 2) + z(3\\sqrt [3]{\\frac {abc}{xyz} }- 3})\\ge0$\r\nHence \r\n\r\n$ a + b + c\\ge x + y + z$\r\n\r\nNow \r\n\r\n${{{ \\frac {a}{b}.\\frac {b}{c}.\\frac {c}{a} = (\\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\\frac 1{x}. (\\frac {b^{x} + a^{x}}{c^{x} + a^{x}}})^\\frac 1{x}. (\\frac {c^{x} + b^{x}}{a^{x} + b^{x}}})^\\frac 1{x} = 1$\r\n\r\nCan help me to prove \r\n\r\n${{{{ \\max\\{{\\frac {a}{b},\\frac {b}{c},\\frac {c}{a}}\\}\\ge\\max\\{(\\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\\frac 1{x},(\\frac {b^{x} + a^{x}}{c^{x} + a^{x}}})^\\frac 1{x},(\\frac {c^{x} + b^{x}}{a^{x} + b^{x}}})^\\frac 1{x}}\\}$\r\n\r\nand\r\n\r\n${{{{{ \\min\\{\\frac {a}{b},\\frac {b}{c},\\frac {c}{a}}\\}\\leq \\min\\{(\\frac {a^{x} + c^{x}}{b^{x} + c^{x}}})^\\frac 1{x},(\\frac {b^{x} + a^{x}}{c^{x} + a^{x}}})^\\frac 1{x},(\\frac {c^{x} + b^{x}}{a^{x} + b^{x}}})^\\frac 1{x}}\\}$", "Solution_17": "[quote=\"manlio\"]Dear can_hang2007, can you please post your impressive solution by contradiction method?\n\nThank you very much.\n\nI am very curious to see it.[/quote]\r\nHere is it :)", "Solution_18": "Thank you very much, can_hang2007.\r\n\r\nVery nice inequality!!!", "Solution_19": "[quote=cefer]\n$a^{4}c^{4}+2ab^{3}c^{4}+2a^{3}c^{2}b^{3}\\geq 3a^{2}b^{4}c^{2}+3a^{2}b^{2}c^{4}$\n[/quote]\nYou lost one expression:\n$a^{4}c^{4}+2ab^{3}c^{4}+2a^{3}c^{2}b^{3}+b^6c^2\\geq 3a^{2}b^{4}c^{2}+3a^{2}b^{2}c^{4}$\nInstead of factorizing one may also use AM-GM:\n$a^{4}c^{4}+2ab^{3}c^{4}\\ge 3\\sqrt[3]{a^{4}c^{4}\\cdot a^2b^{6}c^{8}}=3a^2b^2c^4$\n$2a^{3}c^{2}b^{3}+b^6c^2\\geq 3a^2b^4c^2$" } { "Tag": [ "algebra", "polynomial", "MATHCOUNTS", "calculus", "integration", "inequalities", "induction" ], "Problem": "N is a natural number such that $2^x>x^8$ for all x>N. What is the minimum value for N?\r\n\r\n(Source: Mathcounts 2000-2001 Workout 2)", "Solution_1": "I'm assuming that N is an integer?", "Solution_2": "Yeah, N is a natural number", "Solution_3": "[hide]\n44\nI couldn't think of any better way than to just do:\n$\\root 8 \\of {2^{x}} > x$\nAnd then just guess and check:\n$\\root 8 \\of {2^{44}} = 45.25$\n[/hide]", "Solution_4": "Another way to do it:\r\n\r\nTaking log (base 2) of both sides, we have\r\n2^x>x^8 iff\r\nx>8*log_2(x)\r\n\r\nTesting powers of 2s, we find that x is between 32 and 64. After some guessing and checking, we find that \r\n43<8*log_2(43) and \r\n44>8*log_2(44), \r\nso N=43.\r\n\r\nBut what's the non-calculus way to prove that \r\n2^x>x^8 \r\nfor all x>43? I couldn't find a rigorous way.", "Solution_5": "Here's a way.\r\nFirst, assume $2^x>x^8$ for some x (43 for example). Also, $2<(1+\\frac{1}{x})^x<3$. So\r\n$2^x*(1+\\frac{1}{x})^x>x^8*(1+\\frac{1}{x})^x$\r\nAnd $(1+\\frac{1}{x})^x>(1+\\frac{1}{x})^8$ for x>8, which it is in this case.\r\nTherefore $2^x*(1+\\frac{1}{x})^x>x^8*(1+\\frac{1}{x})^8=(x+1)^8$\r\nNow since $2^x>3>(1+\\frac{1}{x})^x$ for x>1, the left side can become\r\n\r\nEdit: I made a mistake. Can anyone fix this?", "Solution_6": "Nice, never thought of that.", "Solution_7": "Oops should be 43", "Solution_8": "Well, you know that N has to be bigger x when $2^x = x^8$, so you could just solve for $2^x = x^8$. But I really don't know where to start.", "Solution_9": "[quote=\"EFuzzy\"]N is a natural number such that $2^x>x^8$ for all x>N. What is the minimum value for N?[/quote]The answer to that actually depends on whether $x$ is defined as a real variable or an integral variable. I'll assume $x\\in\\mathbf{N}$.[quote=\"SnowStorm\"][hide]\n44\nI couldn't think of any better way than to just do:\n$\\root 8 \\of {2^{x}} > x$\nAnd then just guess and check:\n$\\root 8 \\of {2^{44}} = 45.25$\n[/hide][/quote]If you're going to use a calculator you don't need to do any manipulation of the inequality beforehand. Also, I doubt that $\\root 8 \\of {2^{44}}$ is rational, since $\\root 8 \\of {2^{44}}=\\root 8 \\of {(2^8)^5\\cdot2^4}=\\root 8 \\of {(2^5)^8\\cdot2^4}=2^5\\cdot\\root 8 \\of {16}$. The eighth root of sixteen is not rational. Say $\\frac{m}{n}$ is the eighth root of sixteen and $m,n$ are two relatively prime integers. Intuitively, if $n\\nmid{m}$ then $n^8\\nmid{m^8}$, so $\\frac{m^8}{n^8}\\not\\in\\mathbf{Z}\\Rightarrow\\frac{m^8}{n^8}\\ne16$.[quote=\"Wumbate\"]Here's a way.\nFirst, assume $2^x>x^8$ for some x (43 for example). Also, $2<(1+\\frac{1}{x})^x<3$. So\n$2^x*(1+\\frac{1}{x})^x>x^8*(1+\\frac{1}{x})^x$\nAnd $(1+\\frac{1}{x})^x>(1+\\frac{1}{x})^8$ for x>8, which it is in this case.\nTherefore $2^x*(1+\\frac{1}{x})^x>x^8*(1+\\frac{1}{x})^8=(x+1)^8$\nNow since $2^x>3>(1+\\frac{1}{x})^x$ for x>1, the left side can become\n\nEdit: I made a mistake. Can anyone fix this?[/quote]I've seen induction proofs that use $2\\le\\left(1+\\frac{1}{n}\\right)^n<3$ but I can only see how to use this to show that $m^n>n^k$ for all $n\\ge{N}$ for some given $N$ when $m\\ge3$, but unfortunately $m=2$..[quote=\"h_s_potter2002\"]you could just solve for $2^x = x^8$[/quote]I'll assume you meant to solve for $x$. Still, the problem will require some sort of reasoning rather than just solving an equation. I'm not sure that equation can be solved algebraically anyway." } { "Tag": [ "trigonometry" ], "Problem": "There is a right triangle, ABC, where C is the right angle. What is the value of (Cos A squared) plus (Sin A squared)?", "Solution_1": "It's a well-known trig identity. [hide=\"Answer\"] $ \\sin^{2} \\plus{} \\cos^{2}x \\equal{} 1$.[/hide]", "Solution_2": "Or you can prove it by relating to pythagorian theorem..\r\n\r\n@aznotaku13=:lol: Nice problem, my friend. Just out of curiousity................\r\nIsn't this the problem you show this to me in Chinese class?", "Solution_3": "yea, it was, but i was bored and wanted to see who could figure it out besides you since you already know." } { "Tag": [ "research" ], "Problem": "So are these questions that we thought of that can become research questions, or can this also be ideas for new approaches to already stated problems?", "Solution_1": "And ... why would someone post an idea here instead of doing it with themselves or their friends?", "Solution_2": "[quote=\"solafidefarms\"]And ... why would someone post an idea here instead of doing it with themselves or their friends?[/quote]\r\nTo share it with others here. There was a request in the round table for this forum; many wanted it, including me. Why would you keep an idea to yourself?", "Solution_3": "Check http://www.mathlinks.ro/viewtopic.php?t=194500 for the meaning of this forum, before Valentin or anyone else has given an official description. And feel free to contribute ideas.\r\n\r\n darij" } { "Tag": [ "algebra", "polynomial", "calculus", "calculus computations" ], "Problem": "Find the sequence Yn that satisfies the following difference equation\r\nY(n+1) - 2Y(n-1) = n , where n=1,2,3,...", "Solution_1": "$ Y_{n\\plus{}1} \\minus{} 2Y_{n\\minus{}1} \\equal{} n ,\\; n\\in \\mathbb{N}$.\r\n\r\nOK, this shouldn't be a problem. First, solve the homogeneous analogue of the equation. Its solution is $ 2^{n \\minus{} 1}C$, right? (make the characteristic polynomial equation). Then, guess a particular solution. One possibility is to look for it in a form of $ c_1 n \\plus{} c_2$, as the RHS of the equation is $ n$. After plugging it in your equation, you get $ c_1 \\equal{} \\minus{} 1, c_2 \\equal{} \\minus{} 1$, so your solution is:\r\n\r\n$ Y_n \\equal{} 2^{n \\minus{} 1}C \\minus{} n \\minus{} 1$. Hope this is ok, as I haven't done recurrence relations for a long time.", "Solution_2": "hsiljak, you missed that the equation is 2nd order, not first order. The characteristic polynomial is $ x^2\\minus{}2,$ so the solutions of the homogeneous equation are of the form $ c_12^{n/2}\\plus{}c_2(\\minus{}1)^n2^{n/2}.$\r\n\r\nAlso, I get a particular solution of $ \\minus{}n\\minus{}3$ rather than $ \\minus{}n\\minus{}1.$ So the general solution is\r\n\r\n$ Y_n\\equal{}c_12^{n/2}\\plus{}c_2(\\minus{}1)^n2^{n/2}\\minus{}n\\minus{}3$\r\n\r\nwhere $ c_1$ and $ c_2$ are determined by initial conditions (which the original poster didn't give).", "Solution_3": "$ Y_{n \\plus{} 1} \\equal{} 2Y_{n \\minus{} 1} \\plus{} n\\Longleftrightarrow Y_{n \\plus{} 1} \\plus{} n \\plus{} 1 \\plus{} 3 \\equal{} 2(Y_{n \\minus{} 1} \\plus{} n \\minus{} 1 \\plus{} 3)$. Let $ Z_n \\equal{} Y_n \\plus{} n \\plus{} 3$, we have $ Z_{n \\plus{} 1} \\equal{} 2Z_{n \\minus{} 1}$, yielding $ Z_n \\equal{} 2Z_{n \\minus{} 2}\\ (n\\geq 2)$. \r\n\r\nThus For $ n \\equal{} 2k\\ (k\\equal{}0,\\ 1,\\ 2,\\ \\cdots)$, we have $ Z_{2k} \\equal{} 2Z_{2(k \\minus{} 1)}$, yielding $ Z_{2k} \\equal{} Z_0\\cdot 2^k\\ \\therefore Z_n \\equal{} Z_0\\cdot2^{\\frac {n}{2}}$,\r\n\r\nFor $ n \\equal{} 2k \\minus{} 1\\ (k\\equal{}1,\\ 2,\\ 3,\\ \\cdots)$, we have $ Z_{2k \\minus{} 1} \\equal{} 2Z_{2(k \\minus{} 1) \\minus{} 1}$, yielding $ Z_{2k \\minus{} 1} \\equal{} Z_1\\cdot 2^{k \\minus{} 1}\\ \\therefore Z_n \\equal{} Z_1\\cdot 2^{\\frac {n \\minus{} 1}{2}}$.\r\n\r\nAnswer: $ Y_n \\equal{} C\\cdot 2^{\\left[\\frac {n}{2}\\right]} \\minus{} n \\minus{} 3\\ (n\\geq 1)$.", "Solution_4": "Again, the difference equation is second order, not first order. kunny hasn't captured all solutions.\r\n\r\nUsing the same basic idea as kunny, but varying the presentation a little:\r\n\r\nDefine $ Z_n \\equal{} Y_n \\plus{} n \\plus{} 3$ so that $ Y_n \\equal{} Z_n \\minus{} n \\minus{} 3.$\r\n\r\nUpon substituting that into the original equation and simplifying, we get\r\n\r\n$ Z_{n \\plus{} 1} \\equal{} 2Z_{n \\minus{} 1}.$\r\n\r\nBoth $ Z_n \\equal{} \\left(\\sqrt {2}\\right)^n$ and $ Z_n \\equal{} \\left( \\minus{} \\sqrt {2}\\right)^n$ are solutions to this, as are also all linear combinations of those two. kunny only wrote the first of those two possibilities.", "Solution_5": "I believe my answer is right, because I classified the parity of $ n$.", "Solution_6": "Yes, you did. It's only the very last line of your post, which looks like a single formula with a single constant, that looks misleading. There are two degrees of freedom here, and two pieces of information (such as $ Y_0$ and $ Y_1$) would be needed to pin down the solution.", "Solution_7": "I'm sorry. I've been solving the equation $ Y_{n\\plus{}1} \\minus{} 2Y_{n} \\equal{} n ,\\; n\\in \\mathbb{N}$, and not the OP's one... Typo...", "Solution_8": "Oh, that gets easier than first problem.\r\n\r\nWe have at least 3 solutions. :lol:", "Solution_9": "For the second difference equation, $ y_{n\\plus{}1} \\minus{} 2 y_n \\equal{} n$,\r\n\r\n[hide=\"Solution\"]Applying the z-transform to both sides we get\n\n$ \\mathcal{Z}(y_{n\\plus{}1}) \\minus{} 2 \\mathcal{Z}(y_n) \\equal{} \\mathcal{Z}(n)$\n\n$ z[Y(z) \\minus{} y_0] \\minus{} 2 Y(z) \\equal{} \\frac{z}{(z\\minus{}1)^2}$\n\nand solving for Y(z) we get\n\n$ Y(z) \\equal{} y_0 \\plus{} \\frac{2y_0z^2 \\plus{} (1\\minus{}4y_0)z \\plus{} 2y_0}{(z\\minus{}2)(z\\minus{}1)^2} \\equal{} y_0 \\plus{} \\frac{2\\plus{}2y_0}{z\\minus{}2} \\minus{} \\frac{1}{(z\\minus{}1)^2} \\minus{} \\frac{2}{z\\minus{}1}$.\n\nNow let's find the inverse z-transform of each term:\n\n$ y_n \\equal{} \\mathcal{Z} ^{\\minus{}1}(y_0) \\plus{} (2\\plus{}2y_0) \\mathcal{Z} ^{\\minus{}1}\\left(\\frac{1}{z\\minus{}2}\\right) \\minus{} \\mathcal{Z} ^{\\minus{}1} \\left(\\frac{1}{(z\\minus{}1)^2}\\right) \\minus{} 2 \\mathcal{Z} ^{\\minus{}1}\\left(\\frac{1}{z\\minus{}1}\\right)$\n\n$ \\equal{}> y_n \\equal{} y_0 \\delta_n \\plus{} (2\\plus{}2y_0) 2^{n\\minus{}1} u_{n\\minus{}1} \\minus{} (n\\minus{}1) u_{n\\minus{}1} \\minus{} 2 u_{n\\minus{}1}$.\n\nFor n=0, we get $ y_n \\equal{} y_0$, and for $ n \\geq 1$ we have\n\n$ y_n \\equal{} (1\\plus{}y_0)2^n \\minus{} n \\plus{} 1 \\minus{} 2 \\equal{}> y_n \\equal{} (1\\plus{}y_0)2^n \\minus{} n \\minus{}1$.\n[/hide]", "Solution_10": "For the first difference equation, $ y_{n\\plus{}1} \\minus{} 2 y_{n\\minus{}1} \\equal{} n$:\r\n\r\n[hide=\"Solution\"]Start by shifting the index to get $ y_{n\\plus{}2} \\minus{} 2 y_{n} \\equal{} n \\plus{} 1$. Now let0s apply the z-transform to both sides:\n\n$ \\mathcal{Z} (y_{n\\plus{}2}) \\minus{} 2 \\mathcal{Z}(y_n) \\equal{} \\mathcal{Z}(n) \\plus{} \\mathcal{Z}(1)$\n\n$ \\equal{}> z^2[Y(z) \\minus{} a \\minus{} bz^{\\minus{}1}] \\minus{} 2Y(z) \\equal{} \\frac{z}{(z\\minus{}1)^2} \\plus{} \\frac{z}{z\\minus{}1}$\n\n(where $ a \\equal{} y_0$ and $ b \\equal{} y_1$)\n\n$ \\equal{}> z^2Y(z) \\minus{} az^2 \\minus{} bz \\minus{} 2Y(z) \\equal{} \\frac{z^2}{(z\\minus{}1)^2}$,\n\nand solving for Y(z) we get\n\n$ Y(z) \\equal{} a \\plus{} \\frac{A}{z\\minus{}\\sqrt{2}} \\plus{} \\frac{B}{z \\plus{} \\sqrt{2}} \\plus{} \\frac{C}{(z\\minus{}1)^2} \\plus{} \\frac{D}{z\\minus{}1}$,\n\nwhere A, B, C, and D are constants that are expressed in terms of a and b. Now applying the inverse z-transform to this equality yields\n\n$ y_n \\equal{} \\mathcal{Z} ^{\\minus{}1}(a) \\plus{} A \\mathcal{Z} ^{\\minus{}1} \\left(\\frac{A}{z\\minus{}\\sqrt{2}}\\right) \\plus{} B \\mathcal{Z} ^{\\minus{}1} \\left(\\frac{B}{z \\plus{} \\sqrt{2}}\\right) \\plus{} C \\mathcal{Z} ^{\\minus{}1} \\left(\\frac{C}{(z\\minus{}1)^2}\\right) \\plus{} D \\mathcal{Z} ^{\\minus{}1} \\left(\\frac{D}{z\\minus{}1}\\right)$\n\n$ \\equal{}> y_n \\equal{} a \\delta_n \\plus{} A \\cdot (\\sqrt{2})^{n\\minus{}1} u_{n\\minus{}1} \\plus{} B \\cdot (\\minus{}\\sqrt{2})^{n\\minus{}1} u_{n\\minus{}1} \\plus{} C(n\\minus{}1) u_{n\\minus{}1} \\plus{} D u_{n\\minus{}1}$.\n\nFor n = 0 we get $ y_0 \\equal{} a$; for n = 1, we have $ y_1 \\equal{} A \\plus{} B \\plus{} D$. For n > 1, the expression for $ y_n$ reduces to\n\n$ y_n \\equal{} \\frac{A}{\\sqrt{2}} \\cdot 2^{n/2} \\plus{} \\frac{B}{\\minus{} \\sqrt{2}} \\cdot (\\minus{}1)^n 2^{n/2} \\plus{} Cn \\minus{} C \\plus{} D$.\n\nI have derived the (complicated) expressions for A and B, but for C and D there was no more patience. Of course, this would be much more easy if the original poster has given the initial conditions a and b.\n[/hide]", "Solution_11": "[quote=\"Carcul\"]For the second difference equation, $ y_{n \\plus{} 1} \\minus{} 2 y_n \\equal{} n$,\n\n[hide=\"Solution\"]Applying the z-transform to both sides we get\n\n$ \\mathcal{Z}(y_{n \\plus{} 1}) \\minus{} 2 \\mathcal{Z}(y_n) \\equal{} \\mathcal{Z}(n)$\n\n$ z[Y(z) \\minus{} y_0] \\minus{} 2 Y(z) \\equal{} \\frac {z}{(z \\minus{} 1)^2}$\n\nand solving for Y(z) we get\n\n$ Y(z) \\equal{} y_0 \\plus{} \\frac {2y_0z^2 \\plus{} (1 \\minus{} 4y_0)z \\plus{} 2y_0}{(z \\minus{} 2)(z \\minus{} 1)^2} \\equal{} y_0 \\plus{} \\frac {2 \\plus{} 2y_0}{z \\minus{} 2} \\minus{} \\frac {1}{(z \\minus{} 1)^2} \\minus{} \\frac {2}{z \\minus{} 1}$.\n\nNow let's find the inverse z-transform of each term:\n\n$ y_n \\equal{} \\mathcal{Z} ^{ \\minus{} 1}(y_0) \\plus{} (2 \\plus{} 2y_0) \\mathcal{Z} ^{ \\minus{} 1}\\left(\\frac {1}{z \\minus{} 2}\\right) \\minus{} \\mathcal{Z} ^{ \\minus{} 1} \\left(\\frac {1}{(z \\minus{} 1)^2}\\right) \\minus{} 2 \\mathcal{Z} ^{ \\minus{} 1}\\left(\\frac {1}{z \\minus{} 1}\\right)$\n\n$ \\equal{} > y_n \\equal{} y_0 \\delta_n \\plus{} (2 \\plus{} 2y_0) 2^{n \\minus{} 1} u_{n \\minus{} 1} \\minus{} (n \\minus{} 1) u_{n \\minus{} 1} \\minus{} 2 u_{n \\minus{} 1}$.\n\nFor n=0, we get $ y_n \\equal{} y_0$, and for $ n \\geq 1$ we have\n\n$ y_n \\equal{} (1 \\plus{} y_0)2^n \\minus{} n \\plus{} 1 \\minus{} 2 \\equal{} > y_n \\equal{} (1 \\plus{} y_0)2^n \\minus{} n \\minus{} 1$.\n[/hide][/quote]\r\n\r\nHum, your solution is overdone for the problem.\r\n\r\nHere is my solution.\r\n\r\nSolution 1:\r\n\r\n$ y_{n \\plus{} 1} \\equal{} 2(y_n \\plus{} n)\\Longleftrightarrow y_{n \\plus{} 1} \\plus{} (n \\plus{} 1) \\plus{} 1 \\equal{} 2(y_n \\plus{} n \\plus{} 1)$, yielding $ y_n \\plus{} n \\plus{} 1 \\equal{} 2^n(y_0 \\plus{} 0 \\plus{} 1)$.\r\n\r\n$ \\therefore y_n \\equal{} (y_0 \\plus{} 1)2^n \\minus{} n \\minus{} 1\\ (n\\geq 0)$.\r\n\r\nSolution 2:\r\n\r\n$ y_{n \\plus{} 2} \\equal{} 2y_{n \\plus{} 1} \\plus{} (n \\plus{} 1)$\r\n$ y_{n \\plus{} 1} \\equal{} 2y_n \\plus{} n$\r\n\r\nSubtracting both sides gives $ y_{n \\plus{} 2} \\minus{} y_{n \\plus{} 1} \\equal{} 2(y_{n \\plus{} 1} \\minus{} y_n) \\plus{} 1\\Longleftrightarrow y_{n \\plus{} 2} \\minus{} y_{n \\plus{} 1} \\plus{} 1 \\equal{} 2(y_{n \\plus{} 1} \\minus{} y_n \\plus{} 1)$, yielding\r\n\r\n$ y_{n \\plus{} 1} \\minus{} y_n \\plus{} 1 \\equal{} 2^n(y_1 \\minus{} y_0 \\plus{} 1)$, since $ y_1 \\equal{} 2y_0 \\plus{} 0$, we have $ (2y_n \\plus{} n) \\minus{} y_n \\plus{} 1 \\equal{} 2^n(y_0 \\plus{} 1)$.\r\n\r\n$ \\therefore y_n \\equal{} (y_0 \\plus{} 1)2^n \\minus{} n \\minus{} 1\\ (n\\geq 0)$." } { "Tag": [ "Support", "MATHCOUNTS" ], "Problem": "The requesting, posting, and sharing of copyrighted files of books on this forum must stop. From now own, we will be banning anyone we see doing this. Sharing publicly available files is ok, but asking for books that are clearly under copyright is not. Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge. Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general). The authors are not getting rich off this material - believe me, they could make much more money elsewhere. But they and their families do need to eat.", "Solution_1": "[quote=\"rrusczyk\"]The requesting, posting, and sharing of copyrighted files of books on this forum must stop. From now own, we will be banning anyone we see doing this. Sharing publicly available files is ok, but asking for books that are clearly under copyright is not. Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge. Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general). The authors are not getting rich off this material - believe me, they could make much more money elsewhere. But they and their families do need to eat.[/quote]\r\n\r\nOk! We are agree :wink:", "Solution_2": "Again I repeat, do not share files on copyright on the forum without the author's permission. If you wanna do that in private that's your business (although still wrong), but don't do it over the forums please!", "Solution_3": ":blush: I don't think I am wrong, sorry!", "Solution_4": "Sorry i edited my post this will not happen again", "Solution_5": "isn't mathcounts copyrighted too?", "Solution_6": "[quote=\"rrusczyk\"]The requesting, posting, and sharing of copyrighted files of books on this forum must stop. From now own, we will be banning anyone we see doing this. Sharing publicly available files is ok, but asking for books that are clearly under copyright is not. Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge. Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general). The authors are not getting rich off this material - believe me, they could make much more money elsewhere. But they and their families do need to eat.[/quote]\nYou're right.", "Solution_7": "[quote=\"rrusczyk\"]The requesting, posting, and sharing of copyrighted files of books on this forum must stop. From now own, we will be banning anyone we see doing this. Sharing publicly available files is ok, but asking for books that are clearly under copyright is not. Authors work very hard to share their knowledge, at a very small price compared to the value of that knowledge. Your theft of this work makes it harder for people to be able to afford producing any more new materials (or to support free materials, like this site in general). The authors are not getting rich off this material - believe me, they could make much more money elsewhere. But they and their families do need to eat.[/quote]\n\nGood decision." } { "Tag": [ "trigonometry", "factorial", "search", "algebra", "polynomial", "function", "geometric sequence" ], "Problem": "I'm writing next year's Math Team Selection Test for our school and I need to know the difficulty of the two variations of the following problem. I also want to know if I should have it as a multiple choice or an open answer. Any help would be appreciated. Also, please post how long it took you to do it. Thanks!\r\n\r\n- tZf\r\n\r\nNO CALCULATORS:\r\nVariation 1: Compute $\\lfloor100 sin(\\frac{\\pi}{31})\\rfloor$\r\n\r\nVariation 2: Compute $\\lfloor1000 sin(\\frac{\\pi}{31})\\rfloor$", "Solution_1": "I can't think of a smart way to do this and Taylor approximation without a calculator would be horrid. :( I think you should focus on problems with elegant and clever solutions.", "Solution_2": "well, you could do multiple choice and part of the problem would be looking at the answers and eliminating the wrong answer, i don't see how you could reasonably be able to figure out that trig value without a calculator", "Solution_3": "well if you had to and had way too much time on your hands, you could find an approximation using the half angle identities.", "Solution_4": "zuton force, how were you planning to solve that problem on your own?", "Solution_5": "It's actually somewhat trivial\r\nFor small angles, sinx is approximately equal to x\r\n\r\n$\\pi/31$ is about 0.1013\r\n$sin(\\pi/31)$ is about 0.1012\r\n\r\nSo the answer would be 10 for variation 1 and 101 for variation 2\r\n\r\nMeh. I guess I'll trash it", "Solution_6": "oh..right :blush: \r\n\r\ni knew that.", "Solution_7": "that is not really a rigirous answer, you cannot know that it will not deviate enough for the hundredths digit to remain constant", "Solution_8": "change the problem slightly to be the sum of a gemoetric series or something, at least try to find a trick..", "Solution_9": "[quote=\"t0rajir0u\"]I can't think of a smart way to do this and Taylor approximation without a calculator would be horrid. :( I think you should focus on problems with elegant and clever solutions.[/quote]\r\n\r\nWell actually if you think about it, you only need the first two terms. The third term is about $\\frac{0.1^5}{5!}$, which is obviously too insignificant to affect anything in the first three terms to the right of the decimal. (Even the second term is completely insignificant, but that's going back to saying sinx is nearly x.)\r\n\r\nOf course, I wouldn't expect people to use taylor series expansions anyway, so I guess I'll scrap it, especially since I've accumulated a large pool of non-trivial questions :| \r\n\r\nThanks for the feedback everyone! :)", "Solution_10": "You really need the Lagrange Remainder term of the taylor expansion to say anything about significant figures here. Just because you feel intuitively that all the other terms in the series won't matter...\r\n\r\nThis is basically a pointless problem that doesn't measure anything, especially since it's not even clear that you want an approximate value of the function.", "Solution_11": "Your school has an entrance exam for the math team? Anyone can go to ours (TJ's), and at my base school, I imagine they would be begging people to join due to lack of intrest.", "Solution_12": "[b]Lord[/b] - Unlike TJ, public schools are filled with people that don't know anything that join random clubs for the sake of adding extra-curricular activities to their resumees.\r\n\r\nI don't know about what it's like down there in VA, but in our school is filled with resumee-freaks that purposely burn away to achieve nothing but a 10000 page resumee.\r\n\r\nNow, (arguably thanks to Bush) we're facing severe shortages in money and the math team is basically now running on it's own money. We can't afford to pay for people that don't have their head in it, and we don't want to start charging obscene ammounts of money to people that join for the math.\r\n\r\n\r\n[b]Kalle[/b] - I know what you're driving at. However, intuition is a very important skill, especially since that's all that some of the people come in with. At most American public schools, math is taught extremely mechanically, so even for people that have covered trig and such material, many of them have not had much practice with creative mathematical thinking.\r\n\r\nAbout the \"Lagrange Remainder term\" - I don't really know what this is. But the terms are obviously decreasing faster than a geometric progression. Pretend that after the second term, the terms decrease geometrically. If we draw out an infinite geometric progression based on terms 2 and 3, it's still insignificant. So in the real expansion with the factorials, it would be even more insignificant.", "Solution_13": "I am starting a search for the problem with the most creative solution. Take it to mean what you want, buty PM me with the question/answer if you think you have a good problem.", "Solution_14": "[quote=\"The Zuton Force\"]\nAbout the \"Lagrange Remainder term\" - I don't really know what this is. But the terms are obviously decreasing faster than a geometric progression. Pretend that after the second term, the terms decrease geometrically. If we draw out an infinite geometric progression based on terms 2 and 3, it's still insignificant. So in the real expansion with the factorials, it would be even more insignificant.[/quote]\r\n\r\nThe Lagrange Remainder for the $k^{th}$ order Taylor polynomial $T_k(x)$ of an infinitely differentiable function $f(x)$ is given by $\\frac{ f^{(k+1)} (z) x^{k+1}}{(k+1)!}$, where $f^{(k+1)} (z)$ is the maximum value of $f^{(k+1)} (x)$ (and $f^{(k+1)} (x) = \\frac{d}{dx} f^{k} (x), f^{(0)} (x) = f(x)$.\r\n\r\nNow, for $\\sin x$ the term simply becomes $\\frac{x^{k+1}}{(k+1)!}$, which, for $x \\approx 0.1$ only definitely has error less than $\\frac{1}{1000}$ for $k \\ge 3$. Hence it is only necessary to take $\\sin x \\approx x - \\frac{x^3}{3!}$ to calculate an approximation.\r\n\r\n\r\n\r\nThe point is that Lagrange is a way to say, [i]rigorously[/i], that any new terms are insignificant. You can't just say it.", "Solution_15": "do you have this memorized??? :o", "Solution_16": "Well, I have it, uh... learned. I guess I have an advantage over the other people in my Calc class in that I taught myself Taylor series sometime in 8th grade due to ridiculous boredom, so I didn't have to deal with how poorly the book explains it. :)\r\n\r\n(Oh, I forgot to add that the Lagrange Remainder Term gives $\\epsilon$ such that $| f(x) - T_k (x) | \\le \\epsilon$.)", "Solution_17": "how did you teach it to yourself? sat down with dad's books at home? :huh:", "Solution_18": "Thanks for the explanation :)", "Solution_19": "[quote=\"me@home\"]how did you teach it to yourself? sat down with dad's books at home? :huh:[/quote]\r\n\r\nInternet. Actually, I didn't clearly remember what I read about Taylor series on the internet, so I basically ended up re-deriving the concept myself... minus all those qualms about convergence :)" } { "Tag": [ "geometry", "perimeter", "inradius", "vector", "linear algebra" ], "Problem": "Greetings,\r\n\r\nGot a question. Does anybody how to [b]prove that the perimeter of a right triangle is equal to twice the area[/b]? \r\n \r\nThis was asked as an extra credit in a brain teasers contest here and the organizer didn't have the solution, and nobody got it either...\r\n\r\nTodd", "Solution_1": "Hmm..try a right triangle with sides 1,1, $\\sqrt{2}.$ Perimeter is $2+\\sqrt{2}$ and area is $\\frac 12.$ Obviously this is a contradiction.", "Solution_2": "yes, yes, that's an example.\r\n\r\n But the question is how to PROVE it, much like \r\nproving the pythagorean (sp?) theorem. \r\n\r\nI think it starts with writing the perimeter as a+b+c and then\r\nsubstituting in a^2+b^2 for c, then a bit of magic occurs, \r\nand you end up with it equaling ab/2.", "Solution_3": "I see the contradiction, so maybe I'm not remembering the qualifiers on what kind of triangle completely. But isn't there a perimeter to area relationship for triangles?", "Solution_4": "Uh, no, that was a counterexample. The perimeter is $2+\\sqrt{2}$ whereas twice the area is 1. $2+\\sqrt{2}\\neq 1.$ Therefore [b]the perimeter of every right triangle is not necessarily equal to its twice its area.[/b]\r\n\r\n[edit: ok, i just saw your post (i was writing this while you posted)]", "Solution_5": "Perhaps it was $A=rs$ where $r$ is the radius of the inscribed circle and $s$ is half the perimeter?\r\n\r\nIf that's it I can write up a proof for you.", "Solution_6": "Maybe it has to have whole number sides... but then 20 21 29 wouldn't work because you have the perimeter=90 and the area=210... *thinks about it* *5 seconds later* *gets bored* *gives up*", "Solution_7": "Time to disprove it.\r\n\r\nI'm assuming it's true and working from there. x is the length of one leg, and y is the length of the other one.\r\n\r\nx + y + sqrt(x :^2: + y :^2: ) = xy/2\r\nsqrt(x :^2: + y :^2: ) = xy/2 - x - y\r\nsqrt(x :^2: + y :^2: ) = xy - 2x - 2y\r\nx :^2: + y :^2: = (xy - 2x - 2y) :^2:\r\nx :^2: + y :^2: = x :^2: y :^2: - 2x :^2: y - 2xy :^2: - 2x :^2: y + 4x :^2: + 4xy - 2xy :^2: + 4xy + 4y :^2:\r\nx :^2: + y :^2: = x :^2: y :^2: - 4x :^2: y - 4xy :^2: + 8xy + 4x :^2: + 4y :^2:\r\nx :^2: y :^2: - 4x :^2: y - 4xy :^2: + 8xy + 3x :^2: + 3y :^2: = 0\r\nx :^2: y :^2: + 8xy + 3x :^2: + 3y :^2: = 4x :^2: y + 4xy :^2:\r\n\r\n\r\n(unfinished)", "Solution_8": "I dont really get what this thread is talking about, but oh well...\r\n\r\nA=rs=rp/2=2p => r=4. So the perimeter is only twice the area when the inradius is 4. Can't find a specific example though...", "Solution_9": "Treething, to disprove something a simple counterexample is all that is needed. I already provided one so I don't see why you are doing that.", "Solution_10": "The problem could be to prove that there is only one right triangle with integer sides whose perimeter is equal to twice its area.", "Solution_11": "Well, everyone, this problem is posted in the [b]Game And Fun Factory[/b]\r\nSo i think we could try to find some interesting ways to prove that it is [b]true[/b] for [b]Every right triangles[/b]!", "Solution_12": "Use Heron's. Direct disproval. \r\n\r\n$2\\sqrt{(\\frac{a+b+c}{2})(\\frac{-a+b+c}{2})(\\frac{a-b+c}{2})(\\frac{a+b-c}{2})}$ is not equal to $ab$.", "Solution_13": "Solafide, those two expressions [i]are[/i] equal for any right triangle! Just because it isn't obvious that two things are equal doesn't mean they aren't. (I get this very much when grading linear algebra problems ... \"Well, these vectors are not necessarily the same, so their span is different.\" If you want to show something doesn't work, simply give a counter-example.", "Solution_14": "[quote=\"Treething\"]Time to disprove it.\n\nI'm assuming it's true and working from there. x is the length of one leg, and y is the length of the other one.\n\nx + y + sqrt(x :^2: + y :^2: ) = xy/2\nsqrt(x :^2: + y :^2: ) = xy/2 - x - y\nsqrt(x :^2: + y :^2: ) = xy - 2x - 2y\nx :^2: + y :^2: = (xy - 2x - 2y) :^2:\nx :^2: + y :^2: = x :^2: y :^2: - 2x :^2: y - 2xy :^2: - 2x :^2: y + 4x :^2: + 4xy - 2xy :^2: + 4xy + 4y :^2:\nx :^2: + y :^2: = x :^2: y :^2: - 4x :^2: y - 4xy :^2: + 8xy + 4x :^2: + 4y :^2:\nx :^2: y :^2: - 4x :^2: y - 4xy :^2: + 8xy + 3x :^2: + 3y :^2: = 0\nx :^2: y :^2: + 8xy + 3x :^2: + 3y :^2: = 4x :^2: y + 4xy :^2:\n\n\n(unfinished)[/quote]\r\nYou made a mistake between lines two and three of your proof.", "Solution_15": "There are infinitely many such triangles. If one leg is [b]a[/b], then the other leg is $2(\\frac{a-1}{a-2})$.\r\nand" } { "Tag": [ "vector", "analytic geometry", "trigonometry", "geometry unsolved", "geometry" ], "Problem": "1. Find two unit vectors in $ R^2$ that are perpendicular to $ ( \\minus{} 3,4)$\r\n2. If $ \\vec{a}$ and $ \\vec{b}$ are vectors in 3-space, where $ \\vec{a} \\equal{} t\\vec{b}$, show that $ (\\frac {\\vec{a}\\cdot\\vec{b}}{\\vec{b}\\cdot\\vec{b}})\\vec{b} \\equal{} \\vec{a}$\r\n3. Calculate the angles that $ (3,4,1)$ makes with the coordinate axes.\r\n4. In triangle ABC, E is the midpoint of AC and BE is extended to D so that ED=BE. Prove that $ \\vec{AB} \\equal{} \\vec{DC}$.\r\n5. In the quadrilateral ABCD, the midpoints of the sides AB, BC, CD, and DA are P, Q, R, and S, respectively. Prove that PR and QS bisect each other. Do the points A, B, C, D have to lie in a plane for the result to hold?\r\n6. A man weighing 70 kg lies in a hammock whose ropes make angles of 30 and 45 degrees with the horizontal. What is the tension in each rope?\r\n7. Two tugs are towing a ship. The smaller tug is 10 degrees off the port bow and the larger tug is 20 degrees off the starboard bow. The larger tug pulls twice as hard as the smaller tug. In what direction will the ship move?\r\n8. Find the work done by the force $ \\vec{F}$ exerted on an object moving along the vector $ \\vec{s}$. Assume that the force is measured in newtons and the distance is in meters. ( $ \\vec{F} \\equal{} (3,2)$, $ \\vec{s} \\equal{} (4,1)$ )\r\n9. Find a vector perpendicular to both $ \\vec{u}$ and $ \\vec{v}$. ( $ \\vec{u} \\equal{} ( \\minus{} 2,3,1)$, $ \\vec{v} \\equal{} ( \\minus{} 1,4,0)$ )", "Solution_1": "[quote=\"smellypoopy\"]1. Find two unit vectors in $ R^2$ that are perpendicular to $ ( - 3,4)$\n2. If $ \\vec{a}$ and $ \\vec{b}$ are vectors in 3-space, where $ \\vec{a} = t\\vec{b}$, show that $ (\\frac {\\vec{a}\\cdot\\vec{b}}{\\vec{b}\\cdot\\vec{b}})\\vec{b} = \\vec{a}$\n3. Calculate the angles that $ (3,4,1)$ makes with the coordinate axes.\n4. In triangle ABC, E is the midpoint of AC and BE is extended to D so that ED=BE. Prove that $ \\vec{AB} = \\vec{DC}$.\n5. In the quadrilateral ABCD, the midpoints of the sides AB, BC, CD, and DA are P, Q, R, and S, respectively. Prove that PR and QS bisect each other. Do the points A, B, C, D have to lie in a plane for the result to hold?\n6. A man weighing 70 kg lies in a hammock whose ropes make angles of 30 and 45 degrees with the horizontal. What is the tension in each rope?\n7. Two tugs are towing a ship. The smaller tug is 10 degrees off the port bow and the larger tug is 20 degrees off the starboard bow. The larger tug pulls twice as hard as the smaller tug. In what direction will the ship move?\n8. Find the work done by the force $ \\vec{F}$ exerted on an object moving along the vector $ \\vec{s}$. Assume that the force is measured in newtons and the distance is in meters. ( $ \\vec{F} = (3,2)$, $ \\vec{s} = (4,1)$ )\n9. Find a vector perpendicular to both $ \\vec{u}$ and $ \\vec{v}$. ( $ \\vec{u} = ( - 2,3,1)$, $ \\vec{v} = ( - 1,4,0)$ )[/quote]\r\n\r\n1. $ (\\frac{4}{5}, \\frac{3}{5}), (-\\frac{4}{5}, -\\frac{3}{5})$\r\n2. $ (\\frac {\\vec{a}\\cdot\\vec{b}}{\\vec{b}\\cdot\\vec{b}})\\vec{b} = \\frac{t \\cdot \\vec{a} \\cdot \\vec{a}}{t^2 \\cdot \\vec{a} \\cdot \\vec{a}} \\cdot t \\cdot \\vec{a} = \\vex{a}$. (That is when $ \\vec{a}^2 \\neq 0 \\iff \\vec{a} = (0,0)$)\r\n3. What do you mean?\r\n4. $ E$ is the midpoint of $ A$ and $ C$ so $ E = \\frac{A+C}{2}$. $ E$ is the midpoint of $ B$ and $ D$ so $ E = \\frac{B+D}{2} \\iff D = 2E - B = A+C-B$. Now $ AB = B-A$ and $ DC = C-D = C-A-C+B = B-A$ so $ AB = DC$.\r\n5. $ P = \\frac{A+B}{2}$ and so on. Note that the midpoint of $ PR$ is $ \\frac{P+R}{2} = \\frac{A+B+C+D}{4}$. The same with $ QS$ and we are done. No they do not.\r\n\r\nThe rest seems like physics, which I'll not waste my time with.", "Solution_2": "3.it is actually to find [url=http://mathworld.wolfram.com/DirectionCosine.html]direction cosines[/url] of vector $ \\vec{u}\\equal{}(3,4,1)$\r\nformula is:\r\nIf $ \\alpha$ is angle made by vector$ (a,b,c)$ with X axis,then:\r\n$ cos {\\alpha}\\equal{}\\frac{a}{\\sqrt{a^2\\plus{}b^2\\plus{}c^2}}$\r\n\r\n6 and 7:Please put in physics forum(Hint:just break each force in horizontal and vertical components and add)\r\n\r\n8.It also belongs to physics forum.\r\n$ W\\equal{}\\vec{F}\\cdot \\vec{s}$\r\nso,$ W\\equal{}3\\cdot 4\\plus{}2\\cdot 1 \\equal{}14N$\r\n\r\n9.A vector perpendicular to both $ \\vec{u}$ and $ \\vec{v}$ is given by $ \\vec{u} \\times \\vec{v}$\r\nPlease check [url=http://en.wikipedia.org/wiki/Cross_product]cross products[/url]" } { "Tag": [ "MATHCOUNTS" ], "Problem": "Suppose that S is a set of three positive integers. At most, how many of the following four properties could the set S have?\r\n\r\nI. S has mean 10\r\nII. S has median 8\r\nIII. S has sum 45\r\nIV. S contains the number 35", "Solution_1": "[quote=\"ch1n353ch3s54a1l\"]Suppose that S is a set of three positive integers. At most, how many of the following four properties could the set S have?\n\nI. S has mean 10\nII. S has median 8\nIII. S has sum 45\nIV. S contains the number 35[/quote]\r\n\r\n[hide]3 properties.[/hide]", "Solution_2": "[hide]3 because if the third 1 is true the first one can't be, but if the first one is true the third and fourth one can't and the second is possible no matter what.[/hide]", "Solution_3": "[hide]3 :P \n\nNumber one was used to trick all the mathcounts noobs that want to blow their brains out....\n\nSorry :D [/hide]", "Solution_4": "3 :jump: :harhar: 1st one isn't right", "Solution_5": "[hide]\nAt best, either 1 or 3 isnt true.\nso $3$[/hide]", "Solution_6": "[hide]3 II, III, and IV can all be true.[/hide]" } { "Tag": [ "trigonometry", "function" ], "Problem": "Do the properties for trig functions hold for inverse trig function? ie, ${\\sin^{-1}x=\\cos^{-1}({\\frac{\\pi}{2}}-x})$", "Solution_1": "[quote=\"eryaman\"]Do the properties for trig functions hold for inverse trig function? ie, ${\\sin^{-1}x=\\cos^{-1}({\\frac{\\pi}{2}}-x})$[/quote]\r\n\r\nNo, it should be $\\sin^{-1}x + \\cos^{-1}x = \\frac{\\pi}{2}$" } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Find all natural pairs $ (x,y)$, such that $ x$ and $ y$ are relative prime and satisfy equality: $ 2x^2 \\plus{} 5xy \\plus{} 3y^2 \\equal{} 41x \\plus{} 62y \\plus{} 21$.", "Solution_1": "$ 2x^2 \\plus{} 5xy \\plus{} 3y^2 \\equal{} (x \\plus{} y)(2x \\plus{} 3y)$\r\n$ 41x \\plus{} 62y \\plus{} 21 \\equal{} 21(2x \\plus{} 3y) \\minus{} (x \\plus{} y) \\plus{} 21$\r\n$ 2x^2 \\plus{} 5xy \\plus{} 3y^2 \\equal{} 41x \\plus{} 62y \\plus{} 21$ iff $ (x \\plus{} y \\minus{} 21)(2x \\plus{} 3y \\plus{} 1) \\equal{} 0$. x, y>0, hence we must have x+y=21. If \"reciprocal prime\" is a mistake (what can it possible mean?...), then the possible solutions are only (2,19) and (19,2).", "Solution_2": "I tried to translate, but may be I made mistake. I meant (x,y)=1. Can you say how it is in English? And there are not 2, but about 12 solutions.", "Solution_3": "[quote=bambaman]$ 2x^2 \\plus{} 5xy \\plus{} 3y^2 \\equal{} (x \\plus{} y)(2x \\plus{} 3y)$\n$ 41x \\plus{} 62y \\plus{} 21 \\equal{} 21(2x \\plus{} 3y) \\minus{} (x \\plus{} y) \\plus{} 21$\n$ 2x^2 \\plus{} 5xy \\plus{} 3y^2 \\equal{} 41x \\plus{} 62y \\plus{} 21$ iff $ (x \\plus{} y \\minus{} 21)(2x \\plus{} 3y \\plus{} 1) \\equal{} 0$. x, y>0, hence we must have x+y=21. If \"reciprocal prime\" is a mistake (what can it possible mean?...), then the possible solutions are only (2,19) and (19,2).[/quote]\n\nNice sol! I did the same thing, but the solutions are (1, 20), (2, 19), (4, 17), (5, 16), (8, 13), (10, 11), and it\u2019s permutations, which gives a total of 12 solutions." } { "Tag": [ "probability", "induction", "modular arithmetic", "floor function", "limit", "strong induction" ], "Problem": "Nathan and Abi are playing a game. Abi always goes first. The players take turns changing a positive integer to a smaller one and then passing the smaller number back to their opponent. On each move, a player may either subtract one from the integer or halve it, rounding down if necessary. Thus, from $ 28$ the legal moves are to $ 27$ or to $ 14$; from $ 27$, the legal moves are to $ 26$ or to $ 13$. The game ends when the integer reaches $ 0$. The player who makes the last move wins. For example, if the starting integer is $ 15$, Abi might move to $ 7$, Nathan to $ 6$, Abi to $ 3$, Nathan to $ 2$, Abi to $ 1$, and now Nathan moves to $ 0$ and wins. (However, in this sample game Abi could have played better!)\r\n\r\n[list]a) Assuming both Nathan and Abi play according to the best possible strategy, who will win if the starting integer is $ 1000$? $ 2000$? [i]Prove your answer.[/i]\n\nb) As you might expect, for some starting integers Abi will win and for others Nathan will win. If we pick a starting integer at random from all the integers from $ 1$ to $ n$ inclusive, we can consider the probability of Nathan winning. This probability will fluctuate as $ n$ increases, but what is its limit as $ n$ tends to infinity? [i]Prove your answer.[/i][/list]", "Solution_1": "[hide=\"a\"]Let the starting integer be $ n\\equal{}2^{k}\\cdot a$, where $ a$ is an odd positive integer and $ k$ is a \nnonnegative integer. We claim that Abi wins if and only if $ k$ is even, and Nathan if and only if $ k$ is odd. Thus, since $ 1000\\equal{}2^{3}\\cdot125$ and $ 2000\\equal{}2^{4}\\cdot125$, Nathan wins when we start with 1000, and Abi wins when we start with 2000.\n\nWe prove this by strong induction on $ n$. The claim is true for $ n\\equal{}1$: we have $ 1\\equal{}2^{0}\\cdot1$, and Abi wins here because she can begin by moving from 1 to 0.\n\nNow, assume the claim holds for all $ n\\le x\\minus{}1$ for some integer $ x\\ge2$. We prove it for $ n\\equal{}x$. We have a few cases to consider. First, $ k\\equal{}0$, that is, $ x$ is odd. If $ n\\equiv1\\pmod{4}$, let $ x\\equal{}4i\\plus{}1$. We want to show that Abi wins. She has two choices for her first move: she can either move to $ 4i$ or $ 2i$. We can write these in the form $ 2^{k}\\cdot a$; note that exactly one of these has $ k$ odd. Then, if Abi moves to the one where $ k$ is odd, she has reduced the game to starting with this new number, either $ 4i$ or $ 2i$, which by the inductive hypothesis Nathan wins. However, this game is reversed, since Nathan is \"starting\". Thus, Abi will win. Now, if $ n\\equiv3\\pmod{4}$, let $ x\\equal{}4i\\plus{}3$. Again, we want to show that Abi wins. She can either move to $ 4i\\plus{}2$ or $ 2i\\plus{}1$. If she does the former, she reduces the game to one in which we start with $ 4i\\plus{}2$, but reversed because Nathan starts. But Nathan wins this game, because this is the $ k\\equal{}1$ case (as 2 divides $ 4i\\plus{}2$, but 4 doesn't). Thus, in the reversed version, Abi wins.\n\nNow, consider $ k>0$ and even (note that $ k\\ge2$). We want to show that Abi wins. In her first move, we can have Abi divide the number by 2, in which we get a number with $ k$ odd, a game that Nathan would win. However, the game is now reversed, so Abi wins. Finally, consider $ k>0$ and odd. We want to show that no matter what move Abi moves, she will set up Nathan in a winning position. If she subtracts 1, we will end up with an odd integer (since we start from an even), so $ k\\equal{}0$. This gives Nathan a position in which Abi would win, but the game is now reversed, so Nathan will win. Now, if she divides by 2, we will end up with an integer with $ k$ even, in which Nathan also wins (because Abi wins in the non-reversed version). We have thus exhausted all cases, completing the proof.[/hide]\n\n[hide=\"b\"]The answer is $ \\dfrac{1}{3}$.\n\nAs proven in part (a), Nathan wins if and only if $ k$ is odd. An equivalent statement is that Nathan wins with starting integer $ z$ if and only if $ z\\equiv2^{i}\\pmod{2^{i\\plus{}1}}$ for some odd positive integer $ i$. As $ n$ goes to infinity, the probability that $ z\\equiv2^{i}\\pmod{2^{i\\plus{}1}}$ is $ \\dfrac{1}{2^{i\\plus{}1}}$, since the residues modulo $ 2^{i\\plus{}1}$ are periodic with period $ 2^{i\\plus{}1}$. Thus, we want to sum this over all odd positive integers $ i$; we can make the transformation $ i\\equal{}2k\\minus{}1$, so that we are summing over all positive integers $ k$. Thus, the probability is\n\n$ \\displaystyle\\sum_{k\\equal{}1}^{\\infty}(\\dfrac{1}{2})^{2k}\\equal{}\\displaystyle\\sum_{k\\equal{}1}^{\\infty}(\\dfrac{1}{4})^{k}\\equal{}\\dfrac{\\dfrac{1}{4}}{1\\minus{}\\dfrac{1}{4}}\\equal{}\\dfrac{1}{3}$,\n\nas desired.[/hide]", "Solution_2": "[hide]Define the sequence $ \\{a_k\\}_{k \\ge 1}$ such that for every positive integer $ i$, if a player can always win if he starts with the number $ i$, then $ a_i = 1$, and if the player must always lose if he starts with $ i$, then $ a_i = 0$. In other words, if having $ i$ is a winning position, then $ a_i = 1$, and if $ i$ is a losing position, $ a_i = 0$. \n\nNote that $ i$ is a winning position if either $ \\lfloor i/2 \\rfloor$ or $ i-1$ is a losing position, because then the opponent can be given a losing position. In addition, $ i$ is a losing position if both $ \\lfloor i/2 \\rfloor$ and $ i-1$ are winning positions, because then the opponent always gets a winning position. Thus, using the language of Boolean logic, \\[ a_i = (\\neg a_{\\lfloor i/2 \\rfloor }) \\vee (\\neg a_{i-1}).\\] Trivially, $ a_1 = 1$, because either operation will give 0.\n\nLemma: $ a_{2n-1} = 1$ and $ a_{2n} = \\neg a_n$ for all positive integers $ n$. \n\nProof: We proceed by induction. The base case is clearly satisfied: $ a_1 = 1$ and $ a_2 = 0$. Assume that that the inductive hypothesis is true for $ n=k$. Then $ a_{2(k+1) - 1} = a_{2k+1} = (\\neg a_k) \\vee (\\neg a_{2k}) = a_{2k} \\vee (\\neg a_{2k}) = 1$, and $ a_{2(k+1)} = a_{2k+2} = (\\neg a_{k+1}) \\vee (\\neg a_{2k+1}) = (\\neg a_{k+1}) \\vee 0 = \\neg a_{k + 1}$, as desired, so the induction is complete.\n\nCorollary: For any positive integer $ n$, let $ 2^p$ be the highest power of $ 2$ dividing $ n$. Then if $ p$ is odd, $ a_n = 0$, and if $ p$ is even, $ a_n = 1$.\n\n\nPART A\nFrom our lemma, it is clear that $ a_{1000} = \\neg a_{500} = a_{250} = \\neg a_{125} = \\neg 1 = 0$, so since Abi goes first, he will lose. Finally, $ a_{2000} = \\neg a_{1000} = 1$, so Abi will win.\n\nPART B Let $ \\{S_k \\}_{k \\ge 1}$ be a sequence of strings of 1s and 0s such that for every positive integer $ i$, we have $ S_i = a_1 a_2 a_3 \\dots a_{2^{i-1}}$, so $ S_1 = 1$, $ S_2 = 10$, $ S_3 = 1011$, $ S_4 = 10111010$, and so on. \n\nThen by our lemma, it is clear that $ S_{n+1}$ is obtained by replacing each 1 in $ S_n$ with the string 10, and replacing each 0 in $ S_n$ with the string $ 11$: if $ a_n=1$, then $ a_{2n-1}$ and $ a_{2n}$ are 1 and 0, and if $ a_n = 0$, then $ a_{2n-1}$ and $ a_{2n}$ are 1 and 1.\n\nThus, if we let $ p(n)$ be the probability that a randomly chosen digit from $ S_n$ is a 1, then we have $ p(n) = \\frac{1}{2} \\cdot p(n-1) + 1 \\cdot (1 - p(n-1))$, or \\[ p(n) = 1 - \\frac{1}{2} p(n-1).\\]\n\nWe will prove that $ p(n) = -\\frac{2}{3} (-\\frac{1}{2})^n + \\frac{2}{3}.$ The base case $ p(1) = 1$ is clear. If this formula works for $ n=k$, then\n\\begin{eqnarray*}\np(k+1) &=& 1 - \\frac{1}{2} p(k) \\\\\n&=& 1 - \\frac{1}{2} \\left(-\\frac{2}{3} \\left(-\\frac{1}{2} \\right)^{k} + \\frac{2}{3} \\right)\\\\\n&=& 1 + \\frac{1}{3} \\left( -\\frac{1}{2} \\right)^{k} - \\frac{1}{3} \\\\\n&=& \\frac{1}{3} \\left( -\\frac{1}{2} \\right)^{k} + \\frac{2}{3} \\\\\n&=& -\\frac{2}{3} \\left(-\\frac{1}{2} \\right)^{k+1} + \\frac{2}{3},\n\\end{eqnarray*}\nso the formula works for $ n=k+1$ and the induction is complete. Thus, we have\n\n\\[ \\lim_{n \\rightarrow \\infty} p(n) = \\lim_{n \\rightarrow \\infty} \\left( -\\frac{2}{3} \\left(-\\frac{1}{2} \\right)^n + \\frac{2}{3} \\right) = \\frac{2}{3}.\\]\n\nSince Abi always goes first and Nathan goes second, the number we want (the probability that Nathan wins) is $ 1 - \\frac{2}{3} = \\boxed{\\frac{1}{3}}$[/hide]" } { "Tag": [ "topology", "real analysis", "real analysis unsolved" ], "Problem": "Let K be a planar, compact and convex set in $\\mathbb{R}^n$. Prove that the boundary countains four points which are the verticies of a arbitrary square.", "Solution_1": "This seems pretty badly stated. If \"planar\" really means \"2-manifold with boundary\", we shouldn't bother with $\\mathbb{R}^n$; convexity means we can restrict to a plane anyway. If it means something else, it should still obviously reduce to a plane problem- and then you have to do something to exclude a line segment.\r\nAlso, the topological boundary of a \"2-dimensional\" set in $\\mathbb{R}^n$ for $n> 2$ is the whole set.\r\n\r\nIn $\\mathbb{R}^2$, the statement is much simpler: compact, convex, nonempty interior.", "Solution_2": "adapt the solution of this topic : http://www.mathlinks.ro/Forum/viewtopic.php?highlight=area&t=39780\r\n(post No 2). It works fine here." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "inequalities", "calculus", "search", "function" ], "Problem": "Is there a list of theorems that you are allowed to use without proof on the USAMO? (yeah, it would be pretty long) I know there are like 12 inequalities that you are allowed to use\r\n\r\nAlso, if you give a calculus solution that's correct, will you get full credit?", "Solution_1": "You can pretty much use any theorem, as long as it doesn't trivialize a problem, and as long as you cite it properly. The more bizarre or brutish a theorem you use, the more and more scrutiny it is put under for whether it is cited properly. If you give a calculus solution that's correct, it gets full credit, but it's generally frowned upon, and like citing bizarre or brutish theorems, or generally using any brute force or dumb(you know the last three letters) solution such as calculus on an olympiad problem, the smallest mistake can knock off all of your points. (this question's been asked a lot before, so I'd use the search function next time).", "Solution_2": "What do you mean by cite it properly?", "Solution_3": "Give the name, conductions, and result of the theorem.", "Solution_4": "[quote=\"JSteinhardt\"]Give the name, conductions, and result of the theorem.[/quote]\r\n\r\nIs that for all theorems? Even for like Pythagorean Theorem?\r\n\r\nI thought I read somewhere that for very-well known theorems you only have to state the theorem, I could be wrong though.", "Solution_5": "You can generally gloss over stuff like the Pythagorean theorem if it's pretty obvious. The main point of citing theorems is to let the graders know that you know what you are doing. It is true, however, that some theorems are more equal than others. A good example of this is Nesbitt's inequality, which is not a fundamental inequality per se, but is nevertheless encountered by vast numbers of students, as it is a good instructional problem.\r\n\r\nAnyhow, better than any advice : look at examples. Try working problems, and then look at the official solutions to see where you could have been more direct, what different approaches you could have used, and whether you are being needlessly verbose or abstemiously stingy.", "Solution_6": "Are there any olympiads with problems of the difficulty of AIME problems ranging from #6-15 (medium to hard AIME problems)" } { "Tag": [ "algebra", "polynomial", "number theory", "Rational Root Theorem", "number theory solved" ], "Problem": "Show that sqrt(2) + sqrt(3) is irrational.", "Solution_1": "Solution\r\n\r\nLet $ x \\equal{}\\sqrt{2}\\plus{}\\sqrt{3}$\r\n\r\n$ \\Rightarrow x^{2}\\equal{} 2\\plus{}2\\sqrt{6}\\plus{}3 \\equal{} 5\\plus{}2\\sqrt{6}$\r\n\r\n$ \\sqrt{6}$ is irrational so this makes $ x^{2}$ is irrational. Hence $ x$ is irrational. If this were not the case then $ x$ would be rational, and can be expressed as $ x \\equal{}\\frac{a}{b}$ where $ a$ and $ b$ are integers. This would make $ x^{2}\\equal{}\\frac{a^{2}}{b^{2}}$ which is rational and this contradicts the fact that $ x^{2}$ is irrational.", "Solution_2": "Rationality/irrationality is more number theory than it is analysis. sludgethrower's proof does qualify as number theory. I'm moving the topic.", "Solution_3": "Although it is quite simple to prove the assertion by very elementary tools, you can also prove that the number $ \\sqrt{2}\\plus{}\\sqrt{3}$ is irrational, like this:\r\n Suppose, by the way of contradiction, that $ \\sqrt{2}\\plus{}\\sqrt{3}$ is rational. Since $ \\sqrt{2}$ and $ \\sqrt{3}$ are algebraic integers (because for example $ \\sqrt{2}$ is root of a monic polynomial with integer coefficients, in our case $ x^{2}\\minus{}2$). So, $ \\sqrt{2}\\plus{}\\sqrt{3}$ is algebraic integer (the sum of two algebraic integers is an algebraic integer). It follows that the number $ \\sqrt{2}\\plus{}\\sqrt{3}$ is actually an integer (because a rational number which is algebraic integer, is an integer number). But, it is easy to see that $ 3 <\\sqrt{2}\\plus{}\\sqrt{3}< 4$, contradiction, because there is no integer in the interval $ (3,4)$.", "Solution_4": "[quote=\"Cezar Lupu\"]Since $ \\sqrt{2}$ and $ \\sqrt{3}$ are algebraic integers [/quote]\r\n\r\nWhat do you mean please by [size=150][b]algebraic integer[/b][/size]??", "Solution_5": "An algebraic number is the root of a polynomial with integer coefficients. An algebraic integer is the root of a [i]monic[/i] polynomial with integer coefficients. Here $ \\sqrt{2}$ is the root of the polynomial $ x^{2}\\minus{}2$, and $ \\sqrt{3}$ is the root of the polynomial $ x^{2}\\minus{}3$.\r\n\r\nThe algebraic numbers are an interesting and important set because they are the [i]algebraic closure[/i] of the rationals: in other words,\r\n\r\n- every rational number is an algebraic number,\r\n- the sum of any two algebraic numbers is an algebraic number,\r\n- the product of any two algebraic numbers is an algebraic number,\r\n- the root of any polynomial with algebraic coefficients is an algebraic number.\r\n\r\nIt turns out that the algebraic integers are also closed under addition and multiplication. Hence, Cezar Lupu's reasoning is as follows: based on some results in algebraic number theory, because $ \\sqrt{2},\\sqrt{3}$ are roots of monic polynomials with integer coefficients, $ \\sqrt{2}\\plus{}\\sqrt{3}$ is also the root of a monic polynomial with integer coefficients. This is not hard to show explicitly:\r\n\r\n$ x \\equal{}\\sqrt{2}\\plus{}\\sqrt{3}\\implies$\r\n$ x^{2}\\equal{} 5\\plus{}2\\sqrt{6}\\implies$\r\n$ x^{2}\\minus{}5 \\equal{} 2\\sqrt{6}\\implies$\r\n$ (x^{2}\\minus{}5)^{2}\\equal{} 24$\r\n\r\nwhich is a monic quartic polynomial with integer coefficients. Now by the Rational Root Theorem we conclude that if $ x$ is rational it must be an integer (moreover, an integer that divides $ 25\\minus{}24 \\equal{} 1$), absurd.", "Solution_6": "$ \\sqrt 2 \\equal{}1.41..,\\ \\sqrt 3 \\equal{}1.73...$, therefore $ \\sqrt 2\\plus{}\\sqrt 3\\equal{} 3.14...\\equal{}\\pi$ is irrational. :D", "Solution_7": "[quote=\"t0rajir0u\"]An algebraic number ...[/quote]\r\n\r\nThanks a lot for your explanation\r\n\r\nIm in class 12 but I've never heard about algeabric numbers, it does'nt exist in Morrocan system :mad: :lol:", "Solution_8": "[quote=\"Rust\"]$ \\sqrt 2 \\equal{} 1.41..,\\ \\sqrt 3 \\equal{} 1.73...$, therefore $ \\sqrt 2\\plus{}\\sqrt 3 \\equal{} 3.14... \\equal{}\\pi$ is irrational. :D[/quote]\r\nI don't think that because $ \\pi$ is a advance number but $ \\sqrt2\\plus{}\\sqrt3$ is an algebraic number \r\nSo $ \\sqrt2\\plus{}\\sqrt3\\equal{}\\pi$ not true.\r\nHave a same problem\r\n$ Let a,b\\in Z$ satisfy\r\n$ \\sqrt[3]{3}a\\plus{}\\sqrt[3]{9}b$ is a rational .\r\nProve that \r\n$ a\\equal{}b\\equal{}0$\r\nb)Find the polynomial $ P[x]\\in Z[x]$has least degree such that \r\n$ P(\\sqrt2\\plus{}\\sqrt3)\\equal{}0$", "Solution_9": "[quote=\"TTsphn\"][quote=\"Rust\"]$ \\sqrt 2 \\equal{} 1.41..,\\ \\sqrt 3 \\equal{} 1.73...$, therefore $ \\sqrt 2\\plus{}\\sqrt 3 \\equal{} 3.14... \\equal{}\\pi$ is irrational. :D[/quote]\nI don't think that because $ \\pi$ is a advance number but $ \\sqrt2\\plus{}\\sqrt3$ is an algebraic number \nSo $ \\sqrt2\\plus{}\\sqrt3 \\equal{}\\pi$ not true.\n[/quote]\r\n\r\nHe was just joking :lol:", "Solution_10": "[quote=\"t0rajir0u\"]Rational Root Theorem[/quote]\r\n\r\n[b]Nobody knows the name of this theorem in French please.[/b]", "Solution_11": "[b]Theorem:[/b] If $ r$ is a rational root of a polynomial $ P(x)$ of degree $ n$ with integer coefficients with leading term $ a_{n}$ and constant term $ a_{0}$, then $ r \\equal{}\\frac{p}{q}$ where $ gcd(p, q) \\equal{} 1, p | a_{0}, q | a_{n}$. \r\n\r\nIn other words, $ (qx\\minus{}p) | P(x)$. For monic polynomials, $ q \\equal{} 1$, hence $ r$ is an integer.", "Solution_12": "[quote=\"TTsphn\"] $ \\pi$ is a [color=red][b]advance[/b][/color] number but $ \\sqrt2\\plus{}\\sqrt3$ is an algebraic number \nSo $ \\sqrt2\\plus{}\\sqrt3 \\equal{}\\pi$ not true.\n[/quote]\r\n\r\nWhat is advance number please?", "Solution_13": "I think, he mean ''Transcendental Number ''. :wink:", "Solution_14": "[quote=\"t0rajir0u\"][b]Theorem:[/b] If $ r$ is a rational root of a polynomial $ P(x)$ of degree $ n$ with integer coefficients with leading term $ a_{n}$ and constant term $ a_{0}$, then $ r \\equal{}\\frac{p}{q}$ where $ gcd(p, q) \\equal{} 1, p | a_{0}, q | a_{n}$. \n\nIn other words, $ (qx\\minus{}p) | P(x)$. For monic polynomials, $ q \\equal{} 1$, hence $ r$ is an integer.[/quote]\r\n\r\nOk, now I see, Its a very interresting theorem.\r\n\r\nThanks :)", "Solution_15": "Yes prove it quite easy\r\nA problem is same from Viet Nam Olympiad\r\n$ \\frac{x^{3}\\plus{}1}{y\\plus{}1}\\plus{}\\frac{y^{3}\\plus{}1}{x\\plus{}1}$ is a integer \r\nProve that\r\n$ \\frac{x^{2007}\\plus{}1}{y\\plus{}1}$ is an integer.", "Solution_16": "[quote=\"TTsphn\"]Yes prove it quite easy\nA problem is same from Viet Nam Olympiad\n$ \\frac{x^{3}\\plus{}1}{y\\plus{}1}\\plus{}\\frac{y^{3}\\plus{}1}{x\\plus{}1}$ is a integer \nProve that\n$ \\frac{x^{2007}\\plus{}1}{y\\plus{}1}$ is an integer.[/quote]\r\n\r\nYour problem isn't same to allrighty'problem , TTsphn :(", "Solution_17": "No I want to apply theory\r\nIf $ x_{0}$ is a rational root a monic polynomial then $ x_{0}\\in Z$", "Solution_18": "[quote=\"Rust\"]$ \\sqrt 2 \\equal{} 1.41\\ldots\\ ,\\ \\sqrt 3 \\equal{} 1.73\\ldots$. Therefore $ \\left(\\sqrt 2 \\plus{} \\sqrt 3\\right)\\approx\\pi\\ .$[/quote]\r\n[color=darkblue][b]Fine remark and a nice joke, Rust ![/b] \n\n[b]Exercise.[/b] What is greater between $ \\left(\\sqrt 2 \\plus{} \\sqrt 3\\right)$ and $ \\pi$ (without computer) ?[/color]", "Solution_19": "[quote=\"allrighty\"]Show that sqrt(2) + sqrt(3) is irrational.[/quote]\r\nThis a good link http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 :wink:", "Solution_20": "Let $ x\\equal{}\\sqrt{2}\\plus{}\\sqrt{3}.$\r\nSuppose,by the way of contradiction,that $ x$ is rational. Then $ x^2$ is rational. But $ x^2\\equal{}5\\plus{}2\\sqrt{6}$ implies that $ \\sqrt{6}\\equal{}\\dfrac{x^2\\minus{}5}{2}$ is rational, contradiction ($ \\sqrt{6}$ is irrational!). :ninja:", "Solution_21": "[quote=\"TTsphn\"]Yes prove it quite easy\nA problem is same from Viet Nam Olympiad\n$ \\frac {x^{3} \\plus{} 1}{y \\plus{} 1} \\plus{} \\frac {y^{3} \\plus{} 1}{x \\plus{} 1}$ is a integer \nProve that\n$ \\frac {x^{2007} \\plus{} 1}{y \\plus{} 1}$ is an integer.[/quote]\r\n\r\nWhats the year of this problem?\r\nAre $ x,y$ integers??", "Solution_22": "Year :2007, $ x,y\\in\\mathbb{Z}$.", "Solution_23": "Thank you TUAN :wink: \r\n\r\nIt seems to be an interresting problem, Ill try to solve it :roll:" } { "Tag": [ "integration" ], "Problem": "In the picture, let $ V_{A}\\minus{}V_{b}\\equal{}5$ volts, $ E\\equal{}10$ volts, $ C1 \\equal{} 1\\mu F,C2 \\equal{} 2\\mu F$. What's the voltage over $ C1$ and $ C2$ respectively?", "Solution_1": "Shouldn't it be $ V_{1}\\plus{}10\\plus{}V_{2}\\equal{} 5$ and $ V_{2}\\equal{} 2V_{1}$?", "Solution_2": "Solution\r\n\r\nI assume $ A$ is on the left.\r\n\r\nLet the voltages across the capacitors $ C_{1}$ and $ C_{2}$ be $ V_{1}$ and $ V_{2}$ respectively. Voltages are measured from the subtracting the potential on the left from the potential on the right. The total potential difference, $ \\minus{}5\\,V$, is equal to the sum of the potential differences of the various components. \r\n\r\n$ \\Rightarrow V_{1}\\plus{}V_{E}\\plus{}V_{2}\\equal{}\\minus{}5\\Rightarrow V_{1}\\plus{}10\\plus{}V_{2}\\equal{}\\minus{}5\\Rightarrow V_{1}\\plus{}V_{2}\\equal{}\\minus{}15$ \r\n\r\nNote that the sign of $ V_{E}$ is positive since the positive terminal is on the right.\r\n\r\nThe charge across a capacitor is given by $ Q \\equal{} CV$. The current in this portion $ AB$ is the same at at any point. If the capacitors initally have the same charge then the charge on the capcitors will always be the same. This follows from:\r\n\r\n$ Q(t) \\equal{}\\int I(t)\\,dt$\r\n\r\nThus: $ C_{1}V_{1}\\equal{} C_{2}V_{2}$\r\n\r\n$ \\Rightarrow V_{1}\\equal{} 2V_{2}$\r\n\r\nSubstituting this into the other equation:\r\n\r\n$ 2V_{2}\\plus{}V_{2}\\equal{}\\minus{}15\\Rightarrow 3V_{2}\\equal{}\\minus{}15\\Rightarrow V_{2}\\equal{}\\minus{}5\\,V$\r\n\r\n$ \\Rightarrow V_{1}\\equal{}\\minus{}10\\,V$" } { "Tag": [ "FTW" ], "Problem": "hello hello helllo", "Solution_1": "hello hello hello.\r\n\r\nThere are no stategies in FTW, just read as fast as possible, and get the answer as fast as possible. Or mem.\r\n\r\n:P:P:P", "Solution_2": "this should be moved.\r\n\r\n[size=75][color=red]Moved. [/color][/size]", "Solution_3": "Here are some strategies if you are low on rating:\r\n1. play a lot of 2/5 games\r\n2. bomb your rating (pretending that you're bad at FTW) so that when you play high-rated people, you gain rating epicly with such low rating..\r\n3. Try to mem (this is the \"foolish\" player's option)", "Solution_4": "most of the time....\r\n\r\nWhenever long questions appear, I just skip to the last sentence.", "Solution_5": "Have paper and pen or pencil by you at all times during a game.", "Solution_6": "please don't purposefully try to mem just to gain rating...\r\nit takes out the learning part of FTW and makes it a memming game", "Solution_7": "[quote=\"agentcx\"]Here are some strategies if you are low on rating:\n1. play a lot of 2/5 games\n[/quote]\nThis tends to be 1. very annoying and 2. you don't learn anything because you don't have enough time to work the problem through.\n[quote=\"agentcx\"]3. Try to mem (this is the \"foolish\" player's option)[/quote]\r\nPlease don't mem. This makes the game very unfair for other players, and, once again, you don't learn anything. (as vallon22 said)\r\n\r\nI would recommend making sure that you are familiar with formulas. Practice. Take a course if you need to improve on any one subject. AoPS has many courses available. Try to pinpoint what you are having trouble with (geo, algebra, etc), and work to get better at these subjects. :)", "Solution_8": "[quote=\"agentcx\"]Here are some strategies if you are low on rating:\n1. play a lot of 2/5 games\n2. bomb your rating (pretending that you're bad at FTW) so that when you play high-rated people, you gain rating epicly with such low rating..\n3. Try to mem (this is the \"foolish\" player's option)[/quote]\r\n\r\n1. People won't join after awhile, except memmers who will take away your rating.\r\n2. Uhm, this is pretty dumb...losing a game then winning a game will give you less rating than just winning a game without losing one first :wink:\r\n3. Good luck[size=0]kids[/size] with that.", "Solution_9": "Do you mean like strategies to get higher rating or strategies for answering questions faster? \r\n\r\nIf you want to get higher rating, there's really no strategy for it. You just have to get better.\r\n\r\nAnd for answering questions faster, here's my strategy.\r\n\r\nI skim the passage and then read the last sentence thoroughly because that's usually the sentence that tells you what its asking for. (It kind of makes me confused sometimes when the last sentence isn't the question.)\r\nSkim the passage as fast as you can but comprehending it at the same time.", "Solution_10": "[quote=\"AIME15\"][quote=\"agentcx\"]Here are some strategies if you are low on rating:\n1. play a lot of 2/5 games\n2. bomb your rating (pretending that you're bad at FTW) so that when you play high-rated people, you gain rating epicly with such low rating..\n3. Try to mem (this is the \"foolish\" player's option)[/quote]\n\n1. People won't join after awhile, except memmers who will take away your rating.\n2. Uhm, this is pretty dumb...losing a game then winning a game will give you less rating than just winning a game without losing one first :wink:\n3. Good luck[size=0]kids[/size] with that.[/quote]\r\n\r\nstrategy 2 is good for dumb people!", "Solution_11": "Well thats why I said it was the foolish players option. I don't try to purposefully mem, and I don't want anyone else to try to mem on purpose, so I labeled it the foolish players option. The player is foolish because memming is strongly discouraged (even though it's not cheating). \r\nWell if you've come to the point where you've started memming most of the problems, then I suggest you go to other resources for learning (like Alcumus).", "Solution_12": "The main reason that memming is \"foolish\" is because memming will not help you with the real CD. FTW was developed to help users do better at the real CD. :)", "Solution_13": "[cue=\"to moderators\"]Hi, I think this entire thread has became quite foolish, don't you?[/cue]" } { "Tag": [ "MIT", "college", "Princeton", "Harvard", "Stanford", "geometry", "geometric transformation" ], "Problem": "I'm looking for colleges that have really strong math programs. I know the obvious schools (MIT CalTech) and I've heard that Princeton, University of Chicago, and UC Berkeley are also really good, but I'm just wondering if there are some less known schools that I don't know about. I want to major in math so this is my biggest criterion for colleges. Are there any schools with less harsh acceptance rates that are just as good? Also are there any other Ivy league schools with particularly good math departments?", "Solution_1": "Some strong math colleges:\r\n\r\nHarvard\r\nHarvey Mudd\r\nU of Mich", "Solution_2": "Are Caltech and Stanford good for math (compared to Harvard, MIT, and Princeton)?\r\n\r\nBy the way, mathkid, how did the hurricanes hit your area? (my area was very closely missed 3 times - Frances, Ivan, Jeanne)", "Solution_3": "I live in Vero Beach. We got the grunt of the storm during both Frances and Jeanne, so our town was, and still is, in bad shape. .. scary stuff.", "Solution_4": "Generally, if you want a very strong math department, even as a undergraduate, you should go to a school which has a Ph.D. program in mathematics. The availability of a large breadth of advanced undergraduate courses, not to mention seminars and colloquia, will generally be much better than a school which doesn't have a graduate program.\r\n\r\nHaving said that, the American Mathematical Society has a ranking of Ph.D. programs at:\r\n\r\n[url]http://www.ams.org/employment/groups_des.html[/url]\r\n\r\nAs with all \"rankings\", take this list with a huge grain of salt. In fact, let me quote from the AMS's disclaimer:\r\n\r\n[i]Please keep in mind that these groupings are used for statistical reporting purposes only and do not necessarily accurately reflect current program quality at individual departments.[/i]\r\n\r\nBut if you're looking for a list of strong math programs, this list is not a bad place to start. In particular, the list of \"Group I Public\" are all very strong programs at large state universities.", "Solution_5": "Hi - I put a post (very personal opinion) on this subject before... and anyway here is some good discussion on this subject . It took place some time ago in this form. If you have not seen it, please look at\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=9170\r\n******\r\n\r\nThere are many good schools. You must look at them, (or talk to friends who have gone there). and then make up your mind. \r\n\r\nAlso remember, among good schools, there is no such thing as a 'bad choice' \r\n\r\nOn personal level I (with my sons) looked at many schools. Apart from the schools I mentioned in that post, you may like to check out Washington University (St Louis). ... (Duke and Washingto U are one of a few schools that go out of their way to recruit good students.) \r\n\r\nNow both of my son's are in college. One went to Duke (his other choices were were Caltech and MIT) and other went to MIT (Other choices Duke and WashU). \r\n\r\n(Though I pushed a little, they did not even apply to IIT's - their mom put the foot down, \"her babies were not going to go to a school so far away\") \r\n\r\nBoth are very happy with their decesion (and so am I).\r\n\r\nIs any one here thinking of going to I IT's?", "Solution_6": "[quote=\"DPatrick\"]Generally, if you want a very strong math department, even as a undergraduate, you should go to a school which has a Ph.D. program in mathematics. The availability of a large breadth of advanced undergraduate courses, not to mention seminars and colloquia, will generally be much better than a school which doesn't have a graduate program.[/quote]\r\nThat's a very reasonable point of view. I'd like to put another point of view up here, just as something to think about. I will say up front that I do have an ax to grind, a bias which should be obvious from my signature. I will also say that I am not talking to everyone who reads this board - there are certainly many for whom the above advice does apply.\r\n\r\nThe only thing is that the business of Ph.D. granting institutions is mathematical research and that Ph.D. program. Teaching loads are low (which means from your point of view relatively few chances to have any particular professor for a class), research pressures are high, and the graduate students may have the first claim on the faculty's time. Those seminars and colloquia? Some of them are wonderfully stimulating, but many of them are highly technical research mathematics, accessible only to those who already have graduate-level knowledge in those fields. The question about those seminars is this: will [i]you[/i] be attending them while you are still an undergraduate student? More broadly, where will you be in your academic development in 3 or 4 years? How much will the availability of graduate courses matter to you? (There are some other possible questions that concern finances.)\r\n\r\nThose of you who are reading this are talking about starting an undergraduate program - and while you may be 90% sure you're going to major in math, you do know that students change their minds. If you go to a school without a Ph.D. program, you may see a lot more of the faculty, for whom your undergraduate education is a high priority. You may have more opportunities to spend time with one or two particular professors. Even institutions with less selective admissions have their star students; you shouldn't take for granted that you would be one yourself without working for it.\r\n\r\nThere are two kinds of colleges without Ph.D programs that I'm talking about here. One is small private liberal arts colleges. There are some, particularly in the Midwest, that have excellent reputations in mathematics. The other is second-tier state universities, such as the one I represent, and as masters-granting institutions, we do have an array of first-year graduate courses available to keep you moving forward should you reach that point. Give us a chance. Weigh the benefits of the personal attention against the fame, and decide what is best for you.", "Solution_7": "[quote=\"Gyan\"]Is any one here thinking of going to I IT's?[/quote]\r\n\r\nI am definitely curious about the Indian Institute of Technology (IIT) schools. I was just looking up information about the joint entrance examination the other day. We have no Indian ethnic heritage, but that is the whole point--going to IIT would be a great multicultural experience for a half-European-American, half-Taiwanese young man. \r\n\r\nWhat is the process for applying to IIT as a foreign student?", "Solution_8": "http://www.jee.iitb.ac.in/Mmaths.htm\r\n\r\nThere's a sample mathematics entrance exam from IIT.", "Solution_9": "[quote=\"mathkid\"]http://www.jee.iitb.ac.in/Mmaths.htm\n\nThere's a sample mathematics entrance exam from IIT.[/quote]\r\n\r\nThank you for letting everybody know what aspiring Indian mathematicians have to do test on to get into the top colleges in India. My question is more directed to the issue of how foreigners can sit for those tests. Does one have to travel to India from abroad to take the preliminary and main JEE exams, or is there a way to take them overseas? Or is there a modified examination procedure for foreigners (as there definitely is for universities in China or in Taiwan)?", "Solution_10": "Tokenadult:\r\nIIT's require three exams if you are going into a tech field such as Engineering or math: physics, chemistry, and math. To get into medical school, you need to take physics, biology, and chemistry. No writing is needed at all, simply because the primary focus of the IIT's are science/tech/math stuff and medicine.\r\nFor internationals, I belive you can substitute each of the Indian tests with a SAT II (2C for math) for whatever field you want to get into....... got this from my relatives (4 in IIT's). \r\nAlso, there won't be much culture shock for your son if he chooses to attend, because, aside from the fact nearly everyone is Indian, the professors and students speak English near flawlessly, and English is the only language used in instruction. In dorms, as well, English is spoken around 50% of the time simply because India has hundreds of languages, and English is the one everyone knows. I am confident that your son will not have to know a single word of Hindi to pass the 4 years of college, simply because he won't have to venture outside the campus.\r\nAnother thing: Ragging (hazing) is prevalent in these dorms: stick head in toilet, sing, servant to upperclassemen, etc. Your son will be able to escape all that since he is Chinese- American, and people will be intimidated of him for that.\r\nWhat IIT is he planning to apply to?\r\n\r\n\r\nHope that helps.", "Solution_11": "Sorry if this is long ---\r\nI can talk a little about IIT. I went to IIT Kanpur a long time ago but, yes, I would still be very biased \uf04a ) . I found that the quality of students, and faculty was excellent. There was diversity, although majority were Indian students, there were a few foreign students (including Americans) . And my time at IIT was simply the best time I had in any school. Although IITs are primary Engineering schools, the sciences and math programs are also excellent . (I majored in Physics) \r\n\r\nIITs have kept their standards, and enjoy a very good reputation. For example when I applied to graduate schools I got into every school I applied to (including Caltech, MIT) I did not even take a GRE.. There was a CBS segment 60-minutes on IITs and it said some thing like Put Princeton, Harvard and MIT together .. and you start getting some idea about the prestige it has in India.. MIT prof Keniston once joked \r\n[url=http://www.kamalsinha.com/iit/news/mafia/article.html]joke is that MIT is run by IIT Kanpur Miafia ..[/url] \r\n\r\nFor admissions and other info, you may like to see http://www.iit.org or the link given above by mathkid which has more details and the syllabus for the tests etc.\r\n\r\nThere was a special channel called DASA (Direct admission for students at Abroad) for American (and other foreign) students which used SAT (minimum 2100 in SAT IIs combined in Math, Physics and Chemistry to be considered for application) and other scores to form comparative merit lists but I think it is no longer used. The criteria for all (Indians or Foreign students) now is JEE (Joint Entrance Exam used by IITs and a few other schools) JEE has two parts: A three hour screening test ( 10th April 2005) (About 15% of about 200,000 students gets to the next stage) and a six hour Main-test (3x2hous for Physics, Chemistry , Mathematics) (15% of above or about 2% of total gets selected) . \r\n\r\nDifficulty level *very rough - my estimate only* - If you can qualify for USAMO you will have no difficulty)\r\n\r\nMost of American High Schools are accredited but you have to have enough courses in Physics, Chemistry and Mathematics (see the syllabus in mathkids link). \r\n\r\nJust few points which Samus raised - IIT does not have medical school . Though some now have a few graduate business related degrees but primary (as in 99%) focus is math/science/engineering. \r\n\r\nHazing, I also heard, could be problem, but honestly I have never encountered it (or even seen it ) and if you stay away from frat-party type scene you should be okay. I believe there was a bad case (some one committed suicide) so I believe that administration is now very strict. \r\n\r\nIITs are not very expensive. (Tution+ living expenses+ say one trip per year) may be around $10,000-12,000. JEE is for common admissons (to all IIT's)\r\n\r\nEnglish is the medium, but it really does not matter, (The JEE exams are offered in many other languages) It is really easy to learn a new language while you are studding it in a IIT type atmosphere.\r\n\r\nOne important tip for parents here (in US): Check out the atmosphere in the school ( say while you are in Junior year summer) to see if think if you like it. IIT alumini chapters are many places in USA and they can be a good resource too... \r\n\r\nHope this information is useful.\r\n\r\n\r\n\r\n\r\n .", "Solution_12": "[quote]My question is more directed to the issue of how foreigners can sit for those tests. [/quote]\r\n\r\nAs I said in the previous post, at present only JEE (which is same for all - Foreigners and Indians) is the criteria. DASA (For foreign students), I belive, i[b]s no longer (starting from 2003) an option. \n[/b]\r\nI will find out for sure and the locations where one can take JEE. (Kids I knew (from US) did trave to india but they had family in india and arranged that any way - Indian Embessy is/was possibliity but I am not sure at all so I will post it here when I find out something definitive)", "Solution_13": "[quote=\"samus\"]Tokenadult: \nWhat IIT is he planning to apply to?\n[/quote]\r\n\r\nThank you for the very helpful answer. So far my son is hardly thinking about this at all, but I like to consider lots of possibilities for his education. We have been attending information sessions for United States colleges (the usual \"top\" math schools) this month. I liked the years I spent in Taiwan after college, during which I was a student part of the time, and I think international experience is a good idea. But my son doesn't know even the names of the IIT campuses yet, although one girl in his car pool to his math program is an abcd. \r\n\r\nYes, I expected higher education in India to be in the medium of English for the reasons you mentioned. When my son was in Ireland this summer he found out that English isn't spoken everywhere the way it is spoken in Minnesota [grin]. Also in Taiwan most of the textbooks are exactly the same English-language textbooks used in the top schools in the United States--although of course some students buy cram books written in Chinese. \r\n\r\nWe have a lot to think about in the next few years. I first became aware of the international reputation of the IITs when my mother watched a Sixty Minutes report on one of the IIT schools a year or so ago.", "Solution_14": "Sorry for the confusion. I was thinking of other schools in India that have both engineering and med schools. Also, the general math skills of peope in IIT are of Olympiad (international to national)/pre-Olympiad levels (from my cousin in India, who is a national olympiad qualifier, similar to (better than?) colleges such as MIT CalTech. Finally, a word of caution: the IIT's probably do not offer students a well-rounded education.\r\n\r\nSorry if I'm rambling :( .", "Solution_15": "can we limit the search just a bit more to great small (about maybe less than 50 students to 1 teacher ratio or more....) colleges with great math departments for undergrad looking forward to pure math and great teachers (teachers there to teach and not for the reseach) to teach them?", "Solution_16": "Well, at Yale, I'd say we have a pretty good math department. Although I think that a lot of the ivies have very similar math departments, ie. princeton, harvard, yale, cornell. So you can't really go wrong with any of those choices, though I feel that our ratio between faculty and students is relatively low. The only thing that you might have to consider is you might want to look into certain distributional requirements you need to meet, if you are only interested in doing straight math and no humanities or social sciences. But like I said, you can't really go wrong with any of the choices, though i'd strongly suggest yale (guess i'm biased? :) ).", "Solution_17": "Yes, it's the US News ranking. But anyways, it may still yield some useful info. BTW, this is for graduate school so liberal arts colleges are out. \r\n\r\n*** THE SCIENCES (Ranked in 2002) ***\r\n\r\nThe Sciences->Applied Mathematics->Top Applied Mathematics Programs\r\n\r\nCategories:\r\nRank.|School|Average assessment score (5 = highest)\r\n\r\n1.|Massachusetts Institute of Technology|4.7\r\n |New York University|4.7\r\n3.|California Institute of Technology|4.6\r\n4.|Stanford University (CA)|4.5\r\n5.|Brown University (RI)|4.4\r\n |Princeton University (NJ)|4.4\r\n |University of California-Berkeley|4.4\r\n |University of California-Los Angeles|4.4\r\n9.|University of Minnesota-Twin Cities|4.3\r\n10.|Cornell University (NY)|4.2\r\n11.|Carnegie Mellon University (PA)|4.0\r\n |University of Maryland-College Park|4.0\r\n |University of Texas-Austin|4.0\r\n14.|Northwestern University (IL)|3.9\r\n |Rice University (TX)|3.9\r\n |University of Washington|3.9\r\n |University of Wisconsin-Madison|3.9\r\n18.|Georgia Institute of Technology|3.8\r\n |University of Chicago|3.8\r\n |University of Michigan-Ann Arbor|3.8\r\n21.|Harvard University (MA)|3.7\r\n |Rensselaer Polytechnic Institute (NY)|3.7\r\n |Rutgers State University-New Brunswick (NJ)|3.7\r\n |SUNY-Stony Brook|3.7\r\n |University of Arizona|3.7\r\n |University of Colorado-Boulder|3.7\r\n27.|Duke University (NC)|3.6\r\n |Purdue University-West Lafayette (IN)|3.6\r\n |University of Illinois-Urbana-Champaign|3.6\r\n |Yale University (CT)|3.6\r\n31.|North Carolina State University|3.5\r\n |University of California-San Diego|3.5\r\n33.|Columbia University (NY)|3.4\r\n |Indiana University-Bloomington|3.4\r\n |Johns Hopkins University (MD)|3.4\r\n |Penn State University-University Park|3.4\r\n |University of California-Davis|3.4\r\n |Virginia Tech|3.4\r\n39.|Ohio State University|3.3\r\n |Texas A&M University-College Station|3.3\r\n |University of North Carolina-Chapel Hill|3.3\r\n |University of Utah|3.3\r\n43.|Boston University|3.2\r\n44.|Arizona State University|3.1\r\n |University of California-Santa Barbara|3.1\r\n |University of Delaware|3.1\r\n47.|Claremont Graduate University (CA)|3.0\r\n |CUNY Graduate School and University Center|3.0\r\n |Iowa State University|3.0\r\n |University of California-Irvine|3.0\r\n |University of Pennsylvania|3.0\r\n |University of Southern California|3.0\r\n |Washington University in St. Louis|3.0\r\n54.|Florida State University|2.9\r\n |Michigan State University|2.9\r\n |University of Illinois-Chicago|2.9\r\n |University of Pittsburgh|2.9\r\n |University of Virginia|2.9\r\n59.|University of Houston|2.8\r\n |University of Iowa|2.8\r\n61.|University of Florida|2.7\r\n |University of Massachusetts-Amherst|2.7\r\n |University of Tennessee-Knoxville|2.7\r\n64.|Brandeis University (MA)|2.6\r\n |New Jersey Institute of Technology|2.6\r\n |SUNY-Buffalo|2.6\r\n |University of Notre Dame (IN)|2.6\r\n |Vanderbilt University (TN)|2.6\r\n69.|Case Western Reserve University (OH)|2.5\r\n |Clemson University (SC)|2.5\r\n |Colorado School of Mines|2.5\r\n |Colorado State University|2.5\r\n |Dartmouth College (NH)|2.5\r\n |Emory University (GA)|2.5\r\n |Oregon State University|2.5\r\n |Polytechnic University (NY)|2.5\r\n |University of Georgia|2.5\r\n |University of Missouri-Columbia|2.5\r\n |Washington State University|2.5\r\n\r\n\r\nThe Sciences->Mathematics->Top Mathematics Programs\r\n\r\nCategories:\r\nRank.|School|Average assessment score (5 = highest)\r\n\r\n1.|Massachusetts Institute of Technology|5.0\r\n2.|Harvard University (MA)|4.9\r\n |Princeton University (NJ)|4.9\r\n |Stanford University (CA)|4.9\r\n |University of California-Berkeley|4.9\r\n6.|University of Chicago|4.8\r\n7.|Yale University (CT)|4.7\r\n8.|California Institute of Technology|4.6\r\n |University of Michigan-Ann Arbor|4.6\r\n10.|Cornell University (NY)|4.4\r\n |New York University|4.4\r\n |University of California-Los Angeles|4.4\r\n13.|Columbia University (NY)|4.3\r\n |University of Wisconsin-Madison|4.3\r\n15.|University of Texas-Austin|4.2\r\n16.|Rutgers State University-New Brunswick (NJ)|4.1\r\n |University of Illinois-Urbana-Champaign|4.1\r\n |University of Maryland-College Park|4.1\r\n |University of Minnesota-Twin Cities|4.1\r\n |University of Pennsylvania|4.1\r\n21.|Brown University (RI)|4.0\r\n |Northwestern University (IL)|4.0\r\n |SUNY-Stony Brook|4.0\r\n |University of California-San Diego|4.0\r\n25.|Duke University (NC)|3.9\r\n26.|Indiana University-Bloomington|3.8\r\n |Johns Hopkins University (MD)|3.8\r\n |Pennsylvania State University-University Park|3.8\r\n |Purdue University-West Lafayette (IN)|3.8\r\n |Rice University (TX)|3.8\r\n |University of Washington|3.8\r\n32.|Ohio State University|3.7\r\n |University of North Carolina-Chapel Hill|3.7\r\n34.|Brandeis University (MA)|3.6\r\n |Carnegie Mellon University (PA)|3.6\r\n |CUNY Graduate School and University Center|3.6\r\n37.|Georgia Institute of Technology|3.5\r\n |Michigan State University|3.5\r\n |University of Illinois-Chicago|3.5\r\n |University of Utah|3.5\r\n |Washington University in St. Louis|3.5\r\n42.|Texas A&M University-College Station|3.4\r\n |University of Arizona|3.4\r\n |University of California-Davis|3.4\r\n |University of Colorado-Boulder|3.4\r\n |University of Virginia|3.4\r\n47.|University of California-Irvine|3.3\r\n |University of California-Santa Barbara|3.3\r\n |University of Notre Dame (IN)|3.3\r\n |University of Southern California|3.3\r\n51.|Boston University|3.2\r\n |Dartmouth College (NH)|3.2\r\n |University of Florida|3.2\r\n |University of Georgia|3.2\r\n55.|North Carolina State University|3.1\r\n |University of Iowa|3.1\r\n |University of Massachusetts-Amherst|3.1\r\n |University of Oregon|3.1\r\n |Virginia Tech|3.1\r\n60.|Arizona State University|3.0\r\n |Iowa State University|3.0\r\n |Northeastern University (MA)|3.0\r\n |University of California-Riverside|3.0\r\n |University of Missouri-Columbia|3.0\r\n |University of Rochester (NY)|3.0\r\n |Vanderbilt University (TN)|3.0\r\n67.|Louisiana State University-Baton Rouge|2.9\r\n |SUNY-Buffalo|2.9\r\n |University of Kansas|2.9\r\n |University of Nebraska-Lincoln|2.9\r\n |University of Pittsburgh|2.9\r\n |University of Tennessee-Knoxville|2.9\r\n73.|Emory University (GA)|2.8\r\n |Florida State University|2.8\r\n |Tulane University (LA)|2.8\r\n |University of California-Santa Cruz|2.8\r\n |University of Delaware|2.8\r\n |University of Kentucky|2.8\r\n |University of Oklahoma|2.8\r\n80.|Claremont Graduate University (CA)|2.7\r\n |Colorado State University|2.7\r\n |Oregon State University|2.7\r\n |Syracuse University (NY)|2.7\r\n |Tufts University (MA)|2.7\r\n |University of Connecticut|2.7\r\n |University of Houston|2.7\r\n87.|Auburn University (AL)|2.6\r\n |Case Western Reserve University (OH)|2.6\r\n |Kansas State University|2.6\r\n |Oklahoma State University|2.6\r\n |Polytechnic University (NY)|2.6\r\n |SUNY-Albany|2.6\r\n |Temple University (PA)|2.6\r\n |University of New Mexico|2.6\r\n |University of South Carolina|2.6\r\n |Washington State University|2.6\r\n97.|New Mexico State University|2.5\r\n |SUNY-Binghamton|2.5", "Solution_18": "[quote=\"samus\"]Also, the general math skills of peope in IIT are of Olympiad (international to national)/pre-Olympiad levels (from my cousin in India, who is a national olympiad qualifier, similar to (better than?) colleges such as MIT CalTech.[/quote]\r\n\r\nThis is highly unlikely, as the best US math students (who go almost completely to Harvard, Princeton, MIT, Caltech, Duke, Stanford, etc.) are better than the best Indian students, as far as IMO, etc, are concerned.\r\n\r\nIn fact, India didn't have a single gold medal at the IMO this year, and only one in the past 3 years.\r\n\r\nIt is quite possible that the best Canadian schools, even, attract better math students than IIT.", "Solution_19": "[quote=\"mathkid\"]http://www.jee.iitb.ac.in/Mmaths.htm\n\nThere's a sample mathematics entrance exam from IIT.[/quote]\r\n\r\nmathkid, thanks a lot for posting that.\r\n\r\nit's rare that I get to see good calculus/linear algebra questions like that.", "Solution_20": "You may like to check out NY Times (registration is required) article about \r\n[url=http://www.nytimes.com/2004/10/20/nyregion/20ranking.html?oref=login]Top Colleges, Rated by Those Who Chose Them[/url]\r\n\r\nAs expected Harvard and Yale ranked high in general. For Math/Science students MIT, Caltech, Stanford and Princeton ranked high.\r\n\r\nMIT and Caltech, among science related subjects had top appeal. \r\n\r\nSeveral \"top\" schools (as ranked from usual sources) did not do as well. \r\n\r\nFor example, Washington University in St. Louis, which I liked when I checked it out for my son, was ranked around 10 in this year's U.S. News , was the 62nd rated choice among students' preferences . Places like the University of Pennsylvania, the University of Chicago and Duke placed at least eight spots lower on the preference ranking than on the U.S. News list.\r\n\r\nSlightly related topic - \r\n\r\nIMO, Duke (and Wash U to some extent) has another attraction, it does have prestigious programs - like A.B. Duke Scholarship - (which opens up lot of priviledged oppurtunities, not to mention that those programs provide full tution scholarships too) for really bright students... So the select few \"recruited \" Math students, more often than not, , choose Duke over schools like Harvard... (No wonder that some Ivy college compliain that Duke 'steals' their good potential students by bribing them :) )", "Solution_21": "Frankly, I think that's a incredibly pointless rating system. I have no basis whatsoever to rate my school against other schools, because I've only been to one of them. I think those schools which drop in the rankings suffer (with the exception of UPenn) from geographic problems -- the places that send the most students to private colleges are California and the north-East, especially the New York City tristate region (that is, New Jersey, New York and Conneticut). Students from those areas are much less likely to wander out into the midwest or down south. It doesn't surprise me at all that Duke, Chicago and Washington in St.L ranked lower -- I mean, who wants to go to school in North Carolina, for crying out loud?\r\n\r\n\r\nAnd who wants to go to Penn, anyhow? ;)", "Solution_22": "[quote=\"JBL\"]I mean, who wants to go to school in North Carolina, for crying out loud?[/quote]\r\n\r\nNorth Carolinians (and South Carolinians) presumably are happy to go to school in the Carolinas, in at least some instances. It is still relevant, for comparing colleges, to find out whether a North Carolinian accepted to Duke and Harvard will prefer Harvard or not, and whether a New Yorker accepted to Harvard, to Yale, and to Princeton will prefer one school over the other two. \r\n\r\nThe \"[url=http://post.economics.harvard.edu/faculty/hoxby/papers/revealedprefranking.pdf]A Revealed Preference Ranking of\r\nU.S. Colleges and Universities[/url]\" is pretty interesting and has some good discussion of methodological issues.", "Solution_23": "[quote=\"tokenadult\"]It is still relevant, for comparing colleges, to find out whether a North Carolinian accepted to Duke and Harvard will prefer Harvard or not, and whether a New Yorker accepted to Harvard, to Yale, and to Princeton will prefer one school over the other two. [/quote]\r\n\r\nInteresting, maybe. I don't see how it is relevant. Students who have not yet attended college are not in the best position to choose what the best college for them would be, honestly. What if all my friends at Harvard would really have been happier at Washington University? There's absolutely no way to include that kind of measure in the ranking. What it tells you is that, when presented with a choice between two schools, students tended to choose one over the other -- I'm not convinced this would really represent what school would be best for any student. I think there are much more telling pieces of information, like quality-of-life measurements or perceived enjoyment of already-enrolled students, that could build a much better picture of what students want. Given the basic types of data they collected, the thing that would most warm me up to it is if they also broke it down according to what reasons students gave for choosing the schools they did -- it would give a picture of why students chose schools which would be much more helpful to me, as a student, picking where I wanted to go.", "Solution_24": "But no one has shown that students who decide which offer of admission to accept lack access to currently enrolled students, who presumably would tell the newly admitted what they think about each school--plausibly or not.", "Solution_25": "Don't many of the best students also go to large public institutions? Considering financial and \"staying close-to-home\" concerns...\r\n\r\nIt seems like nany great scientists did not all go to the best institutions. \r\n\r\nFor example, [[Dr. David Goodstein]] only went to state schools. There are plenty of others.\r\n\r\nBut hmmm...\r\n\r\nDo the majority of the best students only go on to state schools? Hmmm... How many people are at age 18? Tkae 1 % of that; and take the addition of the freshman pops of the top 15 universities. This should give an impression...", "Solution_26": "Simfish -- thousands more students go to large public universities than to selective private colleges or universities, but that wouldn't necessarily move those public schools up in the ranking, because it compared where people chose to go once accepted among their various options. Relatively few people who apply to both large public universities and selective private universities and get in to both go to the public schools nowadays -- most of the students you are talking about would never have bothered applying to the private colleges in the first place.\r\n\r\n\r\nTokenadult -- my entire position can be summed up in the following two sentences:\r\nHaving been, relatively recently, among a large group of college-bound high school students, I am not convinced that tracing the patterns of where such students choose to go to school gives sufficient evidence of school quality to declare that the most popular schools are anything other than most popular. In particular, I feel that the relative rankings of any two colleges is essentially meaningless.", "Solution_27": "[quote]I mean, who wants to go to school in North Carolina, for crying out loud?[/quote] :)\n\nReminds me of a T-shirt I saw in Cambridge:\n\n[quote]Harvard: Because every one cant get into MIT ![/quote]\n:D \nMy son is at Duke. He chose Duke over MIT (and Caltech) and I know quite a few kids (My sons friends, and my friends children) who have chosen North Carolina over Harvard and Yale, .and some of them even live closer to Connecticut or Massachusetts ! I think, last year, all (or almost all) who were offered AB Duke scholarships ( this I mentioned it the previous post) did go to Duke (one can be sure that they were also accepted in other top colleges in US).. BTW Duke has done pretty good in Putnam competition, so may be some of those kids who like math wants to go to North Carolina. :)\n\nSeriously, IMO, far from pointless, the survey is has lot of points. When we were visiting colleges from my sons , one of the most often asked question, (when we were getting perspective from other students, and indeed my friends asks me who are trying to get my perspective) is/was which other colleges you were seriously considering? and why you choose x rather than y. This type of survey gives some statistical perspective to get answer to those types of questions. Sure, you can have a lot of data points, you can see rankings, you can talk to your friends, professors, and other students, you can visit colleges and attend classes there, you can also get perspective on why a person would choose college x from college y specially if your choice also happens to be between x and y. Nothing pointless about this. Just my thoughts.\n\n\n\nBlahblahblah - IITs are mainly engineering schools, and some would say that math is not their top field but as I said before, these are, IMO, excellent even for Math and Physics etc. I dont think many typical US students go to IIT (unless some reason like - their parents went to IIT) but I do know some Indian Kids who chose IIT over top US schools. In final analysis, its hard do define the best school and in reality one would not go wrong if one chose any one of the top schools. \n\nBTW, I may be wrong as I dont know how popular AMC/AIME/USAMO type contests are in India today, but when I was there, some of the best math students(and their schools) did not even know of those (nmo etc) tests and contestants for IMO did not represent the top math talent from Indian High schools. \n\n[quote]But no one has shown that students who decide which offer of admission to accept lack access to currently enrolled students, who presumably would tell the newly admitted what they think about each school--plausibly or not..[/quote] \r\nAgree ..IMO they are very well informed", "Solution_28": "This information about foreign students may be of interesti to some:\r\n[url=http://www.iitb.ac.in:8080/examples/acad/foreignstudents.jsp] link.\r\n [quote]General Information - Admissions for foreign students\n\n 1. Foreign student is one who holds foreign passport, or has Indian passport but is studying outside India and wants to do partial studies at IIT Bombay\n\n\n 2. Admission for a full programme with degree from IIT Bombay\n * For B. Tech : Admissions are made through JEE\n * For M. Tech / Ph.D. and other post-graduate degrees\n o Admissions are made based on academic performance\n o Please send your application (brochures / forms are available at our website http://www.iitb.ac.in ) to the department of your interest\n o Offer will be made based on recommendations received from the departments\n o The fees at present are US $ 6000.00 per semester (likely to be reduced). You will be given hostel accommodation on the campus. Your other expenses are expected to be about US $ 200 - 300 pm.\n o You will need to get student visa from the Indian Embassy in your country\n o The schedule for admission is available at our web-site\n\n 3. Admissions for visiting / exchange students\n 1. You may wish to study a semester or carry out your project / research work at IIT Bombay for a limited duration (2 months upto 1 year)\n\n\n 2. If you wish to come as exchange student under memorandum of your Institute with IIT Bombay, the fees / charges will be as per the terms of Memorandum. If you wish to do a semester of course work here as a visiting student, the fee will be US $ 6000.00 per semester (likely to be reduced), and you will have to bear your local expenses. If you are planning to do a short R\\&D project here, the fees will be about US $ 500.\n\n\n 3. The admission procedure will be as follows :\n * If you are an exchange student, send your application through your Dean to the Dy. Registrar (Academic) at IIT Bombay\n * If you are a visiting student, send your application giving full details of your education and academic performance, to the Head of the Department, or the Professor with whom you wish to do the project work.\n * Admission offers will be made based on the recommendation of the department / faculty.\n * You will need to obtain student visa for the duration of your visit.\n\n 4. In case you plan to take transfer of credits for your course work done here, you must plan the courses (in consultation with your faculty advisor at your university) beforehand and inform the Head of the Department here. You should ensure that you have the right background to do the course work here.\n[/quote][/url]", "Solution_29": "Thanks for all the info...still trying to decide how much to use the rankings...btw does anyone know how Carnegie Mellon is in more theoretical math (as opposed to applied). A friend was recommending it and I was just wondering if anyone knew any specifics. Thanks again for all the replies.", "Solution_30": "[quote]BTW, I may be wrong as I dont know how popular AMC/AIME/USAMO type contests are in India today, but when I was there, some of the best math students(and their schools) did not even know of those (nmo etc) tests and contestants for IMO did not represent the top math talent from Indian High schools. [/quote]\r\n\r\nBy the way Gyan, my brother-in-law, who qualified for the Indian National Olympiad had several other friends do that too at IIT-(Dehli?) when he attended from 1996-2000.", "Solution_31": "[quote=\"stargazer\"]Thanks for all the info...still trying to decide how much to use the rankings...btw does anyone know how Carnegie Mellon is in more theoretical math (as opposed to applied). A friend was recommending it and I was just wondering if anyone knew any specifics. Thanks again for all the replies.[/quote]\r\n\r\nI am a proud graduate of Carnegie Mellon, with a B.S. and an M.S. in mathematics. The program prepared me well enough to get a Ph.D. at MIT, so that says something I hope.\r\n\r\nAs you note, the department at CMU is heavily tilted towards applied mathematics. However, they have a very strong undergraduate \"pure\" mathematics course sequence, called Mathematical Studies, and offer a program to get a master's degree in 4 years (which I did).\r\n\r\nSee [url]http://www.math.cmu.edu/ug/ungrd/special.html[/url] for more specifics, and feel free to post and/or email me if you have some specific questions.", "Solution_32": "Isn\u2019t institute of advanced studies good for mathematics?", "Solution_33": "The Institute for Advanced Studies, as its name implies, is not an institution you can apply to for an undergraduate education." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Find the smallest constant k>0 s.t. \r\n\r\n \\frac{ab}{a+b+2c}+\\frac{bc}{b+c+2a}+\\frac{ca}{c+a+2b} \\leq k(a+b+c)\r\n\r\nfor every positive a,b,c", "Solution_1": "[quote=\"maruemu\"]Find the smallest constant k>0 s.t. \n\n $ \\frac{ab}{a\\plus{}b\\plus{}2c}\\plus{}\\frac{bc}{b\\plus{}c\\plus{}2a}\\plus{}\\frac{ca}{c\\plus{}a\\plus{}2b} \\leq k(a\\plus{}b\\plus{}c)$\n\nfor every positive a,b,c[/quote]\r\n\r\n $ LHS\\equal{}\\sum {\\frac{ab}{(a\\plus{}c)\\plus{}(b\\plus{}c)}}\\le \\sum {\\frac{1}{4}(\\frac{ab}{a\\plus{}c}\\plus{}\\frac{ab}{b\\plus{}c})}\\equal{}\\frac{a\\plus{}b\\plus{}c}{4}$\r\n \r\n therefore $ k\\equal{}\\frac{1}{4}$" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "A curve has parametric equations\r\n\r\n$ x\\equal{}2t \\plus{} t^{2}$, $ y\\equal{} 2t^{2} \\plus{} t^{3}$.\r\n\r\n(i) Express $ \\frac{dy}{dx}$ in terms of $ t$ and find the gradient of the curve at the point $ (3, \\minus{}9)$.\r\n\r\n(ii) By considering $ \\frac{y}{x}$, find a cartesian equatiooj of the curve, giving your answer in a form not involving fractions.", "Solution_1": "$ \\frac {dx}{dt} \\equal{} 2 \\plus{} 2t$\r\n\r\n$ \\frac {dy}{dt} \\equal{} 4t \\plus{} 3t^2$\r\n\r\n\r\n$ \\frac {dy}{dx} \\equal{} \\frac {dy/dt}{dx/dt} \\equal{} \\frac {4t \\plus{} 3t^2}{2 \\plus{} 2t}$\r\n\r\n\r\nvalue of t satisfying x = 3 and y = -9 is $ t \\equal{} \\minus{} 3$\r\n\r\n\r\nso the gradient of the curve at the point (3, -9) is $ \\frac { \\minus{} 15}{4}$\r\n\r\n\r\n\r\nThe second part ...\r\n\r\n$ \\frac {y}{x} \\equal{} \\frac {2t^2 \\plus{} t^3}{2t \\plus{} t^2} \\equal{} t$ \r\n\r\nso, what will be the Cartesian form ? is it a straight line ?", "Solution_2": "Hi Nora.91.\r\n\r\nI have \r\n\r\n$ \\frac{y}{x} \\equal{} t$\r\n\r\n$ \\Rightarrow\\, x\\equal{} 2(\\frac{y}{x}) \\plus{} (\\frac{y}{x})^{2}$\r\n\r\n$ \\Rightarrow\\, x^{3} \\equal{} 2xy \\plus{} y^{2}$.", "Solution_3": "[i]perfect ... [/i] Thanks." } { "Tag": [ "quadratics", "superior algebra", "superior algebra unsolved" ], "Problem": "prove that for a quadratic field K, p is ramified in K if and only if p divides the discriminant of K. (where p is a prime in Z).", "Solution_1": "This is true for all number fields in general.\r\n\r\nFor this special case, not much is to do:\r\nLet $ d$ be the discriminant of that extension.\r\nHandle the case $ p\\equal{}2$ seperately. Now let $ p$ be odd and let the extension be generated by adjoining the square root of $ a$. Show that $ d\\equal{}2^s a$, thus we have to show:\r\nAn odd prime $ p$ ramifies iff it divides $ a$.\r\nNow show that if the ideal $ (p)$ factors (into two ideals), then $ x^2 \\equiv a \\mod p$ has a solution $ x$, and $ (p)\\equal{} (p,\\sqrt a\\plus{}x)(p,\\sqrt a\\minus{}x)$. Now just show that these factors are equal iff $ p |a$." } { "Tag": [], "Problem": "From Mathcamp's website:\r\n\r\nfor houshold income between $\\$40,000 and $\\$100,000:\r\n\r\ndiscount: $\\frac{100,000-x}{50}$\r\n\r\npayment: $\\frac{91,700+x}{50}$\r\n\r\nThe sum is $\\$3,834 while the full payment is $\\$3,195. What is the real value of the discount/fee?", "Solution_1": "No clue how the person who updated the website arrived at that second formula. Certainly it should be $(59750+X)/50." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "For a given k, what is the minimum number of integers we have to take such that it always possible to choose k of them such that their sum is divisible by k ?", "Solution_1": "$ 2k\\minus{}1$, see http://www.mathlinks.ro/viewtopic.php?p=389030 .", "Solution_2": "Thankyou ZetaX" } { "Tag": [ "geometry", "3D geometry", "number theory", "prime numbers", "prime factorization" ], "Problem": "Find the largest integer $ n$ such that $ n$ has exactly 4 positive divisors and $ n$ divides $ 100!$.", "Solution_1": "Even after reading the solution, I couldn't understand this problem.", "Solution_2": "When two prime numbers multiply together, you get a number with $ 4$ factors. Call one prime $ a$ and the other $ b$. The prime factorization is $ a^{1}\\cdot b^{1}$. The number of factors it has is $ (1\\plus{}1)(1\\plus{}1)\\equal{}4$. The largest $ 2$ prime factors of $ 100!$ are $ 97$ and $ 89$. Multiplying these together, we get the answer of $ \\boxed{8633}$.", "Solution_3": "@jxl28: However, a prime^3 also yields 4 factors! (1, prime, prime^2,prime^3). The floor of 100/3 is 33, and the greatest prime less than 33 is 31. So the correct answer should be 29791, because that is greater than 8633.\n\nHope that helped :)\n\nWatermelon99", "Solution_4": "Hmm. I disagree with these answers. 97*89 may have 4 divisors, but it is not the greatest factor of 100! with 4 divisors. Cubes of prime divisors of 100! will also work, 1, the number being cubed, the number squared, and number cubed will be the divisors. So using this fact we get 31^3 as our answer which is 29791.", "Solution_5": "yup after ready everybody's posts and i'm pretty sure j_f_c_w is correct and 29,791 is the answer but remember to hide your answers please and thank you! ", "Solution_6": "Please do not revive old topics.", "Solution_7": "[quote=Aditya11]Please do not revive old topics.[/quote]\n\nI think he gets the point", "Solution_8": "Let's hope he does.", "Solution_9": "what's with the 4 solutions, no hidden ones" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Positive reals $a,b,c$ satisfy $abc=1$. Prove that \r\n$\\displaystyle 1+ \\frac{3}{a+b+c} \\ge \\frac{6}{ab+bc+ca}$.", "Solution_1": "old and easy:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=14765\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=32861\r\n\r\n dg", "Solution_2": "[quote=\"k2c901_1\"]Positive reals $a,b,c$ satisfy $abc=1$. Prove that \n$ 1+\\frac{3}{a+b+c}\\ge \\frac{6}{ab+bc+ca}$.[/quote]\r\nWe have: \\[\\left(1-\\sqrt{3 \\over a+b+c}\\right)^{2}\\ge 0 \\\\ \\implies 1+{3 \\over a+b+c}\\ge{6 \\over \\sqrt{3(a+b+c)}}={6 \\over \\sqrt{3abc(a+b+c)}}\\ge{6 \\over ab+bc+ca}.\\]", "Solution_3": "Let $a,$ $b$ and $c$ are positive numbers such that $abc=1$. Prove that \r\n$1+\\frac{8}{a+b+c}\\ge \\frac{11}{ab+bc+ca}$\r\nIs it true?", "Solution_4": "The stronger inequality holds:\r\n \r\n$ 1+\\frac{7}{a+b+c}\\ge \\frac{10}{ab+bc+ca}$.", "Solution_5": "wrong :(", "Solution_6": "[quote=\"silouan\"]\nBut $xy+yz+zx\\leq \\sqrt{3(x+y+z)}$ \n[/quote]\r\nBut $xy+yz+zx\\geq \\sqrt{3(x+y+z)}$ :wink:", "Solution_7": "Sorry I was wrong :(" } { "Tag": [ "inequalities", "geometry", "area of a triangle", "Heron\\u0027s formula", "inequalities proposed" ], "Problem": "Prove that in a triangle with sides a, b and c, semiperimeter s and area A,\r\n\r\nA^4 <= s^2((a^2+b^2+c^2-s^2)/3)^3", "Solution_1": "Prove that in a triangle with sides a, b and c, semiperimeter s and area A we have:\r\n\r\n$A^{4} \\leq s^{2} (\\frac{\\sum a^{2}-s^{2}}{3} )^{3}$$.$\r\n\r\nI have a solution using duality!\r\n\r\n$a=y+z$; $b=z+x$ and $c=x+y$ then the semiperimeter $s=x+y+z$ and the area $A=\\sqrt{(x+y+z)xyz}$. Thus our inequality will become:\r\n\r\n$ (x+y+z)^{2} (\\frac{2\\sum x^{2}+2\\sum xy - \\sum x^{2}-2 \\sum xy}{3} )^{3}=(x+y+z)^{2}(\\frac{\\sum x^{2}}{3} )^{3}$ so our inequality is equivalent to: \r\n\r\n$(x+y+z)^{2}(\\frac{\\sum x^{2}}{3} )^{3} \\geq (x+y+z)^{2}(xyz)^{2}$ so it will become: $ (\\sum x^{2})^{3} \\geq 27x^{2}y^{2}z^{2}$ which is true AM- GM.\r\n\r\nhave fun. what is your solution? does anybody have another approach?\r\n\r\ncheers! :D :D", "Solution_2": "I found a neat identity yesterday, so I thought I'd turn it into an inequality:\r\n\r\nif s=(x+y+z)/2, then\r\n\r\nx^2+y^2+z^2=s^2+(s-x)^2+(s-y)^2+(s-z)^2 (1)\r\n\r\nSo by AM-GM, we can get the weak result x^2+y^2+z^2>4A (using Heron's formula), or the stronger one mentioned above if we apply the AM-GM only to the last three terms on the right and then apply Heron's formula.\r\n\r\nWhat's neat is that (1) generalizes:\r\n\r\nfor B+3A^2-4A=0,\r\nx^2+y^2+z^2=Bs^2+(As-x)^2+(As-y)^2+(As-z)^2\r\n\r\nMaybe someone else can make a nicer problem out of that equation..." } { "Tag": [ "function", "abstract algebra", "linear algebra", "matrix", "integration", "calculus", "circumcircle" ], "Problem": "[i]Math Contest-Memorialul \"Hegy Lajos\" January 2004 Targu-Mures[/i]\r\n\r\n[b]9th grade[/b]\r\n\r\n[b]1[/b]. Proove that the ecuation x^2-x(n^2+1)/n-(n^2+1)/n has real roots and they are irational for every n \\in N*.\r\n\r\n[b]2[/b]. Proove that for every x,y,z>0 we have:\r\nsqrt((x+y)/(x+z))+sqrt((x+z)/(x+y)) <=(y+z)/sqrt(yz).\r\n\r\n[b]3[/b].In the convex quadangle ABCD we denote with G the weight centre of triangle BCD and with H the ortocenter of triangle ACD. Proove that the points A,B,G,H represent in this order the points of a paralelogram(i don't know the english version for this word:( ) if and only if G is the center of the circumscribed circle of triangle ACD.\r\n\r\n[b]4[/b].We have a number with 2 digits. Between their digits we write another digit and we obtain another number. AM of these two numbers is equal with the number with 2 digits obtained if we inverse the order of the digits of the initial number. Find the initial number.\r\n\r\n[b]10th grade[/b]\r\n\r\n[b]1[/b]. Solve in R the ecuations:\r\n\ta) 2^(x+2)+4^(x+1) = 3x^2;\r\n\tb) 2^(x+2)-4^(x+1) = x^2+2x+2;\r\n\r\n[b]2[/b].The sequence (x_n)n>=1 satisfie the relation\r\n\tx_(n+1)+x_n = 2n^2, for every n \\in N*, x_1 \\in Z.\r\na)Exists in the sequence 3 consecutive termns to be in geometric progression.\r\nb)If x_1 = -2004, then which is the value of x_2004?\r\n\r\n[b]3[/b]. Find x,y,z>1 wich for we have\r\nlog{x}(y^2+z^2)+log{y}(z^2+x^2)+log{z}(x^2+y^2)=7 and \\sum log{x}2 = 1.\r\n(i denoted with log{p}(q) = ln(q)/ln(p) )\r\n\r\n[b]4[/b]. The measurements of the angles A, B, D, C of the convex quadangle ABCD are consecutive terms, in this order, of an increasing arithmetic progesion. AC and BD made between them a \\pi/2 angle. The measurements of A,B,D of the triangle ABD are too in an arithmetic progression(in this order).\r\nCompute AC/BD.\r\n\r\n[b]11th grade[/b]\r\n\r\n[b]1[/b].Let (a_n)n>=0 be a sequence satisfying the relation\r\n(n+1)a_(n+2)-(n+2)a_(n+1)+a_n = 0, a_0=1, a_1 = 3.\r\na)Determine the general term of the sequence a_n and proove the convergence of (a_n)n.\r\nb)Compute the limit lim{n->oo}x_n, where x_n=(2+2e-a_n)n!.\r\n\r\n[b]2[/b].Compute lim{n->oo}(nln(n))/([n/1]+[n/2]+[n/3]+...[n/n]).\r\ni denote with [x] the integer part of x.\r\n\r\n[b]3[/b].Let (a_n)n>0 \\in R*. For each n \\in N*\\{1}, we consider e_n : a_nx+a_(n-1)y=0.\r\nWe denote with (b_n)n respectively (c_n)n those (a_n)n for wich e_n and e_(n+1) are perpendicular respectively parallel.\r\nStudy the convergence of (b_n)n and (c_n)n.\r\n\r\n[b]4[/b]. Find all the matrices X \\in M2(R) s.t. \r\nX^2+2X = \r\n(5 5)\r\n(10 10).\r\n\r\n[b]12th grade[/b]\r\n\r\n[b]1[/b]. Let (G,*) be a group with the property that there exists an injective function f:G->G,s.t. f(x*f(x)) = x²*f(x*y) for every x,y \\in G. Proove that (G,*) is an abelian group.\r\n\r\n[b]2[/b]. If F is a primitive of f:R->R, f(x) = ex² for evry x \\in R, compute lim{x->oo}(xF(x))/f(x).\r\n\r\n[b]3[/b].Let A \\in Mn(R), n>=2, invertible. Proove that (A*)* = (detA)A.\r\n\r\n[b]4[/b]. Proove that from 20 integer numbers we can choose 2 s.t. their difference is divisible by 19.\r\n\r\n\r\n[i]What is your impression about this contest? i think it had a very low level... and it was an interdistrict contest :([/i]", "Solution_1": "Can you give us more precision \r\n\r\n\r\n12th grade \r\n\r\n1. ... f(x*f(x)) = x²*f(x*y) ....\r\n\r\nWhat do you mean by x&2sup\r\n\r\n2. ... ex²... \r\n\r\n3.... (A*)* = (detA)R, f(x) = ex² for evry x R, \r\ncompute lim{x->oo}(xF(x))/f(x).\r\n\r\nF(x)= \\int _[1,x]exp(t^2)dt integration by part gives\r\nF(x)= exp(x^2)/2x - e/2 + \\int _[1,x]exp(t^2)/(2t^2) dt \r\nthis leads to \r\nxF(x)/f(x) = 1/2 - x/exp(x^2).e/2 + \r\nx/exp(x^2).1/2. \\int _[1,x]exp(t^2)/t^2 dt ---> 1/2 when x-->+oo\r\n use again integartion by part \r\n\\int _[1,x]exp(t^2)/t^2 dt = \\int _[1,x] (2t.exp(t^2))/(2t^3)dt =\r\nexp(x^2)/(2x^3) - e/2 + 3/2.\\int _[1,x]exp(t^2)/t^3dt\r\n\r\nnice problem for 12th grade, but it could be CE1 grade :D", "Solution_2": "maybe it's ^(x^2). this forum cannot display this style.", "Solution_3": "sorryyy... i'l edit my post... i wrote all the contest in notepad and after that i paste it here and i saw that instead \" 2 \" is something else... i modified them but i think i missed some :D", "Solution_4": "please do not reply anymore in this thread. \r\nput one problem / post ! :)" } { "Tag": [ "calculus", "AoPS Books" ], "Problem": "I don't know if this has already made the rounds, but in this clip Art Benjamin makes his case for changing math education in the US.\r\n\r\n[url]http://metroplexmathcircle.wordpress.com/2009/10/23/arthur-benjamins-formula-for-changing-math/[/url]", "Solution_1": "I partially agree with him. What he says is to take the focus off calculus and onto discrete math. I think that the math curriculum should not focus on a particular subject, but rather let the students try different topics in the wide range of mathematics. Every branch of mathematics is important and useful.\r\n\r\nAnother big point, that I didn't see him mention, is memorization versus understanding, formula versus methods, etc. As a student taking both high school and middle school math classes, as well as the son of two engineering professors, I can say that students in middle and high school are taught insufficient math and lots, including students of many subjects (because math is a foundation for many thing), are not able to take the leap between high school and college/university. This both the teachers' and students' fault, but more so the curriculum and textbooks used, who make math sound like memorization and computation, when it is really deep and creative thinking and problem solving. Mathematicians are not given a worksheet of problems and find theorems and identities, they think.\r\n\r\nTo fix this, we have to make the idea of \"Calculus\" not sound hard and cold. We have to make students really understand things rather than memorizing. For example, in one of my classes, we are given a sheet of formulas and theorems to use on tests and homework, while most students don't understand them and haven't seen them proved. To fix the problem, we have to introduce a new curriculum, such as the way AoPS books work, because they emphasize on the understanding, and show the formula once the student has understood it or derived it by himself/herself by answering questions and solving problems.", "Solution_2": "we know him, he is nice, we went to dinner with him at Davidson in baltimore , and my mom went to school with him :lol:" } { "Tag": [ "Asymptote", "LaTeX" ], "Problem": "Hi.\r\n\r\nCan someone tell me how to install Ghostscript onto my computer? Because in the AoPS wiki webpage about Asymptote, it just has a link that says \"download GS here\". I tried to download AFPL GhostScript 8.54 onto my computer (I clicked on the AFPL link after clicking on the AoPS wiki link) through sourceforge, but they just gave me a bunch of winRar documents (I had to download each part of AFPL separately). \r\n\r\nAnd now, after \"downloading\", I still can't get Ghostscript to work. In fact, my computer says that Ghostscript was not downloaded at all. So, is there any particular site I can go to to download AFPL GS 8.54 all at one time? And, to make my computer accept GS as downloaded? Thanks!\r\n\r\nMr.Ocax", "Solution_1": "Unfortunately you don't tell us\r\n1. which operating system you are using;\r\n2. why you chose the AFPL version* which is the license for older versions. Why not choose the GPL version 8.60 on the same page as Asymptote directed you to http://pages.cs.wisc.edu/~ghost/ ?\r\n\r\nSo assuming you are happy with GPL** then the links will lead you to [url=https://sourceforge.net/project/showfiles.php?group_id=1897&package_id=108733&release_id=529280]here[/url] and there you can download the relevant file for your system eg Windows 32 bit is [i]gs860w32.exe[/i] which, when run, is the installer. \r\n\r\nHowever, if you are using Linux (which flavour?) then it would be better to use your installer/package manager for your system.\r\n\r\n* Aladdin Free Public License\r\n** GNU General Public License", "Solution_2": "[quote=\"stevem\"]Unfortunately you don't tell us\n1. which operating system you are using;\n2. why you chose the AFPL version* which is the license for older versions. Why not choose the GPL version 8.60 on the same page as Asymptote directed you to http://pages.cs.wisc.edu/~ghost/ ?\n\nSo assuming you are happy with GPL** then the links will lead you to [url=https://sourceforge.net/project/showfiles.php?group_id=1897&package_id=108733&release_id=529280]here[/url] and there you can download the relevant file for your system eg Windows 32 bit is [i]gs860w32.exe[/i] which, when run, is the installer. \n\nHowever, if you are using Linux (which flavour?) then it would be better to use your installer/package manager for your system.\n\n* Aladdin Free Public License\n** GNU General Public License[/quote]\r\nThe wiki page actually leads you to: http://pages.cs.wisc.edu/~ghost/doc/AFPL/index.htm", "Solution_3": "Ah, that explains it (the link probably needs updating), though that page leads to http://sourceforge.net/projects/ghostscript/ then http://sourceforge.net/project/showfiles.php?group_id=1897 where there is a choice and the latest version is GPL Ghostscript 8.60. So you still end up on the same pages as going via http://pages.cs.wisc.edu/~ghost/.\r\n[quote=\"Mr.Ocax\"]they just gave me a bunch of winRar documents (I had to download each part of AFPL separately)....So, is there any particular site I can go to to download AFPL GS 8.54 all at one time?[/quote]As far as I'm aware there are no WinRar files and only [i]single[/i] compressed [i]alternative[/i] downloads with tar.bz2, tar.gz, .zip and .exe files, all but the last being marked as source files, so need compiling. Unless Linux users want to compile the program themselves, binary files for Linux are best found using the relevant Linux package manager rather than using the Ghostscript site.", "Solution_4": "Thank you. no, I do not have a linux; I have windows, it makes things more straight forward. Thanks for the updated links and the GPL info,\r\n\r\nMr.Ocax" } { "Tag": [ "inequalities", "geometry", "geometric transformation", "reflection" ], "Problem": "Prove that the cinrcumradius of a triangle is at least twice the inradius.", "Solution_1": "[hide=\"One simple way\"]Prove $ \\sum_{cyc}cosA \\equal{} 1\\plus{}\\frac{r}{R}\\leq\\frac{3}{2}$[/hide]", "Solution_2": "[hide=\"method 1\"]Bloundon's[/hide]\n[hide=\"method 2\"]Euler's Formula[/hide]\n[hide=\"method 3\"]The nine-point circle's radius is at least the incircle's radius.[/hide]\n[hide=\"method 4\"]Ptolomey's Inequality on OBMC in triangle ABC with O being the reflection of A across BC.[/hide]" } { "Tag": [ "search", "combinatorics unsolved", "combinatorics" ], "Problem": "Set $ S\\equal{}\\{1, 2, 3,\\dots 2004\\}.$ We denote by $ d_{1}$ the number of subsets of $ S$ such that the sum of elements of the subset has remainder $ 7$ when divided by $ 32$. We denote by $ d_{2}$ the number of subset of $ S$ such that the sum of elements of subset has reaminder $ 14$ when divided by $ 16.$ Compute $ \\frac{d_{1}}{d_{2}}.$", "Solution_1": "Let $ A \\equal{} {1, 2, 4, 8, 16}$ and $ B \\equal{} S \\minus{} A$. For each subset $ X$ of $ B$, there is exactly one subset $ Y$ of $ A$ for which\r\n\r\n$ \\sum_{x \\in A \\bigcup B} \\equiv 7 mod 32$\r\n\r\nIn fact, this is equivalent with $ \\sum_{x \\in A} x \\equiv 7 \\minus{} \\sum_{x \\in B}x mod 32$ and this presents a solution, because $ \\sum_{x \\in A} \\equiv t mod 32$ always has a solution. This follows from the case $ 0 \\le t < 32$, in which our choice gives the binary representation of $ t$.\r\n\r\nThus, $ d_1 \\equal{} 2^{|B|} \\equal{} 2^{1999}$. Similarly, we get $ d_2 \\equal{} 2^{2000}$.\r\n\r\nSee also http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1867594669&t=39550." } { "Tag": [ "MATHCOUNTS" ], "Problem": "It gets annoying to be smart sometimes because everybody asks you what you got after every test. They say, \"what did you get this time - a 100?\" and start to scoff. It is soooooooooo annoying. I was wondering if people in your school ever bothered you solely because you knew more than they did. Also, I can't have a moments peace because they keep asking you questions based on the assumption that you will always be able to provide them with the correct answer. I don't brag, but they just are annoying like that.\r\n\r\n\r\n\r\nP.S. Visit the AMC MATHCOUNTS counting forum!", "Solution_1": "=)\r\n\r\nYou know, it's actually nice to be asked questions. Even when you can't give them satisfactory answers (which is often the case with me). But you learn so much more when you try teaching others.", "Solution_2": "Yay math-team-captains-teaching!\r\n\r\n(Sorry for the unrelated nature of this post.)", "Solution_3": "Cus of my race they automatically assume that I am smart, and plus the people at my school ask for helparhm...arhm.. help like I said, but they don't really care. They know it ticks me off so they do it.", "Solution_4": "argh. Nobody sees it my way... oh well.", "Solution_5": "I see it ur way I hate being annoyed by them. They ask me for my phone number so they can call me if they ever need help and it drives me up a wall. I politely tell them no Im not a 24\\7 homework hotline. I hate it when they bother me when Im trying to get some reading done in homeroom and they assume that I can answer it and I dont mind because Im not busy as they put it. They assume that Im smart because the teachers voted me president of the NJHS and I made the mistake of telling one friend my report card and they brodcasted it all over the school. Now everyone comes up and asks me what did u make on ur report card and then smirk probably a 100 like u always do. I get real tired of it so I go to the library during homeroom now and sit in the comfy chairs and read at least Im not bothered their unless someone wants to check out a book.", "Solution_6": "Why? You're going to miss that attention when you lose it anyway, fame is like that. People did that to me and I was uncomfortable with it but now that it's over I look back and go *sigh* You'll see, you'll see...", "Solution_7": "I wouldnt miss the attention for that but if people stopped asking me for advice then I would miss that (I mean emotional advice.) If I dont know the answer I simply say go ask the teacher (sometimes I do that anyways even if I do know the answer hehehe...) I just dont like it when they bother me because it seems as if when I do help them they are ungrateful Ill talk some more tommorow Im in trouble its past my bedtime.", "Solution_8": "Yeah some are but then you can't win 'em all. I know that most I helped have helped me back in some way whether it was teaching me how to teach better :lol: or how to explain something better. You can always learn something by teaching others (and better understand it yourself).\r\n\r\nBS? Does that mean make something up?", "Solution_9": "People try to annoy me, but never succeed. Hey, I mean, why tease me, it proves them stupid and makes them look stupider than they actually are (which for some of them is hard to imagine :twisted:)", "Solution_10": "I'm rather being ignored than being annoyed. I'm just not popular even when I do good in school subjects and in other math contests. I am unpopular everywhere... I'm fated to be lonesome.[/img]", "Solution_11": "That is not necessarily true, you are lonesome because you believe you are lonesome when in fact if you were to open your eyes you'd see you have a lot of friends...that I figured out on my own :cry:\r\n\r\nYou don't have to be talented in order to be popular, just kind-hearted and helpful and just be nice.", "Solution_12": "bleumoose wrote:[hide]BS IT!!! they will never know the difference.[/hide]\n\nim joking, of course\n\n\n\nAh, ha, ha. Sure, you've never done anything remotely like that before. \n\n\n\nPersonally, I say \"yes\" to half of the question and \"no\" to the other half. When someone says \"What did you get? 100? *smirks*\" or something like that, I'm irritated. Also, sometimes people think they're complimenting you (like \"you're way too smart\"), but they're actually also making you feel \"different\". That sort of thing gets annoying.\n\n\n\nHowever, when people ask for help, I generally help them because 1) helping someone can be fun as long as they're not completely inept and 2) I want them to understand and see the beauty of math/science/whatever. Plus, it gives you a warm feeling like hot cocoa on a snowy day.", "Solution_13": "No, Tare. They actually ignore me and I have no doubt about that. When I was in Taiwan a few years ago, I was repected. Now that I'm here, I'm feeling like there's a gap between Americans and me. I can seldom find people with commonalities. However, that doesn't mean I don't have friends, I'm just not popular, that's all.", "Solution_14": "MY problem is that people tease me way to much about it and when I get a grade like a A- thry go crazy and keep rubbing salt in my wounds. Also it's annoying now because people keep calling me \"MATH-ew\". :cry:", "Solution_15": "People used to tease me all the time but now they dont because if they do they figure they arent going to pass (I tutor them in math.) Plus it doesnt bother me anymore and even though Im smart Im what they consider I guess \"cool\" at my school. I dont really care what social status Im in at school. I used to be a zero. I may seem like the type that is goodie goodie but shhhhhh Im not. Im loud, noisy, late to class, and skipping down the hallway all the time outside the classroom that is. Yet I seem to make straight A's so the teachers like me. Thats all that matters is the teachers. The kids arent going to be the ones that enter my essay in a contest or grant me scholarships its the teachers so Im friends with all of the teachers except that of a few coaches who have I have grudges agianst (long story) and do not like me. Oh well just in the end dont worry about what the kids think of you just worry about you teachers and what they think.", "Solution_16": "Well among normal kids I'm cool, but when it comes to smart kids that I'm smarter than they kill me.", "Solution_17": "Heh, we were having a discussion in English about whether being the most popular or the smartest was better, I have now voted that the ones who said popular are utmostly stupid, so they have no clue what it means to be smart.", "Solution_18": "Here's a question: how do you know that you know what it is to be smart?", "Solution_19": "NOOOOOOOOO, NOT ANOTHER DEBATE!!!!!!!", "Solution_20": "I'm not trying to start a debate, but if you're afraid I am, there are a whole bunch of problems to be done in the Intermediate forum and a couple in AMC.", "Solution_21": "In my experience, a lot of smart people don't want to be popular. For a long time, I didn't understand that. In 7th and 8th grade I used to wish I were at least somewhat popular so that I wouldn't have to worry about being made fun of. Now, I realize that being popular won't make you any less vulnerable - in fact, it'll probably make you more vulnerable.\n\n\n\nThe only reason guys want to be popular is because of girls, and after [hide]breaking up with my girlfriend I realized how shallow must girls are (no offense to raccoon, aalindsey, rippledance, or any other girls)[/hide]\n\n\n\nSo a word of advice to anyone: don't worry about popularity at all.", "Solution_22": "[quote=\"Tare\"]However the second part I disagree, popularity does not necessarily mean a lot of girls it just means it gives you a lot of options to choose from when you want to make new friends.[/quote]\r\n\r\nActually, it almost always does mean \"a lot of girls\", at least temporarily. And no, it doesn't give you more friends to choose from, because many of those \"friendships\" aren't true friendships. When you're popular, everyone wants to be your friend, but once you become unpopular, you'll be deserted. You're more likely to have real friends who like you because of who you are if you're not the most popular kid in the crowd.", "Solution_23": "Well wouldnt u rather have a girl like u for who u are? See I just dont get guys. They are so good with their friends but when it comes to girls they are clueless. Girls genrally dont care about how u look most of the time anyways (at least I dont.) But u seem to think we do. Yall are so complicated I just dont understand. By the way I like one of these people on here but Im going to keep it a secret so u can guess all u want but u aint going to ever know the answer.", "Solution_24": "[quote=\"aalindsey100\"]Well wouldnt u rather have a girl like u for who u are? See I just dont get guys. They are so good with their friends but when it comes to girls they are clueless. Girls genrally dont care about how u look most of the time anyways (at least I dont.) But u seem to think we do. Yall are so complicated I just dont understand. By the way I like one of these people on here but Im going to keep it a secret so u can guess all u want but u aint going to ever know the answer.[/quote]\r\nYes, I would, and yes, she does. I must contest your statement that guys are so good with their friends. If you would like to be called names, punched in the shoulder, and then would like to go make something explode, all in good fun, then please, let the guys in your life know about that. Most girls would much rather have long talks and walks on the beach.", "Solution_25": "Talking about girls... they don't really like smart people who are nerdy like. If you are truly smart, and you hide it to others and act normally around school, they don't annoy you. They don't think of you as different, or aloof from the main group of people. I like someone at my school, but she probably doesn't like me just because im different. That is when it sucks to be different.", "Solution_26": "[quote=\"white_horse_king88\"]Talking about girls... they don't really like smart people who are nerdy like. If you are truly smart, and you hide it to others and act normally around school, they don't annoy you. They don't think of you as different, or aloof from the main group of people. I like someone at my school, but she probably doesn't like me just because im different. That is when it sucks to be different.[/quote]\r\nThey don't? This is news.", "Solution_27": "I will tell them this but thats not what I prefer. What I like to do is something most guys would find interesting themselves. For fun Id rather go to a nascar race, play tackle football, basketball, do math and science, and most of all just hang out. I love it when there is no stress to impress just when the guys are themselves. I think guys have got girls all wrong. Yea sure some girls are romantic but most are rough around the edges like I am. Actually my brothers idea of a romantic date is candle light dinners with a real burning fire and my idea of a romantic date would be to go to an amusment park, bike riding, camping, hikeing, or like rock climbing. Now does that totally change ur perspective on girls. I guess thats a small town girl who doesnt like the big city just likes to get away. I just am not like most girls or what u guys think most girls are like. By the way this guy he knows I like him or at least I think he does. Well I used to like him alot but now I like him more as a friend.\r\n\r\nPs. some of us like smart guys I know Ive never liked a real dumb guy before theyve all been smart. The person I liked last year looked sort of nerdy but he was cute and I wasnt the only one that thought so either. He ended up likeing me at the end of the year and then I told and he got mad. I told u I didnt understand guys.", "Solution_28": "[quote=\"confuted\"]If you would like to be called names, punched in the shoulder, and then would like to go make something explode, all in good fun, then please, let the guys in your life know about that. Most girls would much rather have long talks and walks on the beach.[/quote]\r\n\r\nBlowing things up is fun! Hey, you can be a girl and still like boy things. Besides, I detest one-on-one talks for a long time. It puts too much pressure on both people and eventually you run out of things to say. And you can't walk on the beach here anyway. I'm landlocked.", "Solution_29": "[quote]I just remebered that lots of girls like guys who can dance really good.. and I'm not one of them. Is that true? \n[/quote]\n\nMan I cant even dance Im sure some girls do though.[/quote]", "Solution_30": "i can't dance. don't even try making me. the only time i did was for PE. it was required. \r\n\r\ni like guys for their minds, not how popular they are. Most popular people are stuck-up and think that they are the center of teh universe. I actually start to strongly dislike ( hate, but i feel that word is too strong) them.\r\n\r\n[quote]Talking about girls... they don't really like smart people who are nerdy like. If you are truly smart, and you hide it to others and act normally around school, they don't annoy you. They don't think of you as different, or aloof from the main group of people.[/quote]\r\n\r\n\r\nthat was a very sterotypical comment.", "Solution_31": "Dancing...in PE!? Boy that's even weirder than us (which is pretty weird: fencing :?:)\r\n\r\n*sigh* The sad truth is, popularity=beautiful/cute=stuck-up=bad personality yet still you're attracted to them since they look so good...another reason how we have betrayed evolution...", "Solution_32": "yes, dancing in PE. \r\n\r\nsixth: square dancing\r\nseventh: slow dancing or folk dancing/line dancing ( this depends on what teacher you have)\r\neighth: cha-cha and salsa\r\n\r\nmy school is very screwed up.", "Solution_33": "I agree heartily, where do you live???", "Solution_34": "california", "Solution_35": "[quote=\"aalindsey100\"]By the way I like one of these people on here but Im going to keep it a secret so u can guess all u want but u aint going to ever know the answer.[/quote]\r\n\r\n*checks memberlist for people from texas*\r\n\r\nirollupstairs? juggle7? JRBoyd? gmann? brian_31407? Krenium? mafps? Smartjock256? Rep123max? wow, there are a lot of people from Texas on AoPS.\r\n\r\nIt could be Fiery. Or jojobird. Or Neal. Granted, none of them live in Texas, but it's a guess.", "Solution_36": "Popular people annoy me. when they talk to me I can't figure out whether they are trying to make me look bad, if they are doing it as a joke, or if they are just doing it to annoy me. Yes. Popular people are really stuck-up. They think they are above everybody else, and that annoys me the most.", "Solution_37": "You know that's contradictory to your signature :lol:\r\n\r\nThough I have to agree, when I look back at how I was when I was popular, I cry out in disgust. I was actually saying stuff like \"Yeah I am better at math than anybody else in the world\" :-P *scrubs tongue with soap*" } { "Tag": [ "floor function", "calculus", "function" ], "Problem": "Find the max value of $K$, such that:\r\n\r\n$\\lfloor \\frac{x^2+6x+14}{x^3+27} \\rfloor \\geq K$; if $x \\in [-2;2]$", "Solution_1": "Calculus ?", "Solution_2": "[hide]\nSo we pretty much just need the floor of the smallest value of $\\frac{x^2+6x+14}{x^3+27}, x \\in [-2,2]$\nThe numerator has no real roots, and it thus always positive. The denominator is negative for $x < -3$, thus positive on the interval. Then the expression is always positive. However, for $x=0$, the expression is $\\frac{14}{27}$. So the minimum value $m$ must satisfy $0 < m \\le \\frac{14}{27}$, and we have the floor of $m$ is 0.\n[/hide]", "Solution_3": "[quote=\"deej21\"][hide]\nSo we pretty much just need the floor of the smallest value of $\\frac{x^2+6x+14}{x^3+27}, x \\in [-2,2]$\nThe numerator has no real roots, and it thus always positive. The denominator is negative for $x < -3$, thus positive on the interval. Then the expression is always positive. However, for $x=0$, the expression is $\\frac{14}{27}$. So the minimum value $m$ must satisfy $0 < m \\le \\frac{14}{27}$, and we have the floor of $m$ is 0.\n[/hide][/quote]\r\n\r\nwe want max value ;)", "Solution_4": "[quote=\"Altheman\"]we want max value ;)[/quote]\r\n\r\nYou want the maximum of $K$ -- he showed that the given expression is never negative (so that $K \\geq 0$) but that $\\frac{x^2+6x+14}{x^3+27}$ is sometimes less than 1, so that $\\lfloor \\frac{x^2+6x+14}{x^3+27} \\rfloor$ is sometimes 0, so that $K \\leq 0$.\r\n\r\nThe problem seems odd to me, because it's not particularly a close question: the given function is [i]always[/i] 0 over the interval $[-2, 2]$.", "Solution_5": "Let ABC be a triangle and the rectangular MNPQ is inscribed ABC(M belong to AB, N belong to AC, P and Q belong BC).Prove that\r\n$\\ S_{ABC}\\geq\\ 2S_{MNPQ}$ ;) \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nMath is my life :rotfl:", "Solution_6": ":blush: Oh,sorry I have a mistake, I have pressed the post reply buttom but I have to press new topic.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nMath is my life :!:" } { "Tag": [ "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $ h : \\mathbb{R} \\rightarrow \\mathbb{R}^\\plus{}$ be a decreasing function. Prove that $ ( \\int_0^{\\infty} \\sqrt{h(x)}dx)^2 \\geq 2 \\int_0^{\\infty} xh(x) dx .$", "Solution_1": "Maybe try Riemann sums and use the following: If $ a_1 \\geq a_2 \\geq ... \\geq a_n\\geq0$, then $ (\\sum a_i)^2 \\equal{} \\sum a_i^2 \\plus{} 2\\sum_{i2^k$ is analogous).\n\n\nWe will prove now that for any such matrix $ n\\times n$ with $ n\\geq 2^k\\plus{}1$ there exists $ i$ such that $ \\{a_{i,1},\\cdots,a_{i,i\\minus{}1}\\}$ and $ \\{a_{i,i\\plus{}1},\\cdots,a_{i,n}\\}$ are [b]not[/b] disjoint.\n\nBecause the matrix is symmetrical, that is equivalent to $ \\{a_{i,1},\\cdots,a_{i,i\\minus{}1}\\}$ e $ \\{a_{i\\plus{}1,i},\\cdots,a_{n,i}\\}$ having a common element.\n\nConsider the following sets:\n\n$ \\{a_{2,1}\\}$\n$ \\{a_{3,1},a_{3,2}\\}$\n$ \\cdots$\n$ \\{a_{n,1},\\cdots,a_{n,n\\minus{}1}\\}$\n\nSince there are $ n\\minus{}1>2^k\\minus{}1$ sets and $ \\{1,\\cdots,k\\}$ only has $ 2^k\\minus{}1$ non-empty subsets, by the pigeonhole principle we have that two of those sets are equal. Let they be $ \\{a_{i1},\\cdots,a_{i,i\\minus{}1}\\}$ e $ \\{a_{j1},\\cdots,a_{j,j\\minus{}1}\\}$, $ j>i$. Then $ a_{j,i}$ is in both sets (because they are equal), so $ \\{a_{i1},\\cdots,a_{i,i\\minus{}1}\\}$ e $ \\{a_{i\\plus{}1,i},\\cdots,a_{j,i},\\cdots,a_{n,i}\\}$ are not disjoint, as desired.[/hide]" } { "Tag": [ "graph theory", "combinatorics proposed", "combinatorics" ], "Problem": "Let $G = (V_1, V_2, E)$ be a connected bipartite graph whose vertices are marked with integers. Thus with $G$ we have a mapping $f: V_1 \\cup V_2 \\mapsto \\mathbb{Z}$ and with $x \\in V_1 \\cup V_2$ let $f(x)$ be the number assigned to the vertex $x$. Let $S_1 = \\sum_{x \\in V_1} f\\left(x\\right)$ and $S_2 = \\sum_{x \\in V_2} f\\left(x\\right)$.\n\nDefine $B$ as a step in which the numbers/markings $f(x)$ and $f(y)$ of the vertices $x,y$ of an edge $\\{x,y\\} \\in E$ are substituted by the values $f(x) + a$ and $f(y) + a$, respectively, where $a \\in \\{-1,1\\}$.\n\nDefine $Z_2$ as the state where all the vertices of the bipartite graph have the number/marking 0.\n\nUnder the assumption that we have $S_1 = S_2$ prove that state $Z_2$ can be reached by a finite application of steps $B$.\n\nA related problem goes as follows: Let $m,n \\geq 2$ be integers. To each of the fields of a $m \\times n$ checker board a non-negative integer is assigned. A elementary operation is defined by adding an integer to two neighboring numbers such that in both fields the result is non-negative. Provide a necessary and sufficient condition that by a finite number of elementary operations we obtain a state where 0 is assigned to every field on the checker board. Two fields are called neighbors if they have an edge in common.", "Solution_1": "some things seem to be missing....\r\nprobably you also need the graph to be connected (for instance in a graph with no edges,no steps are possible).", "Solution_2": "[quote=\"seshadri\"]some things seem to be missing....\nprobably you also need the graph to be connected (for instance in a graph with no edges,no steps are possible).[/quote]\r\n\r\nIt was obvious that I forgot to add the graph is connected but now the statement should be complete. :)", "Solution_3": "It is very simple, isn't it?\r\nBy removing some edges from $G$ we can obtain tree. Choose arbitrary vertex $x$ of degree 1 and by means of elementary operations make it 0, now exclude it from consideration. We again have tree and can repeat such procedure, and so on.\r\nFinally, we will obtain tree with only vertex. If $S_1=S_2$ then its marking should be 0.", "Solution_4": "Nice idea Myth. :) \r\n\r\nPierre.", "Solution_5": "In order to illustrate the connections between those checker board problems and graphs and the mapping among them I show an example. Depending on the definition of neighborhood we have:\r\n\r\na.) shows the mapping of checker boards to a graph if neighborhood is defined as the property two fields have an edge or a vertex in common\r\n\r\nb.) shows the mapping of checker boards to a bipartite graph if neighborhood is defined as the property two fields have an edge in common\r\n\r\nc.) shows the mapping of checker boards to a bipartite graph if neighborhood is defined as the property two fields have a vertex (diagonals) in common", "Solution_6": "attachment c.) And sorry I was too lazy to prepare a properly drawn diagram. Maybe next year. :P", "Solution_7": "In order to see the content of the images better just store it to your hard disk or open the image in a new window (right mouse button -> open in new window)." } { "Tag": [], "Problem": "Solve the following by substitution: $ \\frac {1}{x(x \\plus{} 2)} \\minus{} \\frac {1}{(x \\plus{} 1)^2} \\equal{} \\frac {1}{12},\\;x\\in\\mathbb{Z}^ \\plus{}$.", "Solution_1": "[quote=\"i_like_pie\"]Solve the following by substitution: $ \\frac {1}{x(x \\plus{} 2)} \\plus{} \\frac {1}{(x \\plus{} 1)^2} \\equal{} \\frac {1}{12},\\;x\\in\\mathbb{Z}^ \\plus{}$.[/quote]\r\n[hide] Say that $ y \\equal{} (x \\plus{} 1)^2$. Thus, we have that\n\\[ \\frac {1}{y \\minus{} 1} \\plus{} \\frac {1}{y} \\equal{} \\frac {1}{12}\n\\]\n\n\\[ \\frac {2y \\minus{} 1}{y(y \\minus{} 1)} \\equal{} \\frac {1}{12}\n\\]\n\n\\[ 24y \\minus{} 12 \\equal{} y^2 \\minus{} y\n\\]\n\n\\[ y^2 \\minus{} 25y \\plus{} 12 \\equal{} 0\n\\]\n\n\\[ y \\equal{} \\frac {25\\pm \\sqrt {625 \\minus{} 48}}{2}\n\\]\n\n\\[ y \\equal{} \\frac {25\\pm \\sqrt {577}}{2}\n\\]\n\n\\[ \\boxed{x \\equal{} \\minus{} 1\\pm \\sqrt {\\frac {25\\pm \\sqrt {577}}{2}}}\n\\]\n[/hide]\nEDIT: Ah, okay, that makes life quite a bit easier...\n[hide=\"Revised Solution\"]\nAgain, we substitute $ (x \\plus{} 1)^2 \\equal{} y$ to obtain that\n\\[ \\frac {1}{y \\minus{} 1} \\minus{} \\frac {1}{y} \\equal{} \\frac {1}{12}\n\\]\n\n\\[ \\frac {1}{y(y \\minus{} 1)} \\equal{} \\frac {1}{12}\n\\]\n\n\\[ y^2 \\minus{} y \\equal{} 12\n\\]\n\n\\[ (y \\minus{} 4)(y \\plus{} 3) \\equal{} 0\n\\]\n\n\\[ y \\equal{} 4, \\minus{} 3\n\\]\ny cannot be -3 or else x would be imaginary. Thus,\n\\[ (x \\plus{} 1)^2 \\equal{} 4\n\\]\n\n\\[ x \\plus{} 1 \\equal{} 2, \\minus{} 2\n\\]\n\n\\[ \\boxed{x \\equal{} 1, \\minus{}3}\n\\]\n[/hide]", "Solution_2": "Sorry, the plus was supposed to be a minus. :ninja:", "Solution_3": "[quote=\"i_like_pie\"]Solve the following by substitution: $ \\frac {1}{x(x \\plus{} 2)} \\minus{} \\frac {1}{(x \\plus{} 1)^2} \\equal{} \\frac {1}{12},\\;x\\in\\mathbb{Z}^ \\plus{}$.[/quote]\r\n[hide=\"solution\"]\nLet $ s$ = $ x \\plus{} 1$\n$ \\frac {1}{(s \\minus{} 1)(s \\plus{} 1)} \\minus{} \\frac {1}{s^2} \\equal{} \\frac {1}{12}, \\;s \\ne \\pm 1$\n$ \\frac {1}{s ^ 2 \\minus{} 1} \\minus{} \\frac {1}{s^2} \\equal{} \\frac {1}{12}$\n$ 12(s^2)(s^2 \\minus{} 1) \\cdot \\left[ \\frac {1}{s ^ 2 \\minus{} 1} \\minus{} \\frac {1}{s^2} \\equal{} \\frac {1}{12} \\right]$\n$ 12s^2 \\minus{} 12(s^2 \\minus{} 1) \\equal{} s^2(s^2 \\minus{} 1)$\n$ 12s^2 \\minus{} 12s^2 \\plus{} 12 \\equal{} s^4 \\minus{} s^2$\n$ s^4 \\minus{} s^2 \\minus{} 12 \\equal{} 0$\n$ (s^2 \\minus{} 4)(s^2 \\plus{} 3) \\equal{} 0$\nThe only real roots are in $ (s^2 \\minus{} 4) \\equal{} 0$\n$ s^2 \\minus{} 4 \\equal{} 0$\n$ (s \\plus{} 2)(s \\minus{} 2) \\equal{} 0$\n$ s \\equal{} \\pm2$\n$ s \\equal{} 2$ $ OR$ $ s \\equal{} \\minus{} 2$\n$ x \\plus{} 1 \\equal{} 2$ $ OR$ $ x \\plus{} 1 \\equal{} \\minus{} 2$\n$ x \\equal{} 1$ OR $ x \\equal{} \\minus{} 3$\n$ \\boxed{x \\equal{} \\minus{} 3,1}$\n[/hide]", "Solution_4": "ProtestanT's solution is correct, except for one little mistake...\r\n\r\n[hide=\"Solution\"]Given: $ \\frac {1}{x(x \\plus{} 2)} \\minus{} \\frac {1}{(x \\plus{} 1)^2} \\equal{} \\frac {1}{12}$.\n\nExpand the denominators to get $ \\frac {1}{x^2 \\plus{} 2x} \\minus{} \\frac {1}{x^2 \\plus{} 2x \\plus{} 1} \\equal{} \\frac {1}{12}$\n\nSubstitute $ u \\equal{} x^2 \\plus{} 2x$. The equation now becomes $ \\frac {1}{u} \\minus{} \\frac {1}{u \\plus{} 1} \\equal{} \\frac {1}{12}$.\n\n$ 12(u \\plus{} 1 \\minus{} u) \\equal{} u(u \\plus{} 1)\\quad\\to\\quad 12 \\equal{} u^2 \\plus{} u\\quad\\to\\quad (u \\minus{} 3)(u \\plus{} 4) \\equal{} 0$\n\n$ u$ is either $ \\minus{} 4$ or $ 3$. The former does not yield real solutions for $ x$, but the latter gives $ x \\equal{} \\{ \\minus{} 3,1\\}$.\n\nRemembering that $ x\\in\\mathbb{Z}^ \\plus{}$, we rule out $ x \\equal{} \\minus{} 3$.\n\nAnswer: $ x \\equal{} \\boxed{1}$.[/hide]", "Solution_5": "[quote=\"i_like_pie\"]ProtestanT's solution is correct, except for one little mistake...\n\n[hide=\"Solution\"]Given: $ \\frac {1}{x(x \\plus{} 2)} \\minus{} \\frac {1}{(x \\plus{} 1)^2} \\equal{} \\frac {1}{12}$.\n\nExpand the denominators to get $ \\frac {1}{x^2 \\plus{} 2x} \\minus{} \\frac {1}{x^2 \\plus{} 2x \\plus{} 1} \\equal{} \\frac {1}{12}$\n\nSubstitute $ u \\equal{} x^2 \\plus{} 2x$. The equation now becomes $ \\frac {1}{u} \\minus{} \\frac {1}{u \\plus{} 1} \\equal{} \\frac {1}{12}$.\n\n$ 12(u \\plus{} 1 \\minus{} u) \\equal{} u(u \\plus{} 1)\\quad\\to\\quad 12 \\equal{} u^2 \\plus{} u\\quad\\to\\quad (u \\minus{} 3)(u \\plus{} 4) \\equal{} 0$\n\n$ u$ is either $ \\minus{} 4$ or $ 3$. The former does not yield real solutions for $ x$, but the latter gives $ x \\equal{} \\{ \\minus{} 3,1\\}$.\n\nRemembering that $ x\\in\\mathbb{Z}^ \\plus{}$, we rule out $ x \\equal{} \\minus{} 3$.\n\nAnswer: $ x \\equal{} \\boxed{1}$.[/hide][/quote]\r\nI didn't get what the uppercase Z meant. Could you explain?", "Solution_6": "$ \\mathbb{Z}$ means the set of integers\r\n\r\n$ \\mathbb{R}$ means the set of real numbers\r\n\r\n$ \\mathbb{Z}^\\plus{}$ means positive integers\r\n\r\n$ \\mathbb{Z}^\\minus{}$ means negative integers.\r\n\r\netc...", "Solution_7": "$ \\mathbb{Z^\\plus{}}\\equal{}\\mathbb{N}$ isnt it ?", "Solution_8": "Yes and no. $ \\mathbb{N}$ generally does mean the same as $ \\mathbb{Z}^\\plus{}$, but there are some people who define $ \\mathbb{N}$ to be $ \\mathbb{Z}^* \\equal{} \\mathbb{Z}^\\plus{}\\cup\\{0\\}$ - the nonnegative integers. Use $ \\mathbb{Z}^\\plus{}$ and $ \\mathbb{Z}^*$ rather than $ \\mathbb{N}$ for clarity.", "Solution_9": "[quote=\"binomial_4eva\"]$ \\mathbb{Z^ \\plus{} } \\equal{} \\mathbb{N}$ isnt it ?[/quote]\r\n\r\nSomewhere $ \\mathbb{N}\\equal{}\\mathbb{Z}^\\plus{}$, and elsewhere $ \\mathbb{N}\\equal{}\\mathbb{Z}^\\plus{}_0$\r\n\r\nIn the system A, the set B is denoted $ \\mathbb{N}_0$, and in the system B, the set A is denoted $ \\mathbb{N}^*$\r\n\r\nEDIT: See above" } { "Tag": [ "geometry", "geometric transformation", "reflection", "HCSSiM" ], "Problem": "All right, we are starting a new game. But before we do, I have some rule change proposals:\r\n\r\nProposal A: Once a biohazard suit is bought, it is permanent.\r\nProposal B: Shields will protect against conventional attacks for the same duration, and can prevent the first missile attack that occurs in the duration.\r\nProposal C: Each state start with 35 points. \r\nProposal D: Each state has a maximum of 1 player.\r\nProposal E: Nuclear Defense Systems and SAM Rocket-Intercepts cost 35 instead of 50.\r\nProposal F: Nukes cost 40.\r\n\r\nHere is most of my reasoning.\r\n\r\n[quote=\"chenhsi\"]Anyway, some general comments on the game:\n\nShields are pointless: Conventional attacks are almost never used, and missiles are used instead. Maybe make the shield protect against rocket attacks as well, but only prevent a limited amount of damage?\n\nBiohazard Suits: Too expensive. Spending 25 points to protect against 5 damage? This should be changed so that once you buy the suit, you can NEVER get any nuke splash damage. This is still balance, since nukes are so rarely used.\n\nNuke Defense + Rocket Intercept: Rarely used. Consider making cheaper.\n\nNukes: Rarely used, since V2s can take out an undefended state, and a maxxed out alliance can do 9 damage with conventional attacks, for a total of 34 damage.\n\nMy recommendation: Start with more points, maybe 40. It is too easy to destroy an state with no players. 1 V2 eliminates a state. You saw how I used the combined forces of my alliance to destroy 6 states in 1 turn? Yeah.\n\nAlso, why don't you make it 1 person/state max? There isn't that many players. This way, we also don't have to check whether there are 3 players in a state, we only need to check for 1.[/quote]", "Solution_1": "A, accept\r\nB, partial accept. Once a missile hits a shield the shield should dissipate.\r\nC, accept\r\nD, accept\r\nE, cost should be 40.\r\nF, reject: nukes KILL a state instantly + possibly 25-40 extra splash damage.\r\n\r\nthose are my votes, waiting for others to check proposal...\r\n\r\nNew Proposals:\r\n\r\nProposal G: You may only buy one item per post [e.g. no buying 3 V2 Rockets to blow up states in one post]\r\nProposal H: 1-hour post limit, no double-posting\r\nProposal I: V2 cost = 25, V1 cost = 12\r\nProposal J: Kills give 3 pts to killer", "Solution_2": "B. Yes, I did mean that when a missile hits a shield the shield disappears. Sorry if I wasn't obvious.\r\nE. Sure.\r\nF. I meant that it should cost 40 to buy a nuke. How about 40 to buy, kills a state instantly and does 10 splash damage? 25-40 would probably kill any nearby states as well. 10 would severly weaken.\r\n\r\nG. No.\r\nH. No. I disagree with any rule that discourages being on AoPS more.\r\nI. Sure.\r\nJ. Sure.\r\n\r\nEdit: Proposal K: The game wins when only 1 alliance is left standing. If the alliance wants to disband and keep fighting, fine, but the states are not forced to kill each other.", "Solution_3": "How about adding mini-rockets? So that...\r\n\r\nV1 Rocket- 5 cost, 5 dmg\r\nV2 Rocket- 10 cost, 10 dmg\r\nV3 Rocket- 12 cost, 15 dmg\r\nV4 Rocket- 15 cost, 20 dmg\r\nV5 Rocket- 20 cost, 25 dmg\r\nV10 Rocket- 40 cost, 50 dmg", "Solution_4": "Mini-Nuke- 30 cost, 5 dmg everytime you post within 10 hours\r\nNuke- 50 cost, automatically kills a state, -10 to states surrounding the nuked state\r\nUltra-Nuke- 100 cost, automatically kills a state, kills the states surrounding that state", "Solution_5": "Shield- 5 cost, protects you from 1 point damage attack\r\nStrong Shield- 8 cost, protects you from 2 point damage attacks\r\nDay Shield- 10 cost, protects you from 3 point damage attacks\r\nRocket Shield- 15 cost, protects you from 1 rocket attack\r\nStrong Rocket Shield- 25 cost, protects you from 2 rocket attacks\r\nDay Rocket Shield- 40 cost, protects you from 3 rocket attacks\r\nBiohazard Defense System- 60 cost, protects you from nukes\r\nRocket Stealer- 60 cost, when you are attacked by a rocket, take that rocket and fire it back at the state that fired that rocket. works twice.", "Solution_6": "yay typing.\r\n\r\nA: accept\r\nB: accept\r\nC: accept\r\nD: neutral\r\nE: Accept\r\nF: neutral\r\nG: reject; if someone wants to do that, we can all gang up on him/her or his/her alliance and destroy\r\nH: partial reject; how about half-hour limits?\r\nI: neutral\r\nJ: accept\r\nK: accept\r\nErnie: mini rockets??? no two versions is enough. mini nuke is bad because someone can dish out 100 dmg max. Ultra nuke is a bit powerful, but it does come with a heavy cost. The only drawback is if there is a four person alliance with all 100 health, they can split the cost and kill 8 states max. They can do it two more times if they are crazy and G is rejected. Your shields, um some of them might not be rejected, but if a state has lots of hp, it can buy a biohazard defense system and a rocket stealer and pwn everybody if there is no expiration date.", "Solution_7": "I don't like ernie's items.", "Solution_8": "A: 3 accepts \r\nB: 3 accepts\r\nC: 3 accepts \r\nD: neutral, 2 accepts\r\nE: 3 accepts \r\nF: neutral, accept\r\nG: 2 rejects, accept\r\nH: partial reject, full reject, accept\r\nI: neutral, 2 accepts\r\nJ: 3 accepts\r\nK: 2 accepts\r\nErnie's Items: 2 Rejects\r\n\r\nJames, you need to vote on Item F (I know you rejected it, but I think you misunderstood the item), Item K, and Ernie's Items.\r\n\r\nErnie, you need to vote on all of them.\r\n\r\nI would like it if stevenmeow voted, since he was in the other game.", "Solution_9": "Partial reject to Prop F. 10 pts change for nuke price is not going to change the outcome of the war all that much.\r\n\r\nOn ernie's new items: normal shields are fine, rocket shields don't look fine, stealer is terrible. Mini-nuke: Don't get it. Nuke: Accept. Ultra-Nuke: Reject. WAY too powerful. V3-10 Rockets :too repetitive. Two versions are enough. Also if Prop G is axed out, then you could just \"stack\" multiple rockets to form one big rocket. \r\n\r\n[Ernie's Items final proposal: partial reject]\r\n\r\nProposal K: Neutral. Seriously I don't care if they want to keep going or not.\r\n\r\n\r\nNew Proposal L: Biohazard suits: They should ameliorate splash damage, not eliminate it. [Say maybe lower splash damage from 10 to 2?]\r\n\r\nThis is to reflect the fact that Biohazard suits don't 100% block out radiation.\r\n\r\n1 hr post limit => Weak Accept\r\n\r\n[code]\nProp.A..a..N..r..R..Description\nA 3 - - - - Permanent Biohazard Suits\nB 3 - - - - Missile-Preventing Shields\nC 3 - - - - 35pts start\nD 2 - 1 - - 1 player states\nE 3 - - - - 35pt cost for Rocket/Nuke defense\nF 1 - 1 1 - 40pt Nukes\nG 1 - - - 2 1 item per post\nH - 1 - 1 1 1hr post limit\nI 2 - 1 - - 12/25 cost for V1/V2 rockets\nJ 3 - - - - Kills give 3pts to killer\nK 2 - 1 - - Game ends at single alliance\nL 1 - - - - [NEW] Weaker Biohazard Suits\nE.I - - - 1 2 Addition of new items\n\nNote: E.I = Ernie's Items\nA = Accept, a = Weak/Partial Accept, N = Neutral, r = Weak/Partial Reject, R = Reject\n\n[/code]", "Solution_10": "Partial Reject to Proposal L.", "Solution_11": "[code]\nProp.A..a..N..r..R..Description\nA 3 - - - - Permanent Biohazard Suits\nB 3 - - - - Missile-Preventing Shields\nC 3 - - - - 35pts start\nD 2 - 1 - - 1 player states\nE 3 - - - - 35pt cost for Rocket/Nuke defense\nF 1 - 1 1 - 40pt Nukes\nG 1 - - - 2 1 item per post\nH - 1 - 1 1 1hr post limit\nI 2 - 1 - - 12/25 cost for V1/V2 rockets\nJ 3 - - - - Kills give 3pts to killer\nK 2 - 1 - - Game ends at single alliance\nL 1 - - 1 - [NEW] Weaker Biohazard Suits\nE.I - - - 1 2 Addition of new items\n\nNote: E.I = Ernie's Items\nA = Accept, a = Weak/Partial Accept, N = Neutral, r = Weak/Partial Reject, R = Reject\n\n[/code]", "Solution_12": "Erm, when is the Civil war actually going to take place?", "Solution_13": "The states are at peace right now, trying to figure out what weapons to allow and what weapons not to allow.\r\n\r\n :P", "Solution_14": "If chenhsi and ernie both live in CA, it would be more like state. In reality, states just attack when their opponent is still arguing about which items to allow.", "Solution_15": "+4 Califorina, -5 Georgia\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-40 pts\nCalifornia-42 pts\nColorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-25 pts\nFlorida-27 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-43 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_16": "+5pts alaska, -4 delaware\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-45 pts\nCalifornia-42 pts\nColorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-21 pts\nFlorida-27 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-43 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_17": "-9 Delaware\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-45 pts\nCalifornia-42 pts\nColorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-12 pts\nFlorida-27 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-43 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_18": "+9 California\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-45 pts\nCalifornia-51 pts\nColorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-12 pts\nFlorida-27 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-43 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_19": "-2 Delaware, -7 Florida\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-45 pts\nCalifornia-51 pts\nColorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-10 pts\nFlorida-20 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-43 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_20": "-9 Missouri\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-45 pts\nCalifornia-51 pts\nColorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-10 pts\nFlorida-20 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts[/color]\nMissouri-34 pts\n[color=green]Montana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_21": "+9 California\r\n\r\n[hide=\"states\"][color=blue]Alabama-39 pts\nAlaska-45 pts[/color]\n[color=orange]California-60 pts[/color]\n[color=blue]Colorado-55 pts\nConnecticut-39 pts[/color]\n[color=green]Delaware-10 pts\nFlorida-20 pts\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-34 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote][/quote]", "Solution_22": "Pythag already staked CA. pick another state\r\n\r\nAlaska: +9\r\n\r\nthen V2 delaware, florida\r\n\r\n[hide=\"states\"][color=blue]Alabama-41 pts\nAlaska-10 pts\nCalifornia-62 pts\nColorado-57 pts\nConnecticut-41 pts[/color]\n[color=green]\nGeorgia-30 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-34 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide]", "Solution_23": "+3 GA\r\nBuys V1, destroys AK, +3 GA for the kill. \r\nCasualties: \"13 people and some caribou.\"\r\n[hide=\"states\"]\n[color=blue]Alabama-41 pts[/color]\nAlaska-PWNT\n[color=blue]California-62 pts[/color]\n[color=blue]Colorado-57 pts[/color]\n[color=blue]Connecticut-41 pts[/color]\nDelaware-PWNT\nFlorida-PWNT\n[color=green]Georgia-26 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-34 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide]", "Solution_24": "...\r\n\r\nWhy did you just kill our ally???\r\n\r\n-7 Georgia\r\n\r\n[hide=\"states\"]\n[color=blue]Alabama-41 pts[/color]\nAlaska-PWNT\n[color=blue]California-62 pts[/color]\n[color=blue]Colorado-57 pts[/color]\n[color=blue]Connecticut-41 pts[/color]\nDelaware-PWNT\nFlorida-PWNT\n[color=green]Georgia-19 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-34 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote]", "Solution_25": "@Xantos: Sarah Palin will NOT be happy about this...", "Solution_26": "Note: Alaka not dead.\r\n\r\njames4l made a typo. I don't think Alaska had 10 points at the time: they had 45.", "Solution_27": "they bought two V2 rockets...", "Solution_28": "-7 Georgia\r\n\r\n[hide=\"states\"]\n[color=blue]Alabama-41 pts[/color]\nAlaska-PWNT\n[color=blue]California-62 pts[/color]\n[color=blue]Colorado-57 pts[/color]\n[color=blue]Connecticut-41 pts[/color]\nDelaware-PWNT\nFlorida-PWNT\n[color=green]Georgia-12 pts\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-34 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote]", "Solution_29": "Now we only get 7 points per turn :(\r\n\r\nI hate you Xantos...temporarily, anyway\r\n\r\n-2 Georgia, +5 Colorado, buy V1 rocket, fire at Georgia, gain points for kill\r\n\r\n[hide=\"states\"]\n[color=blue]Alabama-42 pts[/color]\n[color=blue]California-63 pts[/color]\n[color=blue]Colorado-53 pts[/color]\n[color=blue]Connecticut-42 pts[/color]\n[color=green]\nHawaii-35 pts\nIdaho-35 pts\nIllinois-35 pts\nIndiana-35 pts\nIowa-35 pts\nKansas-35 pts\nKentucky-35 pts\nLouisiana-35 pts\nMaine-35 pts\nMaryland-35 pts\nMassachusetts-35 pts\nMichigan-35 pts\nMinnesota-35 pts\nMississippi-35 pts\nMissouri-34 pts\nMontana-35 pts\nNebraska-35 pts\nNevada-35 pts\nNew Hampshire-35 pts\nNew Jersey-35 pts\nNew Mexico-35 pts\nNew York-35 pts\nNorth Carolina-35 pts\nNorth Dakota-35 pts\nOhio-35 pts\nOklahoma-35 pts\nOregon-35 pts\nPennsylvania-35 pts\nRhode Island-35 pts\nSouth Carolina-35 pts\nSouth Dakota-35 pts\nTennessee-35 pts\nTexas-35 pts\nUtah-35 pts\nVermont-35 pts\nVirginia-35 pts\nWashington-35 pts\nWest Virginia-35 pts\nWisconsin-35 pts\nWyoming-35 pts[/color][/hide][/quote][/quote][/quote]" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "[i]Author of the problem is Adrian Satja Kurdija (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 3 (Geometry) for 9th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]\r\n\r\nAn acute-angled triangle $ ABC$ is given. Bisector of angle $ \\measuredangle{BAC}$ intersects the side $ \\overline{BC}$ in point $ K$. Point $ L$ is the midpoint of segment $ \\overline{AK}$ and point $ N$ is the intersection of altitude from vertex $ A$ and side $ \\overline{BC}$. On side $ \\overline{AB}$, point $ P$ is such that $ \\measuredangle{KLP} \\equal{} \\measuredangle{BAC}$. Circumcircle of triangle $ ANP$ intersects side $ \\overline{AC}$ in point $ Q$. Prove that $ |AP| \\equal{} |AQ|$.", "Solution_1": "Because $ \\angle{PAL}\\equal{}\\frac{\\angle{BAC}}{2}\\equal{}\\frac{\\angle{PLK}}{2}$, triangle $ PLA$ is isosceles and $ AL\\equal{}PL$. But $ AL\\equal{}LN\\equal{}LK$, so $ L$ is circumcenter of $ \\triangle ANP$ and $ AK$ is diameter of this circle. Therefore $ \\angle{KPA}\\equal{}\\angle{ADK}\\equal{}90^\\circ$, but $ KP\\equal{}KD$, thus $ AP\\equal{}AD$.", "Solution_2": "Note that $ \\angle PLK\\equal{}\\angle APL\\plus{}\\angle PAL$, i.e., $ \\angle APL\\equal{}\\angle PLK\\minus{}\\angle PAL\\equal{}\\angle BAC\\minus{}\\angle PAL\\equal{}\\angle LAC\\equal{}\\angle PAL$. So $ \\angle APL\\equal{}\\angle PAL$ and hence $ AL\\equal{}PL$.\r\n\r\nConsider the circumcircle of triangle $ ANK$. Since $ \\angle ANK\\equal{}90^{\\circ}$, by Thales' Theorem, $ L$ is the center of the circle, and hence $ LK\\equal{}LN\\equal{}LA$. So $ LA\\equal{}LP\\equal{}LN\\equal{}LK$. Thus, the circumcircle of triangle $ APN$ coincides with the circumcircle of triangle $ ANK$. Now we have $ LP\\equal{}LQ$. Consider triangle $ APL$ and triangle $ ALQ$. Since $ LP\\equal{}LQ,LA\\equal{}LA,\\angle PAL\\equal{}\\angle LAQ$, they are congruent, which implies $ AP\\equal{}AQ$." } { "Tag": [], "Problem": "How do I factor P^4+5P^2+6?", "Solution_1": "[quote=\"Interval\"]How do I factor P^4+5P^2+6?[/quote]\r\n\r\n[hide]Assume that $P^{2}=a$, so now we have $a^{2}+5a+6$.\n\nSo this can be written as $(a+3)(a+2)$.\n\nSubstituting, we get $(P^{2}+3)(P^{2}+2)$[/hide]", "Solution_2": "I like the way you factored that question.", "Solution_3": "Thanks, but that was pretty much the only way you could do it." } { "Tag": [ "ratio", "AMC", "USA(J)MO", "USAMO", "email" ], "Problem": "The poll should be answered as a fraction of the height between two consecutive lines on notebook paper (the US kind that has 33 (not 26) lines).\r\n\r\nAlso answer these questions if you feel like it:\r\n-Do people sometimes look at your paper and think it is blank when you have actually written all over it?\r\n-Does your English teacher assign a 2 page written essay to everyone, except you, who are assigned a 5 page essay?\r\n-What is the width of your letter \"b\"s as a fraction of the line height?\r\n-Do you write in the margin?\r\n-Can you normally distinguish between normal and italics while writing?\r\n-How sharp is your pencil, or do you prefer mechanical?\r\n-Do you write your lower case \"a\"s like alphas (computer font) or the other o-with-a-tail way?\r\n-Do you cross any of your 7's, 0's, z's, or q's?\r\n-What is the height of your lower case \"a\"s as a fraction of the height of your letter \"b\"s?\r\n-How dark do you write?", "Solution_1": "I write very very skinny laterally, but generously heightwise. I'd say 2/3 the width of most people. I think the reason is because I write cursive, and I like to make my curves very elliptical. Sometimes they turn sharp which doesn't look good, but whatever, it's fast.", "Solution_2": "well..\r\nIf i'm in a rush, i stretch out all my letters and it looks like scribbles.\r\n\r\nteehee\r\n\r\nif i'm writing nicely, my b's are pretty tall and my handwriting is fat.\r\n\r\ni make my a's not the computer way.\r\ni don't cross any 7s, 0s, zs, or qs.\r\nmy lowercase a's are about 2/3 of my b's.\r\nand i press pretty firmly most of the time, so it's pretty dark.", "Solution_3": "Yay! People responded!\r\n\r\nMy b's have a 5:3 height:width ratio :lol: . The b's height is 1/2 the line height (most people think this is small). The darkness of my writing is inversely proportional to the sharpness of the pencil. I write medium-dark with a pen. I cross my 7's, z's and q's.\r\n\r\nAnother question: Dot i first or write i first? I write the i first.", "Solution_4": "My handwriting is okay. Sometimes my pencil can get pretty dull, but I try to sharpen them regularly. Ideally, my pencil will be somewhat sharp, but not fresh out of the sharpener. I cross my z's, but nothing else. I stay pretty much in the margins.", "Solution_5": "argh i hate my handwriting. my teachers can read it barley, but on serious stuff like USAMO i have to spend SO MUCH TIME trying to get it right. I even had to draw lines to not mess up...", "Solution_6": "[quote=\"tjhance\"]I stay pretty much in the margins.[/quote]\r\n\r\nReally? Wow. :wink:", "Solution_7": "I fit the size of my writing to the space between lines on the page. But generally, a \"b\" takes up about 7/8 of the line height.\r\n\r\nWhen I'm not in a rush, my handwriting is really neat. But when I am, say, for note taking, it's pretty messy.\r\n\r\nI like writing with pens. If I have to use a pencil, I really don't care what kind I use.\r\nI write my lowercase a's in the \"O-with-a-tail\" style.\r\nI cross my z's and sometimes I put a \"tail\" on my q's. I don't cross my 7's or 0's, but in my neat, not-in-a-rush handwriting, my 7's have a \"tail\" in the top-left, like in Roman font.\r\n\r\nWhen it's not lined paper (like the USAMO), I tend to write in the size I think will fit my desperate attempt at a solution. I tend not to draw lines in case I have to draw a diagram. Lined paper can sometimes get in the way of diagrams, depending on how dark the lines are.", "Solution_8": "Seems like no one gets what I mean by crossing your q's.... my Spanish teacher is the only person I know who does it, and I got it from her.", "Solution_9": "I think some of those poll options aren't going to come up. Seriously, who writes b's 1/8 of the height of the space between the lines?", "Solution_10": "[quote=\"K81o7\"]I think some of those poll options aren't going to come up. Seriously, who writes b's 1/8 of the height of the space between the lines?[/quote]\r\n\r\nLike 5/4 as well. I don't think anyone goes above the lines on every single letter.", "Solution_11": "I am about 2/3\r\nbut when I write fast I am about 4/5", "Solution_12": "Wow, nice little bell curve looking thingy going on there.\r\n\r\n[quote=\"moogra\"][quote=\"K81o7\"]I think some of those poll options aren't going to come up. Seriously, who writes b's 1/8 of the height of the space between the lines?[/quote]\n\nLike 5/4 as well. I don't think anyone goes above the lines on every single letter.[/quote]\r\n\r\nReally? It's strange that there aren't many extremes to me. There are two people at my school who write really small. I would say their writing is closer to 1/8 than 1/4. The title/heading of their papers is about 1/4... Both of them are very, very, consistent in their handwriting size.\r\n\r\nThen there's that girl in my Spanish class who has permission from the teacher to write her answers on a separate sheet of notebook paper instead of the worksheets... I'd say her writing is at a 2.5.", "Solution_13": "[quote=\"leoxnlin\"]Seems like no one gets what I mean by crossing your q's.... my Spanish teacher is the only person I know who does it, and I got it from her.[/quote]\r\n\r\nI cross my q's.\r\n\r\nI cross the 7 and put the upper tail, too. My b is about 7/8 or 4/5 (normally. If I want to make a good writing, about 3/5 :rotfl: and if writing with hurry, 5/4). I NEVER cross the 0's, except if I need to write a capitalized O next to a 0. I hate margins, and I hate the test sheets where the line of \"email\" is as short as date line. IT SUCKS!", "Solution_14": "[quote=\"pacman2812\"]\nI cross the 7 and put the upper tail, too.[/quote]\r\n\r\nDo you give your 1's an upper tail? A baseline?", "Solution_15": "my handwriting is fairly even, not too tall, not too fat, i write my g's kind of like those in times new roman font (two loops), and i cross my z's, 7's, q's, and occasionally (rarely) my 0's. \r\n\r\nwhen i'm very rushed, i write in spikey italics. oh. and i never capitalize (unless i'm taking an english test, in which case i still don't capitalize my name)" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Let $a,b,c>1$.Prove that $(log_b a+log_c a).log_a (\\frac{b+c}{2})\\geq2$ ;)", "Solution_1": "This problem seems very cute.\r\n\r\nBy AM-GM, $log_a(\\frac{b+c}{2}) \\geq log_a(\\sqrt{bc}) = \\frac{1}{2}(log_a(b)+log_a(c)).$\r\n\r\nThus, $(log_b(a)+log_c(a))log_a(\\frac{b+c}{2}) \\geq \\frac{1}{2}(log_b(a)+log_c(a))(log_a(b)+log_a(c)) = $\r\n$\\frac{1}{2}(log_b(a)*log_a(b)+log_b(a)*log_a(c)+log_c(a)*log_a(b)+log_c(a)*log_a(c)).$ \r\nBy log rules, we can simplify this to $1+\\frac{1}{2}(log_b(c)+log_c(b)).$ By a final application of AM-GM, this quantity is greater than or equal to $1+\\sqrt{log_b(c)*log_c(b)} = 1+1 = 2.$ In conclusion, $(log_b(a)+log_c(a))log_a(\\frac{b+c}{2}) \\geq 2.$ QED" } { "Tag": [ "linear algebra", "matrix", "complex numbers", "real analysis", "real analysis unsolved" ], "Problem": "I am stuck with something that must be a known result that I have forgotten..\r\n\r\nIs it true that given a_i,b_i (complex in general)\r\n\r\nSum 1_N a_i*exp(b_i*n) = Sum 1_M a'_j*exp(b'_j*n)\r\n\r\nfor n = 0,1,..2*M-1 (so that we have enough equations to compute a'_j, b'_j)\r\n\r\n=> (a'_j,b'_j) = (a_i,b_i) ie the exponents and coefficient need to be the same\r\nand so this is possible only if M>=N and if M>N some (a'_j,b'_j) =(0,0)\r\n\r\nIf so how to prove this?\r\nPlease help\r\nNick", "Solution_1": "You should be careful here: adding $2\\pi i$ to any of $b_j$ won't change the value of the exponent at any integer point. Also, you probably assume that $M\\ge N$ (otherwise you have more variables $a_j, a'_j$ in your linear system than equations). Also, of course, you have stupid identities like $2e^n=e^n+e^n$.\r\nSo, let us restate your question as follows:\r\n\r\nLet $a_k$, $b_k$ ($1\\le j\\le K$) be complex numbers, let $b_j$ be all different and satisfy $-\\pi\\le\\Im b_j<\\pi$. Is it true that the conditions $\\sum_{j=1}^K a_j e^{b_j n}=0$ ($0\\le n\\le K-1$) imply that all $a_j=0$?\r\n\r\nThen the answer is yes. All you need to do is to look at these equations as at a linear system of $K$ equations with $K$ unknowns and recall how to compute the determinant of the matrix of the system (the Vandermonde determinant).", "Solution_2": "Sorry for the double post and thnx for the answer.It was clear but I had posted it before this one and forgot about that.My excuses.Ok??\r\nNick" } { "Tag": [ "geometry", "geometric transformation", "rotation", "linear algebra", "matrix", "induction" ], "Problem": "If $ A_{(\\theta)}$ is a rotational matrix, show that for angles $ \\alpha$ and $ \\beta$ we have\r\n\r\n$ A_{(\\alpha)}\\cdot A_{(\\beta)}\\equal{}A_{(\\alpha\\plus{}\\beta)}$", "Solution_1": "You must mean to restrict this question to $ 2\\times 2$ matrices, right? Otherwise, it wouldn't be clear what $ \\alpha,$ $ \\beta,$ or $ \\alpha \\plus{} \\beta$ would mean.", "Solution_2": "[hide=\"Hint\"]$ A_{(\\theta)} \\equal{} \\begin{bmatrix}cos(\\theta) & \\minus{} sin(\\theta) \\\\\nsin(\\theta) & cos(\\theta)\\end{bmatrix}$\n $ A_{(\\alpha)}\\cdot A_{(\\beta)} \\equal{} \\begin{bmatrix}cos(\\alpha) & \\minus{} sin(\\alpha) \\\\\nsin(\\alpha) & cos(\\alpha)\\end{bmatrix}\\begin{bmatrix}cos(\\beta) & \\minus{} sin(\\beta) \\\\\nsin(\\beta) & cos(\\beta)\\end{bmatrix} \\equal{} \\begin{bmatrix}cos(\\alpha \\plus{} \\beta) & \\minus{} sin(\\alpha \\plus{} \\beta) \\\\\nsin(\\alpha \\plus{} \\beta) & cos(\\alpha \\plus{} \\beta)\\end{bmatrix} \\equal{} A_{\\alpha \\plus{} \\beta}$[/hide]", "Solution_3": "[quote=\"Kent Merryfield\"]You must mean to restrict this question to $ 2\\times 2$ matrices, right? Otherwise, it wouldn't be clear what $ \\alpha,$ $ \\beta,$ or $ \\alpha \\plus{} \\beta$ would mean.[/quote]\n\nYes, didn't think I needed to mention this, sorry.\n\n[quote=\"QuyBac\"][hide=\"Hint\"]$ A_{(\\theta)} \\equal{} \\begin{bmatrix}cos(\\theta) & \\minus{} sin(\\theta) \\\\\nsin(\\theta) & cos(\\theta)\\end{bmatrix}$\n $ A_{(\\alpha)}\\cdot A_{(\\beta)} \\equal{} \\begin{bmatrix}cos(\\alpha) & \\minus{} sin(\\alpha) \\\\\nsin(\\alpha) & cos(\\alpha)\\end{bmatrix}\\begin{bmatrix}cos(\\beta) & \\minus{} sin(\\beta) \\\\\nsin(\\beta) & cos(\\beta)\\end{bmatrix} \\equal{} \\begin{bmatrix}cos(\\alpha \\plus{} \\beta) & \\minus{} sin(\\alpha \\plus{} \\beta) \\\\\nsin(\\alpha \\plus{} \\beta) & cos(\\alpha \\plus{} \\beta)\\end{bmatrix} \\equal{} A_{\\alpha \\plus{} \\beta}$[/hide][/quote]\r\n\r\nI see. If I wanted to prove by induction that $ [A_{(\\alpha)}]^n \\equal{} A_{(n\\alpha)}$ a simple induction would suffice, correct?", "Solution_4": "[quote=\"triplebig\"] \n I see. If I wanted to prove by induction that $ [A_{(\\alpha)}]^n \\equal{} A_{(n\\alpha)}$ a simple induction would suffice, correct?[/quote]\r\n See $ A_{(\\alpha)}\\cdot A_{(\\beta)}\\equal{}A_{(\\alpha\\plus{}\\beta)}$ you can see that with angles $ \\alpha_1,\\alpha_2,...,\\alpha_n$\r\n We have \r\n$ A_{(\\alpha_1)}\\cdot A_{(\\alpha_2)}\\cdot .... \\cdot A_{\\alpha_n}\\equal{}A_{(\\alpha_1\\plus{}\\alpha_2\\plus{}...\\plus{}\\alpha_n)}$\r\nIt is easy see $ \\alpha_1\\equal{}\\alpha_2\\equal{}...\\equal{}\\alpha_n\\equal{}\\alpha$ It is $ [A_{(\\alpha)}]^n \\equal{} A_{(n\\alpha)}$", "Solution_5": "[quote=\"Kent Merryfield\"] Otherwise, it wouldn't be clear what $ \\alpha,$ $ \\beta,$ or $ \\alpha \\plus{} \\beta$ would mean.[/quote]\r\n Do you can define $ B_{\\alpha} \\equal{} \\begin{bmatrix}I_s & 0 \\\\\r\n0 & A_{\\alpha}\\end{bmatrix} \\in M_{s \\plus{} 2}(\\mathbb{K})$ ? ,$ K \\equal{} \\mathbb{C}$ or $ \\mathbb{R}$ , $ I_s$ is $ s$x$ s$ identity matrix.\r\n Such that $ B_{\\alpha}B_{\\beta} \\equal{} B_{\\alpha \\plus{} \\beta}$ for angles $ \\alpha$ and $ \\beta$ \r\nThanks !" } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove that the following three conditions are all equivalent for a real 3x3 symetric \r\nmatrix $A$, whose eigenvalues are $a,b,c$\r\n\r\n1/ $tr(A)$ is not an eigenvalue of $A$ \r\n\r\n2/ $(a+b)(b+c)(a+c)\\neq 0$\r\n\r\n3/ The map $L:S\\rightarrow S$ is an isomorphism where $S$ is the set of 3x3 real skew-symmetric \r\n\r\nmatrices and $L(W)=AW+WA$", "Solution_1": "1)$\\iff$ 2) : is evident: since $tr(A)=a+b+c$ we see that $a+b+c$ not in set ($a,b,c$) $\\iff$ $(a+b)(b+c)(a+c)\\neq 0$\r\nAbout 3) i haven't fully understand it yet: you mean that if $L(W)=AW+WA$ the $L$ send $S$ to $S$, it it is, then it also evident $L(W)^{T}=(AW)^T+(WA)^T=-AW-WA$, i think that you meant something else...?", "Solution_2": "I edited for 3/", "Solution_3": "In this case, all that's left to prove is that all antisymmetric matrices (I don't like \"skew-symmetric\" :)) have the form $AW+WA$. We can use the fact that $A$ is orthogonally diagonalizable, so we can work with $U^tAU$ and $U^tWU$ instead of $A,W$ respectively, where $U$ is an orthogonal matrix which diagonalizes $A$. In other words, we may assume that $A$ is diagonal.\r\n\r\nWe must thus prove that all antisymmetric matrices can be written as $((\\lambda_i+\\lambda_j)a_{ij})_{i,j}\\ (*)$, where $(a_{ij})_{i,j}$ is antisymmetric. This is clear.\r\n\r\nOn the other hand, if we asume $3$, we can derive the other parts by using the same idea (if we have $\\lambda_i+\\lambda_j=0$ for some $i\\ne j$, the the antisymmetric matrices we get from must have some $0$ entries which are not on the main diagonal, and this means that we can't get all of them)." } { "Tag": [ "inequalities", "calculus", "derivative", "calculus computations" ], "Problem": "hello ^^\r\nlets $ x$,$ y$y and $ z$ three numbers for $ \\mathbb{R}$,find the minimum value for:\r\n$ S\\equal{}\\Large{\\frac{x^2}{x\\plus{}yz}\\plus{}\\frac{y^2}{y\\plus{}zx}\\plus{}\\frac{z^2}{z\\plus{}xy}}$", "Solution_1": ".but what if x,y,z were getting closer and closer to -1 from below?(infinitessimal)? then wouldnt the expressionn go to zero since the denominator is always larger than the numerator? here is my logic.\r\n\r\nfor some reason, because of the symetry, i thought that the minimum would be at some point where x=y=z because each can be interchanged with another(not entirely rigorous, please correct if wrong) so i substituted every variable with x, and then simplified to get\r\n\r\n$ S\\equal{}3(x^2/(x^2\\plus{}x))$, which simplifies to \r\n$ S/3\\equal{}x/(x\\plus{}1)$, and of course minimizing $ S$ is the same thing as minimizing $ S/3$.\r\nso....x/(x+1)=1-1/(x+1). if we want to minimize this, we want to max. the second term, which means we should get the denom as close to zero, meaning x should be as close to -1 as possible...same for y and z b/c of symetry.", "Solution_2": "[quote]thought that the minimum would be at some point where x=y=z [/quote]\r\n\r\nCan you give any mathematical explanation for this ?", "Solution_3": "The sum can be made arbitrarily large negative- look at $ x\\equal{}y\\equal{}z\\equal{}\\minus{}1\\plus{}\\epsilon$.\r\n\r\nThis might be more interesting restricted to positive numbers." } { "Tag": [ "induction", "modular arithmetic" ], "Problem": "Prove that for positive integer $ n,\\ 2^{6n \\minus{} 5} \\plus{} 3^{2n}$ is the multiple of 11.", "Solution_1": "[quote=\"kunny\"]Prove that for positive integer $ 2^{6n \\minus{} 5} \\plus{} 3^{2n}$ is the multiple of 11.[/quote]\r\n\r\n$ 11|2^{6n \\minus{} 5} \\plus{} 3^{2n} \\Leftrightarrow 11|2^{6n}\\plus{}3^{2n}(32)$\r\n$ \\Leftrightarrow 11|(\\minus{}3)^{2n}\\plus{}3^{2n}(32)\\equal{}3^{2n}(33)$, which is obviously true.", "Solution_2": "[hide]induction[/hide]", "Solution_3": "how specific :)\r\n\r\nbut that should be easy...", "Solution_4": "Well,\r\n[hide]\n$ 2^{6n \\minus{} 5} \\plus{} 3^{2n}$\n\n$ \\equiv 2 \\cdot 2^{6n \\minus{} 6} \\plus{} 9^n$\n\n$ \\equiv 2 \\cdot 64^{n \\minus{} 1} \\plus{} 9^n$\n\n$ \\equiv 2 \\cdot ( \\minus{} 2)^{n \\minus{} 1} \\plus{} ( \\minus{} 2)^n$\n\n$ \\equiv 2 \\cdot ( \\minus{} 2)^{n \\minus{} 1} \\plus{} ( \\minus{} 2) \\cdot ( \\minus{} 2)^{n \\minus{} 1}$\n\n$ \\equiv 0 \\pmod{11}$[/hide]", "Solution_5": "[quote=\"kunny\"]Prove that for positive integer $ n,\\ 2^{6n - 5} + 3^{2n}$ is the multiple of 11.[/quote]\r\n\r\n\r\nIt is also proved by Binomial exp.\r\n[hide]\n\n$ 2^{6n - 5} + 3^{2n}\\\\\n=2(2^{6n - 6}) + 9^{n}\\\\\n=2(64^{n-1}) +9^{n}\\\\\n=2 (66-2)^{n-1}+(11-2)^{n}\\\\\n=2( {}_{n-1}C_066^{n-1} -\\cdots+(-1)^{n-1}{}_{n-1}C_{n-1}2^{n-1})+({}_{n}C_0 11^{n-1} -\\cdots+(-1)^{n}{}_{n}C_{n} 2^{n})\\\\\n\\\\=11\\cdot N +(-1)^{n-1}{}_{n-1}C_{n-1}2^{n}+(-1)^{n}{}_{n}C_{n} 2^{n}\\\\\n\\\\=11\\cdot N +(-1)^{n-1}2^{n}+(-1)^{n} 2^{n}\\\\\n\\\\=11\\cdot N +(-1)^{n}(-2^{n}+ 2^{n})\\\\\n\\\\=11\\cdot N$\nwhere , $ N$ is some integer.\n[/hide]\r\n\r\n :roll:", "Solution_6": "[quote=\"abcak\"]how specific :)\n\nbut that should be easy...[/quote]\r\n$ *$.For $ n\\equal{}1, 11|11$\r\nSo we can assume that for a $ n\\equal{}k$ we have:\r\n$ 2^{6k\\minus{}2}\\plus{}3^{2k}\\equal{}11m$\r\n\r\n$ **$.For $ n\\equal{}k\\plus{}1$.We have to prove that:\r\n$ 11|2^{6(k\\plus{}1)\\minus{}2}\\plus{}3^{2(k\\plus{}1)}$\r\nBut $ 3^{2(k\\plus{}1)}\\equal{}9 \\cdot 3^{2k}\\equal{}9(11m\\minus{}2^{6k\\minus{}2})$\r\n\r\nSo \r\n$ 2^{6(k\\plus{}1)\\minus{}1}\\plus{}3^{2(k\\plus{}1)}\\equal{}2^{6(k\\plus{}1)\\minus{}2}\\plus{}9(11m\\minus{}2^{6k\\minus{}2})\\equal{}2^6 \\cdot 2^{6k\\minus{}2}\\plus{}99m\\minus{}9 \\cdot 2^{6k\\minus{}2}\\equal{}11(5 \\cdot 2^{6k\\minus{}2}\\plus{}9m)$", "Solution_7": "$ 2^8\\equiv 3\\mod 11$ so\r\n\r\n$ 2^{6n\\minus{}5}\\plus{}3^{2n}\\equiv 2^{6n\\minus{}5}\\plus{}2^{16n}\\equiv 2^{6n\\minus{}5}\\plus{}2^{6n}\\equal{}2^{6(n\\minus{}1)}(2\\plus{}2^6)\\equiv 0\\mod 11$\r\n\r\n(comment: $ 2$ is a generator mod $ 11$)", "Solution_8": "[quote=\"Altheman\"](comment: $ 2$ is a generator mod $ 11$)[/quote]\r\nWhat is a generator (mod 11)?", "Solution_9": "Some number $ a$ such that $ a, a^2, a^3, \\ldots$ forms a complete residue class mod 11.", "Solution_10": "Is it true that $ a$ is a generator $ \\pmod{n}$ if and only if $ \\gcd(a,n) \\equal{} 1$?", "Solution_11": "No? Consider a=5, n=6.\r\n\r\n$ a^{e}\\equiv5\\pmod{6}$ for odd e and $ a^e\\equiv{1}\\pmod{6}$ for even e. \r\n\r\nAren't generators basically equivalent to primitive roots?" } { "Tag": [ "LaTeX", "articles" ], "Problem": "What are the different document classes in LaTeX and how are they used?\r\n\r\n(e.g. \\documentclass{article} can be used to make:\r\n\r\n[code]\n\n\\documentclass[12pt]{article} \n\\usepackage{amsmath, amsthm}\n\\author{name} \n\\date{\\today} \n\\title{Title} \n \n\\begin{document} \n\\maketitle \n\\section{Exercise 3-1}\n$1+2=3$\n\\end{document}[/code]", "Solution_1": "Peter Flynn's [url=ftp://ftp.tex.ac.uk/tex-archive/info/beginlatex/html/chapter3.html#basic]Formatting Information: The Document Class Declaration[/url]", "Solution_2": "Hi,\r\n\r\nhave a look at\r\n[list][*] [url=http://www.public.asu.edu/~rjansen/latexdoc/ltx-22.html]Document Class[/url] (LaTeX Hypertext Help)\n[*] [url=http://www.math.uiuc.edu/~hildebr/tex/packages.html]LaTeX tips: Document classes and packages[/url]\n[*] [url=http://texblog.wordpress.com/2007/07/09/documentclassbook-report-article-or-letter/]\\documentclass{book, report, article or letter}[/url] for another short summary.[/list]\r\nThe standard document classes are described by most LaTeX books and introductions, you could also just choose an online text from this list: [url=http://texblog.net/latex-link-archive/introduction-guide/]Introductions, Guides[/url].\r\n\r\nStefan" } { "Tag": [ "\\/closed" ], "Problem": "When you look at the results of a poll now, the bars that show (relatively) how many people get proportionately wider as they get longer (the bars undergo a geometric dilation/homothecy rather than merely extend, as they should).\r\n\r\nYou can view very many samples of this [url=http://www.artofproblemsolving.com/Forum/index.php?f=19]here[/url], unless this is just me.\r\n\r\nI'm using Internet Explorer (the current one, I think), and Windows XP.\r\n\r\nI'll also include a poll here, so you can look at the bug right in this thread (I've set it to self-destruct in 10 days). Please vote, so we can get a good specimen here.\r\n\r\nEDIT: with one vote, looking at the poll results almost makes my screen look as though I'm using a monitor with 10x15 resolution, or something like that.\r\n\r\nEDIT 2: Hmm. With my first edit, the poll vanished. I shall recreate it...", "Solution_1": "In my opinion, the odd thing is the title of the thread is under the poll box. In the old mathlinks, the title is always on the top.", "Solution_2": "I don't get it. What's the bug? :maybe:", "Solution_3": "I don't see anything strange...I'm using Firefox though.", "Solution_4": "It depends what computer you are on. The polls look fine for the computer I am on now, but for they look odd (as he says) on my moms older laptop. It is an IBM thinkpad, I think.\r\n\r\nThe polls can take up a lot of room on the pages on some computers, and I think this is mainly a problem on older computers, or computers that don't have some plugin (but my mom's laptop can run the classroom).", "Solution_5": "[quote=\"Valentin Vornicu\"]I don't get it. What's the bug? :maybe:[/quote]\r\n\r\nWell... it looks like this (to me):\r\n\r\nTake a look at one of the answers with no responses. See how, between the two colored nubs on either end, there's a layer (the part that changes length according to how many votes the answer gets) one pixel wide, and some number of pixels tall?\r\n\r\nWhen I view the polls, that layer doesn't just get wider as the answer gets more responses relative to the other responses, it gets proportionally taller as well. So, whereas the layer should be six pixels (or however many pixels it is) high, it ends up being six times as high as it is wide; rather than simply extend the layer, it is homothetically dilated.\r\n\r\nI'd post a picture, but the only way I'd know how is by copying it into a MS Word document and posting that as an attachment, and I don't really want to do that with the dial-up I'm using.\r\n\r\nEDIT: Okay, the layer's height is naturally more like 12 pixels, so when it gets to be, say, two inches long on my screen, it also becomes 2 feet long on my screen.", "Solution_6": "How about posting the picture itself as an attachment?", "Solution_7": "[quote=\"Valentin Vornicu\"]How about posting the picture itself as an attachment?[/quote]\r\n\r\nI don't know how. I'm not exactly an expert at computer matters...", "Solution_8": "[quote=\"dts\"][quote=\"Valentin Vornicu\"]I don't get it. What's the bug? :maybe:[/quote]\n\nWell... it looks like this (to me):\n\nTake a look at one of the answers with no responses. See how, between the two colored nubs on either end, there's a layer (the part that changes length according to how many votes the answer gets) one pixel wide, and some number of pixels tall?\n\nWhen I view the polls, that layer doesn't just get wider as the answer gets more responses relative to the other responses, it gets proportionally taller as well. So, whereas the layer should be six pixels (or however many pixels it is) high, it ends up being six times as high as it is wide; rather than simply extend the layer, it is homothetically dilated.\n\nI'd post a picture, but the only way I'd know how is by copying it into a MS Word document and posting that as an attachment, and I don't really want to do that with the dial-up I'm using.\n\nEDIT: Okay, the layer's height is naturally more like 12 pixels, so when it gets to be, say, two inches long on my screen, it also becomes 2 feet long on my screen.[/quote]\r\nSame here. The results, instead of going horizontally, go vertically and wider, too.", "Solution_9": "[quote=\"dts\"][quote=\"Valentin Vornicu\"]How about posting the picture itself as an attachment?[/quote]\n\nI don't know how. I'm not exactly an expert at computer matters...[/quote]If you are using Firefox or Opera there's an Add an Attachment link. If you are using IE you should be able to see a Browse button and an Add Attachment button. \r\n\r\nBrowse to the file, push the add attachment button once you have selected the file and submit the post :)", "Solution_10": "[quote=\"Valentin Vornicu\"][quote=\"dts\"][quote=\"Valentin Vornicu\"]How about posting the picture itself as an attachment?[/quote]\n\nI don't know how. I'm not exactly an expert at computer matters...[/quote]If you are using Firefox or Opera there's an Add an Attachment link. If you are using IE you should be able to see a Browse button and an Add Attachment button. \n\nBrowse to the file, push the add attachment button once you have selected the file and submit the post :)[/quote]\r\n\r\nI know how to do [i]that[/i], I just don't know how to take the picture.", "Solution_11": "[quote=\"dts\"][/quote][quote=\"Valentin Vornicu\"][quote=\"dts\"][quote=\"Valentin Vornicu\"]How about posting the picture itself as an attachment?[/quote]\n\nI don't know how. I'm not exactly an expert at computer matters...[/quote]If you are using Firefox or Opera there's an Add an Attachment link. If you are using IE you should be able to see a Browse button and an Add Attachment button. \n\nBrowse to the file, push the add attachment button once you have selected the file and submit the post :)[/quote][quote=\"dts\"]\n\nI know how to do [i]that[/i], I just don't know how to take the picture.[/quote]\r\nTry pressing the \"Prnt Scrn\" key on your keyboard. It takes a snapshot of what's currently on your monitor and places it in the clipboard. From there, you can do copy and paste to put the picture in some sort of document.", "Solution_12": "Press print screen, then open paint application and go to Edit >> Paste. Then save as .jpg and make the dimensions smaller (like 50% for less than 100kB file).", "Solution_13": "Alright.\r\n\r\nIf it's okay, though, I think I'll wait until we get high-speed internet (should happen next week). I'm using 26k dial-up right now, and I really don't want to think about how long that'll take with this internet connection...", "Solution_14": "[quote=\"dts\"]I'm using 26k dial-up right now, and I really don't want to think about how long that'll take with this internet connection...[/quote]About 8 seconds.", "Solution_15": "[quote=\"dts\"] When I view the polls, that layer doesn't just get wider as the answer gets more responses relative to the other responses, it gets proportionally taller as well. So, whereas the layer should be six pixels (or however many pixels it is) high, it ends up being six times as high as it is wide; rather than simply extend the layer, it is homothetically dilated.[/quote]\r\nit looks like that on my computer too. the bars that show how many people voted for each thing get tall and fat. it's probably not a huge problem though, if they look fine on other computers.", "Solution_16": "Yes, this is an IE6 bug. I will look at it on Monday.\r\n\r\n(You do not have to post a screenshot -- I know what's going on.)", "Solution_17": "Should now be fixed.", "Solution_18": "[quote=\"DPatrick\"]Should now be fixed.[/quote]\r\n\r\nYep, looks fine now. Thanks!", "Solution_19": "Unfortunately, now it looks like each option is doubled..." } { "Tag": [ "Gauss", "induction", "search", "geometry", "3D geometry", "arithmetic sequence", "algebra" ], "Problem": "Find the value of $ 100 \\plus{} 100 \\plus{} 100 \\plus{} 99 \\plus{} 99 \\plus{} 99 \\plus{} 98 \\plus{} 98 \\plus{} 98 \\plus{} \\dots \\plus{} 3 \\plus{} 3 \\plus{} 3 \\plus{} 2 \\plus{} 2 \\plus{} 2 \\plus{} 1 \\plus{} 1 \\plus{} 1$.", "Solution_1": "[quote=\"isabella2296\"]Find the value of $ 100 \\plus{} 100 \\plus{} 100 \\plus{} 99 \\plus{} 99 \\plus{} 99 \\plus{} 98 \\plus{} 98 \\plus{} 98 \\plus{} \\dots \\plus{} 3 \\plus{} 3 \\plus{} 3 \\plus{} 2 \\plus{} 2 \\plus{} 2 \\plus{} 1 \\plus{} 1 \\plus{} 1$.[/quote]\r\n\r\n[hide]\n\nThis is $ 3(100\\plus{}99\\plus{}...\\plus{}2\\plus{}1)$\n\n$ 3\\left(\\frac{100\\cdot 101}{2}\\right)$\n\n$ 3 \\cdot 5050$\n\n$ \\boxed{15150}$[/hide]", "Solution_2": "Yes, that is correct.", "Solution_3": "Rofl\r\n\r\nHow's this factoring?\r\n\r\n[hide=\"So other people can do it and be enlightened\"]Basically, you add the numbers 2 at a time, the first one and the last one, so that ALL the sums equal 101, and then find how many sums there are. There are $ \\frac {100}{2}\\times3$ sums, so the sum is $ 150\\times101 \\equal{} 15150$.[/hide]\r\n\r\nEDIT:Beaten nooo!\r\nEDIT2: YES POINCARE conjecture", "Solution_4": "i liked textangle's explanation. thank you, textangle.", "Solution_5": "You're welcome Dreambold :) .", "Solution_6": "This reminds me of a method used by the mathematician Carl Friedrich Gauss. When he was a child, his teacher asked him to add all the integers from 1 to 100, and he came up with 5050 because he saw that 100 + 1 = 101, 99 + 2 = 101, etc. and multiplied it by the number of those kinds of pairs.", "Solution_7": "Oh, yes, I remember reading that. There were 50 pairs, so he got $ 101 \\times 50 \\equal{} \\boxed{5050}$.", "Solution_8": "You have just to divide dhe sum into three groups that contain the same numbers , since the numbers of each group are in arithmetic progression just use Gauss formula $ {\\frac{n(n + 1)}{2}}$ than multiply it by 3.", "Solution_9": "This problem reminds me of a similar problem:\r\nFind the value of A=100^2 + 99^2 + ... + 2^2 + 1\r\nI need a short solution.\r\nAlso try: A=100^3 + 99^3 + ... + 2^3 + 1", "Solution_10": "\\[ 1^2\\plus{}2^2\\plus{}3^2\\plus{}...\\plus{}n^2\\equal{}\\frac{n(n\\plus{}1)(2n\\plus{}1)}{6}\\]\r\n\\[ 1^3\\plus{}2^3\\plus{}3^3\\plus{}...\\plus{}n^3\\equal{}\\left(\\frac{n(n\\plus{}1)}{2}\\right)^2\\]\r\nHope that helps", "Solution_11": "Can you post a pfoof?. I need a proof.", "Solution_12": "1. if n=1, then equation is correct ($ 1^2 \\equal{} \\frac {1*2*3}{6} \\equal{} 1$)\r\n\r\nIf $ 1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} ... \\plus{} k^2 \\equal{} \\frac {k(k \\plus{} 1)(2k \\plus{} 1)}{6}$, then $ 1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} ... \\plus{} k^2 \\plus{} (k \\plus{} 1)^2 \\equal{} \\frac {k(k \\plus{} 1)(2k \\plus{} 1)}{6} \\plus{}$\r\n\r\n$ \\plus{} (k \\plus{} 1)^2 \\equal{} \\frac {k(k \\plus{} 1)(2k \\plus{} 1) \\plus{} 6(k \\plus{} 1)^2}{6} \\equal{} \\frac {(k \\plus{} 1)(k \\plus{} 2)(2k \\plus{} 3)}{6} \\equal{}$\r\n\r\n$ \\equal{} \\frac {(k \\plus{} 1)(k \\plus{} 2)(2(k \\plus{} 1) \\plus{} 1)}{6}$ by mathematical induction, equation is proved.\r\n\r\n2. if n=1, then equation is correct ($ 1^3 \\equal{} (\\frac {1*2}{2})^2 \\equal{} 1^2 \\equal{} 1$)\r\n\r\nIf $ 1^3 \\plus{} 2^3 \\plus{} 3^3 \\plus{} ... \\plus{} k^3 \\equal{} (\\frac {k(k \\plus{} 1)}{2})^2$, then $ 1^3 \\plus{} 2^3 \\plus{} 3^3 \\plus{} ... \\plus{} k^3 \\plus{} (k \\plus{} 1)^3 \\equal{} (\\frac {k(k \\plus{} 1)}{2})^2 \\plus{} (k \\plus{} 1)^3 \\equal{} \\frac {k^2(k \\plus{} 1)^2}{4} \\plus{} (k \\plus{} 1)^3 \\equal{}$\r\n\r\n$ \\equal{} \\frac {k^2(k \\plus{} 1)^2 \\plus{} 4(k \\plus{} 1)^3}{4} \\equal{} \\frac {(k \\plus{} 1)^2(k \\plus{} 2)^2}{4} \\equal{} (\\frac {(k\\plus{}1)(k \\plus{} 2)}{2})^2$ by mathematical induction, equation is proved.", "Solution_13": "[quote=\"Bachukas\"]1. if n=1, then equation is correct ($ 1^2 \\equal{} \\frac {1*2*3}{6} \\equal{} 1$)\n\nIf $ 1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} ... \\plus{} k^2 \\equal{} \\frac {k(k \\plus{} 1)(2k \\plus{} 1)}{6}$, then $ 1^2 \\plus{} 2^2 \\plus{} 3^2 \\plus{} ... \\plus{} k^2 \\plus{} (k \\plus{} 1)^2 \\equal{} \\frac {k(k \\plus{} 1)(2k \\plus{} 1)}{6} \\plus{}$\n\n$ \\plus{} (k \\plus{} 1)^2 \\equal{} \\frac {k(k \\plus{} 1)(2k \\plus{} 1) \\plus{} 6(k \\plus{} 1)^2}{6} \\equal{} \\frac {(k \\plus{} 1)(k \\plus{} 2)(2k \\plus{} 3)}{6} \\equal{}$\n\n$ \\equal{} \\frac {(k \\plus{} 1)(k \\plus{} 2)(2(k \\plus{} 1) \\plus{} 1)}{6}$ by mathematical induction, equation is proved.\n\n2. if n=1, then equation is correct ($ 1^3 \\equal{} (\\frac {1*2}{2})^2 \\equal{} 1^2 \\equal{} 1$)\n\nIf $ 1^3 \\plus{} 2^3 \\plus{} 3^3 \\plus{} ... \\plus{} k^3 \\equal{} (\\frac {k(k \\plus{} 1)}{2})^2$, then $ 1^3 \\plus{} 2^3 \\plus{} 3^3 \\plus{} ... \\plus{} k^3 \\plus{} (k \\plus{} 1)^3 \\equal{} (\\frac {k(k \\plus{} 1)}{2})^2 \\plus{} (k \\plus{} 1)^3 \\equal{} \\frac {k^2(k \\plus{} 1)^2}{4} \\plus{} (k \\plus{} 1)^3 \\equal{}$\n\n$ \\equal{} \\frac {k^2(k \\plus{} 1)^2 \\plus{} 4(k \\plus{} 1)^3}{4} \\equal{} \\frac {(k \\plus{} 1)^2(k \\plus{} 2)^2}{4} \\equal{} (\\frac {(k \\plus{} 1)(k \\plus{} 2)}{2})^2$ by mathematical induction, equation is proved.[/quote]\r\n\r\nThat's induction methor. We can use that methor to solve another question.\r\neg: $ 1\\plus{}3\\plus{}5\\plus{}7\\plus{}.............\\plus{}(2n\\minus{}1)\\equal{}n^2$.", "Solution_14": "[quote=\"Nguyen Ngoc Linh\"]Can you post a pfoof?. I need a proof.[/quote]\r\nProofs don't belong in classroom math >.>", "Solution_15": "[quote=\"Nguyen Ngoc Linh\"]Can you post a pfoof?. I need a proof.[/quote]\r\n\r\nI believe I derived the formula before. Search the test forums.\r\nAlso sum of cubes is $ (\\frac{n(n\\plus{}1)}{2})^{2}$" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Helly's theorem states that given the set of $n$ convex sets s.t. the intersection of any $3$ of these sets is nonempty we can conclude that that in this case the whole collection has a nonempty intersection; \r\nCould you please give a counterexample in case when the sets are not convex for any $n$.", "Solution_1": "Just pick $n$ copies of the empty set and fill them with elements such that any $3$ copies share one element in common.", "Solution_2": "thanks,Julien\r\nYou example turned out even more evident then i supposed" } { "Tag": [ "factorial", "function", "number theory proposed", "number theory" ], "Problem": "Define in the natural numbers two functions as: $ f(0)\\equal{}0.2$, $ g(0)\\equal{}0.5$ and $ f(n\\plus{}1)\\equal{}(n\\plus{}1.2)f(n)$ and $ g(n\\plus{}1)\\equal{}(n\\plus{}1.5)f(n)$. Find:\r\n\r\na) The least $ k$ such that $ kf(2008)g(2008)$ is an integer.\r\nb) The largest $ n$ such that $ 3^n|kf(2008)g(2008)$.", "Solution_1": "Let $ t(n)\\equal{}f(n)g(n)$, then $ t(0)\\equal{}0.1, t(n\\plus{}1)\\equal{}\\frac{10n^2\\plus{}27n\\plus{}18}{10}t(n)$.\r\ntherefore a)$ k\\equal{}10^{2009}$.\r\nb)\\[ 10^{2009}t(2008)\\equal{}\\prod_{n\\equal{}0}^{2007}(2n\\plus{}3)(5n\\plus{}6)\\equal{}\\]\r\n\\[ \\equal{}\\prod_{3\\not |n, \\ 1\\le n<2007}(2n\\plus{}3)(5n\\plus{}6)3^{1340}\\prod_{i\\equal{}0}^{667}(2i\\plus{}1)(5i\\plus{}2)\\].\r\nLet $ ord_3(\\prod_i (2i\\plus{}1))\\equal{}a,ord_3(\\prod_i(5i\\plus{}2)\\equal{}b$.\r\n$ 3^k|2i\\plus{}1\\to i\\equal{}(l3^k\\minus{}1)/2, 3^k|5i\\plus{}2\\to i\\equal{}l3^k\\minus{}2)/5$. Therefore we get $ a\\equal{}334\\plus{}[334/3]\\plus{}[334/9]\\plus{}...\\equal{}499$, and $ b\\equal{}222\\plus{}[(667\\plus{}4)/9]\\plus{}[(667\\plus{}22)/27]\\plus{}[(667\\plus{}49)/81]\\plus{}[(667\\plus{}49)/243]\\equal{}331$. \r\nIt give $ n\\equal{}1970$.", "Solution_2": "I think that $ f(n)$ must have some power of 5 (because some of the terms end in .5) and that $ g(n)$ must have some power of 2 (because some of the terms end in even decimal). So there must be something smaller than $ 10^{2009}$, I think.", "Solution_3": "Yes. We must calculate \\[ ord_p(\\prod_{n=0}^N(an+b))\\]\r\nfor cases $ (p,ab)=1$: \r\n1. $ a=2,b=3,p=5,N=2007$,\r\n2. $ a=5,b=3,p=2,N=1003$\r\n3. $ a=2,b=1,p=3,N=669$,\r\n4. $ a=5,b=2,p=3,N=669$.\r\nFor it we must found maximal degree $ p^l|(an+b),n\\le N$.\r\n1. $ p=5,l=5,n_0=\\frac{3125-3}{2}=1561=311\\mod 625 =61\\mod 125 =11\\mod 25 =1\\mod 5$, therefore\r\n$ ord_5(\\prod_{n=0}^{2007}(2n+3)=[\\frac{2007 -1}{5}]+1+[\\frac{2007-11}{25}]+1+[\\frac{2007-61}{125}]+1+[\\frac{2007-311}{625}]+1+[\\frac{2007-1561}{3125}]+1=502$.\r\n$ ord_2(\\prod_{n=0}^{2007}(5n+6))=1004+ord_2(\\prod_{n=0}^{1003}(5n+3))$\r\n$ 2048=5*409+3$ is maximal mean, when $ 2^{11}|5n+3,n\\le 1003$ and $ 409=409\\mod 1024 =\\409\\mod 512 =153\\mod 256 =25\\mod 128 =25\\mod 32 =9\\mod 16 =1\\mod 8 =1\\mod 4 =1\\mod 2$.\r\nTherefore $ ord_2(\\prod_{n=0}^{2007}(5n+6)=1004+[\\frac{1003-1}{2}]+1+[\\frac{1003-1}{4}]+1+[\\frac{1003-1}{8}]+1+[\\frac{1003-9}{16}]+1+[\\frac{1003-25}{32}]+1+[\\frac{1003-25}{64}]+1=[\\frac{1003-25}{128}]+1+[\\frac{1003-153}{256}]+1+2+1+1=1004+502+251+126+63+31+16+8+4+4=2009.$\r\nTherefore $ k=\\frac{10^{2009}}{2^{2009}5^{502}}=5^{1507},$" } { "Tag": [ "function", "calculus", "calculus computations" ], "Problem": "$ cosh^{ \\minus{} 1} x \\equal{} ln(x \\plus{} \\sqrt {x^2 \\minus{} 1})$, \r\neg.\r\nExpress $ cosh^{ \\minus{} 1}\\frac{5}{3}$ in natural logarithms.\r\n\r\nLet $ y \\equal{} cosh^{ \\minus{} 1}$\r\n\r\n$ coshY \\equal{} \\frac {5}{3}$\r\n\r\n$ \\frac {e^y \\plus{} e^{ \\minus{} y}}{2} \\equal{} \\frac {5}{3}$\r\n\r\n$ 3e^{2y} \\minus{} 10e^y \\plus{} 3 \\equal{} 0$\r\n\r\n$ (3e^y \\minus{} 1)(e^y \\minus{} 3) \\equal{} 0$\r\n\r\nwhy is $ ln\\frac {1}{3}$ ommited as a soln.?", "Solution_1": "That's because $ \\cosh$ is not injective. Try plotting it. You could say the same about $ x^2$ and $ \\sqrt{x}$." } { "Tag": [ "limit", "real analysis", "real analysis unsolved" ], "Problem": "$ \\lim_{n\\rightarrow inf}\\sum_{r\\equal{}1}^{n}{\\frac{4r}{4r^4\\plus{}1}}$", "Solution_1": "please reply :(", "Solution_2": "[hide=\"Hint 1\"]factor $ 4r^{4}+1$[/hide]\n[hide=\"Hint 2\"]$ 2(r+1)^2-2(r+1)+1=2r^2+2r+1$[/hide]\n[hide=\"Solution\"]${ \\lim_{n\\rightarrow inf}\\sum_{r=1}^{n}{\\frac{4r}{4r^{4}+1}}=\\lim_{n\\rightarrow inf}\\sum_{r=1}^{n}{(\\frac{1}{2r^{2}-2r+1}-\\frac{1}{2r^{2}+2r+1}})=\\lim_{n\\rightarrow inf}1-\\frac{1}{2n^{2}+2n+1}}=1$[/hide]", "Solution_3": "I could not understand the factors :maybe: \r\ncan u explain it ? :(", "Solution_4": "It's called the Sophie Germain identity:\r\n$ 4r^4 + 1 = 4r^4 + 4r^2 + 1 - 4r^2 = (2r^2 + 1)^2 - (2r)^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$\r\n\r\nThen you're looking for $ \\frac {ar + b}{2r^2 - 2r + 1} + \\frac {cr + d}{2r^2 + 2r + 1} = \\frac {4r}{4r^4 + 1}$\r\nSo $ a = c = 0$ and $ b = - d = 1$\r\n\r\nPut $ f(x) = \\frac {1}{2x^2 - 2x + 1}$\r\n\r\n$ \\lim_{n\\rightarrow inf}\\sum_{r = 1}^{n}{\\frac {4r}{4r^{4} + 1}} = \\lim_{n\\rightarrow inf}\\sum_{r = 1}^{n}{(\\frac {1}{2r^{2} - 2r + 1} - \\frac {1}{2r^{2} + 2r + 1}})$\r\n$ = \\lim_{n\\rightarrow inf}\\sum_{r = 1}^{n}{(f(r) - f(r + 1)}) = \\lim_{n\\rightarrow inf}(f(1) - f(n + 1))$\r\n${ = \\lim_{n\\rightarrow inf}(1 - \\frac {1}{2n^{2} + 2n + 1}}) = 1$" } { "Tag": [ "integration", "rotation", "calculus", "calculus computations" ], "Problem": "Find the volume of the solid generated when the region enclosed by $y=x^3$, $y=1$, and $x=0$ is revolved about the line $y=1$. \r\n\r\nI prefer the use of cylindrical shells.", "Solution_1": "$\\int_0^1 \\pi (1-x^3)^2\\ dx$", "Solution_2": "Move the graph down 1 unit, so\r\n$y=x^3-1$\r\n\r\nWe therefore need to find the volume of revolution between $1$ and $0$ when rotated around the $x$ axis.\r\n\r\nSo $V=\\pi\\int_0^1 (x^3-1)^2dx$\r\n$=\\pi[\\frac{x^7}{7}-\\frac{x^4}{2}+x]_0^1$\r\n\r\nSo $V=\\frac{9\\pi}{14} units^3$\r\n\r\nEDIT: Sorry, arithmatic error :oops:", "Solution_3": "Your answer seems something wrong. :?" } { "Tag": [ "geometry", "trigonometry" ], "Problem": "Circles $ C_1$, $ C_2$, and $ C_3$ have radius $ 1$ and centers $ O$, $ P$, and $ Q$, respectively. $ C_1$ and $ C_2$ intersect at $ A$, $ C_2$ and $ C_3$ intersect at $ B$, and $ C_3$ and $ C_1$ intersect at $ C$ in such a way that $ \\angle APB \\equal{} 60^{\\circ}$, $ \\angle BQC \\equal{} 36^{\\circ}$, and $ \\angle COA \\equal{} 72^{\\circ}$. Find $ m\\angle ABC$.\r\n\r\nHmm, this problem doesn't make sense... the three circles are mutually tangent to one another externally (since there's only one point of intersection for every two circles), and since all three circles are congruent, shouldn't the three given angles be congruent, too? Maybe I'm misunderstanding the problem... :maybe:\r\n(a diagram would be helpful)", "Solution_1": "Perhaps the circles are in space, and not sitting on a plane\r\n\r\nThe answer i get is $ \\frac{\\sqrt{5}\\plus{}3}{4}$", "Solution_2": "ocha could you show your solution?", "Solution_3": "Its just an application of the cosine law, but here it is anyway\r\n[hide=\"solution\"]\n$ \\overline{AB}^2 \\equal{} 1^2 \\plus{} 1^2 \\minus{} 2(1)(1)\\cos(60) \\equal{} 1\\qquad(1)$\n\n$ \\overline{BC}^2 \\equal{} 2 \\minus{} 2\\cos(36) \\qquad(2)$\n\n$ \\overline{CA}^2 \\equal{} 2 \\minus{} 2\\cos(72) \\qquad(3)$\n\nNow we use these to find $ \\angle ABC$\n\n$ 2 \\minus{} 2\\cos(72) \\equal{} 1 \\plus{} 2 \\minus{} 2\\cos(36) \\minus{} 2\\sqrt {2 \\minus{} 2\\cos(36)}\\cos(\\angle ABC)$\n\n$ \\therefore 3 \\minus{} 2\\cos^2(36) \\equal{} 3 \\minus{} 2\\cos(36) \\minus{} 2\\sqrt {2 \\minus{} 2\\cos(36)}\\cos(\\angle ABC)$\n\n$ \\therefore \\cos(\\angle ABC) \\equal{} \\frac {\\minus{}\\cos(36)\\sqrt {1 \\minus{} cos(36)}}{\\sqrt {2}} \\equal{} \\frac {\\minus{}1}{4}$\n\n$ \\angle ABC \\equal{} \\cos^{\\minus{}1}\\left( \\frac{\\minus{}1}{4}\\right)$\n\nIt would appear that the answer i gave in my previous post was wrong\n[/hide]\r\n\r\nEDIT: I just altered this post like 5 times in the last 5 minutes, I'll try clicking the preview button from now on :rotfl:", "Solution_4": "[quote=\"ocha\"]$ 2 \\minus{} 2\\cos(72) \\equal{} 1 \\plus{} 2 \\minus{} 2\\cos(36) \\minus{} 2\\sqrt {2 \\minus{} 2\\cos(36)}\\cos(\\angle ABC)$\n\n$ \\therefore 3 \\minus{} 2\\cos^2(36) \\equal{} 3 \\minus{} 2\\cos(36) \\minus{} 2\\sqrt {2 \\minus{} 2\\cos(36)}\\cos(\\angle ABC)$\n\n[/quote]\r\nShouldn't the LHS be $ 4\\minus{}4\\cos^2(36)$?", "Solution_5": "Actually this is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=410870#410870]#7[/url] in The Intermediate Geometry Marathon, but I don't really see how 4everwise's getting 90 degrees as the answer... :huh:", "Solution_6": "Could someone please explain why 4everwise's getting $ 90$? :huh:" } { "Tag": [ "ratio", "function", "geometric series", "real analysis", "real analysis unsolved" ], "Problem": "Consider the sum $ \\sum_{n\\equal{}0}^{\\infty}\\frac{x}{(1\\plus{}|x|)^n}$\r\nShow that the sum converges for all $ x$, but not uniformly.", "Solution_1": "When $ x\\equal{}0$, every term is 0, making the series 0. Otherwise, $ \\frac{1}{1\\plus{}|x|}$ is strictly between 0 and 1, so the geometric series with it as a ratio converges and we have\r\n\\[ x\\sum_{n\\equal{}0}^\\infty \\left( \\frac{1}{1\\plus{}|x|}\\right)^n \\equal{}\\frac{x}{1\\minus{}\\frac{1}{1\\plus{}|x|}}\\equal{}\\frac{x}{|x|}(1\\plus{}|x|)\\]\r\nThe convergence can't be uniform because a uniformly convergent series of continuous functions is continuous, while this series has a discontinuity (and not even removable!) at 0." } { "Tag": [ "number theory" ], "Problem": "1. Apparently sale of the CD with former physics olympiads has been discontinued. If you still wanted to get those exams, what would you do?\r\n\r\n2. What is the next best place or book to find practice physics problems that are similar to the ones on the physics olympiad?\r\n\r\n3. Which of the classes listed in this link http://epgy.stanford.edu/courses/physics/ in the third column are most relevant to physics olympiad topics and why?", "Solution_1": "Dude, there are so many redundant threads about this that it is not even funny.\r\n\r\n1. Past exams are impossible to get. If you want online sources of other good problems, see the thread\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=212424\r\n\r\nand take feynman's suggestion of BAUPC. The thread is lower on the page and is quite ambiguously titled \"Past Exams.\" lol\r\n\r\n2. I would recommend Irodov's Problems in Physics. The thread above also has suggestions.\r\n\r\n3. Well, the first two classes in the third column might help, but you might as well just self-study that material if you're just interested in the physics olympiad, and I don't know about EPGY. But then again, [b]there's more to physics than the physics olympiad[/b]!!", "Solution_2": "3. Right... there's more to math than math olympiad too but ... I dunno it's like if you were making a choice to learn some number theory or statistics, then number theory would just be more useful in the context of a high school math competition... \r\n\r\ni dunno... :| \r\n\r\nAlso, supposing you would prefer not to self study for various, indisputable reasons that won't be discussed on this thread because that is a digression... \r\n\r\nI'm actually a little surprised the the first two would be more relevant than mechanics and electricity/magnetism classes... \r\n\r\n1. Why are the first two more relevant? \r\n\r\n2.Do the first two classes in the \"university\" level column contain material you wouldn't see much of in a high school AP physics class?", "Solution_3": "That's right, what I meant when I said the first two classes are more relevant is that the physics olympiad doesn't require extensive knowledge of advanced topics, like you're not going to need Lagrangian mechanics. But you will need to know about waves/optics/thermodynamics/some modern physics, which aren't typically covered in AP Physics. As long as you have a solid understanding of fundamental concepts like those in Halliday and Resnick, and you can apply them creatively, you'll be fine. Plus, EPGY is kind of expensive.\r\n\r\nTaking other classes in the third column isn't necessary for the physics olympiad, but they might be of interest if you want to know more about physics as a whole. Looking on that website, I find it strange that EPGY requires you to study PDEs concurrently with using Griffith's Introduction to Electrodynamics for their E+M course." } { "Tag": [ "geometry", "3D geometry", "number theory", "greatest common divisor", "number theory unsolved" ], "Problem": "proof that if $\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}$is an integer so $abc$ is a cube .", "Solution_1": "It posted before (then $a=b=c\\Longrightarrow abc \\-cube$).", "Solution_2": "It's not correct that $a=b=c$.\r\nYou see that if $a=1,b=2,c=4$ then $\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}=\\frac12+\\frac24+\\frac41=5$ is integer.\r\n\r\n[i]Proof :[/i]\r\nWLOG suppose that $gcd(a,b,c)=1$. By problem hypothesis $abc|a^{c}+b^{2}a+c^{2}b$. If $p|a$ then $p|c^{2}b$. So one of $b,c$ are divisible by $p$. WLOG suppose $p|b$ then if $d=gcd(a,b)$ then $d^{2}|c^{2}b\\Rightarrow d^{2}|b$. We know that $gcd(\\frac{a}{d},\\frac{b}{d})=1$. Thus $gcd(\\frac{a}{d},d)=1$, also $b|a^{2}c$. Now suppose $\\parallel a\\parallel _{p}=k,\\parallel b\\parallel _{p}=s$ then $2\\min\\{k,s\\}\\leq s$ and $2k\\geq s$. This concludes obviously that $s=2k$. So $\\parallel abc\\parallel _{p}=3k$. So $abc$ is cube." } { "Tag": [], "Problem": "Where can I found the solution of All russian mathematics olympiad 2008 on the web?", "Solution_1": "Search.\r\n\r\n :)", "Solution_2": "Thanks a lot for your help", "Solution_3": "The web is right here on the [url=http://www.mathlinks.ro/resources.php]ML/AoPS resources section[/url] where you also can find the problems of the [url=http://www.mathlinks.ro/resources.php?c=143&cid=61]5th round of the Russian Math Olympiad.[/url]", "Solution_4": "Thanks \"orl\"\r\nI have a good resource of russian olympiad.But I want to have the solution of them too.specially I want to have the solution of Russian olympiad in 2008 and 2007. I am very happy that you answered my message.", "Solution_5": "[quote=\"daplikos\"]Thanks \"orl\"\nI have a good resource of russian olympiad.But I want to have the solution of them too.specially I want to have the solution of Russian olympiad in 2008 and 2007. I am very happy that you answered my message.[/quote]\r\n\r\nIn my posting above you can see the problems for 2004 to 2008. Click on the link for the year you like and then on the left you will find the problem numbers of the specific grades. Press on those numbers (i.e. a link) which leads you to the discussion of the problem. Some of the problems may already have solutions, others may not. You need to check it!" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $x_1\\leq{x_2}\\dots\\leq{x_n}\\leq{y_1}\\leq{y_2}\\dots\\leq{y_n}$ be real numbers. Prove that:\r\n\\[ (x_1+x_2+\\dots+x_n+y_1+y_2+\\dots+y_n)^2\\geq{4n{(x_1y_1+x_2y_2+\\dots+x_ny_n)} }\\]", "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=81252\r\n\r\n :)" } { "Tag": [ "Putnam", "national olympiad" ], "Problem": "I just realized, you can use the Internet Archive to retrieve the Chinese MO problems (and more importantly, the [b]solutions[/b]) off Kalva! :D \r\n\r\nSpecifically, see \r\n\r\n(you'll have to copy and paste, since the I can't get the forum to make a link).\r\nAnything on or before 05 April 2004 will do.\r\n\r\nThough, the website doesn't cache all the pages during each date. So sometimes you'll need to look through different dates to find the page you want. This makes downloading the entire archive to the harddrive very difficult (unless you have a LOT of time on your hands :rotfl:, in which case we would love it if you could provide the community with a zip file containing all the problems/solutions).\r\n\r\nOf course, the same also applies to Tournament of the Towns, and (the cause of all this) Australian Olympiad.", "Solution_1": "what happened? why cant kalva have the cmo or the tournament of the towns problems? because i remember when they were there and then suddenly dissappeared. why?", "Solution_2": "See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=5304\r\n\r\nThis is too cooool.... thanks for the info! :P", "Solution_3": "[quote=\"ThAzN1\"]\nThis is too cooool.... thanks for the info! :P[/quote]\r\n\r\nIt is! :)", "Solution_4": "It's really cool to look at the oldes Kalva pages, you can see how it developed etc. In the beginning, there were just a few Newman problems, IMO problems and Putnam problems! So cool.", "Solution_5": "[quote=\"billzhao\"]This makes downloading the entire archive to the harddrive very difficult (unless you have a LOT of time on your hands :rotfl:, in which case we would love it if you could provide the community with a zip file containing all the problems/solutions).[/quote]\r\n\r\nI don't think it is too difficult for anybody experienced with coding. Unfortunately, all I could do were some experiments with WinHTTrack, and they didn't bring much success, but I think it wouldn't be too hard if one knows how to use WGet. Any volunteers?\r\n\r\n Darij" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ a,b,c$ be positive numbers satisfying $ abc\\equal{}1$. Prove that $ \\frac{a\\plus{}3}{(a\\plus{}1)^2}\\plus{}\\frac{b\\plus{}3}{(b\\plus{}1)^2}\\plus{}\\frac{c\\plus{}3}{(c\\plus{}1)^2}\\ge 3$", "Solution_1": "[url=http://www.mathlinks.ro/viewtopic.php?p=463670#463670]See here[/url] for this problem and for more links to threads where it has been discussed." } { "Tag": [ "geometry", "3D geometry", "tetrahedron", "trigonometry", "AMC", "AIME", "analytic geometry" ], "Problem": "Given all the sides of a tetrahedron whats the quickest way to compute the area? (Not a regular tetrahedron.)", "Solution_1": "Sorry I can't be clearer, but Mildorf told us about a Heron's Formula for tetrahedrons which I promptly forgot. Maybe PM him.", "Solution_2": "Surface area or volume?\r\n\r\nI don't know either :blush:", "Solution_3": "Surface area is simple if it's regular: $s^2\\sqrt{3}$. If it's irregular, use Heron's, and remember that there are only six distinct edges. All four faces have a combination of those, so there's probably a way to simplify the sum of Heron's in terms of those six edges. If it's volume, it's more difficult, but not too hard. Find the height and the area of the base of that height, which should nearly always be possible, if not always pretty. :P", "Solution_4": "[quote=\"JesusFreak197\"] Find the height and the area of the base of that height, which should nearly always be possible, if not always pretty. :P[/quote]\r\n\r\nThe only way I can think of finding the height is Pythagorean Thm. But you only know the hypotenuse(the lateral edge), how can you get the other side? \r\nIt's gonna get really ugly with multiple law of cosines or something.", "Solution_5": "There is a general formula for the volume of a tetrahedron given the sides, but I doubt you will use it for the AIME:\r\n\r\nIf the edge between the $i$-th and $j$-th vertices is $d_{ij}$ and the volume is $V$, then\r\n\r\n$288V^2 = \\left | \\begin{array} [c]{c c c c c} 0&1&1&1&1\\\\ \\\\ 1&0&d_{12}^2&d_{13}^2&d_{14}^2\\\\ \\\\ 1&d_{21}^2&0&d_{23}^2&d_{24}^2\\\\ \\\\ 1&d_{31}^2&d_{32}^2&0&d_{34}^2\\\\ \\\\ 1&d_{41}^2&d_{42}^2&d_{43}^2&0\\\\ \\\\ \\end{array} \\right |$\r\n\r\nCourtesy of Mildorf", "Solution_6": "There is a formula given coordinates as well. If one of the vertices is at the origin and the other three are at the points (x1, y1, z1)=u (x2, y2, z2)=v and (x3, y3, z3)=w then\r\n\r\n$V =\\frac{1}{6} \\left | \\begin{array} [c]{c c c} x_{1}&y_{1}&z_{1} \\\\ \\\\ x_{2}&y_{2}&z_{2} \\\\ \\\\ x_{3}&y_{3}&z_{3}\\\\ \\\\ \\end{array} \\right |$\r\n\r\n\r\nThis was in AOPS volume 2 in the analytic geometry chapter.", "Solution_7": "[quote=\"mkkool64\"]There is a formula given coordinates as well. If one of the vertices is at the origin and the other three are at the points (x1, y1, z1)=u (x2, y2, z2)=v and (x3, y3, z3)=w then\n\n$V =\\frac{1}{6} \\left | \\begin{array} [c]{c c c} x_{1}&y_{1}&z_{1} \\\\ \\\\ x_{2}&y_{2}&z_{2} \\\\ \\\\ x_{3}&y_{3}&z_{3}\\\\ \\\\ \\end{array} \\right |$\n\n\nThis was in AOPS volume 2 in the analytic geometry chapter.[/quote]\r\n\r\n\r\nSo basically, what this is called is the \"box product\" which essentially just means that you're taking the cross product, which is basically taking the area of the base, and then you're dotting it with the third vector to get a volume. \r\n\r\nIf you didn't divide it by 6 you would get the volume of a parallipiped with three vertices as those three vectors. However, when you divide by 6 it gives you the volume of the tetrahedron because you have to divide by 2 once because of the fact that it'll be a triangle and 1/3 because its a volume according to the 1/3(area of base)(h) rule, thus multiplying by 1/6. Its sorta like the 1/2 b*h rule for triangles and rectangles.. or at least thats the way I look at it I guess." } { "Tag": [ "topology", "Support" ], "Problem": "This is exercise 16.5(b) in Munkres if you want to look at his statement of the problem. If T, T' are topologies on X and U, U' are topologies on Y, consider the product topologies V, V' on X x Y induced by T, U and T', U'. If V 0 such that a + b + c = abc, prove that\r\n\r\n(1 + a)^(-1/2) + (1 + b)^(-1/2) + (1 + c)^(-1/2) <= 3/2", "Solution_1": "Indeed, it is nice:\r\n\r\nThe identity tgA+tgB+tgC=tgA*tgB*tgC is well known for a triangle ABC, so we can replace a, b, c with tgA, tgB, tgC where abc is an acute triangle (acute because the tangents of the angles are positive). 1/(1+(tgx)^2)=(cosx)^2, so our sum in the left part is, in fact, cosA+cosB+cosC=1+r/R<=3/2 because r(the inradius) <= R/2 (R is the circumradius). The last part can also be proved by using the fact that the function cosx is concave on [0, Pi/2].", "Solution_2": "Indeed. The fact that cos A + cos B + cos C <= 3/2 can also proved by Jensen's inequality as follows. Since the cosine function is concave on [0, Pi/2] we have\r\n\r\ncos A + cos B + cos C <= 3 cos ((A + B + C)/3) = 3 cos (Pi/3) = 3/2.\r\n\r\nIf a condition a + b + c = abc is given then think of tangents !\r\nSimilarly, if ab + bc + ca = 1 think of cotangents.\r\nThere is a similar expression for cosines, isn't it Grobber :D ?\r\n\r\nIn fact we have \r\n\r\ncos A + cos B + cos C <= 3/2\r\n\r\nalso for obtuse triangles. To prove this you need a slightly more sophisticated argument. (Beware of nonconvexity/nonconcavity in [0, Pi].)", "Solution_3": "Wow, nice...geometry certainly makes it all fit nicely. I proved this far less elegantly with the AM-QM inequality and the fact that a+b+c = abc >= V3.", "Solution_4": "Can you post your solution with AM GM ? Just curious ... 8-)", "Solution_5": "Ah, never mind...I made a little error in the beginning that conveniently messes up the whole proof...oh I love it when that happens =).", "Solution_6": "That's what I expected ... 8-) \r\nMaybe this sounds strange but after finding the trig solution (which is obvious if you've seen something like this before (a + b + c = abc)) I tried for a while to prove this with standard inequalities but it didn't work as far as I know.", "Solution_7": "So if x + y + z + 2xyz = 1 then think of cosines !\r\n(Substitute x = cos A, y = cos B, z = cos C where A + B + C = 180).\r\n\r\nIn some USAMO - 1999 or 2000 or so - the condition x + y + z + xyz = 4 was given. The you could substitute x = 2 cos A, y = 2 cos B, z = 2 cos C !", "Solution_8": "erm. could you prove these properties of tangent/cotangent/cosine please? or at least give some hints as to how to prove it...i.e. should we stick with using the geometric defenitions of the trigonometric functions by dropping a couple of perpendiculars, or should we handle it all algebraically with trigonometric identities...or what?", "Solution_9": "Trig identities ! Just use the following :\r\nsin(180 - A) = sin A\r\ncos(180 - A) = -cos A\r\ntan(180 - A) = -tan A\r\nsin(X + Y) = sin X cos Y + cos X sin Y\r\ncos(X + y) = cos X cos Y - sin X sin Y\r\ntan(X + y) = (tan X + tan Y)/(1 - tan X tan Y)\r\nand some algebra of course ! :D", "Solution_10": "The following problem is also very nice and can be related to the above posts.\r\n\r\nGiven 13 distinct reals, prove that there are two of them, a and b, such that\r\n(a - b)/(1 + ab) <= 2 - 3^(1/2).\r\n\r\nIf you don't know about trig substitution this is too difficult, but now I gave you already the trig hint so it should be easy ! 8-)", "Solution_11": "Oh ok we use the identities. \r\nOK now, if we let a=tan A and b=tan B, that expression is tan (A-B). Taking our 13 reals mod Pi, some two must be less than or equal to Pi/12 from each other (by pigeonhole). Taking these two to be A and B gives tan (0) :le: tan (A-B) :le: tan (:pi:/12), since tangent is continuous, which gives our desired inequality.\r\nCorrect?", "Solution_12": "That's it !\r\n\r\n[quote=\"MysticTerminator\"] Taking these two to be A and B gives tan (0) :le: tan (A-B) :le: tan (:pi:/12), since tangent is continuous, which gives our desired inequality.[/quote]\r\n\r\nDo you mean : tan(0) :le: tan(A-B) :le: tan(:pi:/12), since tangent is INCREASING ? ;-)" } { "Tag": [ "number theory solved", "number theory" ], "Problem": "Prove that: If p is prime than 2^p + 3^p cannot be represented as n^k with natural numbers n and k > 1.", "Solution_1": "We easily verify that 2 p + 3 p is not of the form a n for some n > 1, when p = 2,3,5.\r\nNow suppose that p \\geq 7 is a prime. Then p = 4k+1 or p = 4+3.\r\nMoreover 2 p + 3 p = (2+3) \\sum (-1) i 2 i 3 p-1-i , where the sum is over i = 0,...,p-1.\r\nSince 2 = -3mod[5], it is easy to verify that the sum is equal to p mod[5] when p = 4k+1, and to -p mod[5] when p = 4k+3.\r\nThus, in each case, 2 p + 3 p is divisible by 5 but not by 5 2 . Then, it cannot be a number of the form a n where n>1.\r\n\r\nPierre.", "Solution_2": "further more about 2^p+3^p: prove that if this number is divisible by some prime q < p then q=5 :)" } { "Tag": [ "geometry", "rectangle", "ratio", "percent", "AMC" ], "Problem": "A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is $ 2: 1$. The ratio of the rectangle's length to its width is $ 2: 1$. What percent of the rectangle's area is inside the square?\r\n\r\n$ \\textbf{(A)}\\ 12.5 \\qquad \\textbf{(B)}\\ 25 \\qquad \\textbf{(C)}\\ 50 \\qquad \\textbf{(D)}\\ 75 \\qquad \\textbf{(E)}\\ 87.5$", "Solution_1": "[hide=\"Solution\"]Let the square's side length be $ s$. Then, the width of the rectangle is $ 2s$, and the length of the rectangle is $ 4s$.\n\nThe ratio of the square's area to the rectangle's area is $ \\frac {s^2}{(2s)(4s)} \\equal{} 0.125 \\equal{} 12.5\\% \\Rightarrow \\boxed{A}$[/hide]", "Solution_2": "$\\frac{1}{2\\cdot2^2}=\\boxed{12.5\\%}$" } { "Tag": [], "Problem": "Solve for $ x$: $ 3x\\minus{}(1\\minus{}x)\\equal{}5$. Express your answer as a common fraction.", "Solution_1": "[quote=\"GameBot\"]Solve for $ x$: $ 3x \\minus{} (1 \\minus{} x) \\equal{} 5$. Express your answer as a common fraction.[/quote]\r\n\r\n$ 3x \\minus{} (1 \\minus{} x) \\equal{} 5 \\implies 3x \\minus{} 1 \\plus{} x \\equal{} 5 \\implies 4x \\minus{} 1 \\equal{} 5 \\implies x \\equal{} \\frac{3}{2}$" } { "Tag": [], "Problem": "2(16-(1+3)^2)=2(16-4^2)=2(16-16)\r\n\r\nwhat do i do first? the (1+3) or the (4^2)", "Solution_1": "that is correct...this is too easy for getting started though", "Solution_2": "well which one is it?", "Solution_3": "you add 1+3, then 4^2=16, 16-16=0 2*0=0\r\n\r\nthis should go to middle school, generally, you go parathensis, exponents, mult/div, add/sub", "Solution_4": "You do the $1+3$ first, then $4^2$, then $16-16$, then $2\\times0$", "Solution_5": "First of all, yo calcule 1+3, then, 4^2, then 16-16 and finally 2*0.", "Solution_6": "I learned \r\nPlease Excuse My Dear Aunt Sally\r\nEach word starts with a letter which also stands for\r\nParentheses, Exponents, Multiplication, Division, Addition, Subtraction\r\nIt is still how I remember order of operations.", "Solution_7": "How about [hide][b]P[/b]urple [b]E[/b]ggplants [b]M[/b]ake [b]D[/b]elicious [b]A[/b]fternoon [b]S[/b]nacks.[/hide]\r\n\r\n :rotfl:", "Solution_8": "You start out by following order of operations in each of the expressions. Then once you simplify each thing you get 0=0=0 which is the answer.", "Solution_9": "I like robinhe's one. I always think of it to help me remember pemdas", "Solution_10": "I just remeber PEMDAS and think what each thing in PEMDAS means. I don't use any acrostic device.", "Solution_11": "PEMDAS yo and the answer is 0 yo" } { "Tag": [ "geometry", "geometric transformation", "rotation", "invariant", "symmetry", "number theory unsolved", "number theory" ], "Problem": "Suppose that $n\\in\\mathbb{N}$ is a fix number , Prove that $\\forall k\\in\\mathbb{N}$, \\[ \\sum_{i=1}^n k^{gcd(i,n)} \\equiv 0 (mod n). \\]", "Solution_1": "$\\sum_{i=1}^n k^{gcd(i,n)} = \\sum_{d|n} \\phi(d) k^{n/d}$ is $n$ times the number of neclaces with $n$ beads having $k$ different colors, when two that are identical up to a rotation are considered the same.", "Solution_2": "why?\r\n$\\sum_{i=1}^n k^{gcd(i,n)} =\\sum_{d|n} \\phi(d) k^{n/d}$\r\nplease explain more.", "Solution_3": "$\\gcd(n,i)$ is clearly a divisor of $n$. The question is, how often the exponent $d|n$ occurs in the sum.\r\nWell there are $\\phi(n)$ of them for $d=1$, and similar there are $\\phi(\\frac{n}{d})$ of them in general.\r\nThus the result follows.", "Solution_4": "Now the prove that the formula $\\frac{1}{n}\\sum_{i=1}^n k^{gcd(i,n)} = \\frac{1}{n}\\sum_{d|n} \\phi(d) k^{n/d}$ gives the desired number of necklaces:\r\n\r\nConsider any $d$, $1 \\leq d \\leq n$. There are $k^{\\gcd(n,d)}$ necklaces that are invariant under rotation by $d$ (the rotational symmetry by $d$ reduces to the one by $\\gcd(n,d)$ by Bezout's theorem and we can choose the first $\\gcd(n,d)$ beads as we want).\r\nNow when we add up all those terms, we counted every type (meaning the class of necklaces seen as the same under rotation) necklace exactly $n$ times:\r\nWhen a type has rotational symmetry of exactly $r|n$ (thus $r$ is the smallest divisor of $n$ such that the necklace is left the same when rotating by $r$), then consider the 'equivalences' given by a rotation by multiples of $r$ only: every such equivalence class is counted $n/r$ times (since this is the number of $d$ with $r|d \\leq n$)\r\nThere are $r$ truely different classes (those reached by rotation with $d \\leq r$) that are considered the same when using all rotations, thus we counted it $n$ times.\r\n\r\n\r\nNote that this is also a simple corollary to Burnsides (see http://www.mathlinks.ro/Forum/viewtopic.php?t=60832) and Polyas counting lemmas. Here the group of residues $\\mod n$ acts cannonical on the mappings $\\{1,2,...,n\\} \\to \\{1,2,...,k\\}$, and another use of these can be seen in http://www.mathlinks.ro/Forum/viewtopic.php?t=63678 .", "Solution_5": "Thank you so much ZetaX. :)", "Solution_6": "In case you want some more general fun:\r\n\r\nLet $ K$ be a number field and $ \\mathcal O_K$ it's ring of integers. Let $ k \\in \\mathcal O_K$, let $ \\mathfrak n$ be a nonzero ideal of $ \\mathcal O_K$ and for an ideal $ \\mathfrak m$ of $ \\mathcal O_K$ define the norm $ \\mathcal N(\\mathfrak m) : \\equal{} |\\mathcal O_K / \\mathfrak m|$. Then we have\r\n\\[ \\sum_{i \\in \\mathcal O_K/\\mathfrak n} k^{\\mathcal N((i)\\plus{}\\mathfrak n)} \\equiv 0 \\mod \\mathfrak n .\\]\r\n\r\nNote that for $ K\\equal{}\\mathbb Q$, we get the original one." } { "Tag": [ "inequalities", "AMC", "USA(J)MO", "USAMO", "algebra unsolved", "algebra" ], "Problem": "the quesiton: \r\nthe sum of 5 real numbers is 8 and the sum of their squares is 16. What is the largest possible value for one of the numbers?", "Solution_1": "We have $a_1+a_2+a_3+a_4+a_5=8$ and $a_1^2+\\ldots =16$ so from that we obtain\r\n$a_1^2+a_2^2+a_3^2+a_4^2+a_5^2-2a_1-2a_2-2a_3-2a_4-2a_5=0$\r\n\r\n\\Rightarrow $(a_1-1)^2+(a_2-1)^2+(a_3-1)^2+(a_4-1)^2+(a_5-1)^2=5$\r\n\r\nSo one get's the maximum value if all braces are zeros except one (because all\r\nvalues in braces are positive so one is maximum when others are all 0)\r\n\r\nso $(a_1-1)^2=5$ and from that the maximum value is $\\sqrt5 +1$\r\n\r\nI hope so that this is correct :) :)", "Solution_2": "When one actually has a+b+c+d+e = s, a^2+b^2+c^2+d^2+e^2 = f then by applying AMGM you can get \r\nf - a^2 >= 1/4 ( s-a)^2 so that a satisfies the inequality:\r\n5a^2-2as+s^2-4f <= 0 and in this way you can find the boundaries for each of the variables", "Solution_3": "[quote=\"alishkov\"]When one actually has a+b+c+d+e = s, a^2+b^2+c^2+d^2+e^2 = f then by applying AMGM [/quote]\r\n\r\nU can't use AMGM becouse these are real numbers not $positive$ real", "Solution_4": "you have to get a^2+b^2+c^2+d^2 >= 1/4 (a+b+c+d)^2. Expand it and you get to prove:\r\n3(a^2+b^2+c^2+d^2)>=2(ab+ac+ad+bc+bd+cd). You have that:\r\n3(a^2 + ... ) >= 2( |ab| + |ac| +...) which in turn is grater or equal than 2(ab+ac+ad+bc+bd+cd)\r\nOK ? So in this case AMGM does not need positive numbers", "Solution_5": "m@re your solution is wrong because if you calculate you will find that $(\\sqrt5+1) +1+1+1+1\\not=8$ and neither is the sum of squares 16.(you practicaly find the maximum value of one of the numbers when the sum of the numbers is half the sum of the squares but you forgot that the sums actually have values :8 and 16)", "Solution_6": "let $a+b+c+d+e=8, a^2+b^2+c^2+d^2+e^2=16\\ (a\\leq b\\leq c\\leq d)$\r\nUsing Cauchy-Schwalz's inequality, \r\n$(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)\\geq (1\\cdot a+1\\cdot b+1\\cdot c+1\\cdot d)^2$\r\n\r\n$\\Longleftrightarrow 4(16-e^2)\\geq (8-e^2)\\Longleftrightarrow 0\\leq e\\leq \\frac{16}{5}$\r\n\r\nThus desired answer is $\\frac{16}{5}$.", "Solution_7": "[quote=\"Tellah\"]m@re your solution is wrong because if you calculate you will find that $(\\sqrt5+1) +1+1+1+1\\not=8$ and neither is the sum of squares 16.(you practicaly find the maximum value of one of the numbers when the sum of the numbers is half the sum of the squares but you forgot that the sums actually have values :8 and 16)[/quote]\r\n\r\nI saw that little bit later :) :) :) :)", "Solution_8": "This problem was on MIMC contest!!\r\nI just realived I solved a USAMO without realiving it...", "Solution_9": "By Lagrange metod we have exremal mean $y=a_{1}$, when $a_{2}=a_{3}=a_{4}=a_{5}=x$. Therefore we have $y+4x=8. y^{2}+4x^{2}=16$.\r\nIt give $5y^{2}-16y=0$, or $y=a_{1}=\\frac{16}{5}, a_{2}=..=a_{5}=\\frac{6}{5}.$" } { "Tag": [ "quadratics" ], "Problem": "One solution of the equation x^3 + 5x^2 + 5x - 2 = 0 is - 2. Find the sum of the remaining solutions.", "Solution_1": "Solutions sum to 5, thus the rest sum to 7\r\n\r\nand we don't need to know that one is -2", "Solution_2": "[hide]divide to get x^2+3x-1. The sum of roots is -b/a which equals -3/1= [b]-3[/b][/hide]", "Solution_3": "Divide by what to get x^2+3x-1?", "Solution_4": "He meant divide by x+2. By the way, me@home, the sum of the solutions is -5; not 5. The sum of the roots of $ax^{3}+bx^{2}+cx+d$ is -b/a.", "Solution_5": "Thanks for letting me know.", "Solution_6": "[quote=\"The QuattoMaster 6000\"]He meant divide by x+2. By the way, me@home, the sum of the solutions is -5; not 5. The sum of the roots of $ax^{3}+bx^{2}+cx+d$ is -b/a.[/quote]No, it's only like that for 2nd degree quadratics", "Solution_7": "[quote=\"kstan013\"][quote=\"The QuattoMaster 6000\"]He meant divide by x+2. By the way, me@home, the sum of the solutions is -5; not 5. The sum of the roots of $ax^{3}+bx^{2}+cx+d$ is -b/a.[/quote]No, it's only like that for 2nd degree quadratics[/quote]\r\nThat is not true:\r\n[hide] Say the roots of $ax^{3}+bx^{2}+cx+d$ are m, n, and p, so $x^{3}+\\frac{bx^{2}}{a}+\\frac{cx}{a}+\\frac{d}{a}$ also has roots m, n, and p. This means that $(x-m)(x-n)(x-p)=x^{3}-(m+n+p)x^{2}+(mn+np+mp)x-mnp=x^{3}+\\frac{bx^{2}}{a}+\\frac{cx}{a}+\\frac{d}{a}$. Corresponding the $x^{2}$s, we get that $\\frac{bx^{2}}{a}=-x^{2}(m+n+p)$. This means that m+n+p=-b/a. [/hide]", "Solution_8": "Forget I said anything then. Anyway I suppose it makes sense when you think about the synthetic division process" } { "Tag": [ "LaTeX", "number theory unsolved", "number theory" ], "Problem": "if [b]p >2[/b] is a prime number and[b] t[/b] is an odd number, [b]4-p m=p m . [/b]", "Solution_1": "If there weren't any mistakes,here's solution.\r\n With $t$ different from $-p,-p+1,-p+2$ we have $t$ is odd and $m$ is odd as $(p+t)!$ divisible by $4$\r\n Hence $(p+t-1)!$=$p^\\ {m-1}$ $+...+$ $t^\\ {m-1} $,which is odd,contradiction.", "Solution_2": "Guys, try to learn $\\LaTeX$ (it's simple to learn! -read this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 and test the features in the [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url]), instead of [b]bolding[/b] the equations. :)", "Solution_3": "Ok [b]Valentine[/b],i've edited my post.", "Solution_4": "[quote=\"nalpaction\"]Ok $Valentine$,i've edited my post :D :D :D[/quote]It's great you learned $\\LaTeX$ so quickly, but please leave text out of the formulae (my name for example). ;)" } { "Tag": [], "Problem": "a,b,c are positive reals show that:\r\n\r\n$ \\frac{a}{2} \\plus{} \\frac{b}{3} \\plus{} \\frac{c}{6} \\geq \\sqrt{a}\\sqrt[3]{b}\\sqrt[6]{c}$", "Solution_1": "[hide=\"Hint\"]\n$ \\frac{a}{2}\\plus{}\\frac{b}{3}\\plus{}\\frac{c}{6} \\equal{} \\frac{a\\plus{}a\\plus{}a\\plus{}b\\plus{}b\\plus{}c}{6}$[/hide]", "Solution_2": "Kouichi Nakagawa's hint is very good. Now use AM-GM for 6 variables.", "Solution_3": "Thanks a lot. I'm dumb.", "Solution_4": "[quote=\"whomer\"]Thanks a lot. I'm dumb.[/quote]\r\n\r\nIf you can't solve one problem, it doesn't mean that you are dumb...", "Solution_5": "$ a^{1/2}b^{1/3}c^{1/6} \\equal{} (a^3b^2c)^{1/6}.$ So now we must prove that \r\n$ \\frac{3a \\plus{} 2b \\plus{} c}{6} \\geq (a^3b^2c)^{1/6}.$ Rewriting $ 3a \\plus{} 2b \\plus{} c$ as $ a\\plus{}a\\plus{}a\\plus{}b\\plus{}b\\plus{}c$ we see that this is true by $ AM\\minus{}GM$ in 6 variables.\r\n\r\nQ.E.D.", "Solution_6": "That solution has already been written. I don't see the purpose of writing it again...", "Solution_7": "for fun :rotfl:", "Solution_8": "Post spamming, he is :P" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Show that $ \\forall x,y,z>0$ we have:\r\n$ \\frac{(x+y)(y+z)(z+x)}{4xyz} \\geq \\frac{x+z}{y+z} + \\frac{y+z}{x+z}$.\r\n\r\nmade by Caragea C.\r\n\r\n\r\nHere is what I have found:\r\n\r\n$ \\frac{x+y}{4xyz}= \\frac{1}{4yz} +\\frac{1}{4xz} \\geq \\frac{1}{(y+z)^2}+ \\frac{1}{(x+z)^2}$ and so we have that:\r\n$ \\frac{x+y}{4xyz} \\geq \\frac{1}{(y+z)^2} + \\frac{1}{(x+z)^2}$ and we multiply by $(y+z)(x+z)$ and we have our ineq! :D :D \r\n\r\ndo you have other solutions? cheers! :D :D", "Solution_1": "The following inequality is also true.\nFor all triangle holds: $\\frac{R}{r}+1\\geq\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}$.", "Solution_2": "[quote=\"Lagrangia\"]Show that $ \\forall x,y,z>0$ we have:\n$ \\frac{(x+y)(y+z)(z+x)}{4xyz} \\geq \\frac{x+z}{y+z} + \\frac{y+z}{x+z}$.\n\nmade by Caragea C.[/quote]\nrefinement 1\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=466241\nrefinement 2\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=467679", "Solution_3": "[quote=\"Lagrangia\"]Show that $ \\forall x,y,z>0$ we have:\n$ \\frac{(x+y)(y+z)(z+x)}{4xyz} \\geq \\frac{x+z}{y+z} + \\frac{y+z}{x+z}$.\n[hide]made by Caragea C.\n\n\nHere is what I have found:\n\n$ \\frac{x+y}{4xyz}= \\frac{1}{4yz} +\\frac{1}{4xz} \\geq \\frac{1}{(y+z)^2}+ \\frac{1}{(x+z)^2}$ and so we have that:\n$ \\frac{x+y}{4xyz} \\geq \\frac{1}{(y+z)^2} + \\frac{1}{(x+z)^2}$ and we multiply by $(y+z)(x+z)$ and we have our ineq! :D :D \n\ndo you have other solutions? cheers! :D :D[/hide][/quote]\nThis inequality is equivalent to\n\\[sin\\frac{A}{2}sin\\frac{B}{2}sin\\frac{C}{2}(\\frac{sinB}{sinC}+\\frac{sinC}{sinB})\\le\\frac{1}{4}\\]\\[\\Leftrightarrow \\]\\[\\frac{R}{r}\\ge 2(\\frac{b}{c}+\\frac{c}{b}).\\]", "Solution_4": "Inequality $ \\frac{R}{r}\\ge\\frac{b}{c}+\\frac{c}{b} $ belongs V.Bandila (see GM.2/1985,problem C:474 )\nWith usual notations the following result is true:\n1) $ R(\\frac{1}{r'}+\\frac{1}{R'})\\ge\\frac{a}{a'}+\\frac{b}{b'}+\\frac{c}{c'} $\nvalid for two triangles ABC and A'B'C'\nFor $ a'=a,b'=c,c'=b $ from 1) to obtain Bandila's inequality.\nFor $ a'=b,b'=c,c'=a $ from 1) to obtain $ \\frac{R}{r}\\ge\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}-1 $\nwhich is stronger than Euler's inequality $ \\frac{R}{r}\\ge 2 $\n_________\nSandu Marin", "Solution_5": "http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=525899&p=2981763#p2981763", "Solution_6": "[quote=\"sandumarin\"]Inequality $ \\frac{R}{r}\\ge\\frac{b}{c}+\\frac{c}{b} $ belongs V.Bandila (see GM.2/1985,problem C:474 )\nWith usual notations the following result is true:\n1) $ R(\\frac{1}{r'}+\\frac{1}{R'})\\ge\\frac{a}{a'}+\\frac{b}{b'}+\\frac{c}{c'} $\nvalid for two triangles ABC and A'B'C'\nFor $ a'=a,b'=c,c'=b $ from 1) to obtain Bandila's inequality.\nFor $ a'=b,b'=c,c'=a $ from 1) to obtain $ \\frac{R}{r}\\ge\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}-1 $\nwhich is stronger than Euler's inequality $ \\frac{R}{r}\\ge 2 $\n_________\nSandu Marin[/quote]\n\nMy book P320:\nBF151 (d):\n\\[\\frac{R}{r}+\\frac{r}{R}-\\frac{1}{2}\\geq \\frac{b}{c}+\\frac{c}{b}\\]" } { "Tag": [ "number theory", "greatest common divisor", "function", "search", "group theory", "number theory proposed" ], "Problem": "For each positive integer $ n$ let $ \\sigma(n)$ and $ \\tau(n)$ be the sum of the positive divisors of $ n$ and the number of positive divisors of $ n$ respectively. \r\n\r\nThen \\[ \\sigma(n)\\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1}{\\tau(gcd(n,k))}\\] holds for any positive integer $ n$. \r\n\r\n\r\n:)", "Solution_1": "Using the formula $ \\sum_{a|x}\\phi(a) = x$,\r\n\\begin{eqnarray*}\r\n\\sigma(n) & = & \\sum_{d|n} d \\\\\r\n& = & \\sum_{d|n} \\sum_{e|d} \\phi(e) \\\\\r\n& = & \\sum_{e|n} \\phi(e)\\tau(\\frac {n}{e}) \\\\\r\n& = & \\sum_{k = 0}^{n - 1} \\tau(\\gcd(n,k))\r\n\\end{eqnarray*}", "Solution_2": "Nice proof pluristiq. I know 3 very different solutions including yours, which may be the esiest of them once the initial formula is known. Let's wait some days, if nobody post another solution, I'll post the other two. \r\n\r\nIt would be iluminating to explain a little more the lats equality, I think. \r\n\r\n :)", "Solution_3": "Here are the details I left out.\r\n\r\n$ \\sum_{d|n}\\sum_{e|d}\\phi(e)\\equal{}\\sum_{e|n}\\phi(e)\\tau(\\frac{n}{e})$: In the sum on the LHS, $ \\phi(e)$ occurs as many times as there are values of $ d$ that satisfy $ e|d|n$, and it's not hard to see there are $ \\tau(\\frac{n}{e})$ such values.\r\n\r\n$ \\sum_{e|n}\\phi(e)\\tau(\\frac{n}{e})\\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1}\\tau(\\gcd(n,k))$: Let $ \\gcd(n,k)\\equal{}d$. Clearly $ d|n$, and for each $ d|n$, the number of times that $ \\tau(d)$ occurs in the sum $ \\sum_{k\\equal{}0}^{n\\minus{}1}\\tau(\\gcd(n,k))$ is the number of $ k$, $ 0\\leq k\\leq n\\minus{}1$ for which $ \\gcd(n,k)\\equal{}d$, and it's not hard to see that there are $ \\phi(\\frac{n}{d})$ such values. Thus $ \\sum_{k\\equal{}0}^{n\\minus{}1}\\tau(\\gcd(n,k))\\equal{}\\sum_{d|n} \\phi(\\frac{n}{d})\\tau(d)\\equal{}\\sum_{e|n}\\phi(e)\\tau(\\frac{n}{e})$.", "Solution_4": "Here is one of the solutions...\r\n[hide]Let $ f$ and $ g$ be functions defined from the positive integers to the complex numbers. For each positive integer $ n$ denote $ \\displaystyle\\sum_{d|n}{f(d)g\\left(\\frac {n}{d}\\right)}$ with $ (f\\times g)(n)$, and with $ f\\times g$ the function with those values. \n\nThe product just defined is associative and conmutative. \n\nLet $ id$ be the function with values $ 1$ at the number $ 1$ and $ 0$ elsewhere, $ I_k$ be the function with value $ n^k$ at the positive integer $ n$ and $ \\mu$ the Mobius function. \n\nIt is easy to see that $ \\mu\\times I_0 \\equal{} id$, $ \\tau \\equal{} I_0\\times I_0$, $ \\sigma \\equal{} I_0\\times I_1$ and $ \\phi \\equal{} \\mu\\times I_1$. Also $ id\\times f\\equal{}f$ for any $ f$.\n\nTaking the product it follows that $ \\sigma \\equal{} \\phi\\times \\tau$. From here proceed as in plurestiq solution. [/hide]\r\n\r\n:)", "Solution_5": "The other solution... I think this is very nice, but it isn't completely elementary.\r\n[hide]The idea is a double counting. I guess one could search for a \"don't say the word group\" equivalent proof but I haven't tooken the time for it. \n\nConsider a cyclic group $ G$ of order $ n>0$ and let $ S$ be the set of all its subgroups. Note that if $ 02^{32}-1\\approx 4.29\\cdot10^9$.\r\nThis leaves you two choices: find a longer format to hold your integers, or factor $10^6$.\r\nThe second method: find the number in question mod $2^6$ and mod $5^6$, then solve the chinese remainder problem $x\\equiv a\\mod 2^6,x\\equiv b\\mod 5^6$. Since $5^{12}\\approx 2.45\\cdot 10^7$, these calculations will not cause overflow.\r\n\r\nThe solution to the chinese remainder problem is $x\\equiv 5^6\\cdot a\\cdot m+2^6\\cdot b\\cdot n \\mod 10^6$, where $m\\equiv 5^{-6}\\mod 2^6$ and $n\\equiv 2^{-6}\\mod 5^6$. The calculation $a\\cdot m$ can be performed mod $2^6$, and the calculation $b\\cdot n$ can be performed mod $5^6$; this avoids potential overflow.", "Solution_2": "[quote=\"lynn\"] I use VC++ to compute 5^131313 mod 1000000. [/quote]\r\nWhat a waste of computer power! All you need to do is to note that $5^{131313}\\equiv 5\\mod 2^6$ (because $5^{16}-1=(5-1)(5+1)(5^2+1)(5^4+1)(5^8+1)$ is divisible by $2^6$ and $131313\\equiv 1313\\equiv 1280+33\\equiv 0+1=1 \\mod 16$). So, you need a number divisible by $5^6=15625$ that is $5$ modulo $2^6=64$. But $5^6=125^2\\equiv (-3)^2=9\\mod 64$, so you need to find a multiple of $9$ that is $5$ modulo $64$. Note that $9\\cdot(-7)=-63\\equiv 1\\mod 64$, so $9\\cdot(64-5\\cdot 7)=9\\cdot 29$ is what you are looking for. The final answer is $15625\\cdot 29=15625\\cdot 30-15625=468750-15625=453125$. I hope I haven't made some stupid mistake in my computations. My goal was to convince you that, for this particular problem, even using pen and paper is an overkill. \r\n\r\nAs to the exponentiation algorithm itself, I completely agree with jmerry. As a matter of fact, I once faced a similar overflow problem and had to implement a long integer arithmetic myself. I did it the most stupid way by representing integers as arrays of digits in the system with base $100$. Of course, you'll have to write addition, subtraction, multiplication and division subroutines yourself, which takes time and patience (especially division), but once done, you can use it many times, so it really pays off.\r\n\r\nBTW, in your particular situation there is a quick fix. When you multiply two integers $a$ and $b$ both less than $10^6$ and want to get the remainder of the product modulo $10^6$, instead of c=(a*b)%1000000;, write c=(a*(b%1000)+1000*(a%1000)*(b/1000))%1000000; This will keep you within the range up to $2\\cdot 10^9$, which shouldn't cause an overflow.", "Solution_3": "Thank you very much! It's very helpful to me! :)", "Solution_4": "i implemented some time ago long integer arithmetic in c++.\r\ni used base $2^{16}$ because in this case you can use some asm and make even multiplication and division reasonably fast.\r\nanyway, it took me a long, long time to do it properly :P" } { "Tag": [ "function", "algebra proposed", "algebra" ], "Problem": "find all fonctions $ f \\mathbb{R}\\rightarrow \\mathbb{R}$ of class $ C^2$ such that :\r\n\r\n$ f(x^2)\\equal{}f(x\\sqrt{2})$", "Solution_1": "it is a classical fuctional equation! \r\nfirst for $ y=x\\sqrt{2}$ we see that $ f(-y)=f(\\frac{(-y)^2}{2})=f(\\frac{y^2}{2})=f(y)$ , so $ f$ must be an even fuction.\r\nfor $ x > 0$ we have $ f(x^2)=f(x \\sqrt{2})$=$ f((x \\sqrt{2})^{\\frac{1}{2}} \\sqrt{2})=....=f(x^{\\frac{1}{2^n}}2^{\\frac{1}{2^{n+1}}+\\frac{1}{2^n}+...\\frac{1}{2})}$.\r\n so for $ n \\mapsto +\\infty$ since $ f$ is a continious function and $ x^{\\frac{1}{2^n}} \\mapsto 1$ and $ \\frac{1}{2^{n+1}}+\\frac{1}{2^n}+...+\\frac{1}{2}=1-\\frac{1}{2^{n+1}} \\mapsto 1$ we get $ f$ is a constant function in positive real set and also by continuity at $ 0$ and $ f$ is even , $ f=c$ for some real number $ c$ which it is clear solution to this problem." } { "Tag": [ "algorithm", "quadratics", "continued fraction" ], "Problem": "How do you find the continued fraction expansion of a radical. For example $ \\sqrt{7}$?", "Solution_1": "There is a very straightforward [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion]algorithm[/url] (a simplification of the general continued fraction algorithm) in the case of quadratic surds. Do you understand how to compute the continued fraction expansion in general?" } { "Tag": [ "function", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Let $ V_{a}^{b}f$ be the [url=http://en.wikipedia.org/wiki/Bounded_variation#BV_functions_of_one_variable]bounded variation[/url] of a function.\r\n\r\nNow suppose $ f_{n}\\rightarrow f$ uniformly on an interval $ [a,b]$. Is it necessarily true that $ \\lim_{n\\rightarrow\\infty}V_{a}^{b}f_{n}\\equal{}V_{a}^{b}f$?\r\n\r\nIn our analysis section, we tried to come up with some counterexample and we came up with the function $ f_{n}(t)\\equal{}\\frac{1}{n}\\sin(nt)$, which uniformly approximates the function $ f(t)\\equal{}0$. However, there was a disagreement as to whether or not $ V_{a}^{b}f_{n}\\rightarrow 0$. So does it?", "Solution_1": "this is of course false. You can find a lot of continuous functions $ f$ such that $ \\|f\\|_{\\infty}<\\epsilon$ and the variation of $ f$ over $ [a;b]$ is as large as you want (or even infinite).", "Solution_2": "We assumed it was false. We had some disagreements as to whether or not my function worked.", "Solution_3": "It does: you have about $ \\frac{b\\minus{}a}\\pi n$ humps of height $ 1/n$, so the total variation is at least some fixed constant, which is enough to conclude that it does not tend to $ 0$.", "Solution_4": "And if we make the example $ f_n(x)\\equal{}\\frac1n\\sin(n^2x),$ then we can have $ f_n$ tending uniformly to zero while $ V_a^bf_n\\to\\infty.$ \r\n\r\n(Just making concrete something alekk already said above.)" } { "Tag": [ "probability", "articles", "conditional probability" ], "Problem": "Your friend has two children. You meet one of them and find out that he is a boy. What is the probability that the other child is a girl?\r\n\r\nP.S. - A child can only be a boy or girl", "Solution_1": "[hide=\"Solution\"]Assume WLOG that the probability of having a boy is equal to the probability of having a girl. (Not explicitly stated; you wanted an elaborate solution.)\n\nTherefore are three cases:\n\nCase 1: The family had a boy first, then a girl.\n\nCase 2: The family had a girl first, then a boy.\n\nCase 3: The family had two boys.\n\nNow, each case has the same probability of arising. Therefore, two cases out of three state that the other child is a girl, and one case out of three states that the other child is a boy. Therefore, the probability that the family has a girl is $ \\displaystyle\\frac {2}{3}$.[/hide]", "Solution_2": "See http://en.wikipedia.org/wiki/Conditional_probability and http://en.wikipedia.org/wiki/Boy_or_Girl_paradox", "Solution_3": "[hide=\"Different Solution\"]As mentioned in the above wikipedia article, the probability is actually $ \\frac{1}{2}$! However, if your friend told you that one of his children was a boy, then the probability would be $ \\frac{1}{3}$. This is because finding that the child you met is a boy does not affect the probability that the child you didn't meet is a girl.\n\nIf the probability was not $ \\frac{1}{2}$, then if the child you met said she was a girl, the probability that the other child is a girl would be different than if the child you met said he was a boy. However, it is intuitive that this is not the case.\n\nYou can also think of it this way: there are four possibilities for your friend's children, where the first child is the one you met:\nBB BG GB GG\nSince the child you met said he was a boy, this rules out the last two possibilities, and thus the probability that the other child is a girl is $ \\frac{1}{2}$. However, if your friend told you that one of his children was a boy, only the last possibility would be ruled out, and the probability that the other child is a girl would be $ \\frac{2}{3}$.\n\nFinally, as Walk Around The River suggested, we can use conditional probability to get the answer. Assume you met one of the children. Let $ A$ be the event that the child you met is a boy, and $ B$ be the event that the child you didn't meet is a girl. By the rules of conditional probability:\n$ P(B|A)\\equal{}\\frac{P(B\\&A)}{P(A)}$\nWe know that $ P(A)\\equal{}\\frac{1}{2}$. Furthermore, since $ P(B)\\equal{}\\frac{1}{2}$ and the events $ A$ and $ B$ are not correlated, $ P(B\\&A)\\equal{}\\frac{1}{4}$. Thus\n$ P(B|A)\\equal{}\\frac{1/4}{1/2}\\equal{}\\frac{1}{2}$\n\nThis is a very tricky question, and I may not have explained it very clearly. Feel free to ask if you have any more questions![/hide]" } { "Tag": [ "AMC 10", "AMC 10 B", "AMC" ], "Problem": "Does anyone have the answers to the 2005 AMC 10 B?\r\n\r\nI did the problems in the wiki but I didn't realize that there were no solutions posted. \r\nI just want to see how much I would have scored, so I don't need the solutions.\r\n\r\nThanks in advance.", "Solution_1": "See the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43]Resources Section.[/url]", "Solution_2": "See also:\r\n\r\nhttp://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2005-10a/05itemdiff10.html\r\n\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", "Solution_3": "Thanks for the help! :)" } { "Tag": [ "geometry" ], "Problem": "Five circles $\\omega_{1}, \\omega_{2}, \\ldots \\omega_{5}$ are drawn so that the radius of each circle is equal to $16,8,4,2, \\text{ and }1$ respectively; the centers the of the circles are collinear in the order given; and the center of each circle, other than $\\omega_{1}$, is on the circle which has the closest, but larger, radius than the circle to be drawn ($\\omega_{2}$ is on $\\omega_{1}$, etc.) Find the common area of the five circles.", "Solution_1": "[quote=\"cincodemayo5590\"]Five circles $\\omega_{1}, \\omega_{2}, \\ldots \\omega_{5}$ are drawn so that the radius of each circle is equal to $16,8,4,2, \\text{ and }1$ respectively; the centers the of the circles are collinear in the order given; and the center of each circle, other than $\\omega_{1}$, is on the circle which has the closest, but larger, radius than the circle to be drawn ($\\omega_{2}$ is on $\\omega_{1}$, etc.) Find the common area of the five circles.[/quote]\r\n\r\n[hide]Region between $\\omega_{1}$ and $\\omega_{2}$ has the area of\n$4 \\cdot \\frac16 \\cdot 64 \\pi-2 \\cdot 64 \\cdot \\frac{\\sqrt{3}}{4}$ $=\\frac{128}{3}\\pi-32 \\sqrt{3}$.\n\nBecause the pattern repeats, we can add the area of the circles, and subtract the intersections repeatedly.\n\nFirst the area of the circles sum up to\n$16^{2}\\pi+8^{2}\\pi+4^{2}\\pi+2^{2}\\pi+1^{2}\\pi=341 \\pi$.\n\nThe intersections has the pattern that continues while the area decreases to $\\frac{1}{4}$. So\n$(1+\\frac14+\\frac{1}{16}+\\frac{1}{64})(\\frac{128}{3}\\pi-32 \\sqrt{3})$ $=\\frac{170}{3}\\pi-\\frac{85}{2}\\sqrt{3}$.\n\nSubtract the intersections from the area of the circles:\n$\\frac{853}{3}\\pi+\\frac{85}{2}\\sqrt{3}$.[/hide]\r\n\r\nThat doesn't seem very right though :blush:" } { "Tag": [], "Problem": "1. A book weighing 3N is resting on a table 1 meter above the floor. How much force is the table exerting on the book?\r\n\r\n2. How much work is the table doing on the book?", "Solution_1": "1. If the book is in equilibrium, then the sum of the forces that act on it must be zero. The only forces applied to the book are its weight and the force applied by the table, which then must have the same intensity of the weight (3 N), same direction and opposite sense.\r\n\r\n2. Work is a way of tranferring energy by means of some kind of movement. The force that the table exerts on the book is not making it moving, and so the work is zero." } { "Tag": [ "inequalities", "geometry", "trigonometry", "function", "calculus", "AMC", "USA(J)MO" ], "Problem": "Prove the inequality for acute triangle abc with area K such that,\r\n\r\n27((b^2+c^2-a^2)^2)((a^2+c^2-b^2)^2)((a^2+b^2-c^2)^2) :le: (4K)^6\r\n\r\nGosh, I hope I typed that up right!! Well, have fun with this one.\r\n\r\nIW.OAV", "Solution_1": "[color=cyan]I give you two approaches.\n\nNumber 1: notice that we have an inequality which appears to have an equality case. Also, equality appears to occur only when a=b=c. This makes it unlikely that the triangle inequality will prove useful, and instead implies AM-GM. So, put the area in terms of a, b, and c using Heron's Formula and multiply everything out. All the way. Try to make AM-GM work. An hour or less will tell you whether this really does work out in the end.\n\nNumber 2: b:^2: + c:^2: - a:^2:. Hmm. Look familiar? Think about the Law of Cosines. K, eh? How about abSinC/2 ? Go with it from there.\n\n\nNotes: I got bored of number 1 about a third of the way through, and I'm not positive it will work. And if it doesn't, I'd be really interested in knowing why exactly. Because, well, it should. As for number 2, you end up with a trig relationship that I wasn't able to finish off entirely. But it's definitely very doable that way.[/color]", "Solution_2": "Sir, I have tried those approaches and they have been futile... can anyone else give me the proof or a link to the proof, it's supposedly a famous one. If there is no proof, well... there's my science fair project!\r\n\r\nIW.OAV", "Solution_3": "Well, you can get someting like:\r\n\r\n([(CosA)(CosB)(CosC)]^2)/([(SinA)(SinB)(SinC)]^2) :le: 64/27\r\n\r\nI honestly don't know whether that will get you anywhere.", "Solution_4": "[color=cyan]You should have lost the 64 somewhere, Long. That expression can be turned around rather nicely into \ntan(A)*tan(B)*tan(C) :ge: 3:rt3:.\n\nClearly, the equality case for this is A=B=C=60 degrees. This shouldn't be that hard, I don't think. Perhaps some sort of smoothing argument? Or replacing tan(A)*tan(B)*tan(C) with tan(A) + tan(B) + tan(C)?\n\n\nOr use calculus. That should work. Since C is an easy funtion of A and B.[/color]", "Solution_5": "um, yup yup. I should have lost the 64 somewhere, when i canceled out all the powers of 2. \r\n\r\nI was thinking of doing the tangents, but I don't where to go after that...", "Solution_6": "[color=cyan]Join the club :). Calculus should work, though. If only I remembered how . . .[/color]", "Solution_7": "Did you try smoothing? I didn't, but it might work.", "Solution_8": "[color=cyan]I thought about it, but I haven't tried it. It certainly looks doable. Maybe when it isn't 1 AM . . .\n\nI did do it in the case where one of the angles is equal to 60 degrees, although admittedly that doesn't tell you a whole lot . . .[/color]", "Solution_9": "what's a smoothing argument?", "Solution_10": "For a smoothing argument, you might, for example, replace A by 60+x and show that the expression is minimized when x=0.", "Solution_11": "ahh, I see. \r\n\r\nBut maybe we're spending lost time. Maybe its not supposed to get (tanAtanBtanC)^2....", "Solution_12": "[color=blue]I think smoothing works. First, show it works when one of the angles is 60 degrees. Then, try smoothing out three angles into a triangle with one 60 degree angle. I got it to work last night and didn't quite make it work this morning. But I'm pretty sure it's doable.[/color]", "Solution_13": "where did u learn about smoothing.", "Solution_14": "Come on punks, one of yous guys gots ta be able to solve this problem.", "Solution_15": "Don't think it's insolvable... it's doable. You math hotshots should be able to do this... It's a simple inequality!!! Try using schurs or something, and look at this year's USAMO problem I mean come on, they both have a,b, and c's in em so you should be able to do it... Perhaps this should be an Olympiad problem... IMHO it is of olympiad stature...\n\n\n\nPeace\n\nIW.OAV\n\n\n\n[hide]Chickens taste good on mondays whereas on other days I enjoy spraying - on my brother cuz it's funny. After that my mom takes us for ice cream and we have lots of fun. I like to eat ice cream because it tastes yummy in my tummy! Shizam![/hide]", "Solution_16": "[color=cyan](1) why weren't your last 2 messages the same message?\n(2) why are you insulting people trying to help you out?\n(3) why are you insulting people for not being able to do a problem you can't do either?\n(4) why do you care so much, anyhow?\n(5) why don't you try and use some of the ideas that we have already given? The trig thing looks very very promising.[/color]", "Solution_17": "Whoa buddy, chill out... I just really wanna solve this problem and methods you guys give aren't really cuttin for me... I don mean to insult anyone", "Solution_18": "[quote=\"Idiot_without_a_village\"]Whoa buddy, chill out... I just really wanna solve this problem and methods you guys give aren't really cuttin for me... I don mean to insult anyone[/quote]\r\n\r\nIf the methods that JBL, ComplexZeta, darkquantum, etc. are giving you aren't working out for you, just say so. And if YOU want to solve the problem, why are you insulting other people and asking THEM to solve it for you? If you want to solve it, you should do so yourself.\r\n\r\nAnd if you're kiddin' when you call people \"hotshot\", make sure you let them know it. Otherwise they may percieve it as an insult. It sure does to me.", "Solution_19": "Is it that hard for you all to take a joke?\r\n\r\nPeople like you make me regret posting the question in the first place.\r\n\r\nBy the way this is called Ono's inequality or something like that... \r\n\r\nAnd quite frankly, I don't see the need for you to purport on everyone of my posts when all you really do is make other people feel bad. I was just tryin to get peeps to help me with problem in a manner that wasn't so anal.\r\n\r\nThanks.. I think.\r\nIW.OAV", "Solution_20": "OK guys!!!\r\n\r\nI;ve got it!!!!!!!!!!!!!!!!!!!!!!!!\r\n\r\n...hopefully", "Solution_21": "OK, a,b,c are the sides of hte triangle, and let A, B, C be the respective angles.\r\n\r\nNote that K=(abSinC/)2=(bcSinA)/2=(acSinB)/2\r\n\r\nThus, (4K)^6= [(abSinC/)2]^2*[(bcSinA)/2]^2*[(acSinB)/2]^2 which is equal to 4^3*[abSinC]^2*[bcSinA]^2*[acSinB]^2\r\n\r\nnotice that (a^2+b^2-c^2)^2=(2abCosC)^2\r\n (b^2+c^2-a^2)^2=(2bcCosA)^2\r\n (a^2+c^2-b^2)^2=(2acCosB)^2\r\n\r\nNow, divide the 27 and 4^3 from the left to the right, and the [abSinC]^2*[bcSinA]^2*[acSinB]^2 from the right to the left.\r\n\r\nYou are left with [(CosA*CosB*CosC)^2]/[(SinA*SinB*Sinc)^2] :le: 1/27\r\n\r\nTake the reciprocal and square rooting both sides, we get \r\n\r\ntanA*tanB*tanC :ge: 3 root 3\r\n\r\nWe know that in a triangle, the product of the tangents is equal to the sum of the tangents (this is easily proven by letting the third angle equal 180-(A+B) and then using the tangent sum of angle formulas)\r\n\r\nThus, tanA + tanB + tanC :ge: 3 root 3\r\n\r\nLet tanA + tanB + tanC =x\r\n\r\nBy the AM-GM inequality, \r\n\r\nx/3 :ge: the cube root of tanA*tanB*tanC, \r\n \r\nbut we know tanA*tanB*tanC=x,\r\n\r\ntherefore, we get x/3 :ge: the cube root of x\r\n\r\nSolving for x by some simple algebraic manipulation, we get that x :ge: 3 root 3.\r\n\r\nThus, tanA*tanB*tanC=tanA + tanB + tanC :ge: 3 root 3.\r\n\r\n\r\n\r\nQED", "Solution_22": "Hopefully, if that is right, it eliminates the need for painful smoothing arguments or calculus/jensen's theorem. \r\n\r\nI guess, Joel, that your not at your optimum at 1am! :D", "Solution_23": "[color=cyan]ooh, sexy! Nice job. \n\nIW.OAV, I am withholding my response to you, except to tell you not to be so rude.[/color]" } { "Tag": [ "blogs", "limit", "calculus", "calculus computations" ], "Problem": "f:(-\u03b5, \u03b5)->R, \u03b5>0, f(0)=0 and f'(0) exists, show that \r\nlim(x->0) (1/x)[f(x)+f(x/2)+....+f(x/k)]=(1+1/2+...1/k)f'(0), \r\nk belogs to N*.\r\nAll help appreciated. :)", "Solution_1": "Just to make things readable:\r\n\r\nLet $ f \\, : (\\minus{}\\epsilon,\\epsilon) \\to \\mathbb{R}$ be given, $ f(0) \\equal{} 0$ and $ f'(0)$ exists.\r\n\r\nShow that:\r\n\r\n$ \\lim_{x \\to 0} \\frac1x \\sum\\limits_{n\\equal{}1}^k f(\\frac{x}{n}) \\equal{} f'(0) \\sum\\limits_{n\\equal{}1}^k \\frac1n$, where $ k \\in \\mathbb{N}^*$.", "Solution_2": "It's sufficient to see that $ \\displaystyle \\lim_{x\\rightarrow 0}\\frac{f(x/n)}x\\equal{}\\lim_{x\\rightarrow 0}\\frac{f(x/n)\\minus{}f(0)}{x/n}\\frac 1n\\equal{}f'(0)\\frac 1n$\r\n :cool:", "Solution_3": "Thank you very much Diogene :)" } { "Tag": [ "vector", "linear algebra", "linear algebra unsolved" ], "Problem": "Let H be a subspace of dimension n - 1 of an n-dimensional vector space V. Show that if U is a subspace of V and U $ \\nsubseteq$ H, then H + U = V and dim (H$ \\cap$U) = dim(U) - 1.", "Solution_1": "By $ U\\nsubseteq H$,then there exist $ x \\in U$,but $ x \\notin H$.Then it is not hard to prove $ H\\plus{}x$ spans $ V$(In fact we can find a basis of $ H$,say it $ e_1,\\cdots,e_{n\\minus{}1}$,then $ e_1,\\cdots,e_{n\\minus{}1},x$ is linearly independent,thus it is a basis of $ V$)\r\nFor the next,notice:\r\n$ dim(U \\cap H)\\equal{}dim(U)\\plus{}dim(H)\\minus{}dim(U\\plus{}H)$" } { "Tag": [ "IMO", "IMO 2008" ], "Problem": "Results until now:\r\nIRN1(Kasra): 7 7 - 7 7 -\r\nIRN2(Milad) : 7 7 - 7 7 -\r\nIRN3(Nima) : 7 2 - 7 7 -\r\nIRN4(Mohamad) : 7 7 - 7 7 -\r\nIRN5(Amir) : 7 7 - 7 7 -\r\nIRN6(Mahdi) : 7 7 - 7 - -", "Solution_1": "wow! that's pretty good up till now. congrats!", "Solution_2": "IRN6 on problem5 is 0 or...", "Solution_3": "As same for others (-) means, I don't know the exact mark :wink:", "Solution_4": "Oh, IRN6(mahdi) got 2 in fifth.", "Solution_5": "The total score will be something like 181 or 182.", "Solution_6": "congratulation!!! As usual iran do it again and guys proved their ability. i think iran will be 1 or 2 ? am i right? :lol:", "Solution_7": "Of course not.\r\n\r\nAs I know rank of Iranian team will be 6 or 7.", "Solution_8": "Final Result:\r\nIRN1(Kasra): 7 7 1 7 7 7=36\r\nIRN2(Milad) : 7 7 2 7 7 0=30\r\nIRN3(Nima) : 7 2 3 7 7 0=26\r\nIRN4(Mohamad) : 7 7 0 7 7 2=30\r\nIRN5(Amir) : 7 7 1 7 7 0=29\r\nIRN6(Mahdi) : 7 7 7 7 2 0=30\r\n\r\nThe total score =181", "Solution_9": "CONGRATULATIONS!!!\r\nI think IRAN will be 3 or 4.\r\n\r\nWhat about Uzbekistan teams result?", "Solution_10": "[url]http://www.mathlinks.ro/viewtopic.php?p=1195339#1195339[/url]" } { "Tag": [ "trigonometry", "limit", "calculus", "calculus computations" ], "Problem": "I have 2 problems I got wrong on a quiz and I can't seam to figure them out\r\n\r\n\r\nlim [u]xcos(x)[/u]\r\nx->0 cos(x)-1\r\n\r\nI know I need to use L'Hospital's rule because its 0/0 but I can't get it to work out\r\n\r\nlim [u] 2[/u]\r\nx-> x^(x-1)", "Solution_1": "A hint for you. Observe that\r\n\\[ \\frac{{x\\cos x}}{{1 \\minus{} \\cos x}} \\geqslant \\frac{1}{x}\r\n\\]\r\nprovided $ x>0$ small enough.", "Solution_2": "[hide=\"Hint for limit 1\"]Multiply and divide by (cos x + 1) and use the known limit lim(x->0) (sin x/x) = 1.[/hide]\n\n[hide=\"Answer to limit 1\"]$ \\lim_{x \\to 0} \\frac {x \\cos x}{\\cos x \\minus{} 1} \\equal{} \\lim_{x \\to 0} \\frac {x \\cos x(\\cos x \\plus{} 1)}{ \\minus{} \\sin^2x} \\equal{} \\lim_{x \\to 0} \\minus{} \\left(\\frac {x}{\\sin x}\\right)^2 \\cdot \\frac {1}{x} \\cdot \\cos x(\\cos x \\plus{} 1)$\n\n$ \\equal{} \\minus{} 1^2 \\cdot \\infty \\cdot 2 \\equal{} \\infty$.[/hide]\r\n\r\nRegarding your second limit, x tends to what?", "Solution_3": "X-> 1+ sry for the 2nd one\r\n\r\nbut for the 1st, we had to solve it using L'Hospitals rule", "Solution_4": "Using L'H\u00f4pital:\r\n\r\n$ \\lim_{x \\to 0} \\frac {x \\cos x}{\\cos x \\minus{} 1} \\equal{} \\lim_{x \\to 0} \\frac {\\cos x \\minus{} x \\sin x}{ \\minus{} \\sin x} \\equal{} \\frac {1 \\minus{} 0}{0} \\equal{} \\infty$.\r\n\r\n[quote=\"Amandasmith\"]X-> 1+ sry for the 2nd one[/quote]\r\n\r\n(\"sry\" ?). Then the limit is $ 1^0 \\equal{} 1$.", "Solution_5": "sry = sorry :D" } { "Tag": [ "calculus", "integration", "function", "real analysis", "real analysis unsolved" ], "Problem": "Suppose that $f,g:[0,1]\\rightarrow\\mathbb R$ are integrable and both have integral equal to 1. Prove that there is an interval $[a,b]\\subset[0,1]$ such that in that interval both function have integral equal to $\\frac12$.\r\n\r\nBest,", "Solution_1": "Here is the interesting reformulation of the problem:\r\n\r\nOn a blackjack machine you can win or loose one dollar at a time. Show that if two players were playing once every minute in a period, and both of them won 2N dollars, then there was a period during which each of them won exactly N dollars\r\n\r\nSource: Homepage of Vilmos Totik http://tarski.math.usf.edu/~totik/problems.html" } { "Tag": [ "geometry", "incenter", "circumcircle", "perpendicular bisector", "angle bisector" ], "Problem": "i hope this is my own question and no one else has invented it\r\n\r\n\r\nlet ABC be an acute triangle. M,N are on BC, P,Q are on AC, X,Y are on AB, such that BM=BA, AC=CN, AP=AB, BC=CQ, AX=AC, BC=BY. (M is towards C, N towards B, P towards C, Q towards A, X towards B, Y towards A) prove that the perpendicular bisectors of MN, PQ, and XY are concurrent.", "Solution_1": "[hide]It is very easy to prove by suming up segments that the midpoints of MN, PQ, and XY are the tangency points of the inscribed circle of ABC with the sides BC,AC,AB hence these three perpendicular bisectors concur at the incenter I of ABC[/hide]", "Solution_2": "$BM=AB$ means that the perpendicular bisector of $AM$ is the bisector of $\\angle ABC$; similarly, the angle bisector of $\\angle ACB$ is the perpendicular bisector of $AN$, hence incenter $I$ of the triangle $\\triangle ABC$ is the circumcenter of $\\triangle AMN$; similarly for $\\triangle BPQ, \\triangle CXY$, hence the three perpendicular bisectors pass through $I$.\n\nBest regards,\nsunken rock" } { "Tag": [], "Problem": "[b]Problema B4133:[/b]\r\nFie $ ABCD$ un dreptunghi si punctele $ P\\in(BC), \\ Q\\in(CD)$. \r\nSa se arate ca in cazul in care $ \\triangle{APQ}$ este echilateral, avem: $ S_{AQD}\\plus{}S_{ABP}\\equal{}S_{CPQ}.$\r\n\r\n[b]Problema B4139:[/b]\r\nIn triunghiul ascutitunghic $ ABC$ notam cu $ E$ si $ F$ piciarele inaltimilor duse din varfurile $ B$ si $ C$, iar cu $ P$ si cu $ Q$ punctele de intersectie ale dreptei $ EF$ cu cercul circumscris triunghiului. \r\nAratati ca: $ |AP| \\equal{} |AQ|.$\r\n\r\n[b]Problema B4140:[/b]\r\nIn patrulaterul inscriptibil $ ABCD$ punctele $ E$ si $ F$ sunt mijloacele arcelor mici $ BC$ si respectiv $ AD$, iar cu $ P$ si $ Q$ centrele cercurilor inscrise in triunghiurile $ ABC$ si $ ABD$. \r\nAratati ca: $ PQ\\parallel{}EF$.", "Solution_1": "Semidreptele $ [AO$ si $ [AH$ sunt izogonale in $ \\triangle{ABC}$($ O$-fiind centrul cercului circumscris $ \\triangle{ABC}$)$ \\Rightarrow$ ele sunt izogonale si in $ \\triangle{AEF}$.\r\n$ [AH]$ este insa diametru in cercul circumscris patrulaterului inscriptibil $ AEHF\\Rightarrow AO\\perp{PQ}\\Rightarrow$ punctul $ \\{M\\}\\equal{}PQ\\cap AO$ este mijlocul corzii $ [PQ]\\Rightarrow$ dreapta $ AO$ este mediatoarea segmentului $ [PQ]\\Rightarrow |AP|\\equal{}|AQ|.$", "Solution_2": "Pe scurt, cum $ AO$ si $ AH$ sunt izogonale in $ ABC$, iar $ EF$ e antiparalela cu $ BC$, rezulta ca $ AO \\perp EF$, de unde concluzia.", "Solution_3": "These problems are from the current issue of K\u00f6MaL; they should not be discussed until the deadline (15 January 2008) expires.\r\n\r\nPlease do not destroy our work." } { "Tag": [ "blogs", "pigeonhole principle", "combinatorics proposed", "combinatorics" ], "Problem": "During a certain lecture, each of five mathematicians fell asleep exactly twice. For each pair of these mathematicians, there was some moment when both were sleeping simultaneously. Prove that at some moment, some three were sleeping simultaneously.", "Solution_1": "[hide=\"from my blog\"]\nFor each professor, there are 2 falling asleep moments, making 10 total. Call this set of moments S. For each pair, there is a moment at which both are asleep, making 10 also. Call this set of moments T. If any two moments in T are the same, we are done. If not, then they are all different.\n\nWe can correspond each moment in T with a moment in S because at the beginning of any interval in which two professors are sleeping, one of them must have just fallen asleep. However, the first 2 moments in S have to come before or at the same time as the first moment in T (since two have to fall asleep before two can be sleeping at the same time). Then the remaining 9 moments in T have to correspond with the 8 other moments in S, which means (by Pigeonhole) that 2 moments in T must be the same. Contradiction.\n\nHence there must be some moment at which three professors are asleep.[/hide]" } { "Tag": [ "modular arithmetic", "algebra proposed", "algebra" ], "Problem": "$ \\frac{1}{2},\\frac{1}{3},\\frac{2}{3},\\frac{1}{4},\\frac{2}{4},\\frac{3}{4},...,\\frac{1}{2005},\\frac{2}{2005},...,\\frac{2004}{2005}$ fractions are written by rising sequence. Find next fraction of this $ \\frac{5}{41}$ fraction.", "Solution_1": "[quote=\"zaya_yc\"]$ \\frac {1}{2},\\frac {1}{3},\\frac {2}{3},\\frac {1}{4},\\frac {2}{4},\\frac {3}{4},...,\\frac {1}{2005},\\frac {2}{2005},...,\\frac {2004}{2005}$ fractions are written by rising sequence. Find next fraction of this $ \\frac {5}{41}$ fraction.[/quote]\r\nLet us look for the next fraction with bottom part $ q$ : $ \\frac{p\\minus{}1}q\\le \\frac 5{41}<\\frac pq$ with $ p 3$ prime.\r\nIf $f$ has precisely two non-real roots in the complex numbers, then no $n$-th root of a rational can be a root of $f$.\r\n\r\nThe overkill solution (as proposed by ZetaX) would follow from the following Galois theory fact\r\nIf $f ( X ) \\in \\mathbb{Q}[ X ]$ is an irreducible polynomial over $\\mathbb{Q}$ and of degree $p$ prime, and $f$ has precisely two non-real roots in the complex numbers, then the Galois group of $f$ is $S_p$.\r\nTogether with this classical group theory fact: If $n > 4$, then $S_n$ is not solvable.\r\nI believe this is enough motivation to start learning Galois :)\r\n\r\nJohann : Have you got some background in group theory ?", "Solution_4": "Galois theory gives us that the no radical expression solves it, but when we just look for radicals / $n$-th roots itself it's a lot simpler:\r\nWhen we already know that it is irreducible and not of type $x^n-a$, then we are done. The reason is that any root (meaning 'nontrivial' roots, not roots of unity) has a minimal polynomial of type $x^m-b$ (these are irreducible by Eisenstein), and irreducibility of the original polynomial solves the rest.\r\nIn general, simply factor it into irreducible polynomials (well this isn't that simple to do in reality ;) ).\r\n\r\nAlso: it's enough that the polynomial has at least one nonreal and two real solutions to have a nonsolvable Galois group.\r\n\r\nEdit: some nonsense corrected.", "Solution_5": "[quote=\"ZetaX\"]\nAlso: it's enough that the polynomial has at least one nonreal and two real solutions to have a nonsolvable Galois group.\n[/quote]\r\n\r\nLet $f ( X ) \\in \\mathbb{Q}[ X ]$ be defined by $f ( X ) = X^4 - 1$.\r\nThe galois group $G$ of $f$ is a subgroup of $S_4$\r\n$S_4$ is solvable, therefore $G$ (as a subgroup of a solvable group) is solvable.\r\nIs the mistake on my or your side ?", "Solution_6": "Sorry, I forgot to mention: it has to be irreducible and prime degree as yours.", "Solution_7": "[quote=\"{x}\"]\n\nJohann : Have you got some background in group theory ?[/quote]\r\n\r\nWell, I have almost no serious background in Groups but I read I.N Herstein's book :). it is an optional study in my faculty course, and I studied by myself." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "In any triangle ABC prove that [tex] 18R^3\\geq(a^2+b^2+c^2)R+\\sqrt{3}abc [/tex]", "Solution_1": "I think it is very simple. It suffices to see that $ 9R^2\\geq a^2+b^2+c^2$ and then the AM-GM. Am I missing something3$", "Solution_6": "[quote=\"enndb0x\"]solve for the case when the variables are $ > 3$[/quote]\r\np=r=5, q=a=7", "Solution_7": "it's $ pq\\equal{}ra^2$ with $ (p,q)\\equal{}1$ and $ r$ prime...:)", "Solution_8": "[hide=\"Solution\"]\nWe know that since $ (p,q) \\equal{} 1$, one of $ p,q$ must contain $ a^2$ as a factor - two relatively prime factors of a perfect square are of course also perfect squares, so it's impossible for one to just be a factor of $ a^2$. Suppose that $ p$ does. Then in order for $ p$ not to be a perfect square, $ r|p$. But then $ p$ has to equal $ ra^2$ and then $ q$ is $ 1$, a perfect square.[/hide]", "Solution_9": "no it is $ pq\\equal{}ra$ not $ pq\\equal{}ra^2$", "Solution_10": "Except that $ pq \\equal{} ra$ means the problem is completely wrong. If you're going to insist on that you need to give a valid correction.", "Solution_11": "I took this problem from a book , there's a solution but I am not understanding the solution , so here is the solution.\r\n$ p = p_1^{k_1}p_2^{k_2}\\cdots p_m^{k_m}$,$ q = q_1^{s_1}q_2^{s_2}\\cdots q_t^{s_t}$ , $ a = a_1^{l_1}a_2^{l_2}\\cdots a_u^{l_u}$ are the canonic form of the number $ p$,$ q$ and $ a$ , when $ p$ and $ q$ has no common divisor greater then $ 1$.We have :$ p_1^{k_1}p_2^{k_2}\\cdots p_m^{k_m} \\cdot q_1^{s_1}q_2^{s_2}\\cdots q_t^{s_t} = r \\cdot a_1^{l_1}a_2^{l_2}\\cdots a_u^{l_u}$ when $ r$ is prime number , so two sides of equation (left side and right side)\r\nare the canonic form of the same integer.If $ r = a_i$ then we can write $ r\\cdot a_i^{2l} = a_i^{2l + 1}$. We have :${ p = p_1^{k_1}p_2^{k_2}\\cdots p_m^{k_m} = a_1^{{2l_i}_1}a_2^{{2l_1}_2}\\cdots a_n^{{2l_i}_n} = ({a_i}_1}^{{l_i}_1} {{a_i}_2}^{{l_i}_2} \\cdots {{a_i}_n}^{{l_i}_n})^2$\r\nso one of the numbers $ p$,$ q$ is perfect square.", "Solution_12": "Don't blame yourself for not understanding the solution; it looks like total garbage. They seemed to have in fact abruptly switched the problem to $ pq \\equal{} ra^2$ in the middle, since they act like they said $ r \\cdot a_1^{2l_1} a_2 ^ {2l_2} \\cdots$ when in the second half. I'm also not sure what they were thinking when they made that last statement about $ p$. It's true if you say WLOG that the exponent of $ a_i$ in $ q$ is odd (has to be odd in one of $ p,q$), but you can't just completely omit that.", "Solution_13": "They start with $ pq \\equal{} ra$ , and in the middle they seem to assume $ pq \\equal{} ra^2$ ?? :huh: \r\n\r\nCould anyone please help me understand this subtle/absurd solution? :( \r\n\r\nFrom which book did you take that problem?" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "what is the value of\r\n1- cosx cos2x cos3x cos4x ....cos nx", "Solution_1": "If $ x \\equal{} \\frac {(2k\\plus{}1)\\pi} {2\\ell}$ and $ n$ is large enough, the value is $ 1$. Otherwise ..." } { "Tag": [ "calculus", "integration", "trigonometry", "calculus computations" ], "Problem": "How would I find a reduction formula for the integral of $ \\frac{1}{(a\\plus{}bcos(x))^n}$.\r\nI tried substituting t=tan(x/2) but that doesn't seem to make life any easier.", "Solution_1": "Went back to this question and found it was not that hard just needed some manipulation:\r\nAssume $ b<>0$.\r\nThen divide out the b term and substitute $ u\\equal{}\\frac{a}{b}\\plus{}\\cos x$.\r\nAfter some work you can express the original $ I_n$ in terms of $ I_{n\\minus{}1}$, $ I_{n\\plus{}1}$ and a term involving $ \\frac{1}{(a\\plus{}b\\cos x)^n}$." } { "Tag": [ "logarithms", "inequalities", "triangle inequality" ], "Problem": "Solve the equation \r\n$\\sqrt{\\left(\\log_{3}\\sqrt[3]{3x}+\\log_{x}\\sqrt[3]{3x}\\right).\\log_{3}x^{3}}+\\sqrt{\\left(\\log_{3}\\sqrt[3]{\\frac{x}{3}}+\\log_{x}\\sqrt[3]{\\frac{3}{x}}\\right).\\log_{3}x^{3}}=2$", "Solution_1": "[hide=\"Sketch of solution\"]Let $\\log_{3}x = a$. It reduces to $\\sqrt{a^{2}+2a+1}+\\sqrt{a^{2}-2a+1}= 2$, i.e. $|a+1|+|a-1| = 2$. By the triangle inequality, this is true iff $\\textrm{sign}(a+1) = \\textrm{sign}(1-a)$, i.e. $-1\\leq a \\leq 1$. So $\\frac{1}{3}\\leq x \\leq 3$.\n\nEdit: we need $\\log_{x}3$ te be defined, so $x\\neq 1$. So our solution is $x\\in\\left[\\frac{1}{3},3\\right]\\setminus\\{1\\}$.[/hide]", "Solution_2": "[hide=\"solution\"]\nJust slightly different simplifying and factoring than above.\n\nSimplify, log rules\n$\\sqrt{\\log_{3}x}\\cdot \\left( \\sqrt{ log_{3}x+log_{x}3+2 }+\\sqrt{ log_{3}x+log_{x}3-2}\\right) = 2$\n\nSince $\\log_{3}x$ and $\\log_{x}3$ are reciprocles, the terms under the radicals both factor to squared terms. ((x + 1/x)^2 and (x - 1/x)^2)\n\n$\\sqrt{\\log_{3}x}\\cdot \\left( \\sqrt{ log_{3}x }+\\sqrt{ log_{x}3 }+\\sqrt{ log_{3}x }-\\sqrt{ log_{x}3 }\\right) = 2$\n\nWhich is $2\\log_{3}x = 2$, so $x=3$. (Which means it doesn't matter which term was negative when simplifying the second radical term.)\n\n[/hide]", "Solution_3": "hello, i have $x=3$ or $x=\\frac{1}{27}$.\r\nSonnhard.", "Solution_4": "[quote=\"Cicatriz\"]so $x=3$[/quote]\n\n[quote=\"Dr Sonnhard Graubner\"]hello, i have $x=3$ or $x=\\frac{1}{27}$.\nSonnhard.[/quote]\r\n\r\nIt's true for $x=2$, e.g., I checked it. I think my solution is correct.", "Solution_5": "I agree with Kurt Godel.\r\n\r\nWhen you take the square roots you have to remember that you get absolute values of the expressions. Then you have to do casework to see when you get solutions." } { "Tag": [ "probability", "counting", "distinguishability" ], "Problem": "A teacher with a class of 7 students has 10 pieces of candy. If each piece of candy is given to a student at random, what is the probability that every student gets a piece of candy?", "Solution_1": "3/286, Ill post a solution if this is right", "Solution_2": "[quote=\"usaha\"]3/286, Ill post a solution if this is right[/quote]\r\n\r\ncan you explain your answer so we can deem it correct or incorrect?\r\n\r\nedit: your answer isn't logical because there are more pieces of candy than students so there is a better chance than $\\frac{3}{286}$ that every student gets at least 1 piece.", "Solution_3": "No, I think its correct,\r\nIf every studnet gets a candy, then there are 3 candies remaining for any studnet, so thats 9C6\r\nThe total amount of ways the candy can be distributed is 16C6, and ((9C6)/(16C6) is 3/286", "Solution_4": "[quote=\"usaha\"]No, I think its correct,\nIf every studnet gets a candy, then there are 3 candies remaining for any studnet, so thats 9C6\nThe total amount of ways the candy can be distributed is 16C6, and ((9C6)/(16C6) is 3/286[/quote]\r\nLooks solid.", "Solution_5": "but a piece of candy is randomly given to a random student so the first 2 pieces given can be to the same student...or does that not matter? i suck at probability :blush: \r\n\r\nand by 16C6 do you mean $_{16}\\text{C}_{6}$?", "Solution_6": "[quote=\"7h3.D3m0n.117\"]but a piece of candy is randomly given to a random student so the first 2 pieces given can be to the same student...[/quote]\r\n\r\nIt \"doesn't matter\" what order you give out the candies in. If you treat the candies as being distinguishable then the only thing that matters is the final distribution of candy, so we can assume WLOG that the candy was given out in a way such that $7$ went to all $7$ students first.", "Solution_7": "[quote=\"usaha\"]No, I think its correct,\nIf every student gets a candy, then there are 3 candies remaining for any student, so thats 9C6\nThe total amount of ways the candy can be distributed is 16C6, and ((9C6)/(16C6) is 3/286[/quote]\r\n\r\n[hide=\"For those who do not understand where these numbers came from:\"]\n\n[u]Balls and Urns[/u]\n\nSay we draw a diagram depicting this situation, where each candy is a dot (indistinguishable from the others) and each person is an urn, which will be the space between two divider lines. The diagram looks like this:\n\\[| \\cdot | \\cdot | \\cdot | \\cdot | \\cdot | \\cdot | \\cdot | \\ \\ \\cdot \\cdot \\ \\cdot \\]\nThose last three dots are the extra candies to be distributed. Now, we need to devise a system to count all of the possible places where we can add these three candies, while not counting anything we don't want, like one person not having any candies. The method I use is to make larger elements out of the balls and walls, so that no matter what I move, each person (urn) will have at least one candy. The elements are boxed:\n\\[\\boxed{\\boxed{ | \\cdot }}\\boxed{ | \\cdot }\\boxed{ | \\cdot }\\boxed{ | \\cdot }\\boxed{ | \\cdot }\\boxed{ | \\cdot }\\boxed{ | \\cdot }\\cdot \\cdot \\ \\cdot \\boxed{\\boxed{|}}\\]\nNow, we cannot move the double-boxed elements on the left and right, since they determine the outermost urns. Note also that one of the two double-boxed elements needs one ball, so that the left-most or right-most urn always will have at least one ball, along with the others. The single-boxed elements and the extra balls can move around freely; the number of permutations of these is $\\frac{ 9!}{6! 3!}= \\dbinom{9}{6}$. Now, if do not place the restriction that all children must have candy, the diagram looks thus:\n\\[\\boxed{ \\boxed{|}}\\cdot | \\cdot | \\cdot | \\cdot | \\cdot | \\cdot | \\cdot \\cdot \\cdot \\cdot \\boxed{ \\boxed{ | }}\\]\nThe number of permutations in this situation is $\\frac{ 16!}{10! 6!}= \\dbinom{16}{6}$, and since we want the probability, we divide:\n\\[\\frac{ \\dbinom{9}{6}}{ \\dbinom{16}{6}}= \\boxed{ \\frac{3}{286}}\\]\n[/hide]" } { "Tag": [], "Problem": "Sa se demonstreze ca daca $K_1,K_2,K_3$ sunt subcorpuri ale corpului $K$, diferite de $K$ atunci $K_1\\cup K_2\\cup K_3$ e diferit de $K$", "Solution_1": "Este problema 2, ONM, 1987, propusa de C. Nastasescu.\r\nRecomand cartea \"Olimpiadele Nationale de Matematica, 1954-2003\", care se poate comanda online [url=http://www.librarie.net/carti/39911/Olimpiadele-Nationale-de-Matematica-]aici[/url]. Costa sub 300.000 si are peste 500 pagini. Merit\u0103 :)", "Solution_2": "Se poate demonstra ca un corp nu se poate scrie ca reuniune disjuncta de un numar finit de subcorpuri. \r\n\r\nO sa schitez o demonstratie prin reducere la absurd: \r\n\r\n1) Se trateaza cazul in care corpul $K$ este finit. \r\n2) Daca $K$ e infinit, Consideram cea mai mica multime (in sensul incluziunii) de subcorpuri care au ca reuniune pe $K$. Fie aceasta familie $K_1,K_2,...,K_n$.\r\n3) Se demonstreaza ca intersectia celor $n$ subcorpuri contine un sir infinit de elemente distincte (i.e. este infinita) prin inductie. \r\n4) Se demonstreaza ca intersectia lor are cel mult $n-1$ elemente. Presumpunem ca exista $a_1,...a_n$ apartinand intersectiei si consideram elementele de forma $aa_i+b$ unde $a\\in K_1\\setminus \\cup K_i,b\\in K_2\\setminus K_1$.\r\n\r\nDemonstratia e mai lunga si plina de detalii. Daca e necesar o sa mai completez unde e cazul.", "Solution_3": "[quote=\"xirti\"]Se poate demonstra ca un corp nu se poate scrie ca reuniune disjuncta de un numar finit de subcorpuri. \n[/quote]\r\nPai mi se pare evident. Daca $0$ e in unul din subcorpuri, fiind disjuncte, n-ar putea fi si in altul, deci clar nu ar mai fi subcorp. Cred ca ar trebui sa definesti mai clar ce intelegi prin disjuncte.", "Solution_4": "Nu vroiam sa zic nimic prin disjuncta :D singurele conditii sunt ca subcorpurile sa fie proprii." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "MATHCOUNTS", "ratio" ], "Problem": "Who did you think was going to win nationals.\r\nI thought David was, being the top 7th gradfer last year", "Solution_1": "Same as you, same reason.", "Solution_2": "I thought it was gonna be mark zhang because he got a 46 on state. But I don't know if anyone else got a 46..", "Solution_3": "I thought it would be Neal. Of course.\r\n\r\nActually, I thought it would be Mark Zhang. However, I wasn't very surprised to see Sergei win (my anti-speed, anti-kumon, etc. mind, I consider the written winner the winner). But yeah, Mark Zhang was my bet.", "Solution_4": "Pardha Ponugoti and David from IN did really good at state", "Solution_5": "Patricia, cuz shes from CA and so am i", "Solution_6": "Were there any other 46s? or just those 2?", "Solution_7": "there were no other 46's or 45's besides those three. Sergei got 44", "Solution_8": "[quote=\"aznness\"]Patricia, cuz shes from CA and so am i[/quote]\r\n\r\nsame reason, and you should have seen her own at state, i owned aznness over there 3-0 :D, not bragging or anything, and then she came and owned me 3-0, answering before i had even read the question...", "Solution_9": "[quote=\"aznness\"]Patricia, cuz shes from CA and so am i[/quote]\r\n\r\nsame reason. Go California!! Go for first place!! Oh wait we didn't win!! NO!!! a;sdlkfjapoeribth aeriobtupnaewrbtonaipoanb!!!!", "Solution_10": "I thought it was gonna be Mark Zhang cuz he aws really fast at countdown. Btw NoSoupForYou were you the guy who came over to the delaware table at the lunch on Saturday? or was that bob?\r\n\r\nMy count of AOPSers I met at nats: 2", "Solution_11": "Don't remember, but here is a way to tell us apart; bob is Asian, I'm not.\r\n\r\nAoPSers I met: 10+", "Solution_12": "Who were #10, #11, and #12?", "Solution_13": "Do you mean 10, 11, and 12th written winners?\r\n\r\n10) Pardha Ponugoti, Indiana\r\n\r\nThat's all I know.", "Solution_14": "10. Pardha\r\n11. Mike Jin- MO\r\n12. KAren Kota- KS", "Solution_15": "CA is always in the top 3, and consistently has more than 1 person in CD.", "Solution_16": "Not this year. . .", "Solution_17": "Ahem, we'll ignore that... but they did have a tie for 12th, which is just as good in terms of team score.", "Solution_18": "[quote=\"Phelpedo\"]CA is always in the top 3, and consistently has more than 1 person in CD.[/quote]\r\n\r\n2001, CA was 4th...", "Solution_19": "Boo hoo", "Solution_20": "CA will own next year because I will be on the team next year!!!\r\n[size=75]I wish[/size]", "Solution_21": "CA, as a team, won 5 times - more than any other team by far.", "Solution_22": "Yes and when they don't win they're still up there in the top ranks.", "Solution_23": "[quote=\"Phelpedo\"]TX, IN, and MA all had good years this year. By this I mean an exceptionally strong team. But CA is good nearly every year (although did you know that CA has never had a national champion?)[/quote]\r\n\r\nAs in, 1st place individual? Didn't TianKai Liu (sorry if I misspelled that) get first place in 2000? Or did he lose in CD?\r\n\r\n[b]EDIT:[/b] LynnelleYe... do you have an older sister who goes to North Hollywood High School? If not, it must just be the same last name...", "Solution_24": "[quote=\"JesusFreak197\"][quote=\"Phelpedo\"]TX, IN, and MA all had good years this year. By this I mean an exceptionally strong team. But CA is good nearly every year (although did you know that CA has never had a national champion?)[/quote]\n\nAs in, 1st place individual? Didn't TianKai Liu (sorry if I misspelled that) get first place in 2000? Or did he lose in CD?\n\n[b]EDIT:[/b] LynnelleYe... do you have an older sister who goes to North Hollywood High School? If not, it must just be the same last name...[/quote]\r\n\r\nHe got 2nd after countdown, I think...", "Solution_25": "[quote=\"NoSoupForYou\"]CA, as a team, won 5 times - more than any other team by far.[/quote]\r\nWhat about individual winner...........nope :!: :o ;)", "Solution_26": "Nope, no older sister. An irritatingly large number of people have my last name actually...almost every time they announce AMC perfect scorers or USAMO qualifiers or MathCounts top 25% at state or whatever they will have someone with my last name. Unfortunately it's never followed by my first name.", "Solution_27": "The reason why CA won 5 times as a team is because it has such a large population. That's one unfair thing about nationals. Since California has the largest population, it had the greatest chance of selecting the smartest people. \r\n\r\nIncidentally, a mixed blessing, because if California were to be on a level playing field with say, Wyoming, then it would be divided into 74 different states, all of whom got to bring 4 people to nations. Then, many more of your fellow Californians would have been able to make nationals.", "Solution_28": "The one downside to having such a big population is that students almost never make nationals two years in a row; so, some people from other states can get more experience with the actual national competition because they live in smaller states, and can attend more than one year.", "Solution_29": "Yeah but TX also has a large population; nowhere near as large as CA but certainly still large. Yet, their win ratio is much lower than their population ratio." } { "Tag": [ "counting", "distinguishability" ], "Problem": "John wants to buy a dozen flowers. There are three colors: red, pink, and purple. He wants to have at least one of each color. How many different ways can he choose the dozen of roses?", "Solution_1": "Are you familiar with Balls and Urns? If so, heres what you can do:\r\n$ r$= # of red flowers\r\n$ p$= # of pink flowers\r\n$ u$= # of purple flowers\r\n\r\nSince each one must be at least 1, we have:\r\n$ r \\plus{} p \\plus{} u \\equal{} 9$\r\nWhich by balls and urns has $ \\binom{11}{2} \\equal{} \\boxed{55}$ solutions.\r\n\r\nOR if you dont know balls and urns:\r\nWe have 12 flowers. Lets examine this by cases:\r\nCASE 1;1 red flower: We have $ 11$ flowers left. of those, we can have any number from $ 1 \\minus{} 10$ of the flowers purple (we cant have 11 otherwise there will be no pink flowers). This has a total of $ 10$ ways.\r\nCASE 2; 2 red flowers: We have $ 10$ flowers left. of those, we can have any number from $ 1 \\minus{} 9$ of the flowers purple.This has a total of $ 9$ ways.\r\nCASE 3; 3 red flowers:We have $ 9$ flowers left. of those, we can have any number from $ 1 \\minus{} 8$ of the flowers purple.This has a total of $ 8$ ways.\r\nNow we see a pattern. The solution, based on our pattern, is $ 1 \\plus{} 2 \\plus{} 3 \\plus{} \\dots \\plus{} 10 \\equal{} \\frac {10 \\cdot 11}{2}$(which is also $ \\binom{11}{2}$)$ \\equal{} \\boxed{55}$", "Solution_2": "I should explain the concept first...\r\n\r\nSO, here goes nothing.\r\n\r\nThe formula for balls and urns is $ \\binom {n \\plus{} k \\minus{} 1}{n \\minus{} 1}$ which distributes Indistinguishable to Distinguishable.\r\n\r\nIn your problem John wants to buy a dozen flowers. There are three colors: red, pink, and purple. He wants to have at least one of each color. How many different ways can he choose the dozen of roses?\r\n\r\nSince he must have at least one of each color, there are three colors, and so $ 12 \\minus{} 3 \\equal{} 9$\r\n\r\n$ 9 \\equal{} n$ in this case. \r\n$ 3 \\equal{} k$ \r\n\r\nSoo, by balls and urns is $ \\binom {9 \\plus{} 3 \\minus{} 1}{3 \\minus{} 1}$ $ \\equal{}$ $ \\binom {11}{2}$ $ \\equal{}$ $ \\boxed 55$\r\n\r\nDid this help?", "Solution_3": "Thank you so much for the help. Now I know where to go when I have problems.\r\n\r\nTech." } { "Tag": [ "topology" ], "Problem": "State an example of a space X and a topology T on X that is induced by no metric of X. :D\r\n\r\nHope you like the proposal.\r\n\r\nBest of luck.", "Solution_1": "Take any non-hausdorff space.", "Solution_2": "even better, take any non-normal space.", "Solution_3": "Or best, take any non-metric space.", "Solution_4": "[quote=\"ZetaX\"]Or best, take any non-metric space.[/quote]\r\n\r\nZetaX, could you please tell me what you exactly mean by [i]a non-metric space[/i]? :wink:", "Solution_5": "A space whose topology is not induced by a metric; i.e. a space that fulfills your requirements :wink:", "Solution_6": "Aren't those spaces typically referred to as non-metrizable spaces?\r\n\r\nThe expression non-metric space doesn't make much sense to me...", "Solution_7": "Let $ X$ be a set and $ T$ a collection of subsets of $ X$. We say that $ (X,T)$ is $ \\emph{a topological space}$ if\r\n1. $ T$ is closed under arbitrary unions.\r\n2. $ T$ is closed under finite intersections.\r\n3. $ \\emptyset\\in T$ and $ X\\in T$.\r\n(Actually 3. follows from 1. and 2. using properties of the empty set)\r\nTopological space $ (X,T)$ is said to be $ \\emph{metrizable}$ if there is a metric $ d$ such that $ (X,T) \\equal{} (X,T_d)$. If there is no such $ d$ then it is a non-metrizable topological space. For example the sets $ U(a,t) \\equal{} \\{a\\}\\cup [t,\\infty [,a,t\\in\\mathbb{R}$ forms a basis of a non-metrizable space. This follows from the fact that $ X$ is not Hausdorff but every metrizable space is Hausdorff." } { "Tag": [], "Problem": "In a class election, one candidate received more than $94\\%$ (but less than $100\\%$) of the votes cast. What is the least possible number of votes cast?", "Solution_1": "[quote=\"mathgeniuse^ln(x)\"]In a class election, one candidate received more than $94\\%$ (but less than $100\\%$) of the votes cast. What is the least possible number of votes cast?[/quote][hide] $\\frac{1}{x}<.06=\\frac{3}{50}\\Longrightarrow x>\\frac{50}{3}$. $x=17$. [/hide]", "Solution_2": "[hide]$\\frac{16}{17}$ is the smallest fraction whose decimal representation is bigger than 0.94. so 17[/hide]", "Solution_3": "You are all correct. Good Explanation." } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "perpendicular bisector", "geometry unsolved" ], "Problem": ":( :( :(", "Solution_1": "$ D_aM_a$ is the perpendicular bisector of the side $ BC.$ Perpendicular to $ D_aM_a$ at $ M_a$ is the sideline $ a.$ $ N \\in a$ is foot of perpendicular from $ S$ to $ a.$ $ S$ is the Fermat point $ \\Longrightarrow$ $ \\angle BSC \\equal{} 120^\\circ.$ Inversion with center $ S$ and power $ SN^2$ takes the line $ a$ into a circle $ \\mathcal A'$ with diameter $ SN.$ Unknown rays $ SB, SC$ through the inversion center $ S$ go to themselves, intersecting the known circle $ \\mathcal A'$ at unknown $ B', C'.$ Circumcircle $ \\mathcal P$ of $ \\triangle SBC$ passing through the inversion center $ S$ goes to the line $ p' \\equiv B'C'.$ This unknown line is tangent to a circle $ \\mathcal Q'$ concentric with the circle $ \\mathcal A'$ and diameter $ \\frac {_{SN}}{^2},$ because of the angle $ \\angle B'SC' \\equal{} \\angle BSC \\equal{} 120^\\circ.$ This circle can be constructed. Let $ T$ be reflection of $ S$ in the line $ D_aM_a$ and $ T'$ inversion image of $ T'.$ Since the circle $ \\mathcal P$ goes through $ T,$ the line $ p'$ goes through $ T'.$ Construct tangents from $ T'$ to $ \\mathcal Q'$ and pick one that separates $ S$ and the common center of $ \\mathcal A', \\mathcal Q'.$ This tangent intersects the circle $ \\mathcal A'$ at $ B', C',$ so that $ \\angle B'SC' \\equal{} 120^\\circ,$ it is the line $ p'.$ Rays $ SB', SC'$ cut the line $ a$ at $ B, C.$ Circumcircle $ \\mathcal O$ of $ \\triangle D_aBC$ is also circumcircle of $ \\triangle ABC.$ Internal bisector line of the angle $ \\angle BSC$ cuts $ \\mathcal O$ at $ A,$ such that $ \\angle CSA \\equal{} \\angle ASB \\equal{} \\angle BSC \\equal{} 120^\\circ.$" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "The followig problem is from Bulgarian Winter Mathematical Competition this year for 8th grade. No one of the contestants solved the problem.\r\nFind all integer numbers \"a\" and \"b\" such that $ a^{2} > \\frac {2.4.6\\cdots2008}{1.3.5\\cdots2007} > b^{2}$.\r\nI know some solutions.", "Solution_1": "And of course \"a\" and \"b\" must be consecutive numbers!", "Solution_2": "Any ideas?", "Solution_3": "It remains me Canadian Math Olympiad 1997 problem.", "Solution_4": "Is it the best selection for a and b? or we just find a, b with the suitable coefficients.", "Solution_5": "There are only two pairs of consecutive numbers \"a\" and \"b\" (depends on if the numbers are positive ot negative). We need \"a\" and \"b\" such that they are [b]consecutive[/b] numbers.", "Solution_6": "[quote=\"SAPOSTO\"]The followig problem is from Bulgarian Winter Mathematical Competition this year for 8th grade. No one of the contestants solved the problem.\nFind all integer numbers \"a\" and \"b\" such that $ a^{2} > \\frac {2.4.6\\cdots2008}{1.3.5\\cdots2007} > b^{2}$.\nI know some solutions.[/quote]\r\nwe must find natural N such that : $ N \\le \\sqrt {\\frac {2008!!}{2007!!}}\\le (N \\plus{} 1)$\r\n1.$ (2*4) < 3^2,(4*6) < 5^2 \\cdots (2004*2006) < 2005^2,(2006*2008) < 2007^2$ $ \\equal{} > \\sqrt {\\frac {2008!!}{2007!!}} < (2*2008)^{1/4} < 8$\r\n2.$ 4^2 > (3*5),6^2 > (5*7)\\cdots 2004^2 >$$ (2003*2005),2006^2 > (2005*2007) \\equal{} > \\sqrt {\\frac {2008!!}{2007!!}} > (\\frac {4*2007}{3})^{1/4} > 7$\r\nso : $ 7 < \\sqrt {\\frac {2008!!}{2007!!}}< 8$ :)", "Solution_7": "Good work! The solution I have is almost the same. Finally (7,8) and (-7,-8) are the numbers we wanted to find." } { "Tag": [ "logarithms", "function", "parameterization", "calculus", "analytic geometry", "graphing lines", "slope" ], "Problem": "The number of solutions of the equation $ 2^x(4\\minus{}x) \\equal{} 2x\\plus{}4$ is ______.", "Solution_1": "This is my first time doing a pre-olympiad problem, so my answer might be wrong. \r\n[hide=\"My Solution\"]\n$ 2^x(4 - x) = 2x + 4 \\\\\n2^x(4 - x) = 2(x + 2) \\\\\n\\log{(2^x)(4 - x)} = \\log{(2)(x + 4)} \\\\\n\\log{2^x} + \\log{(4 - x)} = \\log{2} + \\log{(x + 4)}$\nNow we have two log functions being applied on both side to the same base (10) so we can equate the parameters (I think that's what they're called).\n$ 1.) 2^x = 2 \\\\\nx = 1%Error. \"bigspace\" is a bad command.\n\\\\\n2.)2^x = x + 2\\\\\nno\\,solutions? \\\\\n3.)4 - x = 2 \\\\\n- x = - 2 \\\\\nx = 2%Error. \"bigspace\" is a bad command.\n\\\\\n4.)4 - x = x + 2 \\\\\n2x = 2 \\\\\nx = 1$\n\n$ x = 0,1,2$ all work when you plug them back in. As for Case #2, I am not sure how you would find the value of $ x$ for such equations. At most there will be 4 solutions, but if Case#2 has no solutions, then there are 3.\n[/hide]", "Solution_2": "[quote=\"sunehra\"]\n$ 2^x(4 \\minus{} x) \\equal{} 2x \\plus{} 4 \\\\\n2^x(4 \\minus{} x) \\equal{} 2(x \\plus{} 4)$\n[/quote]\r\n\r\n\r\ni think it's rather\r\n\r\n$ 2^x(4 \\minus{} x) \\equal{} 2(x \\plus{} 2)$\r\n\r\nno??", "Solution_3": "Yea it is XD. I just checked. It appears the solutions still work though for some odd reason.", "Solution_4": "[quote=\"sunehra\"][hide=\"My Solution\"]\nNow we have two log functions being applied on both side to the same base (10) so we can equate the parameters (I think that's what they're called).[/hide][/quote]\n\nUm... what? It's not true that $ \\log a \\plus{} \\log b \\equal{} \\log c \\plus{} \\log d$ necessarily implies that $ a \\equal{} c$ or $ a \\equal{} d$. You don't even need to take logs. \n\nI'm not sure how to solve this problem, but here's my partial\n\n[hide=\"Progress\"]Notice that if $ x > 4$, then $ 2^{x} (4 \\minus{} x) < 0$, so $ 2x \\plus{} 4 < 0$. But this is impossible if $ x > 4$.\n\nIf $ x < \\minus{} 2$, then $ 2^{x}(4 \\minus{} x) > 0$, so $ 2x \\plus{} 4 > 0$. But this is impossible for $ x < \\minus{} 2$.\n\nSo we have $ \\minus{} 2\\le x \\le 4$. The solutions are easy to verify: $ x\\equal{}0,1,2$, but I don't know how to get farther than that.[/hide]\r\n\r\nThis problem is like trying to solve $ 2^{x} \\equal{} 3x$. It's pretty annoying (to get a closed form of $ x$), but we can make a guess on how many solutions there are...", "Solution_5": "[hide=\"Calculus outline\"]\nLHS has exactly one inflection point, so use an argument with concavity and slopes of tangent lines to argue that a line can only intersect at most three times, giving the three solutions.[/hide]", "Solution_6": "hello, write your equation in the following form:\r\n$ 2^x(4\\minus{}x)\\equal{}2(x\\plus{}2)$\r\n$ 2^{x\\minus{}1}(4\\minus{}x)\\equal{}x\\plus{}2$\r\n$ 2^{x\\plus{}1}\\minus{}x\\cdot2^{x\\minus{}1}\\equal{}x\\plus{}2$\r\n$ 2^{x\\plus{}1}\\minus{}2\\equal{}x(2^{x\\minus{}1}\\plus{}1)$\r\n$ x\\equal{}\\frac{2^{x\\plus{}1}\\minus{}2}{2^{x\\minus{}1}\\plus{}1}$\r\nNow you can consider the function\r\n$ f(x)\\equal{}x\\minus{}\\frac{2^{x\\plus{}1}\\minus{}2}{2^{x\\minus{}1}\\plus{}1}$.\r\nSonnhard.", "Solution_7": "a simple study of office and you see that this function is zero 3 points" } { "Tag": [ "floor function", "logarithms", "number theory" ], "Problem": "Let u be a fixed positive integer. Prove that the equation $n! = u^{\\alpha} - u^{\\beta}$ has a finite number of solutions $(n, \\alpha, \\beta).$", "Solution_1": "[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows: \r\n\r\n[hide=\"Remark\"]\n- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917\n- second link\netc.\n[/hide]\r\n\r\n[b]Happy Problem-Solving![/b] :)", "Solution_2": "This is not a formal proof (maybe someone would like to formalize it): problem is equivalent to proving that $n!=u^{\\alpha}(u^{\\beta}-1)$ has finite number of solutions. By a well-known lemma if $p^k || n!$, and $p \\perp u$ and $p \\perp u-1$ then $p^k||\\beta$. That actually finishes the problem because $u^{\\beta}-1$ is growing much faster than $n!$ by above observation (that shouldn't be hard to prove).", "Solution_3": "What lemma\u00bf I see problems, e.g. $n=4$, $p=3$, $u=2$.", "Solution_4": "Well, please forget about this nonsense above :blush: I have realized later after I had posted it that it was completely wrong.", "Solution_5": "OK lets try to kill it (the idea is more or less that one of Megus)...\r\n\r\nLemma: when $ p$ is a odd prime and $ p \\nmid a$, then the order of $ 1 + ap^k \\mod p^s$ is $ p^{s - k}$ for all $ s > k$.\r\n(But I will not post the proof as long as noone wants to see it: it's boring calculation and induction)\r\n\r\nNow to the problem:\r\nWe want that $ v_p(n!) = v_p(u^b - 1)$ for all primes $ p$ not dividing $ u$.\r\nLook at any odd prime $ p$ dividing $ u - 1$ (if there is none, so when $ u = 2^k + 1$, just consider any odd prime dividing $ u^s - 1$ for some $ s$).\r\nLet $ p^k \\parallel{} u^s - 1$, so $ u^s = ap^k + 1$, $ p \\nmid a$, and by the above $ u^s$ has order $ p^{t - k}$ seen $ \\mod p^t$.\r\nWhen we also choose $ s$ to be minimal with $ p|u^s - 1$, we get that the order of $ u \\mod p^n$ is $ s \\cdot p^{t - k}$.\r\n\r\nNow lets compute what the order has to be for given $ n$:\r\nThere are $ \\left\\lfloor \\frac {n}{p}\\right\\rfloor > \\frac {n}{p} - 1$ multiples of $ p$ in $ n!$, thus $ v_p(n!) > \\frac {n}{p} - 1$.\r\nWe need that $ u^b \\equiv 1 \\mod p^{v_p(n!)}$, thus $ b$ being a multiple of $ p^{v_p(n!) - k}$.\r\n\r\nSo we have (finally!):\r\n$ n^n \\geq n! = u^a(u^b - 1) > u^{p^{v_p(n!) - k}} - 1 > u^{p^{\\frac {n}{p} - k - 1}} - 1$\r\ngiving:\r\n$ n^n \\geq u^{p^{\\frac {n}{p(p - 1)} - k - 1}}$\r\nwhich by taking logarithms yields:\r\n$ n^2 \\geq n \\cdot \\log(n) \\geq \\log(u) \\cdot p^{\\frac {n}{p(p - 1)} - k - 1}$\r\nor equivalently:\r\n$ \\log(u)^{ - 1} n^2 + \\cdot (k + 1) \\geq n \\cdot \\log(n) \\geq \\left( p^{\\frac {1}{p}\\right)^n}$\r\nBut this can only happen for finetely $ n$ since one the left is a polynomial, on the right is an exponential function.\r\n\r\nSo the problem is dead, dead, dead, any questions ;)\r\n\r\nEdit: simplified it a bit.\r\nEdit number two: changed the error mentioned below.", "Solution_6": "I suppose it should be $p^{s-k}$ instead of $p^{s-k-1}$ in the lemma (check the case for $k=1$: $(1+ap)^{p^{s-2}}\\equiv 1+ap^{s-1} \\mod p^s$) though it doesn't influence the solution which is nice :)", "Solution_7": "Thanks, I corrected it.\r\nBut it's very similar to what you mentioned above :)", "Solution_8": "can i ask something about these lemmas you people keep coming with?\r\ndo you make them to suit the needs for this specific problem or have\r\nyou read them some where and recall them as you notice they would\r\nbe usefull for these problems?", "Solution_9": "The one I used above is another way of expressing something some people call \"Hensels Lemma\" (see the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Formulary[/url]).\r\nIn general, some lemmas reappear from time to time. Others more rarely, and there are also some that you just make up for this special problem.\r\nThus I would say that you can't classify them.", "Solution_10": "Here is an alternative solution based on cyclotomic polynomial :) \r\nTo ZetaX:do you have a nice proof for the lemma below?", "Solution_11": "Here's a new solution with LTE.\n\nFirst, note that for any fixed $n$ there are only finitely many solutions because $\\alpha$ is bounded by considering a prime divisor of $u$ and then $\\beta$ is fixed when $\\alpha$ and $n$ are known. It suffices to show that only finitely many $n$ have solutions, then. We will prove that $n$ is bounded from above. \n\nRewrite the equation as\n \\[ n! = u^{\\beta} (u^{m}-1 ) \\]\n where $m=\\alpha-\\beta.$ Now, consider an odd prime $p$ not dividing $u$ and denote the order of $u$ modulo $p$ by $d$. For sufficiently large $n$, $p$ must divide $n!$ so $d$ must also divide $m$. By the Lifting the Exponent Lemma,\n \\[ \\nu_p(u^m-1) = \\nu_p(u^d-1) + \\nu_p( \\frac{m}{d} ) \\leq \\nu_p( u^d - 1 ) + \\log_p \\left(\\frac{m}{d} \\right) .\\]\n Letting $c=\\nu_p(u^d-1) - \\log_p(d)$, it becomes clear that $\\nu_p(u^n-1) \\leq \\log_p(m) + c$ for a fixed constant $c$. However by Legendre, \n \\[\\nu_p(u^m-1) = \\nu_p(n!) = \\sum_{k=1}^{\\infty} \\left\\rfloor\\frac{n}{p^k} \\right\\rfloor \\geq \\frac{n}{p} - 1. \\]\n From these two inequalities, we deduce that\n \\[ {m} \\geq p^{\\frac{n}{p} - c} \\]\n for a new constant $c$. Then, a chain of inequalities reveals that\n \\[ n^n \\geq n! = u^{\\beta} (u^m - 1 ) \\geq u^{\\log_p{m} + c} \\geq u^{p^{\\frac{n}{p} - c} } . \\]\n Taking logs of each side gives that\n \\[ n \\ln{n} \\geq p^{\\frac{n}{p} - c} \\ln{u} , \\]\n which can only hold for finitely many $n$ since the left hand side is less than a polynomial of $n$ and the right hand is an exponential in $n$. ", "Solution_12": "Not my solution with Zsigmondy's.\n\nRewrite the equation as $n!=u^a(u^b-1)$. Then\n$$a=v_u(n!)=\\sum_{k=1}^{\\infty} \\Biggl\\lfloor\\frac{n}{u^k}\\Biggr\\rfloor < \\sum_{k=1}^{\\infty}\\frac{n}{u^k}\\le n$$\nso $a\\le n-1$. Moreover, by Zsigmondy's theorem there exists a prime number $p$ such that $p|u^b-1$ and $p\\nmid u^m-1$ for any $m\\frac{n}{p}-1$\nNow let $\\nu_p(b-c)\\ge \\frac{n}{p}-k$ for some constant $k$ independent of $n$ and hence $n^n>n!=a^b-a^c \\ge a^{p^{\\frac{n}{p}-k}}$ and by taking logarithms we will be done.", "Solution_14": "[quote=ryan17]Not my solution with Zsigmondy's.\n\nRewrite the equation as $n!=u^a(u^b-1)$. Then\n$$a=v_u(n!)=\\sum_{k=1}^{\\infty} \\Biggl\\lfloor\\frac{n}{u^k}\\Biggr\\rfloor < \\sum_{k=1}^{\\infty}\\frac{n}{u^k}\\le n$$\n[/quote]\n\nIsn't it true for prime u?", "Solution_15": "[quote=SerdarBozdag][quote=ryan17]\n$$a=v_u(n!)=\\sum_{k=1}^{\\infty} \\Biggl\\lfloor\\frac{n}{u^k}\\Biggr\\rfloor < \\sum_{k=1}^{\\infty}\\frac{n}{u^k}\\le n$$\n[/quote]\n\nIsn't it true for prime u?[/quote]\n\nYou're right; Legendre's formula works only when $u$ is a prime. So @ryan17 takes a wrong step here.", "Solution_16": "Suppose that $n, b, c$ are positive integers such that $n! = a^{b} - a^{c}$ for a given $a$. \nThus by Exponent Lifting, we have that for all primes $p$, \\[v_p(a^b - a^c) = v_p(a^c) + v_p(a^{b-c} - 1) = cv_p(a) + v_p(a^{b-c}-1) = v_p(n!) \\]\n\nNote that $a$ and $a^{b-c} - 1$ are relatively prime so let $p_1, p_2, ... ,p_u$ be the primes which divide $a$.\\\\\n\nNow consider $p \\in \\text{\\textbraceleft} p_1, p_2, ... , p_u \\text{\\textbraceright}.$\nThen we know that $cv_p(a) = v_p(n!)$, so it must be true that \n\\[c \\hspace{1mm}|\\hspace{1mm} \\text{gcd}( v_{p_1}(n!), v_{p_2}(n!), ... , v_{p_u}(n!)). \\]\nso $c$ can take on only finitely many values for each $n$. Now let $r$ be the least prime which is greater than every prime which divides $a$ or $a^{b-c}-1$, so $n < r$ and thus $n$ is bounded. For each choice of $n$ and $c$, there is at most one value of $b$ which satisfies $n! = a^{b} - a^{c}$, so we are done. ", "Solution_17": "@above how do we know that there are finitely many primes which divide $a^{b-c} - 1$? Note that $a$ is fixed, but $b-c$ is not.", "Solution_18": "[hide=Solution] The equation rewrites as $n!=u^{\\beta}(u^{d}-1)$ for $d=\\alpha-\\beta$. Suppose that $u \\neq 1$ and take a prime $p \\mid u$; we have $p \\nmid u^d-1$, so $\\beta \\leq \\beta\\nu_p(u)=\\nu_p(n!)<\\frac{n}{p-1}\\leq n$, so $\\betad$ divides $n!$, which yields that $d2p_k$, then $n$ divides $(n-k)!$. [/list]", "Solution_1": "The [url=http://www.kalva.demon.co.uk/apmo/asoln/asol033.html]solution on Kalva[/url] seems incomplete to me... at least I see no justification why we'd only have those 3 cases, e.g. why we cannot have $ q x+164i=4i(x+n+164i)\n =>x+164i=-656+4(x+n)i\n =>x=-656 and 4(x+n)=164\nThen we find n=41-x\n n=697", "Solution_20": "MY FIRST COMPLEX NUMBER AIME PROBLEM! :winner_first: In my birth year :love: \n\nLet $z = a + 164i$, replacing that into the equation yields:\n$$\\frac{a + 164i}{n + a + 164i} = 4i$$\nCross multiplying gets us:\n$$a + 164i = (4(a + n))i - 656$$\nRearrange.\n$$a + 656 = (4(a + n))i - 164i$$\n$$a + 656 = (4(a + n) - 164)i$$\nThis implies that $(4(a + n) - 164)i$ is real, so $4(a + n) - 164 = 0$.\n$$\\implies a = -656$$\n$$\\implies n = \\boxed{697}$$\n", "Solution_21": "[hide=Solution]\nWe let $z=a+164i$.\n\nWe can substitute $z=a+164i$ into $\\frac{z}{z+n}=4i$ to get\n\\begin{align*}\n\\frac{a+164i}{a+n+164i} &= 4i \\\\\na+164i &= -656+4ai+4ni.\n\\end{align*}\n\nWe see that both $a$ and $n$ are real so we can equate real and imaginary parts to get\n\\begin{align*}\na &=-656 \\\\\n164 &= 4a+4n \\\\\n41 &= a+n \\\\\n41 &= -656+n \\\\\n\\boxed{697} &= n. \\\\\n\\end{align*}\n[/hide]", "Solution_22": "[hide = Solution]\nSo we can name $z = a + 164i$ and sub that into the equation. We get..\n$\\frac{a + 164i}{a+ n+ 164i} = 4i.$ If we expand we get.. $a + 164i = 4i(a+n) -656.$ \nWe can equate the real and imaginary parts to create $2$ equations. We get $a = -656$ and $164i = 4i(a+n)$. since $a = -656$ we can sub that into the second equation and we get $n$ to be...\n$164i = 4i(a+n) \\Rightarrow 41 = -656 + n \\Rightarrow n = \\boxed{697}$\n[/hide]", "Solution_23": "p2: Set $z = a+164i$. Then $a+164i=4i((a+n)+164i)$, so $a = 4i \\cdot 164i = -656$ and $a+n=\\frac{164}{4} = 41$, so $n = 41+656 = \\boxed{697}$.", "Solution_24": "I got this wrong in alcumus :(", "Solution_25": "Easy, but complex (see what I did there)\n-----\nLet $z=a+bi$. Our equation is just $$\\frac{a+164i}{a+164i+n}=4i \\rightarrow a=-656$$ from equating coefficients. Resubstituting and equating $\\Im(z)$ gives $4n-2624=164$ from where we simply get $n=\\boxed{697}$.", "Solution_26": "We, let, $z=a+bi$, hence, $\\frac{a+164i}{a+n+164i}=4i \\implies 4ia+4in-656=a+164i$, hence, $a=-656$, also $4(a+n)=164$, hence, $a+n=41$, hence, $n=41+656=\\boxed{697}.$", "Solution_27": "[hide = simple manipulations]\n\\[\\frac {z}{z+n}=4i\\]\n\\[1-\\frac {n}{z+n}=4i\\]\n\\[1-4i=\\frac {n}{z+n}\\]\n\\[\\frac {1}{1-4i}=\\frac {z+n}{n}\\]\n\\[\\frac {1+4i}{17}=\\frac {z}{n}+1\\]\nSince their imaginary part has to be equal,\n\n\\[\\frac {4i}{17}=\\frac {164i}{n}\\]\n\\[n=\\frac {(164)(17)}{4}=697\\]\n\\[n = \\boxed{697}.\\]\n[/hide]" } { "Tag": [ "geometry", "incenter", "geometry solved" ], "Problem": "The angle bisectors of triangle ABC intersect sides BC, CA and AB at points A1, B1 and C1, respectively. On line A1B1, denote by F the perpendicular foot point of C1. Prove that line FC1 bisects angle AFB.", "Solution_1": "Theorem 2 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=20781#20781]http://www.mathlinks.ro/Forum/viewtopic.php?t=6088 post #2[/url], applied to P = incenter of triangle ABC.\r\n\r\n darij", "Solution_2": "$B_1$, $C_1$ and the intersection of $BC$ with the external bisector of $\\widehat{BAC}$ are collinear, hence $F$ lies on the Applollonius circle of $\\triangle ABC$, done.\n\nBest regards,\nsunken rock", "Solution_3": "My solution:\nLet $P$,$Q$, and $R$ be the feet of the perpendiculars from $A$, $B$, and $C$ to $A_1B_1$, respectively. Then since $AP \\parallel C_1F \\parallel BQ$, \n$\\frac{PF}{FQ}=\\frac{AC_1}{C_1B}$\nAlso, since $\\triangle APB_1 \\sim \\triangle CRB_1$,\n$\\frac{AP}{AB_1}=\\frac{CR}{CB_1}$\nSimilarly,\n$\\frac{BQ}{BA_1}=\\frac{CR}{CA_1}$\nDividing the two,\n$\\frac{AP}{BQ}= \\frac{AB_1}{CB_1} \\cdot \\frac{CA_1}{BA_1}=\\frac{AC_1}{C_1B}$\nBy Ceva's theorem.\nThus, since $\\frac{PF}{FQ}=\\frac{AP}{QB}$ and $\\angle APF= \\angle BQF =90$, $\\triangle APF \\sim \\triangle BQF$. Thus, $\\angle PAF=\\angle QBF$. But $\\angle PAF= \\angle AFC_1$ and $\\angle QBF= \\angle BFC_1$, thus $FC_1$ bisects $\\angle AFB$\nNote that for this proof all we need is that $AA_1$, $BB_1$ and $CC_1$ concur, as Darij mentioned, but they do not need to be angle bisectors." } { "Tag": [ "linear algebra", "matrix", "vector", "inequalities" ], "Problem": "A is a hermitian matrix and has orthonormal eigenvectors and real \r\neigenvalues ordered as: lambda1<=lambda2<=lambda3<=....<=lambdan\r\nProve that for any unit vector x, lambda1<= <=lambdan", "Solution_1": "[quote=\"AngelaS\"]A is a hermitian matrix and has orthonormal eigenvectors and real \neigenvalues ordered as: lambda1<=lambda2<=lambda3<=....<=lambdan\nProve that for any unit vector x, lambda1<= <=lambdan[/quote]\r\n\r\n\r\nNo help? :(", "Solution_2": "Diagonalize it: $U$ is a unitary matrix, and $U^{*}AU$ is the matrix $\\begin{pmatrix}\\lambda_{1}&0&\\cdots&0\\\\ 0&\\lambda_{2}&\\cdots&0\\\\ \\vdots&\\vdots&\\ddots&\\vdots\\\\ 0&0&\\cdots&\\lambda_{n}\\end{pmatrix}$\r\nNow $x=Uv$, where $v$ is the unit vector $(c_{1},c_{2},\\cdots,c_{n})^{T}$, and $\\langle x,Ax\\rangle=(Uv)^{*}A(Uv)=v^{*}(U^{*}AU)v= \\lambda_{1}|c_{1}|^{2}+\\lambda_{2}|c_{2}|^{2}+\\cdots+\\lambda_{n}|c_{n}|^{2}$. By elementary inequalities, this is between $\\lambda_{1}|v|^{2}=\\lambda_{1}$ and $\\lambda_{n}|v|^{2}=\\lambda_{n}$. That includes the fact that this inner product is real." } { "Tag": [], "Problem": "So the question is can anyone survive a succesfully detonated bomb in a plane flying 10,000m? \r\nAs it turns out the answer is yes. Check out http://en.wikipedia.org/wiki/Vesna_Vulovi%C4%87 ! :)", "Solution_1": "Yes, well. Without a parachute that...wow. But I'd say most people could survive a 30,000 feet drop with a parachute on. A big parachute.", "Solution_2": "A really, really, big one...", "Solution_3": "Wouldnt you reach maximum velocity at a certain point due to drag?", "Solution_4": "Yeah but even at that velocity you'd die on impact usually... that's quite impressive if I say so myself.", "Solution_5": "it was lucky she landed in snow\r\n\r\nshed be dead otherwise", "Solution_6": "Yes the snow was lucky. It would be good to land in snow or water or fluff or something.", "Solution_7": "Still to survive \r\n\r\n1) a detonating terrorist bomb a couple of feet away \r\n2) plane broke in half -> no oxygen \r\n3) plane crash from 10,000 meters\r\nand 4) to have no permanent injury (after 1 year) ...", "Solution_8": "I think cats have a nonlethal terminal velocity.", "Solution_9": "Are you saying that a cat can survive something like this? :) :?", "Solution_10": "That's interesting. So a cat can survive a fall from 30,000 feet? \r\n\r\nIt would still hurt though. :|", "Solution_11": "[quote]Are you saying that a cat can survive something like this?[/quote]\nIt is well known that this is the case if you tape a buttered toast on the back of the cat. There are two laws:\n[quote]Law 1 - A buttered toast always land on the buttered side.\nand Law 2 -A Cat always land on its feat. [/quote]\nvoila - Obviously , only solution is - Cat will never touch the ground. :)\n\nSeriously - Terminal velocity (among other factors) depends on the size of the object. A small mammal (say size of a mouse) would reach terminal velocity and most likelywill survive if it lands on a soft ground. ( I have known hamsters surviving a fall of 50+ meter) Anything smaller (say a roach) the chances are excellent. For a cat, if the cat orients itself well, the terminal velocity is \"non-fatal\" There was a study I have seen people quoting from the Journal of the American Veterinary Medical Association. [quote]Two vets examined 132 cases of cats that had fallen out of high-rise windows and were brought to the Animal Medical Center, a New York veterinary hospital, for treatment. On average the cats fell 5.5 stories (BTW most will reach terminal velocity from this height) , yet 90 percent survived. (Many did suffer serious injuries.)[/quote] Any mammal larger than a cat and terminal velocity iwill be high enough to be fatal ..\r\n\r\nAbout the person surviving after a fall from the plane - I seem to recall the original story from the newspapers - what was incredible that the person was lucky that some hikers were actually close by who were able to find her and rescue her.", "Solution_12": "I hope Valentin doesn't mind me bumping his old topic back up.\r\nHowever, I find this one much more interesting than \"I have a laptop!\" and other spamtopics that can be found in Round Table nowadays.\r\n\r\nI once did a parachute jump when I was fifteen (a tandem jump). For the first seconds it was very confusing and then I regained my notion of up and down. I'm pretty sure those first seconds are when you are still gaining speed. An average human reaches a terminal velocity of about 60 meters per second, which is theoretically attained after about six seconds (and about 183 meters).\r\nThere is certainly no difference between 1 kilometer and 2 kilometers.\r\n....Except the cold and low air pressure of course. At about four kilometers there wasn't a noticable difference in air pressure, but the altitude combined with the high speed was already enough to make it ice cold! There is no such thing as free fall when you have attained the terminal velocity. You need glasses to protect your eyes, your hands freeze and the wind blows into your face that hard that it becomes hard to breathe.\r\nI was told that unlike jumps from about 4 kilometers, a jump from about 10 kilometers requires a much more advanced suit.\r\n\r\nBut seriously, I had never heard about this attack :o . As I understand it, \"Ustace\" were Croatian catholics, and in the second WWII they fought for the Nazis and against Tito's partizans. What were they trying to accomplish (a flight from Copenhagen that lands in Zagreb, there could be quite a few Croatians on board too right?)", "Solution_13": "On a tangent, if you jumped out of a plane, at say 10,000 feet, would it make any difference at all if you took off your shirt and attempted to use it as a parachute?\r\n\r\nNow that's what you call problem solving :P", "Solution_14": "I don't know penguin,but i s'pose a cape will make a fine one indeed.But i rather not go out in the public without clothes so a cape would be fine for me.\n\nWhy don't you try it;then you'll solve a really really difficult problem for skydivers. :harhar: ", "Solution_15": "Why did you revive a 9 year old topic? ._.", "Solution_16": "LOL @Above", "Solution_17": "It is actually 10 years old, then a year after it died it was revived, then 9 YEARS LATER it was revived AGAIN by Robot7620. Wow.", "Solution_18": "Lol.\n :icecream: ", "Solution_19": "Hahaha.Very funny.(I'm being sarcastic)", "Solution_20": "1. Please read [url=http://www.artofproblemsolving.com/community/c170h1125888_reviving_old_threads_with_contentless_posts]this[/url] before posting (a few years ago, this would have been common sense, but the times change?).\n\n2. Quit replying to spam posts (hint: also spam); just report them and a moderator will take care of it." } { "Tag": [ "linear algebra", "matrix", "vector", "linear algebra unsolved" ], "Problem": "A an $n$x$n$ permutation matrix is defined as a matrix with columns $e_{1}$,$e_{2},\\ldots,e_{n}$ (the standard unit vectors) in any particular order.\r\n\r\nProve that the product of two permutation matrices is again a permutation matrix.", "Solution_1": "This is inmediately clear when you realize that for a vector x, Ax permutes the entries of vector x.\r\nSo for permutation matrices A en B ABx is a product of two permutations, which is again a permutation.", "Solution_2": "Another more formal (boring) way to see it is:\r\nIf $\\sigma$ is the permuation and $A_{\\sigma}$ is the matrix.\r\nDenote by $B=A_{\\sigma}A_{\\sigma'}$, $b_{i,j}=\\sum_{k=1}^{n}\\delta_{i,\\sigma(k)}\\delta_{k,\\sigma(j)}=\\delta_{i,\\sigma \\circ \\sigma '(j)}$.\r\nSo $A_{\\sigma}A_{\\sigma'}=A_{\\sigma \\circ \\sigma '}$" } { "Tag": [ "ratio" ], "Problem": "Does anyone know the proof for the theorem to solve infinate geometric sequences?\r\n\r\ns= t1/ (1- r) with t1 being term 1 and r being the ratio of change", "Solution_1": "(nr^0+nr^1+nr^2+nr^3+nr^4+nr^5...nr^m)/r-1=nr^m+1" } { "Tag": [ "group theory", "superior algebra", "superior algebra solved" ], "Problem": "Need help, \r\nshow that if n is odd then the set of all n-cycle consists of two conjugacy classes of equal size in An\r\n\r\nThx", "Solution_1": "Take $(1 2 ... n)$, his conjugacy class contains elements $(f(1) f(2) ... f(n))$ with $f$ in $A_n$, so this class is cardinal $\\frac{n!}{2}\\times \\frac{1}{n}$ ($\\frac{1}{n}$ since $(12...n)=(n12...)=...$, $n$ representations for one cycle).\r\nNow take $(g(1)g(2)...g(n))$ where $g$ is a transposition, clearly he is not in the first class then you have your second class with same cardinal $\\frac{n!}{2}\\times \\frac{1}{n}$. Union of both classes is cardinal $(n-1)!$ we have all our $n$-cycles ...\r\nI hope I've not done mistakes", "Solution_2": "I'm not sure that what you wrote completely proves that the cardinal of the conjucacy class of an $n$-cycle is $\\frac{(n-1)!}2$. An (another) idea is to check that the commutator of an $n$-cycle is the cyclic group it generates which has $n$ elements, hence the cardinal of its conjucacy class is $(n-1)!/2$." } { "Tag": [ "probability", "rotation", "geometry", "geometric transformation", "reflection", "pigeonhole principle", "conditional probability" ], "Problem": "Robert deals 9 cards in a square from a normal 52 card deck. He then pretends that it is a tic-tac-toe board (black cards against red cards). If the probability that black won is $\\frac{n}{1000}$, compute $n$ to the nearest integer.", "Solution_1": "Well what happens if both win :maybe: like if you have\r\n\r\nXXX\r\nOOO\r\nXOX\r\n\r\nor something like that?", "Solution_2": "Since black and red have the same chance of winning, wouldn't it just be $\\frac{500}{1000}$?", "Solution_3": "What if neither of them win? Or, as [b]A M Y[/b] pointed out, both of them win?", "Solution_4": "A M Y: That counts as a tie.\r\ntOrajirOu: Ties do not count as black winning.\r\nsamath: There is the case where they both win or both lose.", "Solution_5": "[quote=\"miyomiyo\"]A M Y: That counts as a tie.\ntOrajirOu: Ties do not count as black winning.\nsamath: There is the case where they both win or both lose.[/quote]\r\n\r\n\r\nIn order that one can make ANY sense out of this problem, [b]you may at least try to define what do you mean by \"black won\".\n[/b]\r\n[hide]An ordinary tic-tac-toe is played by altering two types (say X and O's).. to see which one (if any) can get three of the same type in a line. \"Random\" 9 cards from a deck need not even have equal (or with a difference of 1) cards of alternate color... \n[/hide]", "Solution_6": "And what about if say...black won twice and red won once?\r\n\r\nBBB\r\nRRR\r\nBBB\r\n\r\nor\r\n\r\nBRB\r\nBRB\r\nBRB", "Solution_7": "Oh ...\r\nI read the problem to mean, assuming one person won, what is the probability that black won.", "Solution_8": "Perhaps the question should be reworded to:\r\n\r\nWhat is the probability that exactly three black cards lie in the same row, column, or diagonal, and that no three of the red cards lie in the same row, column, or diagonal?", "Solution_9": "[quote=\"Gyan\"][quote=\"miyomiyo\"]A M Y: That counts as a tie.\ntOrajirOu: Ties do not count as black winning.\nsamath: There is the case where they both win or both lose.[/quote]\n\n\nIn order that one can make ANY sense out of this problem, [b]you may at least try to define what do you mean by \"black won\".\n[/b]\n[hide]An ordinary tic-tac-toe is played by altering two types (say X and O's).. to see which one (if any) can get three of the same type in a line. \"Random\" 9 cards from a deck need not even have equal (or with a difference of 1) cards of alternate color... \n[/hide][/quote]\n\nUmm... do you know how to play tic-tac-toe?\n\n[quote=\"Hamster1800\"]\nAnd what about if say...black won twice and red won once? \n[/quote]\n\nDid you read my response when A M Y asked the same exact question?\n\n[quote=\"samath\"]\nOh ... \nI read the problem to mean, assuming one person won, what is the probability that black won.\n[/quote]\n\nYeah, I figured that was the problem.\n\n[quote=\"chesspro\"]\nPerhaps the question should be reworded to: \n\nWhat is the probability that exactly three black cards lie in the same row, column, or diagonal, and that no three of the red cards lie in the same row, column, or diagonal?\n[/quote]\r\n\r\nThat is a PERFECT rewording of the question.", "Solution_10": "Approximate Answer:\r\n\r\nI estimate a 50% black and red probability per square. There are 2^9=512 possible combinations.\r\n\r\nBlack has 2 Diagonals: There are 2^4=16 ways this could happen.\r\nBlack has 1 Diagonal:There are 2(2*2^4+2^4)=96 was this could happen\r\nSF: 112. We consider cases without Black diagonals:\r\nBlack has center and 2 adjacent rows: There are 4*2^2=16 ways this could happen.\r\nBlack has center and 1 row: There are 4*2^2=16 ways this could happen (Red having no row crops up first here)\r\nSF: 144\r\nWe consider cases with a Red center:\r\nBlack has 4 rows: 1 way.\r\nBlack has 3 rows: 4 ways.\r\nBlack has 2 rows: 7 ways\r\nBlack has 1 row: 4*2*2=16 ways.\r\nTotal: 172\r\n\r\n172/512=43/128. N=336.", "Solution_11": "1) Yes I did read your response. The question was not EXACTLY the same like you said. The prior question did not account for black winning \"more\" than red. Mine did.\r\n\r\n2) When you ask if someone knows how to play Tic-Tac-Toe, it sounds very demeaning. He was pointing out that it's possible for all 9 cards to be the same color, where as in an \"ordinary\" game of Tic-Tac-Toe, it's not possible for anything other than a 5-4 split.\r\n\r\n3) Yes, I agree, the rewording is great. Try to be as clear as you can when stating a problem and define any corner cases that you may have contained.\r\n\r\n[hide=\"attempt\"]\nSuppose that someone wins. There's a 50% chance of black winning and red winning. So we just count the cases where nobody wins and subtract the probability of hitting one of those.\n\n1) Both have a row:\n6 ways to choose the two rows, 8 ways for the last row: 6*8=48\nBut we overcounted! We counted the cases I described in my last post twice. This is 2*(3 choose 2) = 6 cases. Therefore, there are only 42 of these cases.\n\n2) both have a column. Same as above: 42\n\n3) neither win.\nLet's count manually. This will be exactly the same as cat's games in normal tic-tac-toe. If one player gets 6, he/she can't lose.\nBBR\nRBB\nBRR\n\nRBR\nBBR\nRRB\n\nThese can be rotated and reflected, so there are 8 ways for the top one to occur and 4 for the bottom one, so we remove 24 more.\n\nThe total number of cases we're removing is 108\n\nSo I get $\\frac{2^{9}-108}{2^{1}0}= \\frac{512-108}{1024}= \\frac{404}{1024}= \\frac{202}{512}= \\frac{101}{256}$.\n\nUnfortunately, I just realised that my answer is wrong becacuse as more and more black cards are dealt, red cards become more and more likely. Anyone have a solution that incorporates this? (Mine should work for randomly assigning colors, but not for cards. Also, mine's a bit larger than Bill The Maniac's. Anyone have a third opinion?\n[/hide]", "Solution_12": "There is a 30-point difference between yours and mine. That means that you counted 30 that I didn't.\r\n\r\nI can't find any errors in yours.\r\nI can't find any errors in mine.\r\n\r\nOuch.", "Solution_13": "[quote=\"miyomiyo\"][quote=\"Gyan\"][quote=\"miyomiyo\"]A M Y: That counts as a tie.\ntOrajirOu: Ties do not count as black winning.\nsamath: There is the case where they both win or both lose.[/quote]\n\n\nIn order that one can make ANY sense out of this problem, [b]you may at least try to define what do you mean by \"black won\".\n[/b]\n[hide]An ordinary tic-tac-toe is played by altering two types (say X and O's).. to see which one (if any) can get three of the same type in a line. \"Random\" 9 cards from a deck need not even have equal (or with a difference of 1) cards of alternate color... \n[/hide][/quote]\n\nUmm... do you know how to play tic-tac-toe?\n\n[/quote]\r\n\r\nYes, I believe, I do know the rules of tic-tac-toe! But I [b]don't[/b] believe it is played with \"randomly putting 9 cards from a card deck !\" \r\nWhat's next: \"problem\" about \"pretend poker with X's and O's \" and asking the probability of full house in your next game of tic-tac-toe?\r\n\r\nDon't want to be harsh or insulting, but one has to be a a little precise to pose a mathematical \"problem\" .. the problem as originally posted, IMO, is so vague that it sounds silly and nonsensical. (An obvious example: you can draw nine red cards from a deck, but you are not going to see nine \"X's in a standard tic-tac-toe game) \r\n\r\nHamster1800, Bill The Mack: I did not look very carefully, but the considering your version of the problem statement, the answer is not that easy (or fun), it just requires brute-force counting. \r\n\r\nYou first have to calculate the probability that you will have $n$ read cards out of 9 (Binomial type formula) and then counting in how many ways one can win. (Brute force - more of less)", "Solution_14": "Based on chesspro's use of the word \"exactly\", I will assume that situations with redundant wins will not count.\r\n\r\n[hide=\"Casewise\"]Suppose:\n\n3 black cards are dealt. There are only two ways to win:\nXOO OOX\nOXO and OXO\nOOX XOO\n\n4 black cards are dealt. Same as before, except one of the six reds is replaced with a black in each case, for a total of 12 ways.\n\n5 black cards are dealt. We may either use diagonals or select a row or column.\n\n>> For diagonals, there are 2 diagonals and we must replace two of the red cards so as not to form a row or column. There are two ways to do this keeping them on the same side of the diagonal and 2 ways if they are on different sides, for a total of 4 ways per diagonal, or 8 ways.\n\n>> For rows/columns, there are 6 to pick from, and of the remaining 6 spaces, we must choose 2 to cover with black cards so as not to form a second row/column OR a diagonal. There are 3 rows and columns to be formed, and 2 diagonals, which we must subtract from our original 6*6=9 possibilities per original row/column, to produce 4 ways for each of the 6 rows and columns, or 24 ways.\n\nFor 6 cards, we may either use diagonals or rows/columns.\n\n>> For diagonals, there are only two ways to color the board:\nXOX\nOXX\nXXO as well as its reflection across the diagonal >> 4 ways.\n\n>> For the central row and column, there are only 2 ways to color each >> 4 ways.\n\n>> For a side row or column, there are only two ways to color >> 8 ways.\n\n>> For 7 cards or more cards, we cannot create a diagonal without redundancy.\n\nAlso, if we use a row or column, we must block off each of the other two strips of spaces to prevent a red win. However, with 7 cards, we must use 4, and thus by pigeonhole, at least 2 will line up and create a redundant win. Hence, we are done finding possibilities.\n\nWays to win with 3 black cards: 2\nWays to win with 4 black cards: 12\nWays to win with 5 black cards: 32\nWays to win with 6 black cards: 18\n\nIf three black cards are dealt, then there are $C_{3}^{9}=84$ ways for the cards to be dealt, so the conditional probability of winning given 3 black cards is $P_{3}=2/84=1/42$\n\n$P_{4}=\\frac{12}{C_{4}^{9}}=12/126=2/21$\n$P_{5}=\\frac{32}{C_{5}^{9}}=32/126=16/63$\n$P_{6}=\\frac{18}{C_{6}^{9}}=18/126=1/7$[/hide]\r\n\r\nSomeone can calculate the probabilities of each number of cards being drawn for me and finish the estimation =) I'm going to bed", "Solution_15": "[hide=\"extending my solution\"]\nI actually just realised that counting the ties still works. Whatever the probability that the 9 cards split 4-5, you multiply that by the ways to tie a 4 v 5 game and subtract from 1, then halve it. But like others said, it's still brute force (mine doesn't require casework, though) and not a pretty problem.\n[/hide]", "Solution_16": "On a side note, who here is 100% invincible at tic-tac-toe? (i.e., knows a perfect strategy for going first and one for going second) I'm guessing everyone, because we're all pretty smart here.", "Solution_17": "First, there is no way to win everytime, your opponent can stop you everytime, they just have to know your strategy. Second, you can be delt any number of cards, not just 3, 4, 5, and 6. You didn't include 7, 8, or 9. If you get that many, you obviously have won, but you can get 7, 8, or 9 cards and need to include that." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "$ a > 1$, compute the sum\r\n \r\n$ \\sum_{i \\equal{} 0}^{\\infty} \\frac {2^n}{a^{2^n} \\plus{} 1}$", "Solution_1": "$ \\frac{2^{n}}{a^{2^{n}}\\plus{}1}\\equal{}\\frac{2^{n}}{a^{2^{n}}\\minus{}1}\\minus{}\\frac{2^{n\\plus{}1}}{a^{2^{n\\plus{}1}}\\minus{}1}$\r\n\t\r\nHence the given series becomes\r\n\t\r\n$ \\sum_{n\\equal{}0}^{\\infty }\\frac{2^{n}}{a^{2^{n}}\\plus{}1}\\equal{}\\sum_{n\\equal{}0}^{\\infty }\\left( \\frac{2^{n}}{a^{2^{n}}\\minus{}1}\\minus{}\\frac{2^{n\\plus{}1}}{a^{2^{n\\plus{}1}}\\minus{}1}\\right) \\equal{}\\frac{1}{a\\minus{}1}$" } { "Tag": [], "Problem": "Se considera doua triunghiuri asemenea $A_{1}B_{1}C_{1}$ si $ABC$\r\n iar primul este in interiorul celui de-al doilea. Demonstrati ca exista un punct ce se corespunde lui insusi prin asemanare.", "Solution_1": "Ce faci, Cezare ? Ai pus problema asta si la avansati.", "Solution_2": "O demonstratie foarte rapida ar fi asta:\r\n\r\nAsemanarea care duce triunghiul mare in triunghiul mic e o aplicatie continua de la trinughiul mare in el insusi. Din teorema de punct fix a lui Brouwer are un punct fix, si gata :D.", "Solution_3": "Bai, Alex, bravo, exact la asta m-am gandit si eu. Sau mai poti sa faci si cu contractii, dar o solutie asa mai scolareasca, nu exista? :)", "Solution_4": "eu n`am inteles enuntul . ce punct cautam si ce \"restrictii\" sunt asupra lui ?", "Solution_5": "Cautam un punct fix. Cezar n-a mai mentionat altceva.\r\n\r\nCred ca asemanarile directe sunt date de $z \\mapsto \\alpha z+\\beta$, iar cele inverse de $z \\mapsto \\alpha \\overline z+\\beta$. E destul de usor de dedus ca punctul fix exista daca si numai daca $\\alpha \\neq 1$ sau ($\\alpha = 1$ si $\\beta = 0$). In problema de fata, $|\\alpha| \\neq 1$ ($|\\alpha|$ = \"coeficientul de asemanare\"), pentru ca un triunghi e in interiorul celuilalt." } { "Tag": [ "induction", "ceiling function", "ratio", "combinatorics proposed", "combinatorics" ], "Problem": "Two Players A and B take turns removing chips from a pile that initially contians n chips.A cannot remove all chips on his first turn and must remove atleast one chip.Then onwards a player cannot remove more chips than how much his opponent removed the last turn.The player who removes the last chip is the winner.Find a winning strategy for A.\r\n\r\n\r\nI will post my answer later.\r\n\r\n\r\nAn Alternate:\r\n What if a player is allowed to remove up to twice the number of chips removed by his opponent on the previous turn?", "Solution_1": "Obviously $n>1$. If $n$ is odd, A just takes one chip; then all the moves are forced one-chippers and A takes the last chip. If $n=2(2k+1)$, A takes two chips; if B takes two then A takes two again and so on, winning; if at some point B takes one chip, he leaves an odd number of chips and A can apply the strategy for $n$ odd to win.\r\n\r\nIn general, we claim that A wins iff $n$ is not a power of $2$. Let $n=2^x(2y+1)$ with $y \\geq 1$. We'll prove that A wins by induction on $x$. He begins by taking $2^x$. As long as B takes $2^x$ too, A keeps taking $2^x$ and eventually wins, so suppose that at some point B takes $m<2^x$. The biggest power of $2$ that divides the resulting $n$ is at most $m$, so A can apply the strategy for smaller $x$ and win by induction.\r\n\r\nNow to prove that B wins if $n=2^x$: Say A takes $m$ on his first move, leaving $2^x-m$ chips. If this is a power of two, then it is at most $2^{x-1}$, which means $m \\geq n-m$; then B can take all the remaining chips and win. If $2^x-m$ is not a power of two, then, since $(2^x, 2^x-m)=(2^x, m) \\leq m$ we know that B can take the biggest power of $2$ dividing $n-m$, and then apply A's original strategy.", "Solution_2": "Hey! That's the same solution I had got\r\n\r\n\r\nWhat about the alternate problem?", "Solution_3": "For the alternate:\r\n\r\nWe can check that B wins for $2, 3, 5$ and A wins for $4, 6, 7$.\r\n\r\nA lemma: suppose that $k$ is winning for B and let $n=k+r$ with $r2r$, and in any other case A wins with B's strategy for $k$. This means that if $k$ is winning for B then $k+1, \\ldots, \\lceil \\frac{3k}{2} \\rceil-1$ are winning for A.\r\n\r\nSuppose that $2k$ is winning for B. Then $2k+1, \\ldots, 3k-1$ are winning for A. We want to show that $3k$ is winning for B iff $k$ is (!). Indeed, in the beginning A can't take $k$ or more chips as B would then win immediately, so he will take some $r>1.618$." } { "Tag": [ "calculus", "integration", "function", "algebra", "polynomial", "rational function", "calculus computations" ], "Problem": "I was hoping someone could help me out with 2 substitution problems that appear to run along a smilar vein. \r\n\r\nThe first is to eval. the integral of (x^2 / x+1 )dx and the second is to evaluate the integral of( x / 1+x^4) dx \r\n\r\nI realize the second will yield an arctan with 1 + ( x^2)^2 but I don't know what to do with the x in the numerator.", "Solution_1": "$\\frac{x}{1+x^4}=\\frac{x}{(x^2+\\sqrt{2}x+1)(x^2-\\sqrt{2}x+1)}=\\frac{1}{2\\sqrt{2}}\\left(\\frac{1}{x^2-\\sqrt{2}x+1}-\\frac{1}{x^2+\\sqrt{2}x+1}\\right)$", "Solution_2": "Thinking back on my question, I just realized that (x^2)2 should be u and du should be 2x for the x/1+x^4 integral. That makes things real quick. However, I still am stumped on the other question, which Im sure is very similar.", "Solution_3": "For the other question: Divide through first. Don't try to integrate an improper rational function directly.\r\nYou should get a polynomial part and a fractional part, each of which will be routine.", "Solution_4": "for the firs one if you devide $x^2$ by $x+1$ you'll get $x-1$ and a remeinder of 1 so \r\n$\\int\\frac {x^2} {x+1}dx = \\int\\(x-1dx+\\int\\frac {1} {x+1}dx$\r\nwhich equals to\r\n$\\frac {(x-1)^2} {2} + ln (x+1) $", "Solution_5": "$\\int \\frac{x}{1+x^4}dx = \\frac{1}{2}\\int \\frac{d(x^2)}{1+(x^2)^2}=\\frac{1}{2}\\arctan x^2+c$" } { "Tag": [ "geometry" ], "Problem": "A vein diagram is drawn consisting of two congruent circles of radius 17 units. If each circle passes through the center of the other circle, what is the area of the football shaped region in the middle where the circles intersect?", "Solution_1": "[quote=\"luke(mathnerd)\"]A vein diagram is drawn consisting of two congruent circles of radius 17 units. If each circle passes through the center of the other circle, what is the area of the football shaped region in the middle where the circles intersect?[/quote]\r\n[hide]Draw from the center of the circle to the top arc where they intersect and the bottom arc where they intersect. Then connect the centers, to have two equilateral tiangles with a side length of 17. We need one more thing, the little circular region. That area is equal to the area of the circular sector minus the area of a triangle. Since its an equilateral triangle, the area of the circular sector is $\\frac{289}{6}\\pi$. The area of that equilateral triangle is $\\frac{289\\sqrt{3}}{4}$. So the little area is $\\frac{289}{6}\\pi-\\frac{289\\sqrt{3}}{4}$. But there are four of these sectors so we multiply by 4 to get $\\frac{578}{3}\\pi-289\\sqrt{3}$. Then add in those two equilateral triangles to have our final answer which is $\\frac{578}{3}\\pi-\\frac{289\\sqrt{3}}{2}$.[/hide] \r\nI hope I didn't do anything wrong, I did this all in my head, so theres bound to be something wrong.", "Solution_2": "according to my math, this is incorrect--- BUT i did write this formula myself so there is a good chance my formula could be flawed,, but according to what i came up with today, you are incorrect", "Solution_3": "dude\r\nwho are you?\r\ni need to know\r\nplz", "Solution_4": "[quote=\"ckck\"][quote=\"luke(mathnerd)\"]A vein diagram is drawn consisting of two congruent circles of radius 17 units. If each circle passes through the center of the other circle, what is the area of the football shaped region in the middle where the circles intersect?[/quote]\n[hide]Draw from the center of the circle to the top arc where they intersect and the bottom arc where they intersect. Then connect the centers, to have two equilateral tiangles with a side length of 17. We need one more thing, the little circular region. That area is equal to the area of the circular sector minus the area of a triangle. Since its an equilateral triangle, the area of the circular sector is $\\frac{289}{6}\\pi$. The area of that equilateral triangle is $\\frac{289\\sqrt{3}}{4}$. So the little area is $\\frac{289}{6}\\pi-\\frac{289\\sqrt{3}}{4}$. But there are four of these sectors so we multiply by 4 to get $\\frac{578}{3}\\pi-289\\sqrt{3}$. Then add in those two equilateral triangles to have our final answer which is $\\frac{578}{3}\\pi-\\frac{289\\sqrt{3}}{2}$.[/hide] \nI hope I didn't do anything wrong, I did this all in my head, so theres bound to be something wrong.[/quote]\r\nWhat do you mean there are four regions?, I thought there were only two of those regions, but I may have misunderstood...\r\nedit: nvm, I drew the figure, and I see now. I got the same answer as aznwithabrain", "Solution_5": "[quote=\"Aznwithabrain\"]dude\nwho are you?\ni need to know\nplz[/quote]\r\n\r\nMy name is Luke Wilson\r\nIm from the northern ky MC chapter\r\ngot 1st in N. Ky chapter\r\nI got 17th at state today\r\ngot like 25th in KAAC governors cup math state\r\ni just started liking math this year\r\nthats about it", "Solution_6": "i see\r\nwhat grade are you in?", "Solution_7": "8th(whoohooooo high school next year!)(sorry, ROE)", "Solution_8": "this problem is pretty tough it took me a couple days to figure out how to do it" } { "Tag": [ "search", "number theory unsolved", "number theory" ], "Problem": "Are there finitely or infinitely many pairs of positive integers $(m,n)$ such that \\[2003m-2004n^2+2005(m^2+n)=1\\]", "Solution_1": "Hi, Peng!\r\nI just had one question. Where did you get the problem? It looks so familiar. I remember working on this problem very recently. Was this given at the seminar? :?", "Solution_2": "It's in the PC-PS4 that Dr. Recio gave us by email.", "Solution_3": "[quote=\"pengshi\"]It's in the PC-PS4 that Dr. Recio gave us by email.[/quote]\r\nOh! :idea: You are right! I forgot about them. I should keep working on them... :D", "Solution_4": "If this helps anybody, I think that the answer is that there are finitely many $(m,n)$'s. Indeed, I did a computer search on the first couple of thousand $m$'s and only found the solution $(0,1)$. Then again, I might be wrong." } { "Tag": [], "Problem": "i know what it is. the only use i know of it is when you have something like\r\nhow many sets of 3 single digit positive integers are there that have a sum of 8.\r\n\r\ncan give me a different type of problem involving this?", "Solution_1": "ive never heard of balls and urns could some one explain it.", "Solution_2": "Something like\r\n\r\nDavid wants to eat either chocolate, glazed or rainbow-sprinkled donuts. He can get 6 total and has an ample supply of each. How many different assortments of donuts can he buy?\r\n\r\nThe solution would be:\r\n\r\nLet there be 8 donuts: O O O O O O O O (mmm.... tastey :D )\r\n\r\nYou want to turn 2 of these donuts into \"dividers\" such that each divider starts a new category of donuts. These categories would divide it into chocolate, glazed, or rainbow sprinkled. For example,\r\n\r\nO | O O O | O O Would be 1 chocolate donut, 3 glazed donuts and 2 rainbow sprinkled donuts.\r\n\r\n| O | O O O O O Would be 0 chocolate donuts, 1 glazed donut and 5 rainbow sprinkled donuts.\r\n\r\nSo in the end you just do $ \\binom{8}{2}\\equal{}28$ possible assortments.\r\n\r\nThat would be how to use balls and urns... some people call it stars and bars, and it's a VERY useful method of solving problems (eliminating a lot of annoying casework).", "Solution_3": "and if you had 4 different types of donuts it would be $ \\binom{9} {3}$ right?\r\n\r\nthanks.", "Solution_4": "yeah. it would be like $ \\binom{\\text{total \\plus{} number of different choices} \\minus{} 1}{\\text{number of different choices} \\minus{} 1}$... i think" } { "Tag": [], "Problem": "Catherine has six exterior doors on her dollhouse. In how many distinct ways can Catherine allow her doll to enter and then exit her dollhouse?", "Solution_1": "6 ways to enter and 6 ways to leave\r\n\r\nanswer : 36" }